From 4f924557ad53693c389e1b30d7051c076a88d585 Mon Sep 17 00:00:00 2001 From: jverzani Date: Fri, 26 Apr 2024 18:26:12 -0400 Subject: [PATCH 1/2] many edits --- .gitignore | 3 + Project.toml | 1 + quarto/ODEs/differential_equations.qmd | 2 +- quarto/ODEs/euler.qmd | 2 +- quarto/ODEs/odes.qmd | 2 +- quarto/ODEs/solve.qmd | 1 - quarto/Project.toml | 3 + quarto/README-quarto.md | 14 +-- quarto/_common_code.qmd | 6 +- quarto/_quarto.yml | 4 +- quarto/alternatives/SciML.qmd | 7 +- quarto/alternatives/makie_plotting.qmd | 9 +- quarto/alternatives/plotly_plotting.qmd | 19 --- quarto/alternatives/symbolics.qmd | 21 +++- quarto/derivatives/curve_sketching.qmd | 14 +-- quarto/derivatives/derivatives.qmd | 16 ++- .../derivatives/first_second_derivatives.qmd | 22 ++-- .../derivatives/implicit_differentiation.qmd | 10 +- quarto/derivatives/lhospitals_rule.qmd | 66 +++++------ quarto/derivatives/more_zeros.qmd | 1 + quarto/derivatives/newtons_method.qmd | 10 +- quarto/derivatives/optimization.qmd | 10 +- quarto/derivatives/symbolic_derivatives.qmd | 6 +- .../derivatives/taylor_series_polynomials.qmd | 30 ++--- .../scalar_functions_applications.qmd | 2 +- .../vector_valued_functions.qmd | 5 +- .../vectors.qmd | 108 +++++++++--------- .../div_grad_curl.qmd | 4 +- .../double_triple_integrals.qmd | 21 ++-- .../line_integrals.qmd | 36 +++--- .../stokes_theorem.qmd | 2 - quarto/integrals/area.qmd | 17 +-- quarto/integrals/area_between_curves.qmd | 29 ++--- quarto/integrals/ftc.qmd | 4 +- quarto/integrals/improper_integrals.qmd | 3 +- quarto/integrals/integration_by_parts.qmd | 14 +-- quarto/integrals/partial_fractions.qmd | 6 +- quarto/limits/intermediate_value_theorem.qmd | 12 +- quarto/limits/limits.qmd | 32 ++++-- quarto/limits/limits_extensions.qmd | 2 +- quarto/precalc/calculator.qmd | 11 +- quarto/precalc/functions.qmd | 1 + quarto/precalc/numbers_types.qmd | 1 + quarto/precalc/polynomial_roots.qmd | 32 +++--- quarto/precalc/polynomials_package.qmd | 1 + 45 files changed, 326 insertions(+), 296 deletions(-) create mode 100644 Project.toml diff --git a/.gitignore b/.gitignore index be400c8..af74ee1 100644 --- a/.gitignore +++ b/.gitignore @@ -5,3 +5,6 @@ docs/site test/benchmarks.json Manifest.toml TODO.md +default.profraw +/quarto/default.profraw +/*/*/default.profraw \ No newline at end of file diff --git a/Project.toml b/Project.toml new file mode 100644 index 0000000..81648c0 --- /dev/null +++ b/Project.toml @@ -0,0 +1 @@ +[deps] diff --git a/quarto/ODEs/differential_equations.qmd b/quarto/ODEs/differential_equations.qmd index 4002203..5d8d9e4 100644 --- a/quarto/ODEs/differential_equations.qmd +++ b/quarto/ODEs/differential_equations.qmd @@ -16,7 +16,7 @@ using ModelingToolkit --- -The [`DifferentialEquations`](https://github.com/SciML/DifferentialEquations.jl) suite of packages contains solvers for a wide range of various differential equations. This section just briefly touches on ordinary differential equations (ODEs), and so relies only on `OrdinaryDiffEq` part of the suite. For more detail on this type and many others covered by the suite of packages, there are many other resources, including the [documentation](https://diffeq.sciml.ai/stable/) and accompanying [tutorials](https://github.com/SciML/SciMLTutorials.jl). +The [`DifferentialEquations`](https://github.com/SciML/DifferentialEquations.jl) suite of packages contains solvers for a wide range of various differential equations. This section just briefly touches on ordinary differential equations (ODEs), and so relies only on `OrdinaryDiffEq`, a small part of the suite. For more detail on this type and many others covered by the suite of packages, there are many other resources, including the [documentation](https://diffeq.sciml.ai/stable/) and accompanying [tutorials](https://github.com/SciML/SciMLTutorials.jl). ## SIR Model diff --git a/quarto/ODEs/euler.qmd b/quarto/ODEs/euler.qmd index 510d678..c858cfb 100644 --- a/quarto/ODEs/euler.qmd +++ b/quarto/ODEs/euler.qmd @@ -94,7 +94,7 @@ function make_euler_graph(n) scatter!(p, xs, ys) ## add function - out = dsolve(u'(x) - F(u(x), x), u(x), ics=(u, x0, y0)) + out = dsolve(D(u)(x) - F(u(x), x), u(x), ics=Dict(u(x0) => y0)) plot!(p, rhs(out), x0, xs[end], linewidth=5) p diff --git a/quarto/ODEs/odes.qmd b/quarto/ODEs/odes.qmd index 0ce1d4b..c2ec479 100644 --- a/quarto/ODEs/odes.qmd +++ b/quarto/ODEs/odes.qmd @@ -567,7 +567,7 @@ We enter this into `Julia`: ```{julia} @syms w::positive H::positive y() -eqnc = D2(y)(x) ~ (w/H) * sqrt(1 + y'(x)^2) +eqnc = D2(y)(x) ~ (w/H) * sqrt(1 + D(y(x))^2) ``` Unfortunately, `SymPy` needs a bit of help with this problem, by breaking the problem into steps. diff --git a/quarto/ODEs/solve.qmd b/quarto/ODEs/solve.qmd index 796ceb8..ec4e3c6 100644 --- a/quarto/ODEs/solve.qmd +++ b/quarto/ODEs/solve.qmd @@ -227,7 +227,6 @@ Finally, the author of the post shows how the interface can compose with other p ```{julia} - earth₄ = Problem(y0 = 0.0 ± 0.0, v0 = 30.0 ± 1.0) sol_euler₄ = solve(earth₄) sol_sympl₄ = solve(earth₄, Symplectic2ndOrder(dt = 2.0)) diff --git a/quarto/Project.toml b/quarto/Project.toml index 97a7f11..7525162 100644 --- a/quarto/Project.toml +++ b/quarto/Project.toml @@ -19,10 +19,12 @@ LaTeXStrings = "b964fa9f-0449-5b57-a5c2-d3ea65f4040f" Measures = "442fdcdd-2543-5da2-b0f3-8c86c306513e" Meshing = "e6723b4c-ebff-59f1-b4b7-d97aa5274f73" ModelingToolkit = "961ee093-0014-501f-94e3-6117800e7a78" +MonteCarloMeasurements = "0987c9cc-fe09-11e8-30f0-b96dd679fdca" Mustache = "ffc61752-8dc7-55ee-8c37-f3e9cdd09e70" NonlinearSolve = "8913a72c-1f9b-4ce2-8d82-65094dcecaec" Optimization = "7f7a1694-90dd-40f0-9382-eb1efda571ba" OptimizationOptimJL = "36348300-93cb-4f02-beb5-3c3902f8871e" +OrdinaryDiffEq = "1dea7af3-3e70-54e6-95c3-0bf5283fa5ed" PlotUtils = "995b91a9-d308-5afd-9ec6-746e21dbc043" PlotlyLight = "ca7969ec-10b3-423e-8d99-40f33abb42bf" Plots = "91a5bcdd-55d7-5caf-9e0b-520d859cae80" @@ -31,6 +33,7 @@ Primes = "27ebfcd6-29c5-5fa9-bf4b-fb8fc14df3ae" QuadGK = "1fd47b50-473d-5c70-9696-f719f8f3bcdc" QuizQuestions = "612c44de-1021-4a21-84fb-7261cf5eb2d4" RealPolynomialRoots = "87be438c-38ae-47c4-9398-763eabe5c3be" +RecipesBase = "3cdcf5f2-1ef4-517c-9805-6587b60abb01" Revise = "295af30f-e4ad-537b-8983-00126c2a3abe" Richardson = "708f8203-808e-40c0-ba2d-98a6953ed40d" Roots = "f2b01f46-fcfa-551c-844a-d8ac1e96c665" diff --git a/quarto/README-quarto.md b/quarto/README-quarto.md index e0bcd02..a85900f 100644 --- a/quarto/README-quarto.md +++ b/quarto/README-quarto.md @@ -1,22 +1,18 @@ - # CalculusWithJulia via quarto Short cut. Run first command until happy, then run second to publish ``` quarto render -julia adjust_plotly.jl; quarto publish gh-pages --no-render +#julia adjust_plotly.jl # <-- no longer needed +quarto publish gh-pages --no-render ``` To compile the pages through quarto -* author in `.jmd` files (to be run through pluto) -* run `julia make_qmd.jl` to convert to `.qmd` files - - The files in subdirectories are generated, they should not be edited - - The files in this main directory are quarto specific. - - `_book` and `_freeze` are conveniences +* author in `.qmd` files (to be run through pluto) * run `quarto preview` to develop interactively (kinda slow!) * run `quarto render` to render pages (not too bad) @@ -28,9 +24,7 @@ To compile the pages through quarto * should also push project to github * no need to push `_freeze` the repo, as files are locally rendered for now. -* XXX to get `PlotlyLight` to work the plotly library needs loading **before** require.min.js. This is accomplished by **editing** the .html file and moving up this line: - - +* NO LONGERto get `PlotlyLight` to work the plotly library needs loading **before** require.min.js. This is accomplished by **editing** the .html file and moving up this line: This can be done with this commandline call: julia adjust_plotly.jl diff --git a/quarto/_common_code.qmd b/quarto/_common_code.qmd index f7f9cd3..45327f9 100644 --- a/quarto/_common_code.qmd +++ b/quarto/_common_code.qmd @@ -19,6 +19,11 @@ import Logging Logging.disable_logging(Logging.Info) # or e.g. Logging.Info Logging.disable_logging(Logging.Warn) +``` + +```{julia} +#| eval: false +#| echo: false import SymPy function Base.show(io::IO, ::MIME"text/html", x::T) where {T <: SymPy.SymbolicObject} println(io, " ") @@ -215,5 +220,4 @@ function Base.show(io::IO, m::MIME"text/plain", x::HTMLoutput) println(io, caption) return nothing end - ``` diff --git a/quarto/_quarto.yml b/quarto/_quarto.yml index 8287a86..527694e 100644 --- a/quarto/_quarto.yml +++ b/quarto/_quarto.yml @@ -1,5 +1,5 @@ version: "0.18" -jupyter: julia-1.9 +engine: julia project: type: book @@ -153,5 +153,5 @@ format: # colorlinks: true execute: - error: true + error: false freeze: auto diff --git a/quarto/alternatives/SciML.qmd b/quarto/alternatives/SciML.qmd index 94a6d1a..9bb6a67 100644 --- a/quarto/alternatives/SciML.qmd +++ b/quarto/alternatives/SciML.qmd @@ -147,8 +147,8 @@ Incorporating parameters is readily done. For example to solve $f(x) = \cos(x) - ```{julia} f(x, p) = @. cos(x) - x/p -u0 = (0, pi/2) -p = 2 +u0 = (0.0, pi/2) +p = 2.0 prob = IntervalNonlinearProblem(f, u0, p) solve(prob, Bisection()) ``` @@ -410,6 +410,7 @@ could be similarly approached: y = Area/x # from A = xy P = 2x + 2y @named sys = OptimizationSystem(P, [x], [Area]); +sys = structural_simplify(sys) u0 = [x => 4.0] p = [Area => 25.0] @@ -730,6 +731,6 @@ For a trivial example, we have: ```{julia} f(x, p) = [x[1], x[2]^2] -prob = IntegralProblem(f, [0,0],[3,4], nout=2) +prob = IntegralProblem(f, [0,0],[3,4]) solve(prob, HCubatureJL()) ``` diff --git a/quarto/alternatives/makie_plotting.qmd b/quarto/alternatives/makie_plotting.qmd index a4a8c61..39ca3fb 100644 --- a/quarto/alternatives/makie_plotting.qmd +++ b/quarto/alternatives/makie_plotting.qmd @@ -34,7 +34,6 @@ We begin by loading the main package and the `norm` function from the standard ` ```{julia} using GLMakie import LinearAlgebra: norm - ``` The `Makie` developers have workarounds for the delayed time to first plot, but without utilizing these the time to load the package is lengthy. @@ -317,13 +316,13 @@ xs = 1:5 pts = Point2.(xs, xs) scatter(pts) annotations!("Point " .* string.(xs), pts; - textsize = 50 .- 2*xs, + fontsize = 50 .- 2*xs, rotation = 2pi ./ xs) current_figure() ``` -The graphic shows that `textsize` adjusts the displayed size and `rotation` adjusts the orientation. (The graphic also shows a need to manually override the limits of the `y` axis, as the `Point 5` is chopped off; the `ylims!` function to do so will be shown later.) +The graphic shows that `fontsize` adjusts the displayed size and `rotation` adjusts the orientation. (The graphic also shows a need to manually override the limits of the `y` axis, as the `Point 5` is chopped off; the `ylims!` function to do so will be shown later.) Attributes for `text`, among many others, include: @@ -331,7 +330,7 @@ Attributes for `text`, among many others, include: * `align` Specify the text alignment through `(:pos, :pos)`, where `:pos` can be `:left`, `:center`, or `:right`. * `rotation` to indicate how the text is to be rotated - * `textsize` the font point size for the text + * `fontsize` the font point size for the text * `font` to indicate the desired font @@ -1151,5 +1150,3 @@ current_figure() ``` The slider value is "lifted" by its `value` component, as shown. Otherwise, the above is fairly similar to just using an observable for `h`. - - diff --git a/quarto/alternatives/plotly_plotting.qmd b/quarto/alternatives/plotly_plotting.qmd index 5ba72ab..32b4254 100644 --- a/quarto/alternatives/plotly_plotting.qmd +++ b/quarto/alternatives/plotly_plotting.qmd @@ -1,21 +1,8 @@ ---- -format: - html: - header-includes: | - ---- - # JavaScript based plotting libraries {{< include ../_common_code.qmd >}} -:::{.callout-note} -## Plotly and Quarto -There are some oddities working with `Plotly`, `Julia`, and Quarto that require some hand editing of an HTML file. If the images are not shown below, there was an oversight. A reported issue would be welcome. - -::: - This section uses this add-on package: @@ -23,12 +10,6 @@ This section uses this add-on package: using PlotlyLight ``` -```{julia} -#| echo: false -PlotlyLight.src!(:none) -nothing -``` - To avoid a dependence on the `CalculusWithJulia` package, we load two utility packages: diff --git a/quarto/alternatives/symbolics.qmd b/quarto/alternatives/symbolics.qmd index 0a36132..575b148 100644 --- a/quarto/alternatives/symbolics.qmd +++ b/quarto/alternatives/symbolics.qmd @@ -501,10 +501,11 @@ d, r = polynomial_coeffs(ex, (x,)) r ``` -To find the degree of a monomial expression, the `degree` function is available. Here it is applied to each monomial in `d`: +To find the degree of a monomial expression, the `degree` function is available, though not exported. Here it is applied to each monomial in `d`: ```{julia} +import Symbolics: degree [degree(k) for (k,v) ∈ d] ``` @@ -773,6 +774,11 @@ Symbolics.jacobian(eqs, [x,y]) ### Integration +::: {.callout-note} +#### This area is very much WIP + +The use of `SymbolicNumericIntegration` below is currently broken. +::: The `SymbolicNumericIntegration` package provides a means to integrate *univariate* expressions through its `integrate` function. @@ -904,18 +910,27 @@ v = factor_rational(u) As such, the integrals have numeric differences from their mathematical counterparts: +::: {.callout-note} +#### Errors ahead + +These last commands are note being executed, as there are errors. +::: + ```{julia} -a,b,c = integrate(u) +#| eval: false +a,b,c = integrate(u) # not ``` We can see a bit of how much through the following, which needs a tolerance set to identify the rational numbers of the mathematical factorization correctly: ```{julia} +#| eval: false cs = [first(arguments(term)) for term ∈ arguments(a)] # pick off coefficients ``` ```{julia} -rationalize.(cs; tol=1e-8) +#| eval: false +rationalize.(cs[2:end]; tol=1e-8) ``` diff --git a/quarto/derivatives/curve_sketching.qmd b/quarto/derivatives/curve_sketching.qmd index 0ee843a..fc76aca 100644 --- a/quarto/derivatives/curve_sketching.qmd +++ b/quarto/derivatives/curve_sketching.qmd @@ -179,8 +179,8 @@ Not much to do here if you are satisfied with a graph that only gives insight in ```{julia} -𝒇(x) = ( (x-1)*(x-3)^2 ) / (x * (x-2) ) -plot(𝒇, -50, 50) +f(x) = ( (x-1)*(x-3)^2 ) / (x * (x-2) ) +plot(f, -50, 50) ``` We can see the slant asymptote and hints of vertical asymptotes, but, we'd like to see more of the basic features of the graph. @@ -193,9 +193,9 @@ To identify how wide a viewing window should be, for the rational function the a ```{julia} -𝒇cps = find_zeros(𝒇', -10, 10) -poss_ips = find_zero(𝒇'', (-10, 10)) -extrema(union(𝒇cps, poss_ips)) +cps = find_zeros(f', -10, 10) +poss_ips = find_zero(f'', (-10, 10)) +extrema(union(cps, poss_ips)) ``` So a range over $[-5,5]$ should display the key features including the slant asymptote. @@ -205,7 +205,7 @@ Previously we used the `rangeclamp` function defined in `CalculusWithJulia` to a ```{julia} -plot(rangeclamp(𝒇), -5, 5) +plot(rangeclamp(f), -5, 5) ``` With this graphic, we can now clearly see in the graph the two zeros at $x=1$ and $x=3$, the vertical asymptotes at $x=0$ and $x=2$, and the slant asymptote. @@ -219,7 +219,7 @@ Again, this sort of analysis can be systematized. The rational function type in ```{julia} xᵣ = variable(RationalFunction) -plot(𝒇(xᵣ)) # f(x) of RationalFunction type +plot(f(xᵣ)) # f(x) of RationalFunction type ``` ##### Example diff --git a/quarto/derivatives/derivatives.qmd b/quarto/derivatives/derivatives.qmd index ca67693..e9e7b5c 100644 --- a/quarto/derivatives/derivatives.qmd +++ b/quarto/derivatives/derivatives.qmd @@ -454,7 +454,7 @@ The notation - $\big|$ - on the right-hand side separates the tasks of finding t * Euler used `D` for the operator `D(f)`. This was initially used by [Argobast](http://jeff560.tripod.com/calculus.html). The notation `D(f)(c)` would be needed to evaluate the derivative at a point. - * Newton used a "dot" above the variable, $\dot{x}(t)$, which is still widely used in physics to indicate a derivative in time. This inidicates take the derivative and then plug in $t$. + * Newton used a "dot" above the variable, $\dot{x}(t)$, which is still widely used in physics to indicate a derivative in time. This indicates first taking the derivative and then plugging in $t$. * The notation $[expr]'(c)$ or $[expr]'\big|_{x=c}$would similarly mean, take the derivative of the expression and **then** evaluate at $c$. @@ -493,8 +493,10 @@ We can rearrange $(k(x+h) - k(x))$ as follows: $$ -(a\cdot f(x+h) + b\cdot g(x+h)) - (a\cdot f(x) + b \cdot g(x)) = -a\cdot (f(x+h) - f(x)) + b \cdot (g(x+h) - g(x)). +\begin{align*} +(a\cdot f(x+h) + b\cdot g(x+h)) - (a\cdot f(x) + b \cdot g(x)) &=\\ +\quada\cdot (f(x+h) - f(x)) + b \cdot (g(x+h) - g(x)). & +\end{align*} $$ Dividing by $h$, we see that this becomes @@ -645,8 +647,12 @@ Let's first treat the case of $3$ products: $$ -[u\cdot v\cdot w]' =[ u \cdot (vw)]' = u' (vw) + u [vw]' = u'(vw) + u[v' w + v w'] = -u' vw + u v' w + uvw'. +\begin{align*} +[u\cdot v\cdot w]' &=[ u \cdot (vw)]'\\ +&= u' (vw) + u [vw]'\\ +&= u'(vw) + u[v' w + v w'] \\ +&=u' vw + u v' w + uvw'. +\end{align*} $$ This pattern generalizes, clearly, to: diff --git a/quarto/derivatives/first_second_derivatives.qmd b/quarto/derivatives/first_second_derivatives.qmd index 078c91d..1d6de2b 100644 --- a/quarto/derivatives/first_second_derivatives.qmd +++ b/quarto/derivatives/first_second_derivatives.qmd @@ -180,8 +180,8 @@ Consider the function $f(x) = e^{-\lvert x\rvert} \cos(\pi x)$ over $[-3,3]$: ```{julia} -𝐟(x) = exp(-abs(x)) * cos(pi * x) -plotif(𝐟, 𝐟', -3, 3) +f(x) = exp(-abs(x)) * cos(pi * x) +plotif(f, f', -3, 3) ``` We can see the first derivative test in action: at the peaks and valleys – the relative extrema – the color changes. This is because $f'$ is changing sign as the function changes from increasing to decreasing or vice versa. @@ -191,7 +191,7 @@ This function has a critical point at $0$, as can be seen. It corresponds to a p ```{julia} -find_zeros(𝐟', -3, 3) +find_zeros(f', -3, 3) ``` ##### Example @@ -204,17 +204,17 @@ We will do so numerically. For this task we first need to gather the critical po ```{julia} -𝒇(x) = sin(pi*x) * (x^3 - 4x^2 + 2) -𝒇cps = find_zeros(𝒇', -2, 2) +f(x) = sin(pi*x) * (x^3 - 4x^2 + 2) +cps = find_zeros(f', -2, 2) ``` We should be careful though, as `find_zeros` may miss zeros that are not simple or too close together. A critical point will correspond to a relative maximum if the function crosses the axis, so these can not be "pauses." As this is exactly the case we are screening for, we double check that all the critical points are accounted for by graphing the derivative: ```{julia} -plot(𝒇', -2, 2, legend=false) +plot(f', -2, 2, legend=false) plot!(zero) -scatter!(𝒇cps, 0*𝒇cps) +scatter!(cps, 0*cps) ``` We see the six zeros as stored in `cps` and note that at each the function clearly crosses the $x$ axis. @@ -234,7 +234,7 @@ We will apply the same approach, but need to get a handle on how large the value ```{julia} g(x) = sqrt(abs(x^2 - 1)) -gcps = find_zeros(g', -2, 2) +cps = find_zeros(g', -2, 2) ``` We see the three values $-1$, $0$, $1$ that correspond to the two zeros and the relative minimum of $x^2 - 1$. We could graph things, but instead we characterize these values using a sign chart. A piecewise continuous function can only change sign when it crosses $0$ or jumps over $0$. The derivative will be continuous, except possibly at the three values above, so is piecewise continuous. @@ -244,7 +244,7 @@ A sign chart picks convenient values between crossing points to test if the func ```{julia} -pts = sort(union(-2, gcps, 2)) # this includes the endpoints (a, b) and the critical points +pts = sort(union(-2, cps, 2)) # this includes the endpoints (a, b) and the critical points test_pts = pts[1:end-1] + diff(pts)/2 # midpoints of intervals between pts [test_pts sign.(g'.(test_pts))] ``` @@ -512,14 +512,14 @@ Use the second derivative test to characterize the critical points of $j(x) = x^ ```{julia} j(x) = x^5 - 2x^4 + x^3 -jcps = find_zeros(j', -3, 3) +cps = find_zeros(j', -3, 3) ``` We can check the sign of the second derivative for each critical point: ```{julia} -[jcps j''.(jcps)] +[cps j''.(cps)] ``` That $j''(0.6) < 0$ implies that at $0.6$, $j(x)$ will have a relative maximum. As $j''(1) > 0$, the second derivative test says at $x=1$ there will be a relative minimum. That $j''(0) = 0$ says that only that there **may** be a relative maximum or minimum at $x=0$, as the second derivative test does not speak to this situation. (This last check, requiring a function evaluation to be `0`, is susceptible to floating point errors, so isn't very robust as a general tool.) diff --git a/quarto/derivatives/implicit_differentiation.qmd b/quarto/derivatives/implicit_differentiation.qmd index 1728207..c565b4e 100644 --- a/quarto/derivatives/implicit_differentiation.qmd +++ b/quarto/derivatives/implicit_differentiation.qmd @@ -268,8 +268,8 @@ Setting $a=1$ we have the graph: ```{julia} -𝒂 = 1 -G(x,y) = x^3 + y^3 - 3*𝒂*x*y +a = 1 +G(x,y) = x^3 + y^3 - 3*a*x*y implicit_plot(G) ``` @@ -573,15 +573,15 @@ To illustrate, we need specific values of $a$, $b$, and $L$: ```{julia} -𝐚, 𝐛, 𝐋 = 3, 3, 10 # L > sqrt{a^2 + b^2} -F₀(x, y) = F₀(x, y, 𝐚, 𝐛) +a, b, L = 3, 3, 10 # L > sqrt{a^2 + b^2} +F₀(x, y) = F₀(x, y, a, b) ``` Our values $(x,y)$ must satisfy $f(x,y) = L$. Let's graph: ```{julia} -implicit_plot((x,y) -> F₀(x,y) - 𝐋, xlims=(-5, 7), ylims=(-5, 7)) +implicit_plot((x,y) -> F₀(x,y) - L, xlims=(-5, 7), ylims=(-5, 7)) ``` The graph is an ellipse, though slightly tilted. diff --git a/quarto/derivatives/lhospitals_rule.qmd b/quarto/derivatives/lhospitals_rule.qmd index 8df44f5..5baacff 100644 --- a/quarto/derivatives/lhospitals_rule.qmd +++ b/quarto/derivatives/lhospitals_rule.qmd @@ -229,7 +229,7 @@ Of course, we could have just relied on `limit`, which knows about L'Hospital's ```{julia} -limit(f(x)/g(x), x, a) +limit(f(x)/g(x), x => a) ``` ## Idea behind L'Hospital's rule @@ -272,13 +272,13 @@ function lhopitals_picture_graph(n) plt end -caption = L""" - -Geometric interpretation of ``L=\lim_{x \rightarrow 0} x^2 / (\sqrt{1 + -x} - 1 - x^2)``. At ``0`` this limit is indeterminate of the form +caption = raw""" +Geometric interpretation of +``L=\lim_{x \rightarrow 0} x^2 / (\sqrt{1 + x} - 1 - x^2)``. +At ``0`` this limit is indeterminate of the form ``0/0``. The value for a fixed ``x`` can be seen as the slope of a secant -line of a parametric plot of the two functions, plotted as ``(g, -f)``. In this figure, the limiting "tangent" line has ``0`` slope, +line of a parametric plot of the two functions, plotted as +``(g, f)``. In this figure, the limiting "tangent" line has ``0`` slope, corresponding to the limit ``L``. In general, L'Hospital's rule is nothing more than a statement about slopes of tangent lines. @@ -437,38 +437,38 @@ $$ \lim_{x \rightarrow 1} \frac{x\log(x)-(x-1)}{(x-1)\log(x)} $$ -In `SymPy` we have: +In `SymPy` we have (using italic variable names to avoid a problem with the earlier use of `f` as a function): ```{julia} -𝒇 = x*log(x) - (x-1) -𝒈 = (x-1)*log(x) -𝒇(1), 𝒈(1) +𝑓 = x * log(x) - (x-1) +𝑔 = (x-1) * log(x) +𝑓(1), 𝑔(1) ``` L'Hospital's rule applies to the form $0/0$, so we try: ```{julia} -𝒇′ = diff(𝒇, x) -𝒈′ = diff(𝒈, x) -𝒇′(1), 𝒈′(1) +𝑓′ = diff(𝑓, x) +𝑔′ = diff(𝑔, x) +𝑓′(1), 𝑔′(1) ``` Again, we get the indeterminate form $0/0$, so we try again with second derivatives: ```{julia} -𝒇′′ = diff(𝒇, x, x) -𝒈′′ = diff(𝒈, x, x) -𝒇′′(1), 𝒈′′(1) +𝑓′′ = diff(𝑓, x, x) +𝑔′′ = diff(𝑔, x, x) +𝑓′′(1), 𝑔′′(1) ``` From this we see the limit is $1/2$, as could have been done directly: ```{julia} -limit(𝒇/𝒈, x=>1) +limit(𝑓/𝑔, x=>1) ``` ## The assumptions are necessary @@ -524,7 +524,7 @@ This ratio has no limit, as it oscillates, as confirmed by `SymPy`: ```{julia} -limit(u(x)/v(x), x=> oo) +limit(u(x)/v(x), x => oo) ``` ## Questions @@ -645,8 +645,8 @@ What is $L$? #| hold: true #| echo: false f(x) = (4x - sin(x))/x -L = float(N(limit(f, 0))) -numericq(L) +L = limit(f(x), x=>0) +numericq(float(L)) ``` ###### Question @@ -666,8 +666,8 @@ What is $L$? #| hold: true #| echo: false f(x) = (sqrt(1+x) - 1)/x -L = float(N(limit(f, 0))) -numericq(L) +L = limit(f(x), x => 0) +numericq(float(L)) ``` ###### Question @@ -687,8 +687,8 @@ What is $L$? #| hold: true #| echo: false f(x) = (x - sin(x))/x^3 -L = float(N(limit(f, 0))) -numericq(L) +L = limit(f(x), x=>0) +numericq(float(L)) ``` ###### Question @@ -708,8 +708,8 @@ What is $L$? #| hold: true #| echo: false f(x) = (1 - x^2/2 - cos(x))/x^3 -L = float(N(limit(f, 0))) -numericq(L) +L = limit(f(x), x=>0) +numericq(float(L)) ``` ###### Question @@ -729,8 +729,8 @@ What is $L$? #| hold: true #| echo: false f(x) = log(log(x))/log(x) -L = N(limit(f(x), x=> oo)) -numericq(L) +L = limit(f(x), x=> oo) +numericq(float(L)) ``` ###### Question @@ -750,8 +750,8 @@ What is $L$? #| hold: true #| echo: false f(x) = 1/x - 1/sin(x) -L = float(N(limit(f, 0))) -numericq(L) +L = limit(f(x), x => 0) +numericq(float(L)) ``` ###### Question @@ -770,8 +770,8 @@ What is $L$? ```{julia} #| hold: true #| echo: false -L = float(N(limit(log(x)/x, x=>oo))) -numericq(L) +L = limit(log(x)/x, x=>oo) +numericq(float(L)) ``` ###### Question diff --git a/quarto/derivatives/more_zeros.qmd b/quarto/derivatives/more_zeros.qmd index 054ab4b..90806e4 100644 --- a/quarto/derivatives/more_zeros.qmd +++ b/quarto/derivatives/more_zeros.qmd @@ -355,6 +355,7 @@ There can be problems when the stopping criteria is `abs(b-a) <= 2eps(m))` and t ```{julia} #| hold: true +#| error: true fu(x) = -40*x*exp(-x) chandrapatla(fu, -9, 1, λ3) ``` diff --git a/quarto/derivatives/newtons_method.qmd b/quarto/derivatives/newtons_method.qmd index 1b6f910..d5d3b14 100644 --- a/quarto/derivatives/newtons_method.qmd +++ b/quarto/derivatives/newtons_method.qmd @@ -403,8 +403,8 @@ A plot shows us roughly where the value lies: #| hold: true f(x) = exp(x) g(x) = x^6 -plot(f, 0, 25, label="f") -plot!(g, label="g") +plot(f, 0, 25; label="f") +plot!(g; label="g") ``` Clearly by $20$ the two paths diverge. We know exponentials eventually grow faster than powers, and this is seen in the graph. @@ -858,8 +858,10 @@ The calculation that produces the quadratic convergence now becomes: $$ -x_{i+1} - \alpha = (x_i - \alpha) - \frac{1}{k}(x_i-\alpha - \frac{f''(\xi)}{2f'(x_i)}(x_i-\alpha)^2) = -\frac{k-1}{k} (x_i-\alpha) + \frac{f''(\xi)}{2kf'(x_i)}(x_i-\alpha)^2. +\begin{align*} +x_{i+1} - \alpha &= (x_i - \alpha) - \frac{1}{k}(x_i-\alpha - \frac{f''(\xi)}{2f'(x_i)}(x_i-\alpha)^2) +&= \frac{k-1}{k} (x_i-\alpha) + \frac{f''(\xi)}{2kf'(x_i)}(x_i-\alpha)^2. +\end{align*} $$ As $k > 1$, the $(x_i - \alpha)$ term dominates, and we see the convergence is linear with $\lvert e_{i+1}\rvert \approx (k-1)/k \lvert e_i\rvert$. diff --git a/quarto/derivatives/optimization.qmd b/quarto/derivatives/optimization.qmd index efb80bf..8bd7955 100644 --- a/quarto/derivatives/optimization.qmd +++ b/quarto/derivatives/optimization.qmd @@ -1475,8 +1475,10 @@ Numerically find the value of $a$ that minimizes the length of the line seqment ```{julia} #| hold: true #| echo: false -x(a) = -a - 1/(2a) -d(a) = (a-x(a))^2 + (a^2 - x(a)^2)^2 -a = find_zero(d', 1) -numericq(a) +let + x(a) = -a - 1/(2a) + d(a) = (a-x(a))^2 + (a^2 - x(a)^2)^2 + a = find_zero(d', 1) + numericq(a) +end ``` diff --git a/quarto/derivatives/symbolic_derivatives.qmd b/quarto/derivatives/symbolic_derivatives.qmd index 44bf119..a20fe19 100644 --- a/quarto/derivatives/symbolic_derivatives.qmd +++ b/quarto/derivatives/symbolic_derivatives.qmd @@ -34,7 +34,7 @@ The `Symbolics` package provides native symbolic manipulation abilities for `Jul An expression is an unevaluated portion of code that for our purposes below contains other expressions, symbols, and numeric literals. They are held in the `Expr` type. A symbol, such as `:x`, is distinct from a string (e.g. `"x"`) and is useful to the programmer to distinguish between the contents a variable points to from the name of the variable. Symbols are fundamental to metaprogramming in `Julia`. An expression is a specification of some set of statements to execute. A numeric literal is just a number. -The three main functions from `TermInterface` we leverage are `istree`, `operation`, and `arguments`. The `operation` function returns the "outside" function of an expression. For example: +The three main functions from `TermInterface` we leverage are `isexpr`, `operation`, and `arguments`. The `operation` function returns the "outside" function of an expression. For example: ```{julia} @@ -65,7 +65,7 @@ function arity(ex) end ``` -Differentiation must distinguish between expressions, variables, and numbers. Mathematically expressions have an "outer" function, whereas variables and numbers can be directly differentiated. The `istree` function in `TermInterface` returns `true` when passed an expression, and `false` when passed a symbol or numeric literal. The latter two may be distinguished by `isa(..., Symbol)`. +Differentiation must distinguish between expressions, variables, and numbers. Mathematically expressions have an "outer" function, whereas variables and numbers can be directly differentiated. The `isexpr` function in `TermInterface` returns `true` when passed an expression, and `false` when passed a symbol or numeric literal. The latter two may be distinguished by `isa(..., Symbol)`. Here we create a function, `D`, that when it encounters an expression it *dispatches* to a specific method of `D` based on the outer operation and arity, otherwise if it encounters a symbol or a numeric literal it does the differentiation: @@ -73,7 +73,7 @@ Here we create a function, `D`, that when it encounters an expression it *dispat ```{julia} function D(ex, var=:x) - if istree(ex) + if isexpr(ex) op, args = operation(ex), arguments(ex) D(Val(op), arity(ex), args, var) elseif isa(ex, Symbol) && ex == :x diff --git a/quarto/derivatives/taylor_series_polynomials.qmd b/quarto/derivatives/taylor_series_polynomials.qmd index 6e9a7c2..66027bb 100644 --- a/quarto/derivatives/taylor_series_polynomials.qmd +++ b/quarto/derivatives/taylor_series_polynomials.qmd @@ -592,12 +592,12 @@ To see a plot, we have ```{julia} -𝒇(x) = sin(x) -𝒄, 𝒉, 𝒏 = 0, 1/4, 4 -int_poly = newton_form(𝒇, [𝒄 + i*𝒉 for i in 0:𝒏]) -tp = taylor_poly(𝒇, 𝒄, 𝒏) -𝒂, 𝒃 = -pi, pi -plot(𝒇, 𝒂, 𝒃; linewidth=5, label="f") +f(x) = sin(x) +c, h, n = 0, 1/4, 4 +int_poly = newton_form(f, [c + i*h for i in 0:n]) +tp = taylor_poly(f, c, n) +a, b = -pi, pi +plot(f, a, b; linewidth=5, label="f") plot!(int_poly; color=:green, label="interpolating") plot!(tp; color=:red, label="Taylor") ``` @@ -606,10 +606,10 @@ To get a better sense, we plot the residual differences here: ```{julia} -d1(x) = 𝒇(x) - int_poly(x) -d2(x) = 𝒇(x) - tp(x) -plot(d1, 𝒂, 𝒃; color=:blue, label="interpolating") -plot!(d2; color=:green, label="Taylor") +d1(x) = f(x) - int_poly(x) +d2(x) = f(x) - tp(x) +plot(d1, a, b; linecolor=:blue, label="interpolating") +plot!(d2; linecolor=:green, label="Taylor") ``` The graph should be $0$ at each of the points in `xs`, which we can verify in the graph above. Plotting over a wider region shows a common phenomenon that these polynomials approximate the function near the values, but quickly deviate away: @@ -825,10 +825,10 @@ To try this out to compute $\log(5)$. We have $5 = 2^2(1+0.25)$, so $k=2$ and $m ```{julia} k, m = 2, 0.25 -𝒔 = m / (2+m) -pₗ = 2 * sum(𝒔^(2i)/(2i+1) for i in 1:8) # where the polynomial approximates the logarithm... +s = m / (2+m) +pₗ = 2 * sum(s^(2i)/(2i+1) for i in 1:8) # where the polynomial approximates the logarithm... -log(1 + m), m - 𝒔*(m-pₗ), log(1 + m) - ( m - 𝒔*(m-pₗ)) +log(1 + m), m - s*(m-pₗ), log(1 + m) - ( m - s*(m-pₗ)) ``` @@ -836,7 +836,7 @@ The two values differ by less than $10^{-16}$, as advertised. Re-assembling then ```{julia} -Δ = k * log(2) + (m - 𝒔*(m-pₗ)) - log(5) +Δ = k * log(2) + (m - s*(m-pₗ)) - log(5) ``` The actual code is different, as the Taylor polynomial isn't used. The Taylor polynomial is a great approximation near a point, but there might be better polynomial approximations for all values in an interval. In this case there is, and that polynomial is used in the production setting. This makes things a bit more efficient, but the basic idea remains - for a prescribed accuracy, a polynomial approximation can be found over a given interval, which can be cleverly utilized to solve for all applicable values. @@ -1166,7 +1166,7 @@ Here is one way to get all the values bigger than 1: ex = (1 - x + x^2)*exp(x) Tn = series(ex, x, 0, 100).removeO() ps = sympy.Poly(Tn, x).coeffs() -qs = numer.(ps) +qs = numerator.(ps) qs[qs .> 1] |> Tuple # format better for output ``` diff --git a/quarto/differentiable_vector_calculus/scalar_functions_applications.qmd b/quarto/differentiable_vector_calculus/scalar_functions_applications.qmd index dfc599b..c8a3177 100644 --- a/quarto/differentiable_vector_calculus/scalar_functions_applications.qmd +++ b/quarto/differentiable_vector_calculus/scalar_functions_applications.qmd @@ -1653,7 +1653,7 @@ Here is a snippet of `SymPy` code to verify the above: ```{julia} #| hold: true -@vars y y′ λ C +@syms y y′ λ C ex = Eq(-λ*y′^2/sqrt(1 + y′^2) + λ*sqrt(1 + y′^2), y - C) Δ = sqrt(1 + y′^2) / (y - C) ex1 = Eq(simplify(ex.lhs()*Δ), simplify(ex.rhs() * Δ)) diff --git a/quarto/differentiable_vector_calculus/vector_valued_functions.qmd b/quarto/differentiable_vector_calculus/vector_valued_functions.qmd index 50e6627..314c593 100644 --- a/quarto/differentiable_vector_calculus/vector_valued_functions.qmd +++ b/quarto/differentiable_vector_calculus/vector_valued_functions.qmd @@ -1345,6 +1345,7 @@ e₃(t) = sol(t)[7:9] We fix a small time range and show the trace of the spine and the frame at a single point in time: + ```{julia} a_0, b_0 = 50, 60 ts_0 = range(a_0, b_0, length=251) @@ -1995,10 +1996,10 @@ t0, t1, a = 2PI, PI, PI rp = diff.(r(t), t) speed = 2sin(t/2) -ex = r(t) - rp/speed * integrate(speed, a, t) +ex = r(t) - rp/speed * integrate(speed, t) plot_parametric(0..4pi, r, legend=false) -plot_parametric!(0..4pi, u -> SymPy.N.(subs.(ex, t .=> u))) +plot_parametric!(0..4pi, u -> float.(subs.(ex, t .=> u))) ``` The expression `ex` is secretly `[t + sin(t), 3 + cos(t)]`, another cycloid. diff --git a/quarto/differentiable_vector_calculus/vectors.qmd b/quarto/differentiable_vector_calculus/vectors.qmd index f11a8a2..3003e0d 100644 --- a/quarto/differentiable_vector_calculus/vectors.qmd +++ b/quarto/differentiable_vector_calculus/vectors.qmd @@ -201,13 +201,13 @@ For example, if `xs` is a vector and `ys` a scalar, then the value in `ys` is re ```{julia} -𝐟(x,y) = x + y +f(x,y) = x + y ``` ```{julia} #| hold: true xs = ys = [0, 1] -𝐟.(xs, ys) +f.(xs, ys) ``` This matches `xs` and `ys` to pass `(0,0)` and then `(1,1)` to `f`, returning `0` and `2`. Now consider @@ -216,7 +216,7 @@ This matches `xs` and `ys` to pass `(0,0)` and then `(1,1)` to `f`, returning `0 ```{julia} #| hold: true xs = [0, 1]; ys = [0 1] # xs is a column vector, ys a row vector -𝐟.(xs, ys) +f.(xs, ys) ``` The two dimensions are different so for each value of `xs` the vector of `ys` is broadcast. This returns a matrix now. This will be important for some plotting usages where a grid (matrix) of values is needed. @@ -255,9 +255,9 @@ The dot product is computed in `Julia` by the `dot` function, which is in the `L ```{julia} -𝒖 = [1, 2] -𝒗 = [2, 1] -dot(𝒖, 𝒗) +u = [1, 2] +v = [2, 1] +dot(u, v) ``` :::{.callout-note} @@ -267,15 +267,15 @@ In `Julia`, the unicode operator entered by `\cdot[tab]` can also be used to mir ::: ```{julia} -𝒖 ⋅ 𝒗 # u \cdot[tab] v +u ⋅ v # u \cdot[tab] v ``` Continuing, to find the angle between $\vec{u}$ and $\vec{v}$, we might do this: ```{julia} -𝒄theta = dot(𝒖/norm(𝒖), 𝒗/norm(𝒗)) -acos(𝒄theta) +cos_theta = dot(u/norm(u), v/norm(v)) +acos(cos_theta) ``` The cosine of $\pi/2$ is $0$, so two vectors which are at right angles to each other will have a dot product of $0$: @@ -350,16 +350,16 @@ A [force diagram](https://en.wikipedia.org/wiki/Free_body_diagram) is a useful v ```{julia} -𝗍heta = pi/12 -𝗆ass, 𝗀ravity = 1/9.8, 9.8 +theta = pi/12 +mass, gravity = 1/9.8, 9.8 -𝗅 = [-sin(𝗍heta), cos(𝗍heta)] -𝗉 = -𝗅 -𝖥g = [0, -𝗆ass * 𝗀ravity] +R = [-sin(theta), cos(theta)] +p = -R +Fg = [0, -mass * gravity] plot(legend=false) -arrow!(𝗉, 𝗅) -arrow!(𝗉, 𝖥g) -scatter!(𝗉[1:1], 𝗉[2:2], markersize=5) +arrow!(p, R) +arrow!(p, Fg) +scatter!(p[1:1], p[2:2], markersize=5) ``` The magnitude of the tension force is exactly that of the force of gravity projected onto $\vec{l}$, as the bob is not accelerating in that direction. The component of the gravity force in the perpendicular direction is the part of the gravitational force that causes acceleration in the pendulum. Here we find the projection onto $\vec{l}$ and visualize the two components of the gravitational force. @@ -367,15 +367,15 @@ The magnitude of the tension force is exactly that of the force of gravity proje ```{julia} plot(legend=false, aspect_ratio=:equal) -arrow!(𝗉, 𝗅) -arrow!(𝗉, 𝖥g) -scatter!(𝗉[1:1], 𝗉[2:2], markersize=5) +arrow!(p, R) +arrow!(p, Fg) +scatter!(p[1:1], p[2:2], markersize=5) -𝗉roj = (𝖥g ⋅ 𝗅) / (𝗅 ⋅ 𝗅) * 𝗅 # force of gravity in direction of tension -𝗉orth = 𝖥g - 𝗉roj # force of gravity perpendicular to tension +proj = (Fg ⋅ R) / (R ⋅ R) * R # force of gravity in direction of tension +porth = Fg - proj # force of gravity perpendicular to tension -arrow!(𝗉, 𝗉roj) -arrow!(𝗉, 𝗉orth, linewidth=3) +arrow!(p, proj) +arrow!(p, porth, linewidth=3) ``` ##### Example @@ -551,7 +551,7 @@ As mentioned previously, a matrix in `Julia` is defined component by component w ```{julia} -ℳ = [3 4 -5; 5 -5 7; -3 6 9] +M = [3 4 -5; 5 -5 7; -3 6 9] ``` Space is the separator, which means computing a component during definition (i.e., writing `2 + 3` in place of `5`) can be problematic, as no space can be used in the computation, lest it be parsed as a separator. @@ -568,88 +568,87 @@ In `Julia`, entries in a matrix (or a vector) are stored in a container with a t ```{julia} -@syms x1 x2 x3 -𝓍 = [x1, x2, x3] +@syms xs[1:3] ``` Matrices may also be defined from blocks. This example shows how to make two column vectors into a matrix: ```{julia} -𝓊 = [10, 11, 12] -𝓋 = [13, 14, 15] -[𝓊 𝓋] # horizontally combine +u = [10, 11, 12] +v = [13, 14, 15] +[u v] # horizontally combine ``` Vertically combining the two will stack them: ```{julia} -[𝓊; 𝓋] +[u; v] ``` Scalar multiplication will just work as expected: ```{julia} -2 * ℳ +2 * M ``` Matrix addition is also straightforward: ```{julia} -ℳ + ℳ +M + M ``` Matrix addition expects matrices of the same size. An error will otherwise be thrown. However, if addition is *broadcasted* then the sizes need only be commensurate. For example, this will add `1` to each entry of `M`: ```{julia} -ℳ .+ 1 +M .+ 1 ``` Matrix multiplication is defined by `*`: ```{julia} -ℳ * ℳ +M * M ``` We can then see how the system of equations is represented with matrices: ```{julia} -ℳ * 𝓍 - 𝒷 +M * xs - 𝒷 ``` Here we use `SymPy` to verify the above: ```{julia} -𝒜 = [symbols("A$i$j", real=true) for i in 1:3, j in 1:2] -ℬ = [symbols("B$i$j", real=true) for i in 1:2, j in 1:2] +A = [symbols("A$i$j", real=true) for i in 1:3, j in 1:2] +B = [symbols("B$i$j", real=true) for i in 1:2, j in 1:2] ``` The matrix product has the expected size: the number of rows of `A` ($3$) by the number of columns of `B` ($2$): ```{julia} -𝒜 * ℬ +A * B ``` This confirms how each entry (`(A*B)[i,j]`) is from a dot product (`A[i,:] ⋅ B[:,j]`): ```{julia} -[ (𝒜 * ℬ)[i,j] == 𝒜[i,:] ⋅ ℬ[:,j] for i in 1:3, j in 1:2] +[ (A * B)[i,j] == A[i,:] ⋅ B[:,j] for i in 1:3, j in 1:2] ``` When the multiplication is broadcasted though, with `.*`, the operation will be component wise: ```{julia} -ℳ .* ℳ # component wise (Hadamard product) +M .* M # component wise (Hadamard product) ``` --- @@ -659,7 +658,7 @@ The determinant is found by `det` provided by the `LinearAlgebra` package: ```{julia} -det(ℳ) +det(M) ``` --- @@ -669,14 +668,14 @@ The transpose of a matrix is found through `transpose` which doesn't create a ne ```{julia} -transpose(ℳ) +transpose(M) ``` For matrices with *real* numbers, the transpose can be performed with the postfix operation `'`: ```{julia} -ℳ' +M' ``` (However, this is not true for matrices with complex numbers as `'` is the "adjoint," that is, the transpose of the matrix *after* taking complex conjugates.) @@ -686,14 +685,14 @@ With `u` and `v`, vectors from above, we have: ```{julia} -[𝓊' 𝓋'] # [𝓊 𝓋] was a 3 × 2 matrix, above +[u' v'] # [u v] was a 3 × 2 matrix, above ``` and ```{julia} -[𝓊'; 𝓋'] +[u'; v'] ``` :::{.callout-note} ## Note @@ -782,29 +781,30 @@ In `Julia`, the `cross` function from the `LinearAlgebra` package implements the ```{julia} -𝓪 = [1, 2, 3] -𝓫 = [4, 2, 1] -cross(𝓪, 𝓫) +a = [1, 2, 3] +b = [4, 2, 1] +cross(a, b) ``` There is also the *infix* unicode operator `\times[tab]` that can be used for similarity to traditional mathematical syntax. ```{julia} -𝓪 × 𝓫 +a × b ``` We can see the cross product is anti-commutative by comparing the last answer with: ```{julia} -𝓫 × 𝓪 +b × a ``` Using vectors of size different than $n=3$ produces a dimension mismatch error: ```{julia} +#| error: true [1, 2] × [3, 4] ``` @@ -816,15 +816,15 @@ Let's see that the matrix definition will be identical (after identifications) t ```{julia} @syms î ĵ k̂ -𝓜 = [î ĵ k̂; 3 4 5; 3 6 7] -det(𝓜) |> simplify +M = [î ĵ k̂; 3 4 5; 3 6 7] +det(M) |> simplify ``` Compare with ```{julia} -𝓜[2,:] × 𝓜[3,:] +M[2,:] × M[3,:] ``` --- diff --git a/quarto/integral_vector_calculus/div_grad_curl.qmd b/quarto/integral_vector_calculus/div_grad_curl.qmd index 8966b6c..d3ff059 100644 --- a/quarto/integral_vector_calculus/div_grad_curl.qmd +++ b/quarto/integral_vector_calculus/div_grad_curl.qmd @@ -69,7 +69,7 @@ $$ It has the interpretation of pointing out the direction of greatest ascent for the surface $z=f(x,y)$. -We move now to two other operations, the divergence and the curl, which combine to give a language to describe vector fields in $R^3$. +We move now to two other operations, the *divergence* and the *curl*, which combine to give a language to describe vector fields in $R^3$. ## The divergence @@ -680,7 +680,7 @@ V(v) = V(v...) p = plot([NaN],[NaN],[NaN], legend=false) ys = xs = range(-2,2, length=10 ) zs = range(0, 4, length=3) -CalculusWithJulia.vectorfieldplot3d!(p, V, xs, ys, zs, nz=3) +vectorfieldplot3d!(p, V, xs, ys, zs, nz=3) plot!(p, [0,0], [0,0],[-1,5], linewidth=3) p ``` diff --git a/quarto/integral_vector_calculus/double_triple_integrals.qmd b/quarto/integral_vector_calculus/double_triple_integrals.qmd index d8dcc54..e073557 100644 --- a/quarto/integral_vector_calculus/double_triple_integrals.qmd +++ b/quarto/integral_vector_calculus/double_triple_integrals.qmd @@ -210,8 +210,8 @@ Identifying a formula for this is a bit tricky. Here we use a brute force approa ```{julia} -𝒅(x, y) = sqrt(x^2 + y^2) -function 𝒍(x, y, a) +d(x, y) = sqrt(x^2 + y^2) +function l(x, y, a) theta = atan(y,x) atheta = abs(theta) if (pi/4 <= atheta < 3pi/4) # this is the y=a or y=-a case @@ -226,10 +226,10 @@ And then ```{julia} -𝒇(x,y,a,h) = h * (𝒍(x,y,a) - 𝒅(x,y))/𝒍(x,y,a) +f(x,y,a,h) = h * (l(x,y,a) - d(x,y))/l(x,y,a) 𝒂, 𝒉 = 2, 3 -𝒇(x,y) = 𝒇(x, y, 𝒂, 𝒉) # fix a and h -𝒇(v) = 𝒇(v...) +f(x,y) = f(x, y, 𝒂, 𝒉) # fix a and h +f(v) = f(v...) ``` We can visualize the volume to be computed, as follows: @@ -238,14 +238,14 @@ We can visualize the volume to be computed, as follows: ```{julia} #| hold: true xs = ys = range(-1, 1, length=20) -surface(xs, ys, 𝒇) +surface(xs, ys, f) ``` Trying this, we have: ```{julia} -hcubature(𝒇, (-𝒂/2, -𝒂/2), (𝒂/2, 𝒂/2)) +hcubature(f, (-𝒂/2, -𝒂/2), (𝒂/2, 𝒂/2)) ``` The answer agrees with that known from the formula, $4 = (1/3)a^2 h$, but the answer takes a long time to be produce. The `hcubature` function is slow with functions defined in terms of conditions. For this problem, volumes by [slicing](../integrals/volumes_slice.html) is more direct. But also symmetry can be used, were we able to compute the volume above the triangular region formed by the $x$-axis, the line $x=a/2$ and the line $y=x$, which would be $1/8$th the total volume. (As then $l(x,y,a) = (a/2)/\sin(\tan^{-1}(y,x))$.). @@ -691,8 +691,7 @@ The computationally efficient way to perform multiple integrals numerically woul However, for simple problems, where ease of expressing a region is preferred to computational efficiency, something can be implemented using repeated uses of `quadgk`. Again, this isn't recommended, save for its relationship to how iteration is approached algebraically. -In the `CalculusWithJulia` package, the `fubini` function is provided. For these notes, we define three operations using Unicode operators entered with `\int[tab]`, `\iint[tab]`, `\iiint[tab]`. (Using this, better shows the mechanics involved.) - +For these notes, we define three operations using Unicode operators entered with `\int[tab]`, `\iint[tab]`, `\iiint[tab]`. ```{julia} # adjust endpoints when expressed as a functions of outer variables @@ -1404,11 +1403,11 @@ partition(u,v) = [ u^2-v^2, u*v ] push!(ps, showG(partition)) xlabel!(ps[end], "partition") -l = @layout [a b c; +lyt = @layout [a b c; d e f; g h i] -plot(ps..., layout=l) +plot(ps..., layout=lyt) ``` ### Examples diff --git a/quarto/integral_vector_calculus/line_integrals.qmd b/quarto/integral_vector_calculus/line_integrals.qmd index 1da2715..8c36880 100644 --- a/quarto/integral_vector_calculus/line_integrals.qmd +++ b/quarto/integral_vector_calculus/line_integrals.qmd @@ -906,8 +906,8 @@ Let $S$ be the closed surface bounded by the cylinder $x^2 + y^2 = 1$, the plane ```{julia} -𝐅(x,y,z) = [1, y, z] -𝐅(v) = 𝐅(v...) +F(x,y,z) = [1, y, z] +F(v) = F(v...) ``` The surface has three faces, with different outward pointing normals for each. Let $S_1$ be the unit disk in the $x-y$ plane with normal $-\hat{k}$; $S_2$ be the top part, with normal $\langle-1, 0, 1\rangle$ (as the plane is $-1x + 0y + 1z = 1$); and $S_3$ be the cylindrical part with outward pointing normal $\vec{r}$. @@ -917,32 +917,32 @@ Integrating over $S_1$, we have the parameterization $\Phi(r,\theta) = \langle r ```{julia} -@syms 𝐑::positive 𝐭heta::positive -𝐏hi₁(r,theta) = [r*cos(theta), r*sin(theta), 0] -𝐉ac₁ = 𝐏hi₁(𝐑, 𝐭heta).jacobian([𝐑, 𝐭heta]) -𝐯₁, 𝐰₁ = 𝐉ac₁[:,1], 𝐉ac₁[:,2] -𝐍ormal₁ = 𝐯₁ × 𝐰₁ .|> simplify +@syms R::positive theta::positive +Phi₁(r,theta) = [r*cos(theta), r*sin(theta), 0] +Jac₁ = Phi₁(R, theta).jacobian([R, theta]) +v₁, w₁ = Jac₁[:,1], Jac₁[:,2] +𝐍ormal₁ = v₁ × w₁ .|> simplify ``` ```{julia} -A₁ = integrate(𝐅(𝐏hi₁(𝐑, 𝐭heta)) ⋅ (-𝐍ormal₁), (𝐭heta, 0, 2PI), (𝐑, 0, 1)) # use -Normal for outward pointing +A₁ = integrate(F(Phi₁(R, theta)) ⋅ (-𝐍ormal₁), (theta, 0, 2PI), (R, 0, 1)) # use -Normal for outward pointing ``` Integrating over $S_2$ we use the parameterization $\Phi(r, \theta) = \langle r\cos(\theta), r\sin(\theta), 1 + r\cos(\theta)\rangle$. ```{julia} -𝐏hi₂(r, theta) = [r*cos(theta), r*sin(theta), 1 + r*cos(theta)] -𝐉ac₂ = 𝐏hi₂(𝐑, 𝐭heta).jacobian([𝐑, 𝐭heta]) -𝐯₂, 𝐰₂ = 𝐉ac₂[:,1], 𝐉ac₂[:,2] -𝐍ormal₂ = 𝐯₂ × 𝐰₂ .|> simplify # has correct orientation +Phi₂(r, theta) = [r*cos(theta), r*sin(theta), 1 + r*cos(theta)] +Jac₂ = Phi₂(R, theta).jacobian([R, theta]) +v₂, w₂ = Jac₂[:,1], Jac₂[:,2] +Normal₂ = v₂ × w₂ .|> simplify # has correct orientation ``` With this, the contribution for $S_2$ is: ```{julia} -A₂ = integrate(𝐅(𝐏hi₂(𝐑, 𝐭heta)) ⋅ (𝐍ormal₂), (𝐭heta, 0, 2PI), (𝐑, 0, 1)) +A₂ = integrate(F(Phi₂(R, theta)) ⋅ (Normal₂), (theta, 0, 2PI), (R, 0, 1)) ``` Finally for $S_3$, the parameterization used is $\Phi(z, \theta) = \langle \cos(\theta), \sin(\theta), z\rangle$, but this is over a non-rectangular region, as $z$ is between $0$ and $1 + x$. @@ -953,17 +953,17 @@ This parameterization gives a normal computed through: ```{julia} @syms 𝐳::positive -𝐏hi₃(z, theta) = [cos(theta), sin(theta), 𝐳] -𝐉ac₃ = 𝐏hi₃(𝐳, 𝐭heta).jacobian([𝐳, 𝐭heta]) -𝐯₃, 𝐰₃ = 𝐉ac₃[:,1], 𝐉ac₃[:,2] -𝐍ormal₃ = 𝐯₃ × 𝐰₃ .|> simplify # wrong orientation, so we change sign below +Phi₃(z, theta) = [cos(theta), sin(theta), 𝐳] +Jac₃ = Phi₃(𝐳, theta).jacobian([𝐳, theta]) +v₃, w₃ = Jac₃[:,1], Jac₃[:,2] +Normal₃ = v₃ × w₃ .|> simplify # wrong orientation, so we change sign below ``` The contribution is ```{julia} -A₃ = integrate(𝐅(𝐏hi₃(𝐳, 𝐭heta)) ⋅ (-𝐍ormal₃), (𝐳, 0, 1 + cos(𝐭heta)), (𝐭heta, 0, 2PI)) +A₃ = integrate(F(Phi₃(𝐳, theta)) ⋅ (-Normal₃), (𝐳, 0, 1 + cos(theta)), (theta, 0, 2PI)) ``` In total, the surface integral is diff --git a/quarto/integral_vector_calculus/stokes_theorem.qmd b/quarto/integral_vector_calculus/stokes_theorem.qmd index 394a74c..3f7de1c 100644 --- a/quarto/integral_vector_calculus/stokes_theorem.qmd +++ b/quarto/integral_vector_calculus/stokes_theorem.qmd @@ -685,7 +685,6 @@ The constant $A$ just sets the scale, the parameter $n$ has a qualitative effect ```{julia} #| hold: true -gr() # pyplot doesn't like the color as specified below. n = 2 f(r,theta) = r^n * cos(n*theta) g(r, theta) = r^n * sin(n*theta) @@ -698,7 +697,6 @@ f(v) = f(v...); g(v)= g(v...) xs = ys = range(-2,2, length=50) p = contour(xs, ys, f∘Φ, color=:red, legend=false, aspect_ratio=:equal) contour!(p, xs, ys, g∘Φ, color=:blue, linewidth=3) -#pyplot() p ``` diff --git a/quarto/integrals/area.qmd b/quarto/integrals/area.qmd index 924f596..1e5326f 100644 --- a/quarto/integrals/area.qmd +++ b/quarto/integrals/area.qmd @@ -369,8 +369,10 @@ Any partition $a =x_0 < x_1 < \cdots < x_n=b$ is related to a partition of $[a-c \begin{align*} -f(c_1 -c) \cdot (x_1 - x_0) &+ f(c_2 -c) \cdot (x_2 - x_1) + \cdots + f(c_n -c) \cdot (x_n - x_{n-1}) = \\ -& f(d_1) \cdot(x_1-c - (x_0-c)) + f(d_2) \cdot(x_2-c - (x_1-c)) + \cdots + f(d_n) \cdot(x_n-c - (x_{n-1}-c)). +f(c_1 -c) \cdot (x_1 - x_0) &+ f(c_2 -c) \cdot (x_2 - x_1) + \cdots\\ + &\quad + f(c_n -c) \cdot (x_n - x_{n-1})\\ + &= f(d_1) \cdot(x_1-c - (x_0-c)) + f(d_2) \cdot(x_2-c - (x_1-c)) + \cdots\\ + &\quad + f(d_n) \cdot(x_n-c - (x_{n-1}-c)). \end{align*} @@ -636,15 +638,15 @@ With this, we can easily find an approximate answer. We wrote the function to us ```{julia} -𝒇(x) = exp(x) -riemann(𝒇, 0, 5, 10) # S_10 +f(x) = exp(x) +riemann(f, 0, 5, 10) # S_10 ``` Or with more intervals in the partition ```{julia} -riemann(𝒇, 0, 5, 50_000) +riemann(f, 0, 5, 50_000) ``` (The answer is $e^5 - e^0 = 147.4131591025766\dots$, which shows that even $50,000$ partitions is not enough to guarantee many digits of accuracy.) @@ -719,7 +721,7 @@ We have to be a bit careful with the Riemann sum, as the left Riemann sum will h ```{julia} -𝒉(x) = x > 0 ? x * log(x) : 0.0 +h(x) = x > 0 ? x * log(x) : 0.0 ``` This is actually inefficient, as the check for the size of `x` will slow things down a bit. Since we will call this function 50,000 times, we would like to avoid this, if we can. In this case just using the right sum will work: @@ -910,7 +912,8 @@ $$ $$ ```{julia} -quadgk(x -> x^x, 0, 2) +u(x) = x^x +quadgk(u, 0, 2) ``` $$ diff --git a/quarto/integrals/area_between_curves.qmd b/quarto/integrals/area_between_curves.qmd index dfbef80..c0d7877 100644 --- a/quarto/integrals/area_between_curves.qmd +++ b/quarto/integrals/area_between_curves.qmd @@ -79,14 +79,15 @@ For this problem we need to identify $a$ and $b$. These are found numerically th ```{julia} -a,b = find_zeros(x -> f(x) - g(x), -3, 3) +h(x) = f(x) - g(x) +a,b = find_zeros(h, -3, 3) ``` The answer then can be found numerically: ```{julia} -quadgk(x -> f(x) - g(x), a, b)[1] +first(quadgk(h, a, b)) ``` ##### Example @@ -99,18 +100,18 @@ A plot shows the areas: ```{julia} -𝒇(x) = sin(x) -𝒈(x) = cos(x) -plot(𝒇, 0, 2pi) -plot!(𝒈) +f(x) = sin(x) +g(x) = cos(x) +plot(f, 0, 2pi) +plot!(g) ``` There is a single interval when $f \geq g$ and this can be found algebraically using basic trigonometry, or numerically: ```{julia} -𝒂,𝒃 = find_zeros(x -> 𝒇(x) - 𝒈(x), 0, 2pi) # pi/4, 5pi/4 -quadgk(x -> 𝒇(x) - 𝒈(x), 𝒂, 𝒃)[1] +a, b = find_zeros(x -> f(x) - g(x), 0, 2pi) # pi/4, 5pi/4 +quadgk(x -> f(x) - g(x), a, b)[1] ``` ##### Example @@ -177,15 +178,15 @@ For concreteness, let $f(x) = 2-x^2$ and $[a,b] = [-1, 1/2]$, as in the figure. ```{julia} -𝐟(x) = 2 - x^2 -𝐚, 𝐛 = -1, 1/2 -𝐜 = (𝐚 + 𝐛)/2 +f(x) = 2 - x^2 +a, b = -1, 1/2 +𝐜 = (a + b)/2 -sac, sab, scb = secant(𝐟, 𝐚, 𝐜), secant(𝐟, 𝐚, 𝐛), secant(𝐟, 𝐜, 𝐛) +sac, sab, scb = secant(f, a, 𝐜), secant(f, a, b), secant(f, 𝐜, b) f1(x) = min(sac(x), scb(x)) f2(x) = sab(x) -A1 = quadgk(x -> f1(x) - f2(x), 𝐚, 𝐛)[1] +A1 = quadgk(x -> f1(x) - f2(x), a, b)[1] ``` As we needed three secant lines, we used the `secant` function from `CalculusWithJulia` to create functions representing each. Once that was done, we used the `min` function to facilitate integrating over the top bounding curve, alternatively, we could break the integral over $[a,c]$ and $[c,b]$. @@ -195,7 +196,7 @@ The area of the parabolic segment is more straightforward. ```{julia} -A2 = quadgk(x -> 𝐟(x) - f2(x), 𝐚, 𝐛)[1] +A2 = quadgk(x -> f(x) - f2(x), a, b)[1] ``` Finally, if Archimedes was right, this relationship should bring about $0$ (or something within round-off error): diff --git a/quarto/integrals/ftc.qmd b/quarto/integrals/ftc.qmd index 6894113..bc9c5a6 100644 --- a/quarto/integrals/ftc.qmd +++ b/quarto/integrals/ftc.qmd @@ -202,7 +202,7 @@ $$ \int_{-\pi/2}^{\pi/2} \cos(x) dx = F(\pi/2) - F(-\pi/2) = 1 - (-1) = 2. $$ -### An alternate notation for $F(b) - F(a)$ +### An alternate notation The expression $F(b) - F(a)$ is often written in this more compact form: @@ -617,7 +617,7 @@ We need to figure out when this is $0$. For that, we use some numeric math. ```{julia} F(x) = exp(-x^2) * quadgk(t -> exp(t^2), 0, x)[1] -Fp(x) = -2x*F(x) + 1 +Fp(x) = -2x * F(x) + 1 cps = find_zeros(Fp, -4, 4) ``` diff --git a/quarto/integrals/improper_integrals.qmd b/quarto/integrals/improper_integrals.qmd index a337fc2..0f8f689 100644 --- a/quarto/integrals/improper_integrals.qmd +++ b/quarto/integrals/improper_integrals.qmd @@ -34,8 +34,9 @@ function make_sqrt_x_graph(n) a = 1/2^n xs = range(1/2^8, stop=b, length=250) x1s = range(a, stop=b, length=50) + @syms x f(x) = 1/sqrt(x) - val = N(integrate(f, 1/2^n, b)) + val = N(integrate(f(x), (x, 1/2^n, b))) title = "area under f over [1/$(2^n), $b] is $(rpad(round(val, digits=2), 4))" plt = plot(f, range(a, stop=b, length=251), xlim=(0,b), ylim=(0, 15), legend=false, size=fig_size, title=title) diff --git a/quarto/integrals/integration_by_parts.qmd b/quarto/integrals/integration_by_parts.qmd index 44e3a98..1c5ebdc 100644 --- a/quarto/integrals/integration_by_parts.qmd +++ b/quarto/integrals/integration_by_parts.qmd @@ -440,10 +440,10 @@ Apply the formula to a parameterized circle to ensure, the signed area is proper ```{julia} #| hold: true -@syms 𝒓 t -𝒙 = 𝒓 * cos(t) -𝒚 = 𝒓 * sin(t) --integrate(𝒚 * diff(𝒙, t), (t, 0, 2PI)) +@syms r t +x = r * cos(t) +y = r * sin(t) +-integrate(y * diff(x, t), (t, 0, 2PI)) ``` We see the expected answer for the area of a circle. @@ -461,9 +461,9 @@ Working symbolically, we have one arch given by the following described in a *cl ```{julia} #| hold: true @syms t -𝒙 = t - sin(t) -𝒚 = 1 - cos(t) -integrate(𝒚 * diff(𝒙, t), (t, 0, 2PI)) +x = t - sin(t) +y = 1 - cos(t) +integrate(y * diff(x, t), (t, 0, 2PI)) ``` ([Galileo](https://mathshistory.st-andrews.ac.uk/Curves/Cycloid/) was thwarted in finding this answer exactly and resorted to constructing one from metal to *estimate* the value.) diff --git a/quarto/integrals/partial_fractions.qmd b/quarto/integrals/partial_fractions.qmd index e72ab07..138aca1 100644 --- a/quarto/integrals/partial_fractions.qmd +++ b/quarto/integrals/partial_fractions.qmd @@ -261,15 +261,15 @@ We again just let `SymPy` do the work. A partial fraction decomposition is given ```{julia} -𝒒 = (x^2 - 2x - 3) -apart(1/𝒒) +q = (x^2 - 2x - 3) +apart(1/q) ``` We see what should yield two logarithmic terms: ```{julia} -integrate(1/𝒒, x) +integrate(1/q, x) ``` :::{.callout-note} diff --git a/quarto/limits/intermediate_value_theorem.qmd b/quarto/limits/intermediate_value_theorem.qmd index 84af7ef..d370fed 100644 --- a/quarto/limits/intermediate_value_theorem.qmd +++ b/quarto/limits/intermediate_value_theorem.qmd @@ -300,10 +300,10 @@ The equation $\cos(x) = x$ has just one solution, as can be seen in this plot: ```{julia} -𝒇(x) = cos(x) -𝒈(x) = x -plot(𝒇, -pi, pi) -plot!(𝒈) +f(x) = cos(x) +g(x) = x +plot(f, -pi, pi) +plot!(g) ``` Find it. @@ -313,8 +313,8 @@ We see from the graph that it is clearly between $0$ and $2$, so all we need is ```{julia} -𝒉(x) = 𝒇(x) - 𝒈(x) -find_zero(𝒉, (0, 2)) +h(x) = f(x) - g(x) +find_zero(h, (0, 2)) ``` ##### Example: Inverse functions diff --git a/quarto/limits/limits.qmd b/quarto/limits/limits.qmd index 5195334..df50607 100644 --- a/quarto/limits/limits.qmd +++ b/quarto/limits/limits.qmd @@ -517,6 +517,24 @@ ys = f.(xs) Same story. The numeric evidence supports a limit of $L=0.6$. +::: {.callout-note} +### The `lim` function + +The `CalculusWithJulia` package provides a convenience function `lim(f, c)` to create tables to showcase limits. The `dir` keyword can be `"+-"` (the default) to show values from both the left and the right; `"+"` to only show values to the right of `c`; and `"-"` to only show values to the left of `c`: + +For example: + +```{julia} +lim(f, c) +``` + +The numbers are displayed in decreasing order so the values on the left are read from top to bottom: + +```{julia} +lim(f, c; dir="-") # or even lim(f, c, -) +``` + +::: ##### Example: the secant line @@ -567,18 +585,18 @@ What is the value of $L$, if it exists? A quick attempt numerically yields: ```{julia} -𝒙s = 0 .+ hs -𝒚s = [g(x) for x in 𝒙s] -[𝒙s 𝒚s] +xs = 0 .+ hs +ys = [g(x) for x in xs] +[xs ys] ``` Hmm, the values in `ys` appear to be going to $0.5$, but then end up at $0$. Is the limit $0$ or $1/2$? The answer is $1/2$. The last $0$ is an artifact of floating point arithmetic and the last few deviations from `0.5` due to loss of precision in subtraction. To investigate, we look more carefully at the two ratios: ```{julia} -y1s = [1 - cos(x) for x in 𝒙s] -y2s = [x^2 for x in 𝒙s] -[𝒙s y1s y2s] +y1s = [1 - cos(x) for x in xs] +y2s = [x^2 for x in xs] +[xs y1s y2s] ``` Looking at the bottom of the second column reveals the error. The value of `1 - cos(1.0e-8)` is `0` and not a value around `5e-17`, as would be expected from the pattern above it. This is because the smallest floating point value less than `1.0` is more than `5e-17` units away, so `cos(1e-8)` is evaluated to be `1.0`. There just isn't enough granularity to get this close to $0$. @@ -928,7 +946,7 @@ We know the first factor has a limit found by evaluation: $2/\pi$, so it is real ```{julia} l(x) = cos(PI*x) / (1 - (2x)^2) -limit(l, 1//2) +limit(l(x), x => 1//2) ``` Putting together, we would get $1/2$. Which we could have done directly in this case: diff --git a/quarto/limits/limits_extensions.qmd b/quarto/limits/limits_extensions.qmd index 4050dc0..60a8401 100644 --- a/quarto/limits/limits_extensions.qmd +++ b/quarto/limits/limits_extensions.qmd @@ -153,7 +153,7 @@ Some functions only have one-sided limits as they are not defined in an interval ```{julia} -limit(x^x, x, 0, dir="+") +limit(x^x, x=>0, dir="+") ``` This agrees with the IEEE convention of assigning `0^0` to be `1`. diff --git a/quarto/precalc/calculator.qmd b/quarto/precalc/calculator.qmd index 639fc8f..0219ce0 100644 --- a/quarto/precalc/calculator.qmd +++ b/quarto/precalc/calculator.qmd @@ -570,16 +570,7 @@ In `Julia`, there is a richer set of error types. The value `0/0` will in fact n ```{julia} -sqrt(-1) -``` - -For integer or real-valued inputs, the `sqrt` function expects non-negative values, so that the output will always be a real number. - - -There are other types of errors. Overflow is a common one on most calculators. The value of $1000!$ is actually *very* large (over 2500 digits large). On the Google calculator it returns `Infinity`, a slight stretch. For `factorial(1000)` `Julia` returns an `OverflowError`. This means that the answer is too large to be represented as a regular integer. - - -```{julia} +#| error: true factorial(1000) ``` diff --git a/quarto/precalc/functions.qmd b/quarto/precalc/functions.qmd index 6ef5842..1def502 100644 --- a/quarto/precalc/functions.qmd +++ b/quarto/precalc/functions.qmd @@ -95,6 +95,7 @@ Functions in `Julia` have an implicit domain, just as they do mathematically. In ```{julia} +#| error: true h(-1) ``` diff --git a/quarto/precalc/numbers_types.qmd b/quarto/precalc/numbers_types.qmd index 255beae..49c54a8 100644 --- a/quarto/precalc/numbers_types.qmd +++ b/quarto/precalc/numbers_types.qmd @@ -112,6 +112,7 @@ This could be worked around, as it is with some programming languages, but it is ```{julia} +#| error: true sqrt(-1.0) ``` diff --git a/quarto/precalc/polynomial_roots.qmd b/quarto/precalc/polynomial_roots.qmd index 6eb53d9..a3e3625 100644 --- a/quarto/precalc/polynomial_roots.qmd +++ b/quarto/precalc/polynomial_roots.qmd @@ -427,24 +427,26 @@ SymPy is phasing in the `solveset` function to replace `solve`. The main reason ```{julia} -𝒑 = 8x^4 - 8x^2 + 1 -𝒑_rts = solveset(𝒑) +p = 8x^4 - 8x^2 + 1 +p_rts = solveset(p) ``` -The `𝒑_rts` object, a `FiniteSet`, does not allow immediate access to its elements. For that `elements` will work to return a vector: +The `p_rts` object, a `Set`, does not allow indexed access to its elements. For that `collect` will work to return a vector: ```{julia} -elements(𝒑_rts) +collect(p_rts) ``` -To get the numeric approximation, we compose these function calls: +To get the numeric approximation, we can broadcast: ```{julia} -N.(elements(solveset(𝒑))) +N.(solveset(p)) ``` +(There is no need to call `collect` -- though you can -- as broadcasting over a set falls back to broadcasting over the iteration of the set and in this case returns a vector.) + ## Do numeric methods matter when you can just graph? @@ -455,8 +457,8 @@ For another example, consider the polynomial $(x-20)^5 - (x-20) + 1$. In this fo ```{julia} -𝐩 = x^5 - 100x^4 + 4000x^3 - 80000x^2 + 799999x - 3199979 -plot(𝐩, -10, 10) +p = x^5 - 100x^4 + 4000x^3 - 80000x^2 + 799999x - 3199979 +plot(p, -10, 10) ``` This seems to indicate a root near $10$. But look at the scale of the $y$ axis. The value at $-10$ is around $-25,000,000$ so it is really hard to tell if $f$ is near $0$ when $x=10$, as the range is too large. @@ -466,7 +468,7 @@ A graph over $[10,20]$ is still unclear: ```{julia} -plot(𝐩, 10,20) +plot(p, 10,20) ``` We see that what looked like a zero near $10$, was actually a number around $-100,000$. @@ -476,7 +478,7 @@ Continuing, a plot over $[15, 20]$ still isn't that useful. It isn't until we ge ```{julia} -plot(𝐩, 18, 22) +plot(p, 18, 22) ``` Not that it can't be done, but graphically solving for a root here can require some judicious choice of viewing window. Even worse is the case where something might graphically look like a root, but in fact not be a root. Something like $(x-100)^2 + 0.1$ will demonstrate. @@ -578,6 +580,10 @@ from which it follows that $|x| \leq h$, as desired. For our polynomial $x^5 -x + 1$ we have the sum above is $3$. The lone real root is approximately $-1.1673$ which satisfies $|-1.1673| \leq 3$. + + + + ## Questions @@ -883,10 +889,10 @@ function mag() p = Permutation(0,2) q = Permutation(1,2) m = 0 - for perm in (p, q, q*p, p*q, p*q*p, p^2) - as = perm([2,3,4]) + for perm in (p, q, q*p, p*q, p*q*p)#, p^2) + as = N.(collect(perm([2,3,4]))) fn = x -> x^3 - as[1]*x^2 + as[2]*x - as[3] - rts_ = find_zeros(fn, -10..10) + rts_ = find_zeros(fn, -10,10) a1 = maximum(abs.(rts_)) m = a1 > m ? a1 : m end diff --git a/quarto/precalc/polynomials_package.qmd b/quarto/precalc/polynomials_package.qmd index c5f613d..8353277 100644 --- a/quarto/precalc/polynomials_package.qmd +++ b/quarto/precalc/polynomials_package.qmd @@ -141,6 +141,7 @@ As `r` uses "`x`", and `s` a "`t`" the two can not be added, say: ```{julia} +#| error: true r + s ``` From 83526b02bf1f6c0c51b0fd441bedc64b20ff034d Mon Sep 17 00:00:00 2001 From: jverzani Date: Fri, 26 Apr 2024 18:27:39 -0400 Subject: [PATCH 2/2] clean up cruft --- 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z5zF#P*~vL#^VjlkYZFAMXdKUHt(4tdkSm{W5+t#6Q*)%5zIjo~?7K;*O3d8HfNdC4n2_oHIw*UI?yBd%}$aVRj%9sAhp zUYb!DShx(zS+hy^>CRP)ZRvF?8aN{|m%J@RvNitn?6=_sIkHQD_WiVkV3N3|UzsZ# z$m@#<&7sNvK!G#drqX66EkgsEh3MRwnYT%*xL|-Y`s^L}y`+%pLCulwB2Z5T3Lwpkd~b!@5I)Sj3>&p0Z*IQXvj604M#*qg-=bCaB9~>c zX~|o2nF{k18M6;h$(gkLqf^le!KJU z7`tMbD7ny#X3Mm_Us-tlRJ&?sc-F} z_vUyaq{=Bw^U1kIMFsmffE~~?RUW4&@I8prTDIgxBYOBGFya&$1M5P@{bLG{d;bMo OCw~q7rTM>?{`^08cm{R= diff --git a/CwJ/derivatives/first_second_derivatives.jmd b/CwJ/derivatives/first_second_derivatives.jmd deleted file mode 100644 index 08ca32c..0000000 --- a/CwJ/derivatives/first_second_derivatives.jmd +++ /dev/null @@ -1,998 +0,0 @@ -# The first and second derivatives - -This section uses these add-on packages: - - -```julia -using CalculusWithJulia -using Plots -using SymPy -using Roots -``` - - -```julia; echo=false; results="hidden" -using CalculusWithJulia.WeaveSupport - -const frontmatter = ( - title = "The first and second derivatives", - description = "Calculus with Julia: The first and second derivatives", - tags = ["CalculusWithJulia", "derivatives", "the first and second derivatives"], -); - -nothing -``` - ----- - -This section explores properties of a function, ``f(x)``, that are described by properties of its first and second derivatives, ``f'(x)`` and ``f''(x)``. As part of the conversation two tests are discussed that characterize when a critical point is a relative maximum or minimum. (We know that any relative maximum or minimum occurs at a critical point, but it is not true that *any* critical point will be a relative maximum or minimum.) - - - -## Positive or increasing on an interval - -We start with some vocabulary: - -> A function $f$ is **positive** on an interval $I$ if for any $a$ in $I$ it must be that $f(a) > 0$. - -Of course, we define *negative* in a parallel manner. The intermediate value theorem says a continuous function can not change from positive to negative without crossing $0$. This is not the case for functions with jumps, of course. - -Next, - -> A function, $f$, is (strictly) **increasing** on an interval $I$ if for any $a < b$ it must be that $f(a) < f(b)$. - -The word strictly is related to the inclusion of the $<$ precluding the possibility of a function being flat over an interval that the $\leq$ inequality would allow. - -A parallel definition with $a < b$ implying $f(a) > f(b)$ would be used for a *strictly decreasing* function. - - -We can try and prove these properties for a function algebraically -- -we'll see both are related to the zeros of some function. However, -before proceeding to that it is usually helpful to get an idea of -where the answer is using exploratory graphs. - -We will use a helper function, `plotif(f, g, a, b)` that plots the function `f` over `[a,b]` coloring it red when `g` is positive (and blue otherwise). -Such a function is defined for us in the accompanying `CalculusWithJulia` package, which has been previously been loaded. - - -To see where a function is positive, we simply pass the function -object in for *both* `f` and `g` above. For example, let's look at -where $f(x) = \sin(x)$ is positive: - -```julia; hold=true; -f(x) = sin(x) -plotif(f, f, -2pi, 2pi) -``` - - -Let's graph with `cos` in the masking spot and see what happens: - -```julia; -plotif(sin, cos, -2pi, 2pi) -``` - -Maybe surprisingly, we see that the increasing parts of the sine curve are now -highlighted. Of course, the cosine is the derivative of the sine -function, now we discuss that this is no coincidence. - -For the sequel, we will use `f'` notation to find numeric derivatives, with the notation being defined in the `CalculusWithJulia` package using the `ForwardDiff` package. - - - - -## The relationship of the derivative and increasing - -The derivative, $f'(x)$, computes the slope of the tangent line to the -graph of $f(x)$ at the point $(x,f(x))$. If the derivative is -positive, the tangent line will have an increasing slope. Clearly if -we see an increasing function and mentally layer on a tangent line, it will -have a positive slope. Intuitively then, increasing functions and -positive derivatives are related concepts. But there are some -technicalities. - -Suppose $f(x)$ has a derivative on $I$ . Then - -> If $f'(x)$ is positive on an interval $I=(a,b)$, then $f(x)$ is strictly increasing on $I$. - -Meanwhile, - -> If a function $f(x)$ is increasing on $I$, then $f'(x) \geq 0$. - -The technicality being the equality parts. In the second statement, we -have the derivative is non-negative, as we can't guarantee it is -positive, even if we considered just strictly increasing functions. - -We can see by the example of $f(x) = x^3$ that strictly increasing -functions can have a zero derivative, at a point. - -The mean value theorem provides the reasoning behind the first statement: on -$I$, the slope of any secant line between $d < e$ (both in $I$) is matched by the slope of some -tangent line, which by assumption will always be positive. If the -secant line slope is written as $(f(e) - f(d))/(e - d)$ with $d < e$, -then it is clear then that $f(e) - f(d) > 0$, or $d < e$ implies $f(d) < f(e)$. - -The second part, follows from the secant line equation. The derivative -can be written as a limit of secant-line slopes, each of which is -positive. The limit of positive things can only be non-negative, -though there is no guarantee the limit will be positive. - - -So, to visualize where a function is increasing, we can just pass in -the derivative as the masking function in our `plotif` function, as long as we are wary about places with $0$ derivative (flat spots). - -For example, here, with a more complicated function, the intervals where the function is -increasing are highlighted by passing in the functions derivative to `plotif`: - -```julia; hold=true; -f(x) = sin(pi*x) * (x^3 - 4x^2 + 2) -plotif(f, f', -2, 2) -``` - -### First derivative test - - -When a function changes from increasing to decreasing, or decreasing to increasing, it will have a peak or a valley. More formally, such points are relative extrema. - -When discussing the mean value thereom, we defined *relative -extrema* : - -> * The function $f(x)$ has a *relative maximum* at $c$ if the value $f(c)$ is an *absolute maximum* for some *open* interval containing $c$. -> * Similarly, ``f(x)`` has a *relative minimum* at ``c`` if the value ``f(c)`` is an absolute minimum for *some* open interval about ``c``. - - -We know since [Fermat](http://tinyurl.com/nfgz8fz) that: - -> Relative maxima and minima *must* occur at *critical* points. - -Fermat says that *critical points* -- where the function is defined, but its derivative is either ``0`` or undefined -- are *interesting* points, however: - -> A critical point need not indicate a relative maxima or minima. - -Again, $f(x)=x^3$ provides the example at $x=0$. This is a critical point, but clearly not a -relative maximum or minimum - it is just a slight pause for a -strictly increasing function. - -This leaves the question: - -> When will a critical point correspond to a relative maximum or minimum? - -This question can be answered by considering the first derivative. - -> *The first derivative test*: If $c$ is a critical point for $f(x)$ and -> *if* $f'(x)$ changes sign at $x=c$, then $f(c)$ will be either a -> relative maximum or a relative minimum. -> * It will be a relative maximum if the derivative changes sign from $+$ to $-$. -> * It will be a relative minimum if the derivative changes sign from $-$ to $+$. -> * If $f'(x)$ does not change sign at $c$, then $f(c)$ is *not* a relative maximum or minimum. - -The classification part, should be clear: e.g., if the derivative is positive then -negative, the function $f$ will increase to $(c,f(c))$ then decrease -from $(c,f(c))$ -- so ``f`` will have a local maximum at ``c``. - -Our definition of critical point *assumes* $f(c)$ exists, as $c$ is in -the domain of $f$. With this assumption, vertical asymptotes are -avoided. However, it need not be that $f'(c)$ exists. The absolute -value function at $x=0$ provides an example: this point is a critical -point where the derivative changes sign, but ``f'(x)`` is not defined at exactly -$x=0$. Regardless, it is guaranteed that $f(c)$ will be a relative -minimum by the first derivative test. - -##### Example - -Consider the function $f(x) = e^{-\lvert x\rvert} \cos(\pi x)$ over $[-3,3]$: - -```julia; -𝐟(x) = exp(-abs(x)) * cos(pi * x) -plotif(𝐟, 𝐟', -3, 3) -``` - -We can see the first derivative test in action: at the peaks and -valleys -- the relative extrema -- the color changes. This is because ``f'`` is changing sign as as the function -changes from increasing to decreasing or vice versa. - -This function has a critical point at ``0``, as can be seen. It corresponds to a point where the derivative does not exist. It is still identified through `find_zeros`, which picks up zeros and in case of discontinuous functions, like `f'`, zero crossings: - -```julia -find_zeros(𝐟', -3, 3) -``` - -##### Example - -Find all the relative maxima and minima of the function $f(x) = -\sin(\pi \cdot x) \cdot (x^3 - 4x^2 + 2)$ over the interval $[-2, 2]$. - -We will do so numerically. For -this task we first need to gather the critical points. As each of the -pieces of $f$ are everywhere differentiable and no quotients are -involved, the function $f$ will be everywhere differentiable. As such, -only zeros of $f'(x)$ can be critical points. We find these with - -```julia; -𝒇(x) = sin(pi*x) * (x^3 - 4x^2 + 2) -𝒇cps = find_zeros(𝒇', -2, 2) -``` - -We should be careful though, as `find_zeros` may miss zeros that are not -simple or too close together. A critical point will correspond to a -relative maximum if the function crosses the axis, so these can not be -"pauses." As this is exactly the case we are screening for, we double -check that all the critical points are accounted for by graphing the -derivative: - -```julia; -plot(𝒇', -2, 2, legend=false) -plot!(zero) -scatter!(𝒇cps, 0*𝒇cps) -``` - -We see the six zeros as stored in `cps` and note that at each the -function clearly crosses the $x$ axis. - -From this last graph of the derivative we can also characterize the -graph of $f$: The left-most critical point coincides with a relative minimum -of $f$, as the derivative changes sign from negative to -positive. The critical points then alternate relative maximum, -relative minimum, relative maximum, relative, minimum, and finally relative maximum. - -##### Example - -Consider the function $g(x) = \sqrt{\lvert x^2 - 1\rvert}$. Find the critical -points and characterize them as relative extrema or not. - -We will apply the same approach, but need to get a handle on how large -the values can be. The function is a composition of three -functions. We should expect that the only critical points will occur -when the interior polynomial, $x^2-1$ has values of interest, which is -around the interval $(-1, 1)$. So we look to the slightly wider interval $[-2, 2]$: - -```julia; -g(x) = sqrt(abs(x^2 - 1)) -gcps = find_zeros(g', -2, 2) -``` - -We see the three values $-1$, $0$, $1$ that correspond to the two -zeros and the relative minimum of $x^2 - 1$. We could graph things, -but instead we characterize these values using a sign chart. A -piecewise continuous function can only change sign when it crosses $0$ or jumps over ``0``. The -derivative will be continuous, except possibly at the three values -above, so is piecewise continuous. - -A sign chart picks convenient values between crossing points to test if the function is positive or negative over those intervals. When computing by hand, these would ideally be values for which the function is easily computed. On the computer, this isn't a concern; below the midpoint is chosen: - -```julia; -pts = sort(union(-2, gcps, 2)) # this includes the endpoints (a, b) and the critical points -test_pts = pts[1:end-1] + diff(pts)/2 # midpoints of intervals between pts -[test_pts sign.(g'.(test_pts))] -``` - -Such values are often summarized graphically on a number line using a *sign chart*: - -```julia; eval=false - - ∞ + 0 - ∞ + g' -<---- -1 ----- 0 ----- 1 ----> -``` - -(The values where the function is ``0`` or could jump over ``0`` are shown on the number line, and the sign between these points is indicated. So the first minus sign shows ``g'(x)`` is *negative* on ``(-\infty, -1)``, the second minus sign shows ``g'(x)`` is negative on ``(0,1)``.) - - -Reading this we have: - -- the derivative changes sign from negative to postive at $x=-1$, so $g(x)$ will have a relative minimum. - -- the derivative changes sign from positive to negative at $x=0$, so $g(x)$ will have a relative maximum. - -- the derivative changes sign from negative to postive at $x=1$, so $g(x)$ will have a relative minimum. - -In the `CalculusWithJulia` package there is `sign_chart` function that will do such work for us, though with a different display: - -```julia -sign_chart(g', -2, 2) -``` - -(This function numerically identifies ``x``-values for the specified function which are zeros, infinities, or points where the function jumps ``0``. It then shows the resulting sign pattern of the function from left to right.) - -We did this all without graphs. But, let's look at the graph of the derivative: - -```julia; -plot(g', -2, 2) -``` - -We see asymptotes at $x=-1$ and $x=1$! These aren't zeroes of $f'(x)$, -but rather where $f'(x)$ does not exist. The conclusion is correct - -each of $-1$, $0$ and $1$ are critical points with the identified characterization - but not for the -reason that they are all zeros. - -```julia; -plot(g, -2, 2) -``` - - -Finally, why does `find_zeros` find these values that are not zeros of -$g'(x)$? As discussed briefly above, it uses the bisection algorithm -on bracketing intervals to find zeros which are guaranteed by the -intermediate value theorem, but when applied to discontinuous functions, as `f'` is, will also identify values where the function jumps over ``0``. - - -##### Example - -Consider the function $f(x) = \sin(x) - x$. Characterize the critical points. - -We will work symbolically for this example. - -```julia; -@syms x -fx = sin(x) - x -fp = diff(fx, x) -solve(fp) -``` - -We get values of $0$ and $2\pi$. Let's look at the derivative at these points: - -At $x=0$ we have to the left and right signs found by - -```julia; -fp(-pi/2), fp(pi/2) -``` - -Both are negative. The derivative does not change sign at $0$, so the critical point is neither a relative minimum or maximum. - -What about at $2\pi$? We do something similar: - -```julia; -fp(2pi - pi/2), fp(2pi + pi/2) -``` - -Again, both negative. The function $f(x)$ is just decreasing near -$2\pi$, so again the critical point is neither a relative minimum or maximum. - -A graph verifies this: - -```julia; -plot(fx, -3pi, 3pi) -``` - -We see that at $0$ and $2\pi$ there are "pauses" as the function -decreases. We should also see that this pattern repeats. The critical -points found by `solve` are only those within a certain domain. Any -value that satisfies $\cos(x) - 1 = 0$ will be a critical point, and -there are infinitely many of these of the form $n \cdot 2\pi$ for $n$ -an integer. - - -As a comment, the `solveset` function, which is replacing `solve`, -returns the entire collection of zeros: - -```julia; -solveset(fp) -``` - ----- - -Of course, `sign_chart` also does this, only numerically. We just need to pick an interval wide enough to contains ``[0,2\pi]`` - -```julia -sign_chart((x -> sin(x)-x)', -3pi, 3pi) -``` - - - -##### Example - -Suppose you know $f'(x) = (x-1)\cdot(x-2)\cdot (x-3) = x^3 - 6x^2 + -11x - 6$ and $g'(x) = (x-1)\cdot(x-2)^2\cdot(x-3)^3 = x^6 -14x^5 -+80x^4-238x^3+387x^2-324x+108$. - -How would the graphs of $f(x)$ and $g(x)$ differ, as they share identical critical points? - -The graph of $f(x)$ - a function we do not have a formula for - can have its critical points characterized by the first derivative test. As the derivative changes sign at each, all critical points correspond to relative maxima. The sign pattern is negative/positive/negative/positive so we have from left to right a relative minimum, a relative maximum, and then a relative minimum. This is consistent with a ``4``th degree polynomial with ``3`` relative extrema. - -For the graph of $g(x)$ we can apply the same analysis. Thinking for a -moment, we see as the factor $(x-2)^2$ comes as a power of $2$, the -derivative of $g(x)$ will not change sign at $x=2$, so there is no -relative extreme value there. However, at $x=3$ the factor has an odd -power, so the derivative will change sign at $x=3$. So, as $g'(x)$ is -positive for large *negative* values, there will be a relative maximum -at $x=1$ and, as $g'(x)$ is positive for large *positive* values, a -relative minimum at $x=3$. - -The latter is consistent with a $7$th degree polynomial with positive leading coefficient. It is intuitive that since $g'(x)$ is a $6$th degree polynomial, $g(x)$ will be a $7$th degree one, as the power rule applied to a polynomial results in a polynomial of lesser degree by one. - - -Here is a simple schematic that illustrates the above considerations. - -```julia; eval=false -f' - 0 + 0 - 0 + f'-sign - ↘ ↗ ↘ ↗ f-direction - ∪ ∩ ∪ f-shape - -g' + 0 - 0 - 0 + g'-sign - ↗ ↘ ↘ ↗ g-direction - ∩ ~ ∪ g-shape -<------ 1 ----- 2 ----- 3 ------> -``` - -## Concavity - -Consider the function $f(x) = x^2$. Over this function we draw some -secant lines for a few pairs of $x$ values: - -```julia; echo=false -let - f(x) = x^2 - seca(f,a,b) = x -> f(a) + (f(b) - f(a)) / (b-a) * (x-a) - p = plot(f, -2, 3, legend=false, linewidth=5, xlim=(-2,3), ylim=(-2, 9)) - plot!(p,seca(f, -1, 2)) - a,b = -1, 2; xs = range(a, stop=b, length=50) - plot!(xs, seca(f, a, b).(xs), linewidth=5) - plot!(p,seca(f, 0, 3/2)) - a,b = 0, 3/2; xs = range(a, stop=b, length=50) - plot!(xs, seca(f, a, b).(xs), linewidth=5) - p -end -``` - -The graph attempts to illustrate that for this function the secant -line between any two points $a < b$ will lie above the graph over $[a,b]$. - -This is a special property not shared by all functions. Let $I$ be an open interval. - -> **Concave up**: A function $f(x)$ is concave up on $I$ if for any $a < b$ in $I$, the secant line between $a$ and $b$ lies above the graph of $f(x)$ over $[a,b]$. - -A similar definition exists for *concave down* where the secant lines -lie below the graph. Notationally, concave up says for any $x$ in $[a,b]$: - -```math -f(a) + \frac{f(b) - f(a)}{b-a} \cdot (x-a) \geq f(x) \quad\text{ (concave up) } -``` - -Replacing -$\geq$ with $\leq$ defines *concave down*, and with either $>$ or $<$ -will add the prefix "strictly." These definitions are useful for a -general definition of -[convex functions](https://en.wikipedia.org/wiki/Convex_function). - -We won't work with these definitions in this section, rather we will characterize -concavity for functions which have either a first or second -derivative: - -> * If $f'(x)$ exists and is *increasing* on $(a,b)$, then $f(x)$ is concave up on $(a,b)$. -> * If ``f'(x)`` is *decreasing* on ``(a,b)``, then ``f(x)`` is concave *down*. - -A proof of this makes use of the same trick used to establish the mean -value theorem from Rolle's theorem. Assume ``f'`` is increasing and let -$g(x) = f(x) - (f(a) + M \cdot (x-a))$, where $M$ is the slope of -the secant line between $a$ and $b$. By construction $g(a) = g(b) = -0$. If $f'(x)$ is increasing, then so is $g'(x) = f'(x) + M$. By its -definition above, showing ``f`` is concave up is the same as showing $g(x) \leq -0$. Suppose to the contrary that there is a value where $g(x) > 0$ in -$[a,b]$. We show this can't be. Assuming $g'(x)$ always exists, after -some work, Rolle's theorem will ensure there is a value where $g'(c) = -0$ and $(c,g(c))$ is a relative maximum, and as we know there is at -least one positive value, it must be $g(c) > 0$. The first derivative -test then ensures that $g'(x)$ will increase to the left of $c$ and -decrease to the right of $c$, since $c$ is at a critical point and not -an endpoint. But this can't happen as $g'(x)$ is assumed to be -increasing on the interval. - - -The relationship between increasing functions and their derivatives -- if $f'(x) > 0 $ on $I$, then ``f`` is increasing on $I$ -- -gives this second characterization of concavity when the second -derivative exists: - -> * If $f''(x)$ exists and is positive on $I$, then $f(x)$ is concave up on $I$. -> * If $f''(x)$ exists and is negative on $I$, then $f(x)$ is concave down on $I$. - -This follows, as we can think of $f''(x)$ as just the first derivative -of the function $f'(x)$, so the assumption will force $f'(x)$ to exist and be -increasing, and hence $f(x)$ to be concave up. - -##### Example - -Let's look at the function $x^2 \cdot e^{-x}$ for positive $x$. A -quick graph shows the function is concave up, then down, then up in -the region plotted: - -```julia; -h(x) = x^2 * exp(-x) -plotif(h, h'', 0, 8) -``` - -From the graph, we would expect that the second derivative - which is continuous - would have two zeros on $[0,8]$: - -```julia; -ips = find_zeros(h'', 0, 8) -``` - -As well, between the zeros we should have the sign pattern `+`, `-`, and `+`, as we verify: - -```julia; -sign_chart(h'', 0, 8) -``` - -### Second derivative test - -Concave up functions are "opening" up, and often clearly $U$-shaped, though that is not necessary. At a -relative minimum, where there is a ``U``-shape, the graph will be concave up; conversely -at a relative maximum, where the graph has a downward ``\cap``-shape, the function will be concave down. This observation becomes: - -> The **second derivative test**: If $c$ is a critical point of $f(x)$ -> with $f''(c)$ existing in a neighborhood of $c$, then -> * The value $f(c)$ will be a relative maximum if $f''(c) > 0$, -> * The value $f(c)$ will be a relative minimum if $f''(c) < 0$, and -> * *if* ``f''(c) = 0`` the test is *inconclusive*. - - -If $f''(c)$ is positive in an interval about $c$, then $f''(c) > 0$ implies the function is -concave up at $x=c$. In turn, concave up implies the derivative is increasing -so must go from negative to positive at the critical point. - -The second derivative test is **inconclusive** when $f''(c)=0$. No such -general statement exists, as there isn't enough information. For -example, the function $f(x) = x^3$ has $0$ as a critical point, -$f''(0)=0$ and the value does not correspond to a relative maximum or minimum. On the -other hand $f(x)=x^4$ has $0$ as a critical point, $f''(0)=0$ is a -relative minimum. - -##### Example - -Use the second derivative test to characterize the critical points of $j(x) = x^5 - x^4 + x^3$. - -```julia; -j(x) = x^5 - 2x^4 + x^3 -jcps = find_zeros(j', -3, 3) -``` - -We can check the sign of the second derivative for each critical point: - -```julia; -[jcps j''.(jcps)] -``` - -That $j''(0.6) < 0$ implies that at $0.6$, $j(x)$ will have a relative -maximum. As $''(1) > 0$, the second derivative test says at $x=1$ -there will be a relative minimum. That $j''(0) = 0$ says that only -that there **may** be a relative maximum or minimum at $x=0$, as the second -derivative test does not speak to this situation. (This last check, requiring a function evaluation to be `0`, is susceptible to floating point errors, so isn't very robust as a general tool.) - -This should be consistent with -this graph, where $-0.25$, and $1.25$ are chosen to capture the zero at -$0$ and the two relative extrema: - -```julia; -plotif(j, j'', -0.25, 1.25) -``` - -For the graph we see that $0$ **is not** a relative maximum or minimum. We could have seen this numerically by checking the first derivative test, and noting there is no sign change: - -```julia; -sign_chart(j', -3, 3) -``` - -##### Example - -One way to visualize the second derivative test is to *locally* overlay on a critical point a parabola. For example, consider ``f(x) = \sin(x) + \sin(2x) + \sin(3x)`` over ``[0,2\pi]``. It has ``6`` critical points over ``[0,2\pi]``. In this graphic, we *locally* layer on ``6`` parabolas: - -```julia; hold=true; -f(x) = sin(x) + sin(2x) + sin(3x) -p = plot(f, 0, 2pi, legend=false, color=:blue, linewidth=3) -cps = find_zeros(f', (0, 2pi)) -Δ = 0.5 -for c in cps - parabola(x) = f(c) + (f''(c)/2) * (x-c)^2 - plot!(parabola, c - Δ, c + Δ, color=:red, linewidth=5, alpha=0.6) -end -p -``` - - -The graphic shows that for this function near the relative extrema the parabolas *approximate* the function well, so that the relative extrema are characterized by the relative extrema of the parabolas. - -At each critical point ``c``, the parabolas have the form - -```math -f(c) + \frac{f''(c)}{2}(x-c)^2. -``` - -The ``2`` is a mystery to be answered in the section on [Taylor series](../taylor_series_polynomials.html), the focus here is on the *sign* of ``f''(c)``: - -* if ``f''(c) > 0`` then the approximating parabola opens upward and the critical point is a point of relative minimum for ``f``, -* if ``f''(c) < 0`` then the approximating parabola opens downward and the critical point is a point of relative maximum for ``f``, and -* were ``f''(c) = 0`` then the approximating parabola is just a line -- the tangent line at a critical point -- and is non-informative about extrema. - -That is, the parabola picture is just the second derivative test in this light. - -### Inflection points - -An inflection point is a value where the *second* derivative of $f$ -changes sign. At an inflection point the derivative will change from -increasing to decreasing (or vice versa) and the function will change -from concave up to down (or vice versa). - -We can use the `find_zeros` function to identify potential inflection -points by passing in the second derivative function. For example, -consider the bell-shaped function - -```math -k(x) = e^{-x^2/2}. -``` - -A graph suggests relative a maximum at $x=0$, a horizontal asymptote of $y=0$, -and two inflection points: - -```julia; -k(x) = exp(-x^2/2) -plotif(k, k'', -3, 3) -``` - -The inflection points can be found directly, if desired, or numerically with: - -```julia; -find_zeros(k'', -3, 3) -``` - -(The `find_zeros` function may return points which are not inflection points. It primarily returns points where $k''(x)$ changes sign, but *may* also find points where $k''(x)$ is $0$ yet does not change sign at $x$.) - - - -##### Example - -A car travels from a stop for 1 mile in 2 minutes. A graph of its -position as a function of time might look like any of these graphs: - -```julia; echo=false -let - v(t) = 30/60*t - w(t) = t < 1/2 ? 0.0 : (t > 3/2 ? 1.0 : (t-1/2)) - y(t) = 1 / (1 + exp(-t)) - y1(t) = y(2(t-1)) - y2(t) = y1(t) - y1(0) - y3(t) = 1/y2(2) * y2(t) - plot(v, 0, 2, label="f1") - plot!(w, label="f2") - plot!(y3, label="f3") -end -``` - -All three graphs have the same *average* velocity which is just the -$1/2$ miles per minute (``30`` miles an hour). But the instantaneous -velocity - which is given by the derivative of the position function) -varies. - -The graph `f1` has constant velocity, so the position is a straight -line with slope $v_0$. The graph `f2` is similar, though for first and -last 30 seconds, the car does not move, so must move faster during the -time it moves. A more realistic graph would be `f3`. The position -increases continuously, as do the others, but the velocity changes -more gradually. The initial velocity is less than $v_0$, but -eventually gets to be more than $v_0$, then velocity starts to -increase less. At no point is the velocity not increasing, for `f3`, -the way it is for `f2` after a minute and a half. - -The rate of change of the velocity is the acceleration. For `f1` this -is zero, for `f2` it is zero as well - when it is defined. However, -for `f3` we see the increase in velocity is positive in the first -minute, but negative in the second minute. This fact relates to the -concavity of the graph. As acceleration is the derivative of velocity, -it is the second derivative of position - the graph we see. Where the -acceleration is *positive*, the position graph will be concave *up*, -where the acceleration is *negative* the graph will be concave -*down*. The point $t=1$ is an inflection point, and -would be felt by most riders. - - - -## Questions - - -###### Question - -Consider this graph: - -```julia; -plot(airyai, -5, 0) # airyai in `SpecialFunctions` loaded with `CalculusWithJulia` -``` - -On what intervals (roughly) is the function positive? - -```julia; hold=true; echo=false -choices=[ -"``(-3.2,-1)``", -"``(-5, -4.2)``", -"``(-5, -4.2)`` and ``(-2.5, 0)``", -"``(-4.2, -2.5)``"] -answ = 3 -radioq(choices, answ) -``` - - -###### Question - -Consider this graph: - -```julia; hold=true; echo=false -import SpecialFunctions: besselj -p = plot(x->besselj(x, 1), -5,-3) -``` - -On what intervals (roughly) is the function negative? - -```julia; hold=true; echo=false -choices=[ -"``(-5.0, -4.0)``", -"``(-25.0, 0.0)``", -"``(-5.0, -4.0)`` and ``(-4, -3)``", -"``(-4.0, -3.0)``"] -answ = 4 -radioq(choices, answ) -``` - -###### Question - -Consider this graph - -```julia; hold=true; echo=false -plot(x->besselj(x, 21), -5,-3) -``` - -On what interval(s) is this function increasing? - - -```julia; hold=true; echo=false -choices=[ -"``(-5.0, -3.8)``", -"``(-3.8, -3.0)``", -"``(-4.7, -3.0)``", -"``(-0.17, 0.17)``" -] -answ = 3 -radioq(choices, answ) -``` - -###### Question - - -Consider this graph - -```julia; hold=true; echo=false -p = plot(x -> 1 / (1+x^2), -3, 3) -``` - -On what interval(s) is this function concave up? - - -```julia; hold=true; echo=false -choices=[ -"``(0.1, 1.0)``", -"``(-3.0, 3.0)``", -"``(-0.6, 0.6)``", -" ``(-3.0, -0.6)`` and ``(0.6, 3.0)``" -] -answ = 4 -radioq(choices, answ) -``` - - -###### Question - -If it is known that: - -* A function $f(x)$ has critical points at $x=-1, 0, 1$ - -* at $-2$ an $-1/2$ the values are: $f'(-2) = 1$ and $f'(-1/2) = -1$. - -What can be concluded? - -```julia; hold=true; echo=false -choices = [ - "Nothing", - "That the critical point at ``-1`` is a relative maximum", - "That the critical point at ``-1`` is a relative minimum", - "That the critical point at ``0`` is a relative maximum", - "That the critical point at ``0`` is a relative minimum" -] -answ = 2 -radioq(choices, answ, keep_order=true) -``` - -###### Question - -Mystery function $f(x)$ has $f'(2) = 0$ and $f''(0) = 2$. What is the *most* you can say about $x=2$? - -```julia; hold=true; echo=false -choices = [ -" ``f(x)`` is continuous at ``2``", -" ``f(x)`` is continuous and differentiable at ``2``", -" ``f(x)`` is continuous and differentiable at ``2`` and has a critical point", -" ``f(x)`` is continuous and differentiable at ``2`` and has a critical point that is a relative minimum by the second derivative test" -] -answ = 3 -radioq(choices, answ, keep_order=true) -``` - - -###### Question - -Find the smallest critical point of $f(x) = x^3 e^{-x}$. - -```julia; echo=false -let - f(x)= x^3*exp(-x) - cps = find_zeros(D(f), -5, 10) - val = minimum(cps) - numericq(val) -end -``` - -###### Question - -How many critical points does $f(x) = x^5 - x + 1$ have? - -```julia; echo=false -let - f(x) = x^5 - x + 1 - cps = find_zeros(D(f), -3, 3) - val = length(cps) - numericq(val) -end -``` - -###### Question - -How many inflection points does $f(x) = x^5 - x + 1$ have? - - -```julia; echo=false -let - f(x) = x^5 - x + 1 - cps = find_zeros(D(f,2), -3, 3) - val = length(cps) - numericq(val) -end -``` - -###### Question - -At $c$, $f'(c) = 0$ and $f''(c) = 1 + c^2$. Is $(c,f(c))$ a relative maximum? ($f$ is a "nice" function.) - -```julia; hold=true; echo=false -choices = [ -"No, it is a relative minimum", -"No, the second derivative test is possibly inconclusive", -"Yes" -] -answ = 1 -radioq(choices, answ) -``` - - -###### Question - -At $c$, $f'(c) = 0$ and $f''(c) = c^2$. Is $(c,f(c))$ a relative minimum? ($f$ is a "nice" function.) - -```julia; hold=true; echo=false -choices = [ -"No, it is a relative maximum", -"No, the second derivative test is possibly inconclusive if ``c=0``, but otherwise yes", -"Yes" -] -answ = 2 -radioq(choices, answ) -``` - -###### Question - -```julia; echo=false -let - f(x) = exp(-x) * sin(pi*x) - plot(D(f), 0, 3) -end -``` - -The graph shows $f'(x)$. Is it possible that $f(x) = e^{-x} \sin(\pi x)$? - -```julia; hold=true; echo=false -yesnoq(true) -``` - -(Plot ``f(x)`` and compare features like critical points, increasing decreasing to that indicated by ``f'`` through the graph.) - - -###### Question - -```julia; hold=true; echo=false -f(x) = x^4 - 3x^3 - 2x + 4 -plot(D(f,2), -2, 4) -``` - - -The graph shows $f'(x)$. Is it possible that $f(x) = x^4 - 3x^3 - 2x + 4$? - -```julia; hold=true; echo=false -yesnoq("no") -``` - - - -###### Question - -```julia; hold=true; echo=false -f(x) = (1+x)^(-2) -plot(D(f,2), 0,2) -``` - - -The graph shows $f''(x)$. Is it possible that $f(x) = (1+x)^{-2}$? - -```julia; hold=true; echo=false -yesnoq("yes") -``` - -###### Question - -```julia; hold=true; echo=false -f_p(x) = (x-1)*(x-2)^2*(x-3)^2 -plot(f_p, 0.75, 3.5) -``` - -This plot shows the graph of $f'(x)$. What is true about the critical points and their characterization? - -```julia; hold=true; echo=false -choices = [ -"The critical points are at ``x=1`` (a relative minimum), ``x=2`` (not a relative extrema), and ``x=3`` (not a relative extrema).", -"The critical points are at ``x=1`` (a relative maximum), ``x=2`` (not a relative extrema), and ``x=3`` (not a relative extrema).", -"The critical points are at ``x=1`` (a relative minimum), ``x=2`` (not a relative extrema), and ``x=3`` (a relative minimum).", -"The critical points are at ``x=1`` (a relative minimum), ``x=2`` (a relative minimum), and ``x=3`` (a relative minimum).", -] -answ=1 -radioq(choices, answ) -``` - -##### Question - -You know $f''(x) = (x-1)^3$. What do you know about $f(x)$? - -```julia; hold=true; echo=false -choices = [ -"The function is concave down over ``(-\\infty, 1)`` and concave up over ``(1, \\infty)``", -"The function is decreasing over ``(-\\infty, 1)`` and increasing over ``(1, \\infty)``", -"The function is negative over ``(-\\infty, 1)`` and positive over ``(1, \\infty)``", -] -answ = 1 -radioq(choices, answ) -``` - -##### Question - -While driving we accelerate to get through a light before it turns red. However, at time $t_0$ a car cuts in front of us and we are forced to break. If $s(t)$ represents position, what is $t_0$: - -```julia; hold=true; echo=false -choices = ["A zero of the function", -"A critical point for the function", -"An inflection point for the function"] -answ = 3 -radioq(choices, answ, keep_order=true) -``` - -###### Question - -The [investopedia](https://www.investopedia.com/terms/i/inflectionpoint.asp) website describes: - -"An **inflection point** is an event that results in a significant change in the progress of a company, industry, sector, economy, or geopolitical situation and can be considered a turning point after which a dramatic change, with either positive or negative results, is expected to result." - -This accurately summarizes how the term is used outside of math books. Does it also describe how the term is used *inside* math books? - -```julia; hold=true, echo=false -choices = ["Yes. Same words, same meaning", - """No, but it is close. An inflection point is when the *acceleration* changes from positive to negative, so if "results" are about how a company's rate of change is changing, then it is in the ballpark."""] -radioq(choices, 2) -``` - -###### Question - -The function ``f(x) = x^3 + x^4`` has a critical point at ``0`` and a second derivative of ``0`` at ``x=0``. Without resorting to the first derivative test, and only considering that *near* ``x=0`` the function ``f(x)`` is essentially ``x^3``, as ``f(x) = x^3(1+x)``, what can you say about whether the critical point is a relative extrema? - -```julia; hold=true; echo=false -choices = ["As ``x^3`` has no extrema at ``x=0``, neither will ``f``", - "As ``x^4`` is of higher degree than ``x^3``, ``f`` will be ``U``-shaped, as ``x^4`` is."] -radioq(choices, 1) -``` diff --git a/CwJ/derivatives/implicit_differentiation.jmd b/CwJ/derivatives/implicit_differentiation.jmd deleted file mode 100644 index 1d2df38..0000000 --- a/CwJ/derivatives/implicit_differentiation.jmd +++ /dev/null @@ -1,1014 +0,0 @@ -# Implicit Differentiation - -This section uses these add-on packages: - -```julia -using CalculusWithJulia -using Plots -using ImplicitPlots -using Roots -using SymPy -``` - - - -```julia; echo=false; results="hidden" -using CalculusWithJulia.WeaveSupport - -const frontmatter = ( - title = "Implicit Differentiation", - description = "Calculus with Julia: Implicit Differentiation", - tags = ["CalculusWithJulia", "derivatives", "implicit differentiation"], -); -nothing -``` - ----- - -## Graphs of equations - - -An **equation** in ``y`` and ``x`` is an algebraic expression involving an -equality with two (or more) variables. An example might be ``x^2 + y^2 -= 1``. - -The **solutions** to an equation in the variables ``x`` and ``y`` are all -points ``(x,y)`` which satisfy the equation. - - -The **graph** of an equation is just the set of solutions to the equation represented in the Cartesian plane. - -With this definition, the graph of a function ``f(x)`` is just the graph of the equation ``y = f(x)``. -In general, graphing an equation is more complicated than graphing a -function. For a function, we know for a given value of ``x`` what -the corresponding value of ``f(x)`` is through evaluation of the function. For -equations, we may have ``0``, ``1`` or more ``y`` values for a given ``x`` and -even more problematic is we may have no rule to find these values. - - -There are a few options for plotting equations in `Julia`. We will use `ImplicitPlots` in this section, but note both `ImplicitEquations` and `IntervalConstraintProgramming` offer alternatives that are a bit more flexible. - - -To plot an implicit equation using `ImplicitPlots` requires expressing the relationship in terms of a function, and then plotting the equation `f(x,y) = 0`. In practice this simply requires all the terms be moved to one side of an equals sign. - -To plot the circle of radius ``2``, or the equations ``x^2 + y^2 = 2^2`` we would move all terms to one side ``x^2 + y^2 - 2^2 = 0`` and then express the left hand side through a function: - -```julia; -f(x,y) = x^2 + y^2 - 2^2 -``` - - -This function is then is passed to the `implicit_plot` function, which works with `Plots` to render the graphic: - -```julia; -implicit_plot(f) -``` - - -!!! note - The `f` is a function of *two* variables, used here to express one side of an equation. `Julia` makes this easy to do - just make sure two variables are in the signature of `f` when it is defined. Using functions like this, we can express our equation in the form ``f(x,y) = c`` or, more generally, as ``f(x,y) = g(x,y)``. The latter of which can be expressed as ``h(x,y) = f(x,y) - g(x,y) = 0``. That is, only the form ``f(x,y)=0`` is needed to represent an equation. - -!!! note - There are two different styles in `Julia` to add simple plot recipes. `ImplicitPlots` adds a new plotting function (`implicit_plot`); alternatively many packages add a new recipe for the generic `plot` method using new types. (For example, `SymPy` has a plot recipe for symbolic types. - - -Of course, more complicated equations are possible and the steps are -similar - only the function definition is more involved. For example, -the [Devils -curve](http://www-groups.dcs.st-and.ac.uk/~history/Curves/Devils.html) -has the form - -```math -y^4 - x^4 + ay^2 + bx^2 = 0 -``` - -Here we draw the curve for a particular choice of ``a`` and ``b``. For -illustration purposes, a narrower viewing window is specified below using `xlims` and `ylims`: - -```julia; hold=true -a,b = -1,2 -f(x,y) = y^4 - x^4 + a*y^2 + b*x^2 -implicit_plot(f; xlims=(-3,3), ylims=(-3,3), legend=false) -``` - -## Tangent lines, implicit differentiation - - -The graph ``x^2 + y^2 = 1`` has well-defined tangent lines at all points except -``(-1,0)`` and ``(0, 1)`` and even at these two points, we could call the vertical lines -``x=-1`` and ``x=1`` tangent lines. However, to recover the slope of these tangent lines would -need us to express ``y`` as a function of ``x`` and then differentiate -that function. Of course, in this example, we would need two functions: -``f(x) = \sqrt{1-x^2}`` and ``g(x) = - \sqrt{1-x^2}`` to do this -completely. - -In general though, we may not be able to solve for ``y`` in terms of ``x``. What then? - -The idea is to *assume* that ``y`` is representable by some function of -``x``. This makes sense, moving on the curve from ``(x,y)`` to some nearby -point, means changing ``x`` will cause some change in ``y``. This -assumption is only made *locally* - basically meaning a complicated -graph is reduced to just a small, well-behaved, section of its graph. - - -With this assumption, asking what ``dy/dx`` is has an obvious meaning - -what is the slope of the tangent line to the graph at ``(x,y)``. (The assumption eliminates the question of what a tangent line would mean when a graph self intersects.) - -The method of implicit differentiation allows this question to be -investigated. It begins by differentiating both sides of the equation -assuming ``y`` is a function of ``x`` to derive a new equation involving ``dy/dx``. - -For example, starting with ``x^2 + y^2 = 1``, differentiating both sides in ``x`` gives: - -```math -2x + 2y\cdot \frac{dy}{dx} = 0. -``` - -The chain rule was used to find ``(d/dx)(y^2) = [y(x)^2]' = 2y \cdot dy/dx``. From this we can solve for ``dy/dx`` (the resulting equations are linear in ``dy/dx``, so can always be solved explicitly): - -```math -\frac{dy}{dx} = -\frac{x}{y}. -``` - -This says the slope of the tangent line depends on the point ``(x,y)`` through the formula ``-x/y``. - - -As a check, we compare to what we would have found had we solved for -``y= \sqrt{1 - x^2}`` (for ``(x,y)`` with ``y \geq 0``). We would have -found: ``dy/dx = 1/2 \cdot 1/\sqrt{1 - x^2} \cdot (-2x)``. Which can be -simplified to ``-x/y``. This should show that the method -above - assuming ``y`` is a function of ``x`` and differentiating - is not -only more general, but can even be easier. - -The name - *implicit differentiation* - comes from the assumption that -``y`` is implicitly defined in terms of ``x``. According to the -[Implicit Function Theorem](http://en.wikipedia.org/wiki/Implicit_function_theorem) the -above method will work provided the curve has sufficient smoothness -near the point ``(x,y)``. - -##### Examples - -Consider the [serpentine](http://www-history.mcs.st-and.ac.uk/Curves/Serpentine.html) equation - -```math -x^2y + a\cdot b \cdot y - a^2 \cdot x = 0, \quad a\cdot b > 0. -``` - -For ``a = 2, b=1`` we have the graph: - -```julia;hold=true -a, b = 2, 1 -f(x,y) = x^2*y + a * b * y - a^2 * x -implicit_plot(f) -``` - -We can see that at each point in the viewing window the tangent line -exists due to the smoothness of the curve. Moreover, at a point -``(x,y)`` the tangent will have slope ``dy/dx`` satisfying: - -```math -2xy + x^2 \frac{dy}{dx} + a\cdot b \frac{dy}{dx} - a^2 = 0. -``` - -Solving, yields: - -```math -\frac{dy}{dx} = \frac{a^2 - 2xy}{ab + x^2}. -``` - - -In particular, the point ``(0,0)`` is always on this graph, and the tangent line will have positive slope ``a^2/(ab) = a/b``. - ----- - -The [eight](http://www-history.mcs.st-and.ac.uk/Curves/Eight.html) curve has representation - -```math -x^4 = a^2(x^2-y^2), \quad a \neq 0. -``` - - -A graph for ``a=3`` shows why it has the name it does: - -```julia;hold=true -a = 3 -f(x,y) = x^4 - a^2*(x^2 - y^2) -implicit_plot(f) -``` - -The tangent line at ``(x,y)`` will have slope, ``dy/dx`` satisfying: - -```math -4x^3 = a^2 \cdot (2x - 2y \frac{dy}{dx}). -``` - -Solving gives: - -```math -\frac{dy}{dx} = -\frac{4x^3 - a^2 \cdot 2x}{a^2 \cdot 2y}. -``` - -The point ``(3,0)`` can be seen to be a solution to the equation and -should have a vertical tangent line. This also is reflected in the -formula, as the denominator is ``a^2\cdot 2 y``, which is ``0`` at this point, whereas the numerator is not. - - -##### Example - -The quotient rule can be hard to remember, unlike the product rule. No -reason to despair, the product rule plus implicit differentiation can -be used to recover the quotient rule. Suppose ``y=f(x)/g(x)``, then we -could also write ``y g(x) = f(x)``. Differentiating implicitly gives: - -```math -\frac{dy}{dx} g(x) + y g'(x) = f'(x). -``` - -Solving for ``dy/dx`` gives: - -```math -\frac{dy}{dx} = \frac{f'(x) - y g'(x)}{g(x)}. -``` - -Not quite what we expect, perhaps, but substituting in ``f(x)/g(x)`` for ``y`` gives us the usual formula: - - -```math -\frac{dy}{dx} = \frac{f'(x) - \frac{f(x)}{g(x)} g'(x)}{g(x)} = \frac{f'(x) g(x) - f(x) g'(x)}{g(x)^2}. -``` - -!!! note - In this example we mix notations using ``g'(x)`` to - represent a derivative of ``g`` with respect to ``x`` and ``dy/dx`` to - represent the derivative of ``y`` with respect to ``x``. This is done to - emphasize the value that we are solving for. It is just a convention - though, we could just as well have used the "prime" notation for each. - - -##### Example: Graphing a tangent line - -Let's see how to add a graph of a tangent line to the graph of an -equation. Tangent lines are tangent at a point, so we need a point to -discuss. - -Returning to the equation for a circle, ``x^2 + y^2 = 1``, let's look at -``(\sqrt{2}/2, - \sqrt{2}/2)``. The derivative is ``-y/x``, so the slope -at this point is ``1``. The line itself has equation ``y = b + m \cdot -(x-a)``. The following represents this in `Julia`: - -```julia;hold=true -F(x,y) = x^2 + y^2 - 1 - -a,b = sqrt(2)/2, -sqrt(2)/2 - -m = -a/b -tl(x) = b + m * (x-a) - -implicit_plot(F, xlims=(-2, 2), ylims=(-2, 2), aspect_ratio=:equal) -plot!(tl) -``` - -We added *both* the implicit plot of ``F`` and the tangent line to the graph at the given point. - - -##### Example - -When we assume ``y`` is a function of ``x``, it may not be feasible to -actually find the function algebraically. However, in many cases one -can be found numerically. Suppose ``G(x,y) = c`` describes the -equation. Then for a fixed ``x``, ``y(x)`` solves ``G(x,y(x))) - c = 0``, so -``y(x)`` is a zero of a known function. As long as we can piece together -which ``y`` goes with which, we can find the function. - -For example, the [folium](http://www-history.mcs.st-and.ac.uk/Curves/Foliumd.html) of Descartes has the equation - -```math -x^3 + y^3 = 3axy. -``` - -Setting ``a=1`` we have the graph: - -```julia; -𝒂 = 1 -G(x,y) = x^3 + y^3 - 3*𝒂*x*y -implicit_plot(G) -``` - -We can solve for the lower curve, ``y``, as a function of ``x``, as follows: - -```julia; -y1(x) = minimum(find_zeros(y->G(x,y), -10, 10)) # find_zeros from `Roots` -``` - -This gives the lower part of the curve, which we can plot with: - -```julia; -plot(y1, -5, 5) -``` - -Though, in this case, the cubic equation would admit a closed-form solution, the approach illustrated applies more generally. - - -## Using SymPy for computation - -`SymPy` can be used to perform implicit differentiation. The three -steps are similar: we assume ``y`` is a function of ``x``, *locally*; -differentiate both sides; solve the result for ``dy/dx``. - - -Let's do so for the [Trident of -Newton](http://www-history.mcs.st-and.ac.uk/Curves/Trident.html), which -is represented in Cartesian form as follows: - -```math -xy = cx^3 + dx^2 + ex + h. -``` - - - -To approach this task in `SymPy`, we begin by defining our symbolic expression. For now, we keep the parameters as symbolic values: - -```julia; -@syms a b c d x y -ex = x*y - (a*c^3 + b*x^2 + c*x + d) -``` - -To express that `y` is a locally a function of `x`, we use a "symbolic function" object: - -```julia; -@syms u() -``` - -The object `u` is the symbolic function, and `u(x)` a symbolic expression -involving a symbolic function. This is what we will use to refer to `y`. - - -Assume ``y`` is a function of ``x``, called `u(x)`, this substitution is just a renaming: - -```julia; -ex1 = ex(y => u(x)) -``` - -At this point, we differentiate in `x`: - -```julia; -ex2 = diff(ex1, x) -``` - -The next step is solve for ``dy/dx`` - the lone answer to the linear equation - which is done as follows: - -```julia; -dydx = diff(u(x), x) -ex3 = solve(ex2, dydx)[1] # pull out lone answer with [1] indexing -``` - -As this represents an answer in terms of `u(x)`, we replace that term with the original variable: - -```julia; -dydx₁ = ex3(u(x) => y) -``` - -If `x` and `y` are the variable names, this function will combine the steps above: - -```julia; -function dy_dx(eqn, x, y) - @syms u() - eqn1 = eqn(y => u(x)) - eqn2 = solve(diff(eqn1, x), diff(u(x), x))[1] - eqn2(u(x) => y) -end -``` - - -Let ``a = b = c = d = 1``, then ``(1,4)`` is a point on the curve. We can draw a tangent line to this point with these commands: - -```julia; -H = ex(a=>1, b=>1, c=>1, d=>1) -x0, y0 = 1, 4 -𝒎 = dydx₁(x=>1, y=>4, a=>1, b=>1, c=>1, d=>1) -implicit_plot(lambdify(H); xlims=(-5,5), ylims=(-5,5), legend=false) -plot!(y0 + 𝒎 * (x-x0)) -``` - -Basically this includes all the same steps as if done "by hand." Some effort could have been saved in plotting, had -values for the parameters been substituted initially, but not doing so -shows their dependence in the derivative. - -!!! warning - The use of `lambdify(H)` is needed to turn the symbolic expression, `H`, into a function. - -!!! note - While `SymPy` itself has the `plot_implicit` function for plotting implicit equations, this works only with `PyPlot`, not `Plots`, so we use the `ImplicitPlots` package in these examples. - - -## Higher order derivatives - -Implicit differentiation can be used to find ``d^2y/dx^2`` or other higher-order derivatives. At each stage, the same technique is applied. The -only "trick" is that some simplifications can be made. - -For example, consider ``x^3 - y^3=3``. To find ``d^2y/dx^2``, we first find ``dy/dx``: - -```math -3x^2 - (3y^2 \frac{dy}{dx}) = 0. -``` - -We could solve for ``dy/dx`` at this point - it always appears as a linear factor - to get: - -```math -\frac{dy}{dx} = \frac{3x^2}{3y^2} = \frac{x^2}{y^2}. -``` - -However, we differentiate the first equation, as we generally try to avoid the quotient rule - -```math -6x - (6y \frac{dy}{dx} \cdot \frac{dy}{dx} + 3y^2 \frac{d^2y}{dx^2}) = 0. -``` - -Again, if must be that ``d^2y/dx^2`` appears as a linear factor, so we can solve for it: - -```math -\frac{d^2y}{dx^2} = \frac{6x - 6y (\frac{dy}{dx})^2}{3y^2}. -``` - -One last substitution for ``dy/dx`` gives: - -```math -\frac{d^2y}{dx^2} = \frac{-6x + 6y (\frac{x^2}{y^2})^2}{3y^2} = -2\frac{x}{y^2} + 2\frac{x^4}{y^5} = 2\frac{x}{y^2}(1 - \frac{x^3}{y^3}) = 2\frac{x}{y^5}(y^3 - x^3) = 2 \frac{x}{y^5}(-3). -``` - -It isn't so pretty, but that's all it takes. - - - -To visualize, we plot implicitly and notice that: - -* as we change quadrants from the third to the fourth to the first the - concavity changes from down to up to down, as the sign of the second - derivative changes from negative to positive to negative; - -* and that at these inflection points, the "tangent" line is vertical - when ``y=0`` and flat when ``x=0``. - -```julia; -K(x,y) = x^3 - y^3 - 3 -implicit_plot(K, xlims=(-3, 3), ylims=(-3, 3)) -``` - - -The same problem can be done symbolically. The steps are similar, though the last step (replacing ``x^3 - y^3`` with ``3``) isn't done without explicitly asking. - -```julia; hold=true -@syms x y u() - -eqn = K(x,y) - 3 -eqn1 = eqn(y => u(x)) -dydx = solve(diff(eqn1,x), diff(u(x), x))[1] # 1 solution -d2ydx2 = solve(diff(eqn1, x, 2), diff(u(x),x, 2))[1] # 1 solution -eqn2 = d2ydx2(diff(u(x), x) => dydx, u(x) => y) -simplify(eqn2) -``` - -## Inverse functions - -As [mentioned](../precalc/inversefunctions.html), an [inverse](http://en.wikipedia.org/wiki/Inverse_function) function for ``f(x)`` is a function ``g(x)`` satisfying: -``y = f(x)`` if and only if ``g(y) = x`` for all ``x`` in the domain of ``f`` and ``y`` in the range of ``f``. - -In short, both ``f \circ g`` and ``g \circ f`` are identify functions on their respective domains. -As inverses are unique, their notation, ``f^{-1}(x)``, reflects the name of the related function. - -The chain rule can be used to give the derivative of an inverse -function when applied to ``f(f^{-1}(x)) = x``. Solving gives, -``[f^{-1}(x)]' = 1 / f'(f^{-1}(x))``. - -This is great - if we can remember the rules. If not, sometimes implicit -differentiation can also help. - -Consider the inverse function for the tangent, which exists when the domain of the tangent function is restricted to ``(-\pi/2, \pi/2)``. The function solves ``y = \tan^{-1}(x)`` or ``\tan(y) = x``. Differentiating this yields: - -```math -\sec(y)^2 \frac{dy}{dx} = 1. -``` - -Or ``dy/dx = 1/\sec^2(y)``. - -But ``\sec(y)^2 = 1 + \tan(y)^2 = 1 + x^2``, as can be seen by right-triangle trigonometry. This yields the formula ``dy/dx = [\tan^{-1}(x)]' = 1 / (1 + x^2)``. - -##### Example - -For a more complicated example, suppose we have a moving trajectory ``(x(t), -y(t))``. The angle it makes with the origin satisfies - -```math -\tan(\theta(t)) = \frac{y(t)}{x(t)}. -``` - -Suppose ``\theta(t)`` can be defined in terms of the inverse to some -function (``\tan^{-1}(x)``). We can differentiate implicitly to find ``\theta'(t)`` in -terms of derivatives of ``y`` and ``x``: - -```math -\sec^2(\theta(t)) \cdot \theta'(t) = \frac{y'(t) x(t) - y(t) x'(t)}{x(t))^2}. -``` - -But ``\sec^2(\theta(t)) = (r(t)/x(t))^2 = (x(t)^2 + y(t)^2) / x(t)^2``, so moving to the other side the secant term gives an explicit, albeit complicated, expression for the derivative of ``\theta`` in terms of the functions ``x`` and ``y``: - -```math -\theta'(t) = \frac{x^2}{x^2(t) + y^2(t)} \cdot \frac{y'(t) x(t) - y(t) x'(t)}{x(t))^2} = \frac{y'(t) x(t) - y(t) x'(t)}{x^2(t) + y^2(t)}. -``` - -This could have been made easier, had we leveraged the result of the previous example. - -#### Example: from physics - -Many problems are best done with implicit derivatives. A video showing -such a problem along with how to do it analytically is -[here](http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/unit-2-applications-of-differentiation/part-b-optimization-related-rates-and-newtons-method/session-32-ring-on-a-string/). - -This video starts with a simple question: - - -> If you have a rope and heavy ring, where will the ring position itself -> due to gravity? - - -Well, suppose you hold the rope in two places, which we can take to be -``(0,0)`` and ``(a,b)``. Then let ``(x,y)`` be all the possible positions of -the ring that hold the rope taught. Then we have this picture: - - -```julia; hold=true; echo=false - - P = (4,1) - Q = (1, -3) - scatter([0,4], [0,1], legend=false, xaxis=nothing, yaxis=nothing) - plot!([0,1,4],[0,-3,1]) - a,b = .05, .25 - ts = range(0, 2pi, length=100) - plot!(1 .+ a*sin.(ts), -3 .+ b*cos.(ts), color=:gold) - annotate!((4-0.3,1,"(a,b)")) - plot!([0,1,1],[0,0,-3], color=:gray, alpha=0.25) - plot!([1,1,4],[0,1,1], color=:gray, alpha=0.25) - Δ = 0.15 - annotate!([(1/2, 0-Δ, "x"), (5/2, 1 - Δ, "a-x"), (1-Δ, -1, "|y|"), (1+Δ, -1, "b-y")]) -``` - - - -Since the length of the rope does not change, we must have for any admissible ``(x,y)`` that: - -```math -L = \sqrt{x^2 + y^2} + \sqrt{(a-x)^2 + (b-y)^2}, -``` - -where these terms come from the two hypotenuses in the figure, as computed through -Pythagorean's theorem. - - -> If we assume that the ring will minimize the value of y subject to -> this constraint, can we solve for y? - - -We create a function to represent the equation: - -```julia; -F₀(x, y, a, b) = sqrt(x^2 + y^2) + sqrt((a-x)^2 + (b-y)^2) -``` - -To illustrate, we need specific values of ``a``, ``b``, and ``L``: -```julia; -𝐚, 𝐛, 𝐋 = 3, 3, 10 # L > sqrt{a^2 + b^2} -F₀(x, y) = F₀(x, y, 𝐚, 𝐛) -``` - -Our values ``(x,y)`` must satisfy ``f(x,y) = L``. Let's graph: - -```julia; -implicit_plot((x,y) -> F₀(x,y) - 𝐋, xlims=(-5, 7), ylims=(-5, 7)) -``` - -The graph is an ellipse, though slightly tilted. - -Okay, now to find the lowest point. This will be when the derivative -is ``0``. We solve by assuming ``y`` is a function of ``x`` called `u`. We have already defined symbolic variables `a`, `b`, `x`, and `y`, here we define `L`: - -```julia; -@syms L -``` - -Then - -```julia; -eqn = F₀(x,y,a,b) - L -``` - -```julia; -eqn_1 = diff(eqn(y => u(x)), x) -eqn_2 = solve(eqn_1, diff(u(x), x))[1] -dydx₂ = eqn_2(u(x) => y) -``` - -We are looking for when the tangent line has ``0`` slope, or when -`dydx` is ``0``: - -```julia; -cps = solve(dydx₂, x) -``` - -There are two answers, as we could guess from the graph, but we want the one for the smallest value of ``y``, which is the second. - -The values of `dydx` depend on any pair ``(x,y)``, but our solution must -also satisfy the equation. That is for our value of ``x``, we need to find -the corresponding ``y``. This should be possible by substituting: - -```julia; -eqn1 = eqn(x => cps[2]) -``` - -We would try to solve `eqn1` for `y` with `solve(eqn1, y)`, but -`SymPy` can't complete this problem. Instead, we will approach this -numerically using `find_zero` from the `Roots` package. We make the above a function of `y` alone - -```julia; -eqn2 = eqn1(a=>3, b=>3, L=>10) -ystar = find_zero(eqn2, -3) -``` - -Okay, now we need to put this value back into our expression for the `x` value and also substitute in for the parameters: - -```julia; -xstar = N(cps[2](y => ystar, a =>3, b => 3, L => 3)) -``` - -Our minimum is at `(xstar, ystar)`, as this graphic shows: - -```julia; - -tl(x) = ystar + 0 * (x- xstar) -implicit_plot((x,y) -> F₀(x,y,3,3) - 10, xlims=(-4, 7), ylims=(-10, 10)) -plot!(tl) -``` - - - -If you watch the video linked to above, you will see that the -surprising fact here is the resting point is such that the angles -formed by the rope are the same. Basically this makes the tension in -both parts of the rope equal, so there is a static position (if not -static, the ring would move and not end in the final position). We can -verify this fact numerically by showing the arctangents of the two -triangles are the same up to a sign: - -```julia; -a0, b0 = 0,0 # the foci of the ellipse are (0,0) and (3,3) -a1, b1 = 3, 3 -atan((b0 - ystar)/(a0 - xstar)) + atan((b1 - ystar)/(a1 - xstar)) # ≈ 0 -``` - - - -Now, were we lucky and just happened to take ``a=3``, ``b = 3`` in such a way to -make this work? Well, no. But convince yourself by doing the above for -different values of ``b``. - - ----- - -In the above, we started with ``F(x,y) = L`` and solved symbolically for ``y=f(x)`` so that ``F(x,f(x)) = L``. Then we took a derivative of ``f(x)`` and set this equal to ``0`` to solve for the minimum ``y`` values. - -Here we try the same problem numerically, using a zero-finding approach to identify ``f(x))``. - -Starting with ``F(x,y) = \sqrt{x^2 + y^2} + \sqrt{(x-1)^2 + (b-2)^2}`` and ``L=3``, we have: - -```julia -F₁(x,y) = F₀(x,y, 1, 2) - 3 # a,b,L = 1,2,3 -implicit_plot(F₁) -``` - -Trying to find the lowest ``y`` value we have from the graph it is near ``x=0.1``. We can do better. - -First, we could try so solve for the ``f`` using `find_zero`. Here is one way: - -```julia -f₀(x) = find_zero(y -> F₁(x, y), 0) -``` - -We use ``0`` as an initial guess, as the ``y`` value is near ``0``. More on this later. We could then just sample many ``x`` values between ``-0.5`` and ``1.5`` and find the one corresponding to the smallest ``t`` value: - -```julia -findmin([f₀(x) for x ∈ range(-0.5, 1.5, length=100)]) -``` - -This shows the smallest value is around ``-0.414`` and occurs in the ``33``rd position of the sampled ``x`` values. Pretty good, but we can do better. We just need to differentiate ``f``, solve for ``f'(x) = 0`` and then put that value back into ``f`` to find the smallest ``y``. - -**However** there is one subtle point. Using automatic differentiation, as implemented in `ForwardDiff`, with `find_zero` requires the `x0` initial value to have a certain type. In this case, the same type as the "`x`" passed into ``f(x)``. So rather than use an initial value of ``0``, we must use an initial value `zero(x)`! (Otherwise, there will be an error "`no method matching Float64(::ForwardDiff.Dual{...`".) - -With this slight modification, we have: - -```julia -f₁(x) = find_zero(y -> F₁(x, y), zero(x)) -plot(f₁', -0.5, 1.5) -``` - -The zero of `f'` is a bit to the right of ``0``, say ``0.2``; we use `find_zero` again to find it: - -```julia -xstar₁ = find_zero(f₁', 0.2) -xstar₁, f₁(xstar₁) -``` - -It is important to note that the above uses of `find_zero` required *good* initial guesses, which we were fortunate enough to identify. - - - -## Questions - -###### Question - -Is ``(1,1)`` on the graph of - -```math -x^2 - 2xy + y^2 = 1? -``` - -```julia; hold=true; echo=false -x,y=1,1 -yesnoq(x^2 - 2x*y + y^2 ==1) -``` - -###### Question - -For the equation - -```math -x^2y + 2y - 4 x = 0, -``` - -if ``x=4``, what is a value for ``y`` such that ``(x,y)`` is a point on the graph of the equation? - -```julia; hold=true; echo=false -@syms x y -eqn = x^2*y + 2y - 4x -val = float(N(solve(subs(eqn, (x,4)), y)[1])) -numericq(val) -``` - - -###### Question - -For the equation - -```math -(y-5)\cdot \cos(4\cdot \sqrt{(x-4)^2 + y^2)} = x\cdot\sin(2\sqrt{x^2 + y^2}) -``` - - -is the point ``(5,0)`` a solution? - -```julia; hold=true; echo=false -yesnoq(false) -``` - -##### Question - -Let ``(x/3)^2 + (y/2)^2 = 1``. Find the slope of the tangent line at the point ``(3/\sqrt{2}, 2/\sqrt{2})``. - -```julia; hold=true; echo=false -@syms x y u() -eqn = (x/3)^2 + (y/2)^2 - 1 -dydx = SymPy.solve(SymPy.diff(SymPy.subs(eqn, y, u(x)), x), SymPy.diff(u(x), x))[1] -val = float(SymPy.N(SymPy.subs(dydx, (u(x), y), (x, 3/sqrt(2)), (y, 2/sqrt(2))))) -numericq(val) -``` - - -###### Question - -The [lame](http://www-history.mcs.st-and.ac.uk/Curves/Lame.html) curves satisfy: - -```math -\left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n = 1. -``` - -An ellipse is when ``n=1``. Take ``n=3``, ``a=1``, and ``b=2``. - -Find a *positive* value of ``y`` when ``x=1/2``. - -```julia; hold=true; echo=false -a,b,n=1,2,3 -val = b*(1 - ((1/2)/a)^n)^(1/n) -numericq(val) -``` - -What expression gives ``dy/dx``? - -```julia; hold=true; echo=false -choices = [ -"`` -(y/x) \\cdot (x/a)^n \\cdot (y/b)^{-n}``", -"``b \\cdot (1 - (x/a)^n)^{1/n}``", -"``-(x/a)^n / (y/b)^n``" -] -answ = 1 -radioq(choices, answ) -``` - -###### Question - -Let ``y - x^2 = -\log(x)``. At the point ``(1/2, 0.9431...)``, the graph has a tangent line. Find this line, then find its intersection point with the ``y`` axes. - -This intersection is: - -```julia; hold=true; echo=false -f(x) = x^2 - log(x) -x0 = 1/2 -tl(x) = f(x0) + f'(x0) * (x - x0) -numericq(tl(0)) -``` - - - -###### Question - -The [witch](http://www-history.mcs.st-and.ac.uk/Curves/Witch.html) of [Agnesi](http://www.maa.org/publications/periodicals/convergence/mathematical-treasures-maria-agnesis-analytical-institutions) is the curve given by the equation - -```math -y(x^2 + a^2) = a^3. -``` - -If ``a=1``, numerically find a a value of ``y`` when ``x=2``. - -```julia; hold=true; echo=false -a = 1 -f(x,y) = y*(x^2 + a^2) - a^3 -val = find_zero(y->f(2,y), 1) -numericq(val) -``` - - -What expression yields ``dy/dx`` for this curve: - -```julia; hold=true; echo=false -choices = [ -"``-2xy/(x^2 + a^2)``", -"``2xy / (x^2 + a^2)``", -"``a^3/(x^2 + a^2)``" -] -answ = 1 -radioq(choices, answ) -``` - -###### Question - -```julia; hold=true; echo=false -### {{{lhopital_35}}} -imgfile = "figures/fcarc-may2016-fig35-350.gif" -caption = """ -Image number 35 from L'Hospitals calculus book (the first). Given a description of the curve, identify the point ``E`` which maximizes the height. -""" -ImageFile(:derivatives, imgfile, caption) -``` - - -The figure above shows a problem appearing in L'Hospital's first calculus book. Given a function defined implicitly by ``x^3 + y^3 = axy`` (with ``AP=x``, ``AM=y`` and ``AB=a``) find the point ``E`` that maximizes the height. In the [AMS feature column](http://www.ams.org/samplings/feature-column/fc-2016-05) this problem is illustrated and solved in the historical manner, with the comment that the concept of implicit differentiation wouldn't have occurred to L'Hospital. - -Using Implicit differentiation, find when ``dy/dx = 0``. - -```julia; hold=true; echo=false -choices = ["``y^2 = 3x/a``", "``y=3x^2/a``", "``y=a/(3x^2)``", "``y^2=a/(3x)``"] -answ = 2 -radioq(choices, answ) -``` - -Substituting the correct value of ``y``, above, into the defining equation gives what value for ``x``: - -```julia; hold=true; echo=false -choices=[ -"``x=(1/2) a 2^{1/2}``", -"``x=(1/3) a 2^{1/3}``", -"``x=(1/2) a^3 3^{1/3}``", -"``x=(1/3) a^2 2^{1/2}``" -] -answ = 2 -radioq(choices, answ) -``` - -###### Question - -For the equation of an ellipse: - -```math -\left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2 = 1, -``` - -compute ``d^2y/dx^2``. Is this the answer? - -```math -\frac{d^2y}{dx^2} = -\frac{b^2}{a^2\cdot y} - \frac{b^4\cdot x^2}{a^4\cdot y^3} = -\frac{1}{y}\frac{b^2}{a^2}(1 + \frac{b^2 x^2}{a^2 y^2}). -``` - -```julia; hold=true; echo=false -yesnoq(true) -``` - -If ``y>0`` is the sign positive or negative? - -```julia; hold=true; echo=false -choices = ["positive", "negative", "Can be both"] -answ = 2 -radioq(choices, answ, keep_order=true) -``` - -If ``x>0`` is the sign positive or negative? - - -```julia; hold=true; echo=false -choices = ["positive", "negative", "Can be both"] -answ = 3 -radioq(choices, answ, keep_order=true) -``` - -When ``x>0``, the graph of the equation is... - -```julia; hold=true; echo=false -choices = ["concave up", "concave down", "both concave up and down"] -answ = 3 -radioq(choices, answ, keep_order=true) -``` - - -## Appendix - -There are other packages in the `Julia` ecosystem that can plot implicit equations. - - -### The ImplicitEquations package - -The `ImplicitEquations` packages can plot equations and inequalities. The use is somewhat similar to the examples above, but the object plotted is a predicate, not a function. These predicates are created with functions like `Eq` or `Lt`. - -For example, the `ImplicitPlots` manual shows this function ``f(x,y) = (x^4 + y^4 - 1) \cdot (x^2 + y^2 - 2) + x^5 \cdot y`` to plot. Using `ImplicitEquations`, this equation would be plotted with: - -```julia; hold=true -using ImplicitEquations -f(x,y) = (x^4 + y^4 - 1) * (x^2 + y^2 - 2) + x^5 * y -r = Eq(f, 0) # the equation f(x,y) = 0 -plot(r) -``` - - -Unlike `ImplicitPlots`, inequalities may be displayed: - -```julia; hold=true -f(x,y) = (x^4 + y^4 - 1) * (x^2 + y^2 - 2) + x^5 * y -r = Lt(f, 0) # the inequality f(x,y) < 0 -plot(r; M=10, N=10) # less blocky -``` - - -The rendered plots look "blocky" due to the algorithm used to plot the -equations. As there is no rule defining ``(x,y)`` pairs to plot, a -search by regions is done. A region is initially labeled -undetermined. If it can be shown that for any value in the region the -equation is true (equations can also be inequalities), the region is -colored black. If it can be shown it will never be true, the region is -dropped. If a black-and-white answer is not clear, the region is -subdivided and each subregion is similarly tested. This continues -until the remaining undecided regions are smaller than some -threshold. Such regions comprise a boundary, and here are also colored -black. Only regions are plotted - not ``(x,y)`` pairs - so the results -are blocky. Pass larger values of ``N=M`` (with defaults of ``8``) to -`plot` to lower the threshold at the cost of longer computation times, -as seen in the last example. - - -### The IntervalConstraintProgramming package - -The `IntervalConstraintProgramming` package also can be used to graph implicit equations. For certain problem descriptions it is significantly faster and makes better graphs. The usage is slightly more involved. We show the commands, but don't run them here, as there are minor conflicts with the `CalculusWithJulia`package. - -We specify a problem using the `@constraint` macro. Using a macro -allows expressions to involve free symbols, so the problem is -specified in an equation-like manner: - -```julia; eval=false -S = @constraint x^2 + y^2 <= 2 -``` - - -The right hand side must be a number. - -The area to plot over must be specified as an `IntervalBox`, basically a pair of intervals. The interval ``[a,b]`` is expressed through `a..b`: - -```julia; eval=false -J = -3..3 -X = IntervalArithmetic.IntervalBox(J, J) -``` - -The `pave` command does the heavy lifting: - -```julia; eval=false -region = IntervalConstraintProgramming.pave(S, X) -``` - -A plot can be made of either the boundary, the interior, or both. - -```julia; eval=false -plot(region.inner) # plot interior; use r.boundary for boundary -``` diff --git a/CwJ/derivatives/lhospitals_rule.jmd b/CwJ/derivatives/lhospitals_rule.jmd deleted file mode 100644 index a7df189..0000000 --- a/CwJ/derivatives/lhospitals_rule.jmd +++ /dev/null @@ -1,770 +0,0 @@ -# L'Hospital's Rule - - -This section uses these add-on packages: - -```julia -using CalculusWithJulia -using Plots -using SymPy - -``` - - -```julia; echo=false; results="hidden" -using CalculusWithJulia.WeaveSupport -using Roots - -fig_size=(800, 600) -const frontmatter = ( - title = "L'Hospital's Rule", - description = "Calculus with Julia: L'Hospital's Rule", - tags = ["CalculusWithJulia", "derivatives", "l'hospital's rule"], -); - -nothing -``` - ----- - -Let's return to limits of the form ``\lim_{x \rightarrow c}f(x)/g(x)`` which have an -indeterminate form of ``0/0`` if both are evaluated at ``c``. The typical -example being the limit considered by Euler: - -```math -\lim_{x\rightarrow 0} \frac{\sin(x)}{x}. -``` - -We know this is ``1`` using a bound from geometry, but might also guess -this is one, as we know from linearization near ``0`` that we have ``\sin(x) \approx x`` or, more specifically: - -```math -\sin(x) = x - \sin(\xi)x^2/2, \quad 0 < \xi < x. -``` - -This would yield: - -```math -\lim_{x \rightarrow 0} \frac{\sin(x)}{x} = \lim_{x\rightarrow 0} \frac{x -\sin(\xi) x^2/2}{x} = \lim_{x\rightarrow 0} 1 + \sin(\xi) \cdot x/2 = 1. -``` - -This is because we know ``\sin(\xi) x/2`` has a limit of ``0``, when ``|\xi| \leq |x|``. - -That doesn't look any easier, as we worried about the error term, but -if just mentally replaced ``\sin(x)`` with ``x`` - which it basically is -near ``0`` - then we can see that the limit should be the same as ``x/x`` -which we know is ``1`` without thinking. - - -Basically, we found that in terms of limits, if both ``f(x)`` and ``g(x)`` -are ``0`` at ``c``, that we *might* be able to just take this limit: -``(f(c) + f'(c) \cdot(x-c)) / (g(c) + g'(c) \cdot (x-c))`` which is just -``f'(c)/g'(c)``. - -Wouldn't that be nice? We could find difficult limits just by -differentiating the top and the bottom at ``c`` (and not use the messy quotient rule). - - -Well, in fact that is more or less true, a fact that dates back to -[L'Hospital](http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule) - -who wrote the first textbook on differential calculus - though this result is -likely due to one of the Bernoulli brothers. - -> *L'Hospital's rule*: Suppose: -> * that ``\lim_{x\rightarrow c+} f(c) =0`` and ``\lim_{x\rightarrow c+} g(c) =0``, -> * that ``f`` and ``g`` are differentiable in ``(c,b)``, and -> * that ``g(x)`` exists and is non-zero for *all* ``x`` in ``(c,b)``, -> then **if** the following limit exists: -> ``\lim_{x\rightarrow c+}f'(x)/g'(x)=L`` it follows that -> ``\lim_{x \rightarrow c+}f(x)/g(x) = L``. - - -That is *if* the right limit of ``f(x)/g(x)`` is indeterminate of the -form ``0/0``, but the right limit of ``f'(x)/g'(x)`` is known, -possibly by simple continuity, then the right limit of ``f(x)/g(x)`` -exists and is equal to that of ``f'(x)/g'(x)``. - -The rule equally applies to *left limits* and *limits* at ``c``. Later it will see there are other generalizations. - -To apply this rule to Euler's example, ``\sin(x)/x``, we just need to consider that: - -```math -L = 1 = \lim_{x \rightarrow 0}\frac{\cos(x)}{1}, -``` - -So, as well, ``\lim_{x \rightarrow 0} \sin(x)/x = 1``. - -This is due to ``\cos(x)`` being continuous at ``0``, so this limit is -just ``\cos(0)/1``. (More importantly, the tangent line expansion of -``\sin(x)`` at ``0`` is ``\sin(0) + \cos(0)x``, so that ``\cos(0)`` is why -this answer is as it is, but we don't need to think in terms of -``\cos(0)``, but rather the tangent-line expansion, which is ``\sin(x) -\approx x``, as ``\cos(0)`` appears as the coefficient. - - -!!! note - In [Gruntz](http://www.cybertester.com/data/gruntz.pdf), in a reference attributed to Speiss, we learn that L'Hospital was a French Marquis who was taught in ``1692`` the calculus of Leibniz by Johann Bernoulli. They made a contract obliging Bernoulli to leave his mathematical inventions to L'Hospital in exchange for a regular compensation. This result was discovered in ``1694`` and appeared in L'Hospital's book of ``1696``. - -##### Examples - -- Consider this limit at ``0``: ``(a^x - 1)/x``. We have ``f(x) =a^x-1`` has - ``f(0) = 0``, so this limit is indeterminate of the form ``0/0``. The - derivative of ``f(x)`` is ``f'(x) = a^x \log(a)`` which has ``f'(0) = \log(a)``. - The derivative of the bottom is also ``1`` at ``0``, so we have: - -```math -\log(a) = \frac{\log(a)}{1} = \frac{f'(0)}{g'(0)} = \lim_{x \rightarrow 0}\frac{f'(x)}{g'(x)} = \lim_{x \rightarrow 0}\frac{f(x)}{g(x)} -= \lim_{x \rightarrow 0}\frac{a^x - 1}{x}. -``` - -!!! note - Why rewrite in the "opposite" direction? Because the theorem's result -- ``L`` is the limit -- is only true if the related limit involving the derivative exists. We don't do this in the following, but did so here to emphasize the need for the limit of the ratio of the derivatives to exist. - -- Consider this limit: - -```math -\lim_{x \rightarrow 0} \frac{e^x - e^{-x}}{x}. -``` - -It too is of the indeterminate form ``0/0``. The derivative of the top -is ``e^x + e^{-x}``, which is ``2`` when ``x=0``, so the ratio of -``f'(0)/g'(0)`` is seen to be ``2`` By continuity, the limit of the ratio of the derivatives is ``2``. Then by L'Hospital's rule, the limit above is -``2``. - - -- Sometimes, L'Hospital's rule must be applied twice. Consider this - limit: - -```math -\lim_{x \rightarrow 0} \frac{\cos(x)}{1 - x^2} -``` - -By L'Hospital's rule *if* this following limit exists, the two will be equal: - -```math -\lim_{x \rightarrow 0} \frac{-\sin(x)}{-2x}. -``` - -But if we didn't guess the answer, we see that this new problem is *also* indeterminate -of the form ``0/0``. So, repeating the process, this new limit will exist and be equal to the following -limit, should it exist: - -```math -\lim_{x \rightarrow 0} \frac{-\cos(x)}{-2} = 1/2. -``` - -As ``L = 1/2`` for this related limit, it must also be the limit of the original problem, by L'Hospital's rule. - - -- Our "intuitive" limits can bump into issues. Take for example the limit of ``(\sin(x)-x)/x^2`` as ``x`` goes to ``0``. Using ``\sin(x) \approx x`` makes this look like ``0/x^2`` which is still indeterminate. (Because the difference is higher order than ``x``.) Using L'Hospitals, says this limit will exist (and be equal) if the following one does: - -```math -\lim_{x \rightarrow 0} \frac{\cos(x) - 1}{2x}. -``` - -This particular limit is indeterminate of the form ``0/0``, so we again try L'Hospital's rule and consider - - -```math -\lim_{x \rightarrow 0} \frac{-\sin(x)}{2} = 0 -``` - -So as this limit exists, working backwards, the original limit in question will also be ``0``. - - -- This example comes from the Wikipedia page. It "proves" a discrete approximation for the second derivative. - -Show if ``f''(x)`` exists at ``c`` and is continuous at ``c``, then - -```math -f''(c) = \lim_{h \rightarrow 0} \frac{f(c + h) - 2f(c) + f(c-h)}{h^2}. -``` - -This will follow from two applications of L'Hospital's rule to the -right-hand side. The first says, the limit on the right is equal to -this limit, should it exist: - -```math -\lim_{h \rightarrow 0} \frac{f'(c+h) - 0 - f'(c-h)}{2h}. -``` - -We have to be careful, as we differentiate in the ``h`` variable, not -the ``c`` one, so the chain rule brings out the minus sign. But again, -as we still have an indeterminate form ``0/0``, this limit will equal the -following limit should it exist: - -```math -\lim_{h \rightarrow 0} \frac{f''(c+h) - 0 - (-f''(c-h))}{2} = -\lim_{c \rightarrow 0}\frac{f''(c+h) + f''(c-h)}{2} = f''(c). -``` - -That last equality follows, as it is assumed that ``f''(x)`` exists at ``c`` and is continuous, that is, ``f''(c \pm h) \rightarrow f''(c)``. - -The expression above finds use when second derivatives are numerically approximated. (The middle expression is the basis of the central-finite difference approximation to the derivative.) - - -* L'Hospital himself was interested in this limit for ``a > 0`` ([math overflow](http://mathoverflow.net/questions/51685/how-did-bernoulli-prove-lh%C3%B4pitals-rule)) - -```math -\lim_{x \rightarrow a} \frac{\sqrt{2a^3\cdot x-x^4} - a\cdot(a^2\cdot x)^{1/3}}{ a - (a\cdot x^3)^{1/4}}. -``` - - -These derivatives can be done by hand, but to avoid any minor mistakes -we utilize `SymPy` taking care to use rational numbers for the -fractional powers, so as not to lose precision through floating point -roundoff: - -```julia; -@syms a::positive x::positive -f(x) = sqrt(2a^3*x - x^4) - a * (a^2*x)^(1//3) -g(x) = a - (a*x^3)^(1//4) -``` - -We can see that at ``x=a`` we have the indeterminate form ``0/0``: - -```julia; -f(a), g(a) -``` - -What about the derivatives? - -```julia; -fp, gp = diff(f(x),x), diff(g(x),x) -fp(x=>a), gp(x=>a) -``` - -Their ratio will not be indeterminate, so the limit in question is just the ratio: - -```julia; -fp(x=>a) / gp(x=>a) -``` - -Of course, we could have just relied on `limit`, which knows about L'Hospital's rule: - -```julia; -limit(f(x)/g(x), x, a) -``` - -## Idea behind L'Hospital's rule - -A first proof of L'Hospital's rule takes advantage of Cauchy's -[generalization](http://en.wikipedia.org/wiki/Mean_value_theorem#Cauchy.27s_mean_value_theorem) -of the mean value theorem to two functions. Suppose ``f(x)`` and ``g(x)`` are -continuous on ``[c,b]`` and differentiable on ``(c,b)``. On -``(c,x)``, ``c < x < b`` there exists a ``\xi`` with ``f'(\xi) \cdot (f(x) - f(c)) = -g'(\xi) \cdot (g(x) - g(c))``. In our formulation, both ``f(c)`` and ``g(c)`` -are zero, so we have, provided we know that ``g(x)`` is non zero, that -``f(x)/g(x) = f'(\xi)/g'(\xi)`` for some ``\xi``, ``c < \xi < c + x``. That -the right-hand side has a limit as ``x \rightarrow c+`` is true by the -assumption that the limit of the ratio of the derivatives exists. (The ``\xi`` -part can be removed by considering it as a composition of a function -going to ``c``.) Thus the right limit of the ratio ``f/g`` is -known. - ----- - -```julia; echo=false; cache=true -let -## {{{lhopitals_picture}}} - -function lhopitals_picture_graph(n) - - g = (x) -> sqrt(1 + x) - 1 - x^2 - f = (x) -> x^2 - ts = range(-1/2, stop=1/2, length=50) - - - a, b = 0, 1/2^n * 1/2 - m = (f(b)-f(a)) / (g(b)-g(a)) - - ## get bounds - tl = (x) -> g(0) + m * (x - f(0)) - - lx = max(find_zero(x -> tl(x) - (-0.05), (-1000, 1000)), -0.6) - rx = min(find_zero(x -> tl(x) - (0.25), (-1000, 1000)), 0.2) - xs = [lx, rx] - ys = map(tl, xs) - - plt = plot(g, f, -1/2, 1/2, legend=false, size=fig_size, xlim=(-.6, .5), ylim=(-.1, .3)) - plot!(plt, xs, ys, color=:orange) - scatter!(plt, [g(a),g(b)], [f(a),f(b)], markersize=5, color=:orange) - plt -end - -caption = L""" - -Geometric interpretation of ``L=\lim_{x \rightarrow 0} x^2 / (\sqrt{1 + -x} - 1 - x^2)``. At ``0`` this limit is indeterminate of the form -``0/0``. The value for a fixed ``x`` can be seen as the slope of a secant -line of a parametric plot of the two functions, plotted as ``(g, -f)``. In this figure, the limiting "tangent" line has ``0`` slope, -corresponding to the limit ``L``. In general, L'Hospital's rule is -nothing more than a statement about slopes of tangent lines. - -""" - -n = 6 -anim = @animate for i=1:n - lhopitals_picture_graph(i) -end - -imgfile = tempname() * ".gif" -gif(anim, imgfile, fps = 1) - - -plotly() - ImageFile(imgfile, caption) -end -``` - -## Generalizations - -L'Hospital's rule generalizes to other indeterminate forms, in -particular the indeterminate form ``\infty/\infty`` can be proved at the same time as ``0/0`` -with a more careful -[proof](http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule#General_proof). - -The value ``c`` in the limit can also be infinite. Consider this case with ``c=\infty``: - -```math -\begin{align*} -\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} &= -\lim_{x \rightarrow 0} \frac{f(1/x)}{g(1/x)} -\end{align*} -``` - -L'Hospital's limit applies as ``x \rightarrow 0``, so we differentiate to get: - -```math -\begin{align*} -\lim_{x \rightarrow 0} \frac{[f(1/x)]'}{[g(1/x)]'} -&= \lim_{x \rightarrow 0} \frac{f'(1/x)\cdot(-1/x^2)}{g'(1/x)\cdot(-1/x^2)}\\ -&= \lim_{x \rightarrow 0} \frac{f'(1/x)}{g'(1/x)}\\ -&= \lim_{x \rightarrow \infty} \frac{f'(x)}{g'(x)}, -\end{align*} -``` - -*assuming* the latter limit exists, L'Hospital's rule assures the equality - -```math -\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} = -\lim_{x \rightarrow \infty} \frac{f'(x)}{g'(x)}, -``` - -##### Examples - - -For example, consider - -```math -\lim_{x \rightarrow \infty} \frac{x}{e^x}. -``` - -We see it is of the form ``\infty/\infty``. Taking advantage of the fact that L'Hospital's rule applies to limits -at ``\infty``, we have that this limit will exist and be equal to this one, -should it exist: - -```math -\lim_{x \rightarrow \infty} \frac{1}{e^x}. -``` - -This limit is, of course, ``0``, as it is of the form ``1/\infty``. It is not -hard to build up from here to show that for any integer value of ``n>0`` -that: - -```math -\lim_{x \rightarrow \infty} \frac{x^n}{e^x} = 0. -``` - -This is an expression of the fact that exponential functions grow faster than polynomial functions. - - -Similarly, powers grow faster than logarithms, as this limit shows, which is indeterminate of the form ``\infty/\infty``: - -```math -\lim_{x \rightarrow \infty} \frac{\log(x)}{x} = -\lim_{x \rightarrow \infty} \frac{1/x}{1} = 0, -``` - -the first equality by L'Hospital's rule, as the second limit exists. - - - -## Other indeterminate forms - - -Indeterminate forms of the type ``0 \cdot \infty``, ``0^0``, -``\infty^\infty``, ``\infty - \infty`` can be re-expressed to be in the -form ``0/0`` or ``\infty/\infty`` and then L'Hospital's theorem can be -applied. - - -###### Example: rewriting ``0 \cdot \infty`` - -What is the limit ``x \log(x)`` as ``x \rightarrow 0+``? The form is ``0\cdot \infty``, rewriting, we see this is just: - -```math -\lim_{x \rightarrow 0+}\frac{\log(x)}{1/x}. -``` - -L'Hospital's rule clearly applies to one-sided limits, as well as two -(our proof sketch used one-sided limits), so this limit will equal the -following, should it exist: - -```math -\lim_{x \rightarrow 0+}\frac{1/x}{-1/x^2} = \lim_{x \rightarrow 0+} -x = 0. -``` - -###### Example: rewriting ``0^0`` - -What is the limit ``x^x`` as ``x \rightarrow 0+``? The expression is of the form ``0^0``, which is indeterminate. (Even though floating point math defines the value as ``1``.) We can rewrite this by taking a log: - -```math -x^x = \exp(\log(x^x)) = \exp(x \log(x)) = \exp(\log(x)/(1/x)). -``` - -Be just saw that ``\lim_{x \rightarrow 0+}\log(x)/(1/x) = 0``. So by the -rules for limits of compositions and the fact that ``e^x`` is -continuous, we see ``\lim_{x \rightarrow 0+} x^x = e^0 = 1``. - - - -##### Example: rewriting ``\infty - \infty`` - -A limit ``\lim_{x \rightarrow c} f(x) - g(x)`` of indeterminate form ``\infty - \infty`` can be reexpressed to be of the from ``0/0`` through the transformation: - -```math -\begin{align*} -f(x) - g(x) &= f(x)g(x) \cdot (\frac{1}{g(x)} - \frac{1}{f(x)}) \\ -&= \frac{\frac{1}{g(x)} - \frac{1}{f(x)}}{\frac{1}{f(x)g(x)}}. -\end{align*} -``` - -Applying this to - -```math -L = \lim_{x \rightarrow 1} \big(\frac{x}{x-1} - \frac{1}{\log(x)}\big) -``` - -We get that ``L`` is equal to the following limit: - -```math -\lim_{x \rightarrow 1} \frac{\log(x) - \frac{x-1}{x}}{\frac{x-1}{x} \log(x)} -= -\lim_{x \rightarrow 1} \frac{x\log(x)-(x-1)}{(x-1)\log(x)} -``` - -In `SymPy` we have: - -```julia -𝒇 = x*log(x) - (x-1) -𝒈 = (x-1)*log(x) -𝒇(1), 𝒈(1) -``` - -L'Hospital's rule applies to the form ``0/0``, so we try: - -```julia -𝒇′ = diff(𝒇, x) -𝒈′ = diff(𝒈, x) -𝒇′(1), 𝒈′(1) -``` - -Again, we get the indeterminate form ``0/0``, so we try again with second derivatives: - -```julia -𝒇′′ = diff(𝒇, x, x) -𝒈′′ = diff(𝒈, x, x) -𝒇′′(1), 𝒈′′(1) -``` - -From this we see the limit is ``1/2``, as could have been done directly: - - -```julia -limit(𝒇/𝒈, x=>1) -``` - -## The assumptions are necessary - -##### Example: the limit existing is necessary - -The following limit is *easily* seen by comparing terms of largest growth: - -```math -1 = \lim_{x \rightarrow \infty} \frac{x - \sin(x)}{x} -``` - -However, the limit of the ratio of the derivatives *does* not exist: - -```math -\lim_{x \rightarrow \infty} \frac{1 - \cos(x)}{1}, -``` - -as the function just oscillates. This shows that L'Hospital's rule does not apply when the limit of the the ratio of the derivatives does not exist. - - -##### Example: the assumptions matter - -This example comes from the thesis of Gruntz to highlight possible issues when computer systems do simplifications. - -Consider: - -```math -\lim_{x \rightarrow \infty} \frac{1/2\sin(2x) +x}{\exp(\sin(x))\cdot(\cos(x)\sin(x)+x)}. -``` - -If we apply L'Hospital's rule using simplification we have: - -```julia -u(x) = 1//2*sin(2x) + x -v(x) = exp(sin(x))*(cos(x)*sin(x) + x) -up, vp = diff(u(x),x), diff(v(x),x) -limit(simplify(up/vp), x => oo) -``` - - -However, this answer is incorrect. The reason being subtle. The simplification cancels a term of ``\cos(x)`` that appears in the numerator and denominator. Before cancellation, we have `vp` will have infinitely many zero's as ``x`` approaches ``\infty`` so L'Hospital's won't apply (the limit won't exist, as every ``2\pi`` the ratio is undefined so the function is never eventually close to some ``L``). - -This ratio has no limit, as it oscillates, as confirmed by `SymPy`: - -```julia -limit(u(x)/v(x), x=> oo) -``` - - -## Questions - -###### Question - -This function ``f(x) = \sin(5x)/x`` is *indeterminate* at ``x=0``. What type? - -```julia; echo=false -lh_choices = [ -"``0/0``", -"``\\infty/\\infty``", -"``0^0``", -"``\\infty - \\infty``", -"``0 \\cdot \\infty``" -] -nothing -``` - -```julia; hold=true; echo=false -answ = 1 -radioq(lh_choices, answ, keep_order=true) -``` - -###### Question - -This function ``f(x) = \sin(x)^{\sin(x)}`` is *indeterminate* at ``x=0``. What type? - -```julia; hold=true; echo=false -answ =3 -radioq(lh_choices, answ, keep_order=true) -``` - -###### Question - -This function ``f(x) = (x-2)/(x^2 - 4)`` is *indeterminate* at ``x=2``. What type? - -```julia; hold=true; echo=false -answ = 1 -radioq(lh_choices, answ, keep_order=true) -``` - -###### Question - -This function ``f(x) = (g(x+h) - g(x-h)) / (2h)`` (``g`` is continuous) is *indeterminate* at ``h=0``. What type? - -```julia; hold=true; echo=false -answ = 1 -radioq(lh_choices, answ, keep_order=true) -``` - -###### Question - -This function ``f(x) = x \log(x)`` is *indeterminate* at ``x=0``. What type? - -```julia; hold=true; echo=false -answ = 5 -radioq(lh_choices, answ, keep_order=true) -``` - - -###### Question - -Does L'Hospital's rule apply to this limit: - -```math -\lim_{x \rightarrow \pi} \frac{\sin(\pi x)}{\pi x}. -``` - -```julia; hold=true; echo=false -choices = [ -"Yes. It is of the form ``0/0``", -"No. It is not indeterminate" -] -answ = 2 -radioq(choices, answ) -``` - -###### Question - -Use L'Hospital's rule to find the limit - -```math -L = \lim_{x \rightarrow 0} \frac{4x - \sin(x)}{x}. -``` - -What is ``L``? - -```julia; hold=true; echo=false -f(x) = (4x - sin(x))/x -L = float(N(limit(f, 0))) -numericq(L) -``` - -###### Question - - -Use L'Hospital's rule to find the limit - -```math -L = \lim_{x \rightarrow 0} \frac{\sqrt{1+x} - 1}{x}. -``` - -What is ``L``? - -```julia; hold=true; echo=false -f(x) = (sqrt(1+x) - 1)/x -L = float(N(limit(f, 0))) -numericq(L) -``` - - - -###### Question - - -Use L'Hospital's rule *one* or more times to find the limit - -```math -L = \lim_{x \rightarrow 0} \frac{x - \sin(x)}{x^3}. -``` - -What is ``L``? - -```julia; hold=true; echo=false -f(x) = (x - sin(x))/x^3 -L = float(N(limit(f, 0))) -numericq(L) -``` - - -###### Question - - -Use L'Hospital's rule *one* or more times to find the limit - -```math -L = \lim_{x \rightarrow 0} \frac{1 - x^2/2 - \cos(x)}{x^3}. -``` - -What is ``L``? - -```julia; hold=true; echo=false -f(x) = (1 - x^2/2 - cos(x))/x^3 -L = float(N(limit(f, 0))) -numericq(L) -``` - -###### Question - - -Use L'Hospital's rule *one* or more times to find the limit - -```math -L = \lim_{x \rightarrow \infty} \frac{\log(\log(x))}{\log(x)}. -``` - -What is ``L``? - -```julia; hold=true; echo=false -f(x) = log(log(x))/log(x) -L = N(limit(f(x), x=> oo)) -numericq(L) -``` - - -###### Question - -By using a common denominator to rewrite this expression, use L'Hospital's rule to find the limit - -```math -L = \lim_{x \rightarrow 0} \frac{1}{x} - \frac{1}{\sin(x)}. -``` - -What is ``L``? - -```julia; hold=true; echo=false -f(x) = 1/x - 1/sin(x) -L = float(N(limit(f, 0))) -numericq(L) -``` - -##### Question - -Use L'Hospital's rule to find the limit - -```math -L = \lim_{x \rightarrow \infty} \log(x)/x -``` - -What is ``L``? - -```julia; hold=true; echo=false -L = float(N(limit(log(x)/x, x=>oo))) -numericq(L) -``` - -##### Question - -Using L'Hospital's rule, does - - -```math -\lim_{x \rightarrow 0+} x^{\log(x)} -``` - -exist? - -Consider ``x^{\log(x)} = e^{\log(x)\log(x)}``. - -```julia; hold=true; echo=false -yesnoq(false) -``` - - -##### Question - -Using L'Hospital's rule, find the limit of - -```math -\lim_{x \rightarrow 1} (2-x)^{\tan(\pi/2 \cdot x)}. -``` - -(Hint, express as ``\exp^{\tan(\pi/2 \cdot x) \cdot \log(2-x)}`` and take the limit of the resulting exponent.) - -```julia; hold=true; echo=false -choices = [ -"``e^{2/\\pi}``", -"``{2\\pi}``", -"``1``", -"``0``", -"It does not exist" -] -answ = 1 -radioq(choices, answ) -``` diff --git a/CwJ/derivatives/linearization.jmd b/CwJ/derivatives/linearization.jmd deleted file mode 100644 index 8570a79..0000000 --- a/CwJ/derivatives/linearization.jmd +++ /dev/null @@ -1,806 +0,0 @@ -# Linearization - -This section uses these add-on packages: - -```julia -using CalculusWithJulia -using Plots -using SymPy -using TaylorSeries -using DualNumbers -``` - -```julia; echo=false; results="hidden" -using CalculusWithJulia.WeaveSupport - -const frontmatter = ( - title = "Linearization", - description = "Calculus with Julia: Linearization", - tags = ["CalculusWithJulia", "derivatives", "linearization"], -); - -nothing -``` - ----- - -The derivative of $f(x)$ has the interpretation as the slope of the -tangent line. The tangent line is the line that best approximates the -function at the point. - -Using the point-slope form of a line, we see that the tangent line to the graph of $f(x)$ at $(c,f(c))$ is given by: - -```math -y = f(c) + f'(c) \cdot (x - c). -``` - -This is written as an equation, though we prefer to work with -functions within `Julia`. Here we write such a function as an -operator - it takes a function `f` and returns a function -representing the tangent line. - -```julia; eval=false -tangent(f, c) = x -> f(c) + f'(c) * (x - c) -``` - -(Recall, the `->` indicates that an anonymous function is being generated.) - -This function along with the `f'` notation for automatic derivatives is defined in the -`CalculusWithJulia` package. - - -We make some graphs with tangent lines: - -```julia;hold=true -f(x) = x^2 -plot(f, -3, 3) -plot!(tangent(f, -1)) -plot!(tangent(f, 2)) -``` - -The graph shows that near the point, the line and function are close, -but this need not be the case away from the point. We can express this informally as - -```math -f(x) \approx f(c) + f'(c) \cdot (x-c) -``` - -with the understanding this applies for $x$ "close" to $c$. - -Usually for the applications herein, instead of ``x`` and ``c`` the two points are ``x+\Delta_x`` and ``x``. This gives: - -> *Linearization*: ``\Delta_y = f(x +\Delta_x) - f(x) \approx f'(x) \Delta_x``, for small ``\Delta_x``. - - -This section gives some implications of this fact and quantifies what -"close" can mean. - -##### Example - -There are several approximations that are well known in physics, due to their widespread usage: - -* That $\sin(x) \approx x$ around $x=0$: - -```julia;hold=true -plot(sin, -pi/2, pi/2) -plot!(tangent(sin, 0)) -``` - -Symbolically: - -```julia; hold=true -@syms x -c = 0 -f(x) = sin(x) -f(c) + diff(f(x),x)(c) * (x - c) -``` - - -* That $\log(1 + x) \approx x$ around $x=0$: - -```julia; hold=true; -f(x) = log(1 + x) -plot(f, -1/2, 1/2) -plot!(tangent(f, 0)) -``` - - -Symbolically: - -```julia; hold=true -@syms x -c = 0 -f(x) = log(1 + x) -f(c) + diff(f(x),x)(c) * (x - c) -``` - -(The `log1p` function implements a more accurate version of this function when numeric values are needed.) - - -* That $1/(1-x) \approx x$ around $x=0$: - -```julia; hold=true; -f(x) = 1/(1-x) -plot(f, -1/2, 1/2) -plot!(tangent(f, 0)) -``` - -Symbolically: - -```julia; hold=true -@syms x -c = 0 -f(x) = 1 / (1 - x) -f(c) + diff(f(x),x)(c) * (x - c) -``` - - - - -* That ``(1+x)^n \approx 1 + nx`` around ``x = 0``. For example, with ``n=5`` - -```julia; hold=true; -n = 5 -f(x) = (1+x)^n # f'(0) = n = n(1+x)^(n-1) at x=0 -plot(f, -1/2, 1/2) -plot!(tangent(f, 0)) -``` - - -Symbolically: - -```julia; hold=true -@syms x, n::real -c = 0 -f(x) = (1 + x)^n -f(c) + diff(f(x),x)(x=>c) * (x - c) -``` - ----- - -In each of these cases, a more complicated non-linear function -is well approximated in a region of interest by a simple linear -function. - -## Numeric approximations - -```julia; hold=true; echo=false -f(x) = sin(x) -a, b = -1/4, pi/2 - -p = plot(f, a, b, legend=false); -plot!(p, x->x, a, b); -plot!(p, [0,1,1], [0, 0, 1], color=:brown); -plot!(p, [1,1], [0, sin(1)], color=:green, linewidth=4); -annotate!(p, collect(zip([1/2, 1+.075, 1/2-1/8], [.05, sin(1)/2, .75], ["Δx", "Δy", "m=dy/dx"]))); -p -``` - -The plot shows the tangent line with slope $dy/dx$ and the actual -change in $y$, $\Delta y$, for some specified $\Delta x$. The small -gap above the sine curve is the error were the value of the sine approximated using the drawn tangent line. We can see that approximating -the value of $\Delta y = \sin(c+\Delta x) - \sin(c)$ with the often -easier to compute $(dy/dx) \cdot \Delta x = f'(c)\Delta x$ - for small enough values of -$\Delta x$ - is not going to be too far off provided $\Delta x$ is not too large. - -This approximation is known as linearization. It can be used both in -theoretical computations and in pratical applications. To see how -effective it is, we look at some examples. - -##### Example - -If $f(x) = \sin(x)$, $c=0$ and $\Delta x= 0.1$ then the values for the actual change in the function values and the value of $\Delta y$ are: - -```julia; -f(x) = sin(x) -c, deltax = 0, 0.1 -f(c + deltax) - f(c), f'(c) * deltax -``` - -The values are pretty close. But what is $0.1$ radians? Lets use degrees. Suppose we have $\Delta x = 10^\circ$: - -```julia; -deltax⁰ = 10*pi/180 -actual = f(c + deltax⁰) - f(c) -approx = f'(c) * deltax⁰ -actual, approx -``` - - -They agree until the third decimal value. The *percentage error* is just $1/2$ a percent: - -```julia; -(approx - actual) / actual * 100 -``` - - -### Relative error or relative change - -The relative error is defined by - -```math -\big| \frac{\text{actual} - \text{approximate}}{\text{actual}} \big|. -``` - -However, typically with linearization, we talk about the *relative change*, not relative error, as the denominator is easier to compute. This is - -```math -\frac{f(x + \Delta_x) - f(x)}{f(x)} = \frac{\Delta_y}{f(x)} \approx -\frac{f'(x) \cdot \Delta_x}{f(x)} -``` - -The *percentage change* multiplies by ``100``. - - -##### Example - -What is the relative change in surface area of a sphere if the radius changes from ``r`` to ``r + dr``? - -We have ``S = 4\pi r^2`` so the approximate relative change, ``dy/S`` is given, using the derivative ``dS/dr = 8\pi r``, by - -```math -\frac{8\pi\cdot r\cdot dr}{4\pi r^2} = 2r\cdot dr. -``` - - -##### Example - -We are traveling ``60`` miles. At ``60`` miles an hour, we will take ``60`` minutes (or one hour). How long will it take at ``70`` miles an hour? (Assume you can't divide, but, instead, can only multiply!) - - -Well the answer is $60/70$ hours or $60/70 \cdot 60$ minutes. But we -can't divide, so we turn this into a multiplication problem via some algebra: - -```math -\frac{60}{70} = \frac{60}{60 + 10} = \frac{1}{1 + 10/60} = \frac{1}{1 + 1/6}. -``` - -Okay, so far no calculator was needed. We wrote $70 = 60 + 10$, as we -know that $60/60$ is just $1$. This almost gets us there. If we really -don't want to divide, we can get an answer by using the tangent line -approximation for $1/(1+x)$ around $x=0$. This is $1/(1+x) \approx 1 - -x$. (You can check by finding that $f'(0) = -1$.) Thus, our answer is -approximately $5/6$ of an hour or 50 minutes. - -How much in error are we? - -```julia; -abs(50 - 60/70*60) / (60/70*60) * 100 -``` - -That's about $3$ percent. Not bad considering we could have done all -the above in our head while driving without taking our eyes off the -road to use the calculator on our phone for a division. - -##### Example - -A ``10``cm by ``10``cm by ``10``cm cube will contain ``1`` liter -(``1000``cm``^3``). In manufacturing such a cube, the side lengths are -actually $10.1$ cm. What will be the volume in liters? Compute this -with a linear approximation to $(10.1)^3$. - -Here $f(x) = x^3$ and we are asked to approximate $f(10.1)$. Letting $c=10$, we have: - -```math -f(c + \Delta) \approx f(c) + f'(c) \cdot \Delta = 1000 + f'(c) \cdot (0.1) -``` - -Computing the derivative can be done easily, we get for our answer: - -```julia; -fp(x) = 3*x^2 -c₀, Delta = 10, 0.1 -approx₀ = 1000 + fp(c₀) * Delta -``` - -This is a relative error as a percent of: - -```julia; -actual₀ = 10.1^3 -(actual₀ - approx₀)/actual₀ * 100 -``` - -The manufacturer may be interested instead in comparing the volume of the actual object to the $1$ liter target. They might use the approximate value for this comparison, which would yield: - -```julia; -(1000 - approx₀)/approx₀ * 100 -``` - -This is off by about $3$ percent. Not so bad for some applications, devastating for others. - - -##### Example: Eratosthenes and the circumference of the earth - -[Eratosthenes](https://en.wikipedia.org/wiki/Eratosthenes) is said to have been the first person to estimate the radius (or by relation the circumference) of the earth. The basic idea is based on the difference of shadows cast by the sun. Suppose Eratosthenes sized the circumference as ``252,000`` *stadia*. Taking ``1``` stadia as ``160`` meters and the actual radius of the earth as ``6378.137`` kilometers, we can convert to see that Eratosthenes estimated the radius as ``6417``. - - -If Eratosthenes were to have estimated the volume of a spherical earth, what would be his approximate percentage change between his estimate and the actual? - -Using ``V = 4/3 \pi r^3`` we get ``V' = 4\pi r^2``: - -```julia -rₑ = 6417 -rₐ = 6378.137 -Δᵣ = rₑ - rₐ -Vₛ(r) = 4/3 * pi * r^3 -Δᵥ = Vₛ'(rₑ) * Δᵣ -Δᵥ / Vₛ(rₑ) * 100 -``` - -##### Example: a simple pendulum - -A *simple* pendulum is comprised of a massless "bob" on a rigid "rod" -of length $l$. The rod swings back and forth making an angle $\theta$ -with the perpendicular. At rest $\theta=0$, here we have $\theta$ swinging with $\lvert\theta\rvert \leq \theta_0$ -for some $\theta_0$. - -According to [Wikipedia](http://tinyurl.com/yz5sz7e) - and many -introductory physics book - while swinging, the angle $\theta$ varies -with time following this equation: - -```math -\theta''(t) + \frac{g}{l} \sin(\theta(t)) = 0. -``` - -That is, the second derivative of $\theta$ is proportional to the sine -of $\theta$ where the proportionality constant involves $g$ from -gravity and the length of the "rod." - -This would be much easier if the second derivative were proportional to the angle $\theta$ and not its sine. - -[Huygens](http://en.wikipedia.org/wiki/Christiaan_Huygens) used the -approximation of $\sin(x) \approx x$, noted above, to say that when -the angle is not too big, we have the pendulum's swing obeying -$\theta''(t) = -g/l \cdot t$. Without getting too involved in why, -we can verify by taking two derivatives that $\theta_0\sin(\sqrt{g/l}\cdot t)$ will be a solution to this -modified equation. - -With this solution, the motion is periodic with constant amplitude (assuming frictionless behaviour), as -the sine function is. More surprisingly, the period is found from $T = -2\pi/(\sqrt{g/l}) = 2\pi \sqrt{l/g}$. It depends on $l$ - longer -"rods" take more time to swing back and forth - but does not depend -on the how wide the pendulum is swinging between (provided $\theta_0$ -is not so big the approximation of $\sin(x) \approx x$ fails). This -latter fact may be surprising, though not to Galileo who discovered -it. - -## Differentials - -The Leibniz notation for a derivative is ``dy/dx`` indicating the -change in ``y`` as ``x`` changes. It proves convenient to decouple -this using *differentials* ``dx`` and ``dy``. What do these notations -mean? They measure change along the tangent line in same way -``\Delta_x`` and ``\Delta_y`` measure change for the function. The differential ``dy`` depends on both ``x`` and ``dx``, it being defined by ``dy=f'(x)dx``. As tangent lines locally represent a function, ``dy`` and ``dx`` are often associated with an *infinitesimal* difference. - -Taking ``dx = \Delta_x``, as in the previous graphic, we can compare ``dy`` -- the change along the tangent line given by ``dy/dx \cdot dx`` -- and ``\Delta_y`` -- the change along the function given by ``f(x + \Delta_x) - f(x)``. The linear approximation, ``f(x + \Delta_x) - f(x)\approx f'(x)dx``, says that - -```math -\Delta_y \approx dy; \quad \text{ when } \Delta_x = dx -``` - - - - -## The error in approximation - -How good is the approximation? Graphically we can see it is pretty -good for the graphs we choose, but are there graphs out there for -which the approximation is not so good? Of course. However, we can -say this (the -[Lagrange](http://en.wikipedia.org/wiki/Taylor%27s_theorem) form of a -more general Taylor remainder theorem): - -> Let ``f(x)`` be twice differentiable on ``I=(a,b)``, ->``f`` is continuous on ``[a,b]``, and -> ``a < c < b``. Then for any ``x`` in ``I``, there exists some value ``\xi`` between ``c`` and ``x`` such that -> ``f(x) = f(c) + f'(c)(x-c) + (f''(\xi)/2)\cdot(x-c)^2``. - - -That is, the error is basically a constant depending on the concavity -of $f$ times a quadratic function centered at $c$. - -For $\sin(x)$ at $c=0$ we get $\lvert\sin(x) - x\rvert = \lvert-\sin(\xi)\cdot x^2/2\rvert$. -Since $\lvert\sin(\xi)\rvert \leq 1$, we must have this bound: -$\lvert\sin(x) - x\rvert \leq x^2/2$. - - -Can we verify? Let's do so graphically: - -```julia; hold=true -h(x) = abs(sin(x) - x) -g(x) = x^2/2 -plot(h, -2, 2, label="h") -plot!(g, -2, 2, label="f") -``` - -The graph shows a tight bound near ``0`` and then a bound over this viewing window. - - -Similarly, for $f(x) = \log(1 + x)$ we have the following at $c=0$: - -```math -f'(x) = 1/(1+x), \quad f''(x) = -1/(1+x)^2. -``` - -So, as $f(c)=0$ and $f'(c) = 1$, we have - -```math -\lvert f(x) - x\rvert \leq \lvert f''(\xi)\rvert \cdot \frac{x^2}{2} -``` - -We see that $\lvert f''(x)\rvert$ is decreasing for $x > -1$. So if $-1 < x < c$ we have - -```math -\lvert f(x) - x\rvert \leq \lvert f''(x)\rvert \cdot \frac{x^2}{2} = \frac{x^2}{2(1+x)^2}. -``` - -And for $c=0 < x$, we have - -```math -\lvert f(x) - x\rvert \leq \lvert f''(0)\rvert \cdot \frac{x^2}{2} = x^2/2. -``` - - -Plotting we verify the bound on ``|\log(1+x)-x|``: - -```julia; hold=true -h(x) = abs(log(1+x) - x) -g(x) = x < 0 ? x^2/(2*(1+x)^2) : x^2/2 -plot(h, -0.5, 2, label="h") -plot!(g, -0.5, 2, label="g") -``` - -Again, we see the very close bound near ``0``, which widens at the edges of the viewing window. - -### Why is the remainder term as it is? - -To see formally why the remainder is as it is, we recall the mean value -theorem in the extended form of Cauchy. Suppose $c=0$, $x > 0$, and let $h(x) = f(x) - (f(0) + -f'(0) x)$ and $g(x) = x^2$. Then we have that there exists a $e$ with -$0 < e < x$ such that - -```math -\text{error} = h(x) - h(0) = (g(x) - g(0)) \frac{h'(e)}{g'(e)} = x^2 \cdot \frac{1}{2} \cdot \frac{f'(e) - f'(0)}{e} = -x^2 \cdot \frac{1}{2} \cdot f''(\xi). -``` - -The value of $\xi$, from the mean value theorem applied to $f'(x)$, -satisfies $0 < \xi < e < x$, so is in $[0,x].$ - -### The big (and small) "oh" - -`SymPy` can find the tangent line expression as a special case of its `series` function (which implements [Taylor series](../taylor_series_polynomials.html)). The `series` function needs an expression to approximate; a variable specified, as there may be parameters in the expression; a value ``c`` for *where* the expansion is taken, with default ``0``; and a number of terms, for this example ``2`` for a constant and linear term. (There is also an optional `dir` argument for one-sided expansions.) - - - - -Here we see the answer provided for $e^{\sin(x)}$: - -```julia; -@syms x -series(exp(sin(x)), x, 0, 2) -``` - -The expression $1 + x$ comes from the fact that `exp(sin(0))` is $1$, and the derivative `exp(sin(0)) * cos(0)` is *also* $1$. But what is the $\mathcal{O}(x^2)$? - -We know the answer is *precisely* $f''(\xi)/2 \cdot x^2$ for some ``\xi``, but were we -only concerned about the scale as $x$ goes to zero -that when ``f''`` is continuous that the error when divided by ``x^2`` goes to some finite value (``f''(0)/2``). More generally, if the error divided by ``x^2`` is *bounded* as ``x`` goes to ``0``, then we say the error is "big oh" of ``x^2``. - - -The [big](http://en.wikipedia.org/wiki/Big_O_notation) "oh" notation, -``f(x) = \mathcal{O}(g(x))``, says that the ratio ``f(x)/g(x)`` is -bounded as ``x`` goes to ``0`` (or some other value ``c``, depending -on the context). A little "oh" (e.g., ``f(x) = \mathcal{o}(g(x))``) -would mean that the limit ``f(x)/g(x)`` would be ``0``, as -``x\rightarrow 0``, a much stronger assertion. - -Big "oh" and little "oh" give us a sense of how good an approximation -is without being bogged down in the details of the exact value. As -such they are useful guides in focusing on what is primary and what is -secondary. Applying this to our case, we have this rough form of the -tangent line approximation valid for functions having a continuous second -derivative at ``c``: - -```math -f(x) = f(c) + f'(c)(x-c) + \mathcal{O}((x-c)^2). -``` - -##### Example: the algebra of tangent line approximations - -Suppose $f(x)$ and $g(x)$ are represented by their tangent lines about $c$, respectively: - - -```math -\begin{align*} -f(x) &= f(c) + f'(c)(x-c) + \mathcal{O}((x-c)^2), \\ -g(x) &= g(c) + g'(c)(x-c) + \mathcal{O}((x-c)^2). -\end{align*} -``` - -Consider the sum, after rearranging we have: - -```math -\begin{align*} -f(x) + g(x) &= \left(f(c) + f'(c)(x-c) + \mathcal{O}((x-c)^2)\right) + \left(g(c) + g'(c)(x-c) + \mathcal{O}((x-c)^2)\right)\\ -&= \left(f(c) + g(c)\right) + \left(f'(c)+g'(c)\right)(x-c) + \mathcal{O}((x-c)^2). -\end{align*} -``` - -The two big "Oh" terms become just one as the sum of a constant times $(x-c)^2$ plus a constant time $(x-c)^2$ is just some other constant times $(x-c)^2$. What we can read off from this is the term multiplying $(x-c)$ is just the derivative of $f(x) + g(x)$ (from the sum rule), so this too is a tangent line approximation. - - -Is it a coincidence that a basic algebraic operation with tangent lines approximations produces a tangent line approximation? Let's try multiplication: - -```math -\begin{align*} -f(x) \cdot g(x) &= [f(c) + f'(c)(x-c) + \mathcal{O}((x-c)^2)] \cdot [g(c) + g'(c)(x-c) + \mathcal{O}((x-c)^2)]\\ -&=[(f(c) + f'(c)(x-c)] \cdot [g(c) + g'(c)(x-c)] + (f(c) + f'(c)(x-c) \cdot \mathcal{O}((x-c)^2)) + g(c) + g'(c)(x-c) \cdot \mathcal{O}((x-c)^2)) + [\mathcal{O}((x-c)^2))]^2\\ -&= [(f(c) + f'(c)(x-c)] \cdot [g(c) + g'(c)(x-c)] + \mathcal{O}((x-c)^2)\\ -&= f(c) \cdot g(c) + [f'(c)\cdot g(c) + f(c)\cdot g'(c)] \cdot (x-c) + [f'(c)\cdot g'(c) \cdot (x-c)^2 + \mathcal{O}((x-c)^2)] \\ -&= f(c) \cdot g(c) + [f'(c)\cdot g(c) + f(c)\cdot g'(c)] \cdot (x-c) + \mathcal{O}((x-c)^2) -\end{align*} -``` - -The big "oh" notation just sweeps up many things including any products of it *and* the term $f'(c)\cdot g'(c) \cdot (x-c)^2$. Again, we see from the product rule that this is just a tangent line approximation for $f(x) \cdot g(x)$. - -The basic mathematical operations involving tangent lines can be computed just using the tangent lines when the desired accuracy is at the tangent line level. This is even true for composition, though there the outer and inner functions may have different "$c$"s. - -Knowing this can simplify the task of finding tangent line approximations of compound expressions. - -For example, suppose we know that at $c=0$ we have these formula where $a \approx b$ is a shorthand for the more formal $a=b + \mathcal{O}(x^2)$: - -```math -\sin(x) \approx x, \quad e^x \approx 1 + x, \quad \text{and}\quad 1/(1+x) \approx 1 - x. -``` - -Then we can immediately see these tangent line approximations about $x=0$: - - -```math -e^x \cdot \sin(x) \approx (1+x) \cdot x = x + x^2 \approx x, -``` - -and - -```math -\frac{\sin(x)}{e^x} \approx \frac{x}{1 + x} \approx x \cdot(1-x) = x-x^2 \approx x. -``` - - -Since $\sin(0) = 0$, we can use these to find the tangent line approximation of - -```math -e^{\sin(x)} \approx e^x \approx 1 + x. -``` - - -Note that $\sin(\exp(x))$ is approximately $\sin(1+x)$ but not approximately $1+x$, as the expansion for $\sin$ about $1$ is not simply $x$. - - -### The TaylorSeries package - -The `TaylorSeries` packages will do these calculations in a manner similar to how `SymPy` transforms a function and a symbolic variable into a symbolic expression. - -For example, we have - -```julia -t = Taylor1(Float64, 1) -``` - -The number type and the order is specified to the constructor. Linearization is order ``1``, other orders will be discussed later. This variable can now be composed with mathematical functions and the linearization of the function will be returned: - -```julia -sin(t), exp(t), 1/(1+t) -``` - -```julia -sin(t)/exp(t), exp(sin(t)) -``` - - -##### Example: Automatic differentiation - -Automatic differentiation (forward mode) essentially uses this technique. A "dual" is introduced which has terms ``a +b\epsilon`` where ``\epsilon^2 = 0``. -The ``\epsilon`` is like ``x`` in a linear expansion, so the `a` coefficient encodes the value and the `b` coefficient reflects the derivative at the value. Numbers are treated like a variable, so their "b coefficient" is a `1`. Here then is how `0` is encoded: - -```julia; -Dual(0, 1) -``` - -Then what is ``\(x)``? It should reflect both ``(\sin(0), \cos(0))`` the latter being the derivative of ``\sin``. We can see this is *almost* what is computed behind the scenes through: - -```julia; hold=true -x = Dual(0, 1) -@code_lowered sin(x) -``` - -This output of `@code_lowered` can be confusing, but this simple case needn't be. Working from the end we see an assignment to a variable named `%7` of `Dual(%3, %6)`. The value of `%3` is `sin(x)` where `x` is the value `0` above. The value of `%6` is `cos(x)` *times* the value `1` above (the `xp`), which reflects the *chain* rule being used. (The derivative of `sin(u)` is `cos(u)*du`.) So this dual number encodes both the function value at `0` and the derivative of the function at `0`.) - -Similarly, we can see what happens to `log(x)` at `1` (encoded by `Dual(1,1)`): - -```julia; hold=true -x = Dual(1, 1) -@code_lowered log(x) -``` - -We can see the derivative again reflects the chain rule, it being given by `1/x * xp` where `xp` acts like `dx` (from assignments `%5` and `%4`). Comparing the two outputs, we see only the assignment to `%4` differs, it reflecting the derivative of the function. - - - - -## Questions - -###### Question - -What is the right linear approximation for $\sqrt{1 + x}$ near $0$? - -```julia; hold=true; echo=false -choices = [ -"``1 + 1/2``", -"``1 + x^{1/2}``", -"``1 + (1/2) \\cdot x``", -"``1 - (1/2) \\cdot x``"] -answ = 3 -radioq(choices, answ) -``` - - -###### Question - - -What is the right linear approximation for $(1 + x)^k$ near $0$? - -```julia; hold=true; echo=false -choices = [ -"``1 + k``", -"``1 + x^k``", -"``1 + k \\cdot x``", -"``1 - k \\cdot x``"] -answ = 3 -radioq(choices, answ) -``` - -###### Question - -What is the right linear approximation for $\cos(\sin(x))$ near $0$? - -```julia; hold=true; echo=false -choices = [ -"``1``", -"``1 + x``", -"``x``", -"``1 - x^2/2``" -] -answ = 1 -radioq(choices, answ) -``` - - -###### Question - -What is the right linear approximation for $\tan(x)$ near $0$? - -```julia; hold=true; echo=false -choices = [ -"``1``", -"``x``", -"``1 + x``", -"``1 - x``" -] -answ = 2 -radioq(choices, answ) -``` - - - -###### Question - -What is the right linear approximation of $\sqrt{25 + x}$ near $x=0$? - -```julia; hold=true; echo=false -choices = [ -"``5 \\cdot (1 + (1/2) \\cdot (x/25))``", -"``1 - (1/2) \\cdot x``", -"``1 + x``", -"``25``" -] -answ = 1 -radioq(choices, answ) -``` - - - -###### Question - - -Let $f(x) = \sqrt{x}$. Find the actual error in approximating $f(26)$ by the -value of the tangent line at $(25, f(25))$ at $x=26$. - -```julia; hold=true; echo=false -tgent(x) = 5 + x/10 -answ = tgent(1) - sqrt(26) -numericq(answ) -``` - -###### Question - -An estimate of some quantity was $12.34$ the actual value was $12$. What was the *percentage error*? - -```julia; hold=true; echo=false -est = 12.34 -act = 12.0 -answ = (est -act)/act * 100 -numericq(answ) -``` - - -###### Question - -Find the percentage error in estimating $\sin(5^\circ)$ by $5 \pi/180$. - -```julia; hold=true; echo=false -tl(x) = x -x0 = 5 * pi/180 -est = x0 -act = sin(x0) -answ = (est -act)/act * 100 -numericq(answ) -``` - -###### Question - -The side length of a square is measured roughly to be $2.0$ cm. The actual length $2.2$ cm. What is the difference in area (in absolute values) as *estimated* by a tangent line approximation. - -```julia; hold=true; echo=false -tl(x) = 4 + 4x -answ = tl(.2) - 4 -numericq(abs(answ)) -``` - - -###### Question - -The [Birthday problem](https://en.wikipedia.org/wiki/Birthday_problem) computes the probability that in a group of ``n`` people, under some assumptions, that no two share a birthday. Without trying to spoil the problem, we focus on the calculus specific part of the problem below: - -```math -\begin{align*} -p -&= \frac{365 \cdot 364 \cdot \cdots (365-n+1)}{365^n} \\ -&= \frac{365(1 - 0/365) \cdot 365(1 - 1/365) \cdot 365(1-2/365) \cdot \cdots \cdot 365(1-(n-1)/365)}{365^n}\\ -&= (1 - \frac{0}{365})\cdot(1 -\frac{1}{365})\cdot \cdots \cdot (1-\frac{n-1}{365}). -\end{align*} -``` - -Taking logarithms, we have ``\log(p)`` is - -```math -\log(1 - \frac{0}{365}) + \log(1 -\frac{1}{365})+ \cdots + \log(1-\frac{n-1}{365}). -``` - -Now, use the tangent line approximation for ``\log(1 - x)`` and the sum formula for ``0 + 1 + 2 + \dots + (n-1)`` to simplify the value of ``\log(p)``: - -```julia; hold=true; echo=false -choices = ["``-n(n-1)/2/365``", - "``-n(n-1)/2\\cdot 365``", - "``-n^2/(2\\cdot 365)``", - "``-n^2 / 2 \\cdot 365``"] -radioq(choices, 1, keep_order=true) -``` - - -If ``n = 10``, what is the approximation for ``p`` (not ``\log(p)``)? - -```julia; hold=true; echo=false -n=10 -val = exp(-n*(n-1)/2/365) -numericq(val) -``` - -If ``n=100``, what is the approximation for ``p`` (not ``\log(p)``? - -```julia; hold=true; echo=false -n=100 -val = exp(-n*(n-1)/2/365) -numericq(val, 1e-2) -``` diff --git a/CwJ/derivatives/mean-value.js b/CwJ/derivatives/mean-value.js deleted file mode 100644 index 05550a5..0000000 --- a/CwJ/derivatives/mean-value.js +++ /dev/null @@ -1,23 +0,0 @@ -// https://jsxgraph.uni-bayreuth.de/wiki/index.php?title=Mean_Value_Theorem -var board = JXG.JSXGraph.initBoard('jsxgraph', {boundingbox: [-5, 10, 7, -6], axis:true}); -board.suspendUpdate(); -var p = []; -p[0] = board.create('point', [-1,-2], {size:2}); -p[1] = board.create('point', [6,5], {size:2}); -p[2] = board.create('point', [-0.5,1], {size:2}); -p[3] = board.create('point', [3,3], {size:2}); -var f = JXG.Math.Numerics.lagrangePolynomial(p); -var graph = board.create('functiongraph', [f,-10, 10]); - -var g = function(x) { - return JXG.Math.Numerics.D(f)(x)-(p[1].Y()-p[0].Y())/(p[1].X()-p[0].X()); -}; - -var r = board.create('glider', [ - function() { return JXG.Math.Numerics.root(g,(p[0].X()+p[1].X())*0.5); }, - function() { return f(JXG.Math.Numerics.root(g,(p[0].X()+p[1].X())*0.5)); }, - graph], {name:' ',size:4,fixed:true}); -board.create('tangent', [r], {strokeColor:'#ff0000'}); -line = board.create('line',[p[0],p[1]],{strokeColor:'#ff0000',dash:1}); - -board.unsuspendUpdate(); diff --git a/CwJ/derivatives/mean_value_theorem.jmd b/CwJ/derivatives/mean_value_theorem.jmd deleted file mode 100644 index af71977..0000000 --- a/CwJ/derivatives/mean_value_theorem.jmd +++ /dev/null @@ -1,710 +0,0 @@ -# The mean value theorem for differentiable functions. - -This section uses these add-on packages: - -```julia -using CalculusWithJulia -using Plots -using Roots - -``` - -```julia; echo=false; results="hidden" -using CalculusWithJulia.WeaveSupport -using Printf -using SymPy - -fig_size = (800, 600) - -const frontmatter = ( - title = "The mean value theorem for differentiable functions.", - description = "Calculus with Julia: The mean value theorem for differentiable functions.", - tags = ["CalculusWithJulia", "derivatives", "the mean value theorem for differentiable functions."], -); - -nothing -``` - ----- - -A function is *continuous* at $c$ if $f(c+h) - f(c) \rightarrow 0$ as $h$ goes to $0$. We can write that as ``f(c+h) - f(x) = \epsilon_h``, with ``\epsilon_h`` denoting a function going to ``0`` as ``h \rightarrow 0``. With this notion, differentiability could be written as ``f(c+h) - f(c) - f'(c)h = \epsilon_h \cdot h``. This is clearly a more demanding requirement that mere continuity at ``c``. - -We defined a function to be *continuous* on an interval $I=(a,b)$ if -it was continuous at each point $c$ in $I$. Similarly, we define a -function to be *differentiable* on the interval $I$ it it is differentiable -at each point $c$ in $I$. - -This section looks at properties of differentiable functions. As there is a more stringent definition, perhaps more properties are a consequence of the definition. - -## Differentiable is more restrictive than continuous. - -Let $f$ be a differentiable function on $I=(a,b)$. We see that -``f(c+h) - f(c) = f'(c)h + \epsilon_h\cdot h = h(f'(c) + \epsilon_h)``. The right hand side will clearly go to ``0`` as ``h\rightarrow 0``, so ``f`` will be continuous. In short: - -> A differentiable function on $I=(a,b)$ is continuous on $I$. - -Is it possible that all continuous functions are differentiable? - -The fact that the derivative is related to the tangent line's slope -might give an indication that this won't be the case - we just need a -function which is continuous but has a point with no tangent line. The -usual suspect is $f(x) = \lvert x\rvert$ at $0$. - -```julia; hold=true -f(x) = abs(x) -plot(f, -1,1) -``` - -We can see formally that the secant line expression will not have a -limit when $c=0$ (the left limit is $-1$, the right limit $1$). But -more insight is gained by looking a the shape of the graph. At the origin, the graph -always is vee-shaped. There is no linear function that approximates this function -well. The function is just not smooth enough, as it has a kink. - - -There are other functions that have kinks. These are often associated -with powers. For example, at $x=0$ this function will not have a -derivative: - -```julia; hold=true; -f(x) = (x^2)^(1/3) -plot(f, -1, 1) -``` - - -Other functions have tangent lines that become vertical. The natural slope would be $\infty$, but this isn't a limiting answer (except in the extended sense we don't apply to the definition of derivatives). A candidate for this case is the cube root function: - -```julia; -plot(cbrt, -1, 1) -``` - -The derivative at $0$ would need to be $+\infty$ to match the -graph. This is implied by the formula for the derivative from the -power rule: $f'(x) = 1/3 \cdot x^{-2/3}$, which has a vertical -asymptote at $x=0$. - - -!!! note - The `cbrt` function is used above, instead of `f(x) = x^(1/3)`, as the - latter is not defined for negative `x`. Though it can be for the exact - power `1/3`, it can't be for an exact power like `1/2`. This means the - value of the argument is important in determining the type of the - output - and not just the type of the argument. Having type-stable - functions is part of the magic to making `Julia` run fast, so `x^c` is - not defined for negative `x` and most floating point exponents. - - -Lest you think that continuous functions always have derivatives -except perhaps at exceptional points, this isn't the case. The -functions used to -[model](http://tinyurl.com/cpdpheb) the -stock market are continuous but have no points where they are -differentiable. - - - - - - -## Derivatives and maxima. - -We have defined an *absolute maximum* of $f(x)$ over an interval to be -a value $f(c)$ for a point $c$ in the interval that is as large as any -other value in the interval. Just specifying a function and an -interval does not guarantee an absolute maximum, but specifying a -*continuous* function and a *closed* interval does, by the extreme value theorem. - -> *A relative maximum*: We say $f(x)$ has a *relative maximum* at $c$ -> if there exists *some* interval $I=(a,b)$ with $a < c < b$ for which -> $f(c)$ is an absolute maximum for $f$ and $I$. - -The difference is a bit subtle, for an absolute maximum the interval -must also be specified, for a relative maximum there just needs to -exist some interval, possibly really small, though it must be bigger -than a point. - -!!! note - A hiker can appreciate the difference. A relative maximum would be the - crest of any hill, but an absolute maximum would be the summit. - -What does this have to do with derivatives? - -[Fermat](http://science.larouchepac.com/fermat/fermat-maxmin.pdf), -perhaps with insight from Kepler, was interested in maxima of -polynomial functions. As a warm up, he considered a line segment $AC$ and a point $E$ -with the task of choosing $E$ so that $(E-A) \times (C-A)$ being a maximum. We might recognize this as -finding the maximum of $f(x) = (x-A)\cdot(C-x)$ for some $A < -C$. Geometrically, we know this to be at the midpoint, as the equation -is a parabola, but Fermat was interested in an algebraic solution that -led to more generality. - -He takes $b=AC$ and $a=AE$. Then the product is $a \cdot (b-a) = -ab - a^2$. He then perturbs this writing $AE=a+e$, then this new -product is $(a+e) \cdot (b - a - e)$. Equating the two, and canceling -like terms gives $be = 2ae + e^2$. He cancels the $e$ and basically -comments that this must be true for all $e$ even as $e$ goes to $0$, -so $b = 2a$ and the value is at the midpoint. - -In a more modern approach, this would be the same as looking at this expression: - -```math -\frac{f(x+e) - f(x)}{e} = 0. -``` - -Working on the left hand side, for non-zero $e$ we can cancel the -common $e$ terms, and then let $e$ become $0$. This becomes a problem -in solving $f'(x)=0$. Fermat could compute the derivative for any -polynomial by taking a limit, a task we would do now by the power -rule and the sum and difference of function rules. - - -This insight holds for other types of functions: - -> If $f(c)$ is a relative maximum then either $f'(c) = 0$ or the -> derivative at $c$ does not exist. - -When the derivative exists, this says the tangent line is flat. (If it -had a slope, then the the function would increase by moving left or -right, as appropriate, a point we pursue later.) - - -For a continuous function $f(x)$, call a point $c$ in the domain of -$f$ where either $f'(c)=0$ or the derivative does not exist a **critical** -**point**. - - -We can combine Bolzano's extreme value theorem with Fermat's insight to get the following: - -> A continuous function on $[a,b]$ has an absolute maximum that occurs -> at a critical point $c$, $a < c < b$, or an endpoint, $a$ or $b$. - -A similar statement holds for an absolute minimum. This gives a -restricted set of places to look for absolute maximum and minimum values - all the critical points and the endpoints. - -It is also the case that all relative extrema occur at a critical point, *however* not all critical points correspond to relative extrema. We will see *derivative tests* that help characterize when that occurs. - -```julia;hold=true; echo=false; -### {{{lhopital_32}}} -imgfile = "figures/lhopital-32.png" -caption = L""" -Image number ``32`` from L'Hopitals calculus book (the first) showing that -at a relative minimum, the tangent line is parallel to the -$x$-axis. This of course is true when the tangent line is well defined -by Fermat's observation. -""" -ImageFile(:derivatives, imgfile, caption) -``` - - -### Numeric derivatives - -The `ForwardDiff` package provides a means to numerically compute derivatives without approximations at a point. In `CalculusWithJulia` this is extended to find derivatives of functions and the `'` notation is overloaded for function objects. Hence these two give nearly identical answers, the difference being only the type of number used: - - -```julia;hold=true -f(x) = 3x^3 - 2x -fp(x) = 9x^2 - 2 -f'(3), fp(3) -``` - - -##### Example - -For the function $f(x) = x^2 \cdot e^{-x}$ find the absolute maximum over the interval $[0, 5]$. - -We have that $f(x)$ is continuous on the closed interval of the -question, and in fact differentiable on $(0,5)$, so any critical point -will be a zero of the derivative. We can check for these with: - - -```julia; -f(x) = x^2 * exp(-x) -cps = find_zeros(f', -1, 6) # find_zeros in `Roots` -``` - -We get $0$ and $2$ are critical points. The endpoints are $0$ and -$5$. So the absolute maximum over this interval is either at $0$, $2$, -or $5$: - -```julia; -f(0), f(2), f(5) -``` - -We see that $f(2)$ is then the maximum. - -A few things. First, `find_zeros` can miss some roots, in particular -endpoints and roots that just touch $0$. We should graph to verify it -didn't. Second, it can be easier sometimes to check the values using -the "dot" notation. If `f`, `a`,`b` are the function and the interval, -then this would typically follow this pattern: - -```julia -a, b = 0, 5 -critical_pts = find_zeros(f', a, b) -f.(critical_pts), f(a), f(b) -``` - -For this problem, we have the left endpoint repeated, but in general -this won't be a point where the derivative is zero. - - -As an aside, the output above is not a single container. To achieve that, the values can be combined before the broadcasting: - -```julia -f.(vcat(a, critical_pts, b)) -``` - - -##### Example - -For the function $g(x) = e^x\cdot(x^3 - x)$ find the absolute maximum over the interval $[0, 2]$. - -We follow the same pattern. Since $f(x)$ is continuous on the closed interval and differentiable on the open interval we know that the absolute maximum must occur at an endpoint ($0$ or $2$) or a critical point where $f'(c)=0$. To solve for these, we have again: - -```julia; -g(x) = exp(x) * (x^3 - x) -gcps = find_zeros(g', 0, 2) -``` - -And checking values gives: - -```julia; -g.(vcat(0, gcps, 2)) -``` - -Here the maximum occurs at an endpoint. The critical point $c=0.67\dots$ -does not produce a maximum value. Rather $f(0.67\dots)$ is an absolute -minimum. - -!!! note - **Absolute minimum** We haven't discussed the parallel problem of - absolute minima over a closed interval. By considering the function - $h(x) = - f(x)$, we see that the any thing true for an absolute - maximum should hold in a related manner for an absolute minimum, in - particular an absolute minimum on a closed interval will only occur - at a critical point or an end point. - -## Rolle's theorem - -Let $f(x)$ be differentiable on $(a,b)$ and continuous on -$[a,b]$. Then the absolute maximum occurs at an endpoint or where the -derivative is ``0`` (as the derivative is always defined). This gives rise to: - -> *[Rolle's](http://en.wikipedia.org/wiki/Rolle%27s_theorem) theorem*: For $f$ differentiable on ``(a,b)`` and continuous on ``[a,b]``, if $f(a)=f(b)$, then there exists some $c$ in $(a,b)$ with $f'(c) = 0$. - -This modest observation opens the door to many relationships between a function and its derivative, as it ties the two together in one statement. - -To see why Rolle's theorem is true, we assume that $f(a)=0$, otherwise -consider $g(x)=f(x)-f(a)$. By the extreme value theorem, there must be -an absolute maximum and minimum. If $f(x)$ is ever positive, then the -absolute maximum occurs in $(a,b)$ - not at an endpoint - so at a -critical point where the derivative is $0$. Similarly if $f(x)$ is -ever negative. Finally, if $f(x)$ is just $0$, then take any $c$ in -$(a,b)$. - -The statement in Rolle's theorem speaks to existence. It doesn't give -a recipe to find $c$. It just guarantees that there is *one* or *more* -values in the interval $(a,b)$ where the derivative is $0$ if we -assume differentiability on $(a,b)$ and continuity on $[a,b]$. - -##### Example - -Let $j(x) = e^x \cdot x \cdot (x-1)$. We know $j(0)=0$ and $j(1)=0$, -so on $[0,1]$. Rolle's theorem -guarantees that we can find *at* *least* one answer (unless numeric -issues arise): - -```julia; -j(x) = exp(x) * x * (x-1) -find_zeros(j', 0, 1) -``` - -This graph illustrates the lone value for $c$ for this problem - -```julia; echo=false -x0 = find_zero(j', (0, 1)) -plot([j, x->j(x0) + 0*(x-x0)], 0, 1) -``` - - -## The mean value theorem - -We are driving south and in one hour cover 70 miles. If the speed -limit is 65 miles per hour, were we ever speeding? We'll we averaged -more than the speed limit so we know the answer is yes, but why? -Speeding would mean our instantaneous speed was more than the speed -limit, yet we only know for sure our *average* speed was more than the -speed limit. The mean value tells us that if some conditions are met, -then at some point (possibly more than one) we must have that our -instantaneous speed is equal to our average speed. - -The mean value theorem is a direct generalization of Rolle's theorem. - -> *Mean value theorem*: Let $f(x)$ be differentiable on $(a,b)$ and -> continuous on $[a,b]$. Then there exists a value $c$ in $(a,b)$ -> where $f'(c) = (f(b) - f(a)) / (b - a)$. - - -This says for any secant line between $a < b$ there will -be a parallel tangent line at some $c$ with $a < c < b$ (all provided $f$ -is differentiable on $(a,b)$ and continuous on $[a,b]$). - - -This graph illustrates the theorem. The orange line is the secant -line. A parallel line tangent to the graph is guaranteed by the mean -value theorem. In this figure, there are two such lines, rendered -using red. - - -```julia; hold=true; echo=false -f(x) = x^3 - x -a, b = -2, 1.75 -m = (f(b) - f(a)) / (b-a) -cps = find_zeros(x -> f'(x) - m, a, b) - -p = plot(f, a-1, b+1, linewidth=3, legend=false) -plot!(x -> f(a) + m*(x-a), a-1, b+1, linewidth=3, color=:orange) -scatter!([a,b], [f(a), f(b)]) - -for cp in cps - plot!(x -> f(cp) + f'(cp)*(x-cp), a-1, b+1, color=:red) -end -p -``` - -Like Rolle's theorem this is a guarantee that something exists, not a -recipe to find it. In fact, the mean value theorem is just Rolle's -theorem applied to: - -```math -g(x) = f(x) - (f(a) + (f(b) - f(a)) / (b-a) \cdot (x-a)) -``` - -That is the function $f(x)$, minus the secant line between $(a,f(a))$ and $(b, f(b))$. - -```julia; hold=true; echo=false -# Need to bring jsxgraph into PLUTO -#caption = """ -#Illustration of the mean value theorem from -#[jsxgraph](https://jsxgraph.uni-bayreuth.de/). -#The polynomial function interpolates the points ``A``,``B``,``C``, and ``D``. -#Adjusting these creates different functions. Regardless of the -#function -- which as a polynomial will always be continuous and -#differentiable -- the slope of the secant line between ``A`` and ``B`` is alway#s matched by **some** tangent line between the points ``A`` and ``B``. -#""" -#JSXGraph(:derivatives, "mean-value.js", caption) -nothing -``` - -```=html -

-``` - -```ojs -//| echo: false -//| output: false - -JXG = require("jsxgraph"); - -board = JXG.JSXGraph.initBoard('jsxgraph', {boundingbox: [-5, 10, 7, -6], axis:true}); -p = [ - board.create('point', [-1,-2], {size:2}), - board.create('point', [6,5], {size:2}), - board.create('point', [-0.5,1], {size:2}), - board.create('point', [3,3], {size:2}) -]; -f = JXG.Math.Numerics.lagrangePolynomial(p); -graph = board.create('functiongraph', [f,-10, 10]); - -g = function(x) { - return JXG.Math.Numerics.D(f)(x)-(p[1].Y()-p[0].Y())/(p[1].X()-p[0].X()); -}; - -r = board.create('glider', [ - function() { return JXG.Math.Numerics.root(g,(p[0].X()+p[1].X())*0.5); }, - function() { return f(JXG.Math.Numerics.root(g,(p[0].X()+p[1].X())*0.5)); }, - graph], {name:' ',size:4,fixed:true}); -board.create('tangent', [r], {strokeColor:'#ff0000'}); -line = board.create('line',[p[0],p[1]],{strokeColor:'#ff0000',dash:1}); -``` - -This interactive example can also be found at [jsxgraph](http://jsxgraph.uni-bayreuth.de/wiki/index.php?title=Mean_Value_Theorem). It shows a cubic polynomial fit to the ``4`` adjustable points labeled A through D. The secant line is drawn between points A and B with a dashed line. A tangent line -- with the same slope as the secant line -- is identified at a point ``(\alpha, f(\alpha))`` where ``\alpha`` is between the points A and B. That this can always be done is a conseuqence of the mean value theorem. - - -##### Example - -The mean value theorem is an extremely useful tool to relate properties of a function with properties of its derivative, as, like Rolle's theorem, it includes both ``f`` and ``f'`` in its statement. - - -For example, suppose we have a function $f(x)$ and we know that the -derivative is **always** $0$. What can we say about the function? - -Well, constant functions have derivatives that are constantly $0$. -But do others? We will see the answer is no: If a function has a zero derivative in ``(a,b)`` it must be a constant. We can readily see that if ``f`` is a polynomial function this is the case, as we can differentiate a polynomial function and this will be zero only if **all** its coefficients are ``0``, which would mean there is no non-constant leading term in the polynomial. But polynomials are not representative of all functions, and so a proof requires a bit more effort. - - -Suppose it is known that $f'(x)=0$ on some interval ``I`` and we take any ``a < b`` in ``I``. Since $f'(x)$ always exists, $f(x)$ is always differentiable, and -hence always continuous. So on $[a,b]$ the conditions of the mean -value theorem apply. That is, there is a $c$ in ``(a,b)`` with $(f(b) - f(a)) / (b-a) = -f'(c) = 0$. But this would imply $f(b) - f(a)=0$. That is $f(x)$ is a -constant, as for any $a$ and $b$, we see $f(a)=f(b)$. - -### The Cauchy mean value theorem - -[Cauchy](http://en.wikipedia.org/wiki/Mean_value_theorem#Cauchy.27s_mean_value_theorem) -offered an extension to the mean value theorem above. Suppose both $f$ -and $g$ satisfy the conditions of the mean value theorem on $[a,b]$ with $g(b)-g(a) \neq 0$, -then there exists at least one $c$ with $a < c < b$ such that - -```math -f'(c) = g'(c) \cdot \frac{f(b) - f(a)}{g(b) - g(a)}. -``` - -The proof follows by considering $h(x) = f(x) - r\cdot g(x)$, with $r$ chosen so that $h(a)=h(b)$. Then Rolle's theorem applies so that there is a $c$ with $h'(c)=0$, so $f'(c) = r g'(c)$, but $r$ can be seen to be $(f(b)-f(a))/(g(b)-g(a))$, which proves the theorem. - -Letting $g(x) = x$ demonstrates that the mean value theorem is a special case. - -##### Example - -Suppose $f(x)$ and $g(x)$ satisfy the Cauchy mean value theorem on -$[0,x]$, $g'(x)$ is non-zero on $(0,x)$, and $f(0)=g(0)=0$. Then we have: - -```math -\frac{f(x) - f(0)}{g(x) - g(0)} = \frac{f(x)}{g(x)} = \frac{f'(c)}{g'(c)}, -``` - -For some $c$ in $[0,x]$. If $\lim_{x \rightarrow 0} f'(x)/g'(x) = L$, -then the right hand side will have a limit of $L$, and hence the left -hand side will too. That is, when the limit exists, we have under -these conditions that $\lim_{x\rightarrow 0}f(x)/g(x) = -\lim_{x\rightarrow 0}f'(x)/g'(x)$. - -This could be used to prove the limit of $\sin(x)/x$ as $x$ goes to -$0$ just by showing the limit of $\cos(x)/1$ is $1$, as is known by -continuity. - -### Visualizing the Cauchy mean value theorem - -The Cauchy mean value theorem can be visualized in terms of a tangent -line and a *parallel* secant line in a similar manner as the mean -value theorem as long as a *parametric* graph is used. A parametric -graph plots the points $(g(t), f(t))$ for some range of $t$. That is, -it graphs *both* functions at the same time. The following illustrates -the construction of such a graph: - -```julia; hold=true; echo=false; cache=true -### {{{parametric_fns}}} - - - -function parametric_fns_graph(n) - f = (x) -> sin(x) - g = (x) -> x - - ns = (1:10)/10 - ts = range(-pi/2, stop=-pi/2 + ns[n] * pi, length=100) - - plt = plot(f, g, -pi/2, -pi/2 + ns[n] * pi, legend=false, size=fig_size, - xlim=(-1.1,1.1), ylim=(-pi/2-.1, pi/2+.1)) - scatter!(plt, [f(ts[end])], [g(ts[end])], color=:orange, markersize=5) - val = @sprintf("% 0.2f", ts[end]) - annotate!(plt, [(0, 1, "t = $val")]) -end -caption = L""" - -Illustration of parametric graph of $(g(t), f(t))$ for $-\pi/2 \leq t -\leq \pi/2$ with $g(x) = \sin(x)$ and $f(x) = x$. Each point on the -graph is from some value $t$ in the interval. We can see that the -graph goes through $(0,0)$ as that is when $t=0$. As well, it must go -through $(1, \pi/2)$ as that is when $t=\pi/2$ - -""" - - -n = 10 -anim = @animate for i=1:n - parametric_fns_graph(i) -end - -imgfile = tempname() * ".gif" -gif(anim, imgfile, fps = 1) - -ImageFile(imgfile, caption) -``` - - -With $g(x) = \sin(x)$ and $f(x) = x$, we can take $I=[a,b] = -[0, \pi/2]$. In the figure below, the *secant line* is drawn in red which -connects $(g(a), f(a))$ with the point $(g(b), f(b))$, and hence -has slope $\Delta f/\Delta g$. The parallel lines drawn show the *tangent* lines with slope $f'(c)/g'(c)$. Two exist for this problem, the mean value theorem guarantees at least one will. - - -```julia; hold=true; echo=false -g(x) = sin(x) -f(x) = x -ts = range(-pi/2, stop=pi/2, length=50) -a,b = 0, pi/2 -m = (f(b) - f(a))/(g(b) - g(a)) -cps = find_zeros(x -> f'(x)/g'(x) - m, -pi/2, pi/2) -c = cps[1] -Delta = (0 + m * (c - 0)) - (g(c)) - -p = plot(g, f, -pi/2, pi/2, linewidth=3, legend=false) -plot!(x -> f(a) + m * (x - g(a)), -1, 1, linewidth=3, color=:red) -scatter!([g(a),g(b)], [f(a), f(b)]) -for c in cps - plot!(x -> f(c) + m * (x - g(c)), -1, 1, color=:orange) -end - -p -``` - - - - -## Questions - -###### Question - -Rolle's theorem is a guarantee of a value, but does not provide a recipe to find it. For the function $1 - x^2$ over the interval $[-5,5]$, find a value $c$ that satisfies the result. - -```julia; hold=true; echo=false -c = 0 -numericq(c) -``` - - -###### Question - -The extreme value theorem is a guarantee of a value, but does not provide a recipe to find it. For the function $f(x) = \sin(x)$ on $I=[0, \pi]$ find a value $c$ satisfying the theorem for an absolute maximum. - -```julia; hold=true; echo=false -c = pi/2 -numericq(c) -``` - -###### Question - -The extreme value theorem is a guarantee of a value, but does not provide a recipe to find it. For the function $f(x) = \sin(x)$ on $I=[\pi, 3\pi/2]$ find a value $c$ satisfying the theorem for an absolute maximum. - -```julia; hold=true; echo=false -c = pi -numericq(c) -``` -###### Question - -The mean value theorem is a guarantee of a value, but does not provide a recipe to find it. For $f(x) = x^2$ on $[0,2]$ find a value of $c$ satisfying the theorem. - -```julia; hold=true; echo=false -c = 1 -numericq(c) -``` - -###### Question - -The Cauchy mean value theorem is a guarantee of a value, but does not provide a recipe to find it. For $f(x) = x^3$ and $g(x) = x^2$ find a value $c$ in the interval $[1, 2]$ - -```julia; hold=true; echo=false -c,x = symbols("c, x", real=true) -val = solve(3c^2 / (2c) - (2^3 - 1^3) / (2^2 - 1^2), c)[1] -numericq(float(val)) -``` - - -###### Question - -Will the function $f(x) = x + 1/x$ satisfy the conditions of the mean value theorem over $[-1/2, 1/2]$? - -```julia; hold=true; echo=false -radioq(["Yes", "No"], 2) -``` - -###### Question - -Just as it is a fact that $f'(x) = 0$ (for all $x$ in $I$) implies -$f(x)$ is a constant, so too is it a fact that if $f'(x) = g'(x)$ that -$f(x) - g(x)$ is a constant. What function would you consider, if you -wanted to prove this with the mean value theorem? - -```julia; hold=true; echo=false -choices = [ -"``h(x) = f(x) - (f(b) - f(a)) / (b - a)``", -"``h(x) = f(x) - (f(b) - f(a)) / (b - a) \\cdot g(x)``", -"``h(x) = f(x) - g(x)``", -"``h(x) = f'(x) - g'(x)``" -] -answ = 3 -radioq(choices, answ) -``` - -###### Question - -Suppose $f''(x) > 0$ on $I$. Why is it impossible that $f'(x) = 0$ at more than one value in $I$? - -```julia; hold=true; echo=false -choices = [ -L"It isn't. The function $f(x) = x^2$ has two zeros and $f''(x) = 2 > 0$", -"By the Rolle's theorem, there is at least one, and perhaps more", -L"By the mean value theorem, we must have $f'(b) - f'(a) > 0$ when ever $b > a$. This means $f'(x)$ is increasing and can't double back to have more than one zero." -] -answ = 3 -radioq(choices, answ) -``` - -###### Question - -Let $f(x) = 1/x$. For $0 < a < b$, find $c$ so that $f'(c) = (f(b) - f(a)) / (b-a)$. - -```julia; hold=true; echo=false -choices = [ -"``c = (a+b)/2``", -"``c = \\sqrt{ab}``", -"``c = 1 / (1/a + 1/b)``", -"``c = a + (\\sqrt{5} - 1)/2 \\cdot (b-a)``" -] -answ = 2 -radioq(choices, answ) -``` - -###### Question - -Let $f(x) = x^2$. For $0 < a < b$, find $c$ so that $f'(c) = (f(b) - f(a)) / (b-a)$. - -```julia; hold=true; echo=false -choices = [ -"``c = (a+b)/2``", -"``c = \\sqrt{ab}``", -"``c = 1 / (1/a + 1/b)``", -"``c = a + (\\sqrt{5} - 1)/2 \\cdot (b-a)``" -] -answ = 1 -radioq(choices, answ) -``` - - - - - -###### Question - -In an example, we used the fact that if $0 < c < x$, for some $c$ given by the mean value theorem and $f(x)$ goes to $0$ as $x$ goes to zero then $f(c)$ will also go to zero. Suppose we say that $c=g(x)$ for some function $c$. - -Why is it known that $g(x)$ goes to $0$ as $x$ goes to zero (from the right)? - -```julia; hold=true; echo=false -choices = [L"The squeeze theorem applies, as $0 < g(x) < x$.", -L"As $f(x)$ goes to zero by Rolle's theorem it must be that $g(x)$ goes to $0$.", -L"This follows by the extreme value theorem, as there must be some $c$ in $[0,x]$."] -answ = 1 -radioq(choices, answ) -``` - -Since $g(x)$ goes to zero, why is it true that if $f(x)$ goes to $L$ as $x$ goes to zero that $f(g(x))$ must also have a limit $L$? - -```julia; hold=true; echo=false -choices = ["It isn't true. The limit must be 0", -L"The squeeze theorem applies, as $0 < g(x) < x$", -"This follows from the limit rules for composition of functions"] -answ = 3 -radioq(choices, answ) -``` diff --git a/CwJ/derivatives/more_zeros.jmd b/CwJ/derivatives/more_zeros.jmd deleted file mode 100644 index c6c1f98..0000000 --- a/CwJ/derivatives/more_zeros.jmd +++ /dev/null @@ -1,530 +0,0 @@ -# Derivative-free alternatives to Newton's method - -This section uses these add-on packages: - -```julia -using CalculusWithJulia -using Plots -using ImplicitEquations -using Roots -using SymPy -``` - -```julia; echo=false; results="hidden" -using CalculusWithJulia.WeaveSupport -const frontmatter = ( - title = "Derivative-free alternatives to Newton's method", - description = "Calculus with Julia: Derivative-free alternatives to Newton's method", - tags = ["CalculusWithJulia", "derivatives", "derivative-free alternatives to newton's method"], -); - -nothing -``` - ----- - -Newton's method is not the only algorithm of its kind for identifying zeros of a function. In this section we discuss some alternatives. - -## The `find_zero(f, x0)` function - -The function `find_zero` from the `Roots` packages provides several different algorithms for finding a zero of a function, including some a derivative-free -algorithms for finding zeros when started with an initial -guess. The default method is similar to Newton's method in that only a good initial -guess is needed. However, the algorithm, while possibly slower in terms of -function evaluations and steps, is engineered to be a bit more -robust to the choice of initial estimate than Newton's method. (If it -finds a bracket, it will use a bisection algorithm which is guaranteed to -converge, but can be slower to do so.) Here we see how to call the -function: - -```julia; -f(x) = cos(x) - x -x₀ = 1 -find_zero(f, x₀) -``` - -Compare to this related call which uses the bisection method: - -```julia; -find_zero(f, (0, 1)) ## [0,1] must be a bracketing interval -``` - -For this example both give the same answer, but the bisection method -is a bit less convenient as a bracketing interval must be pre-specified. - -## The secant method - -The default `find_zero` method above uses a secant-like method unless a bracketing method is found. The secant method is historic, dating back over ``3000`` years. Here we discuss the secant method in a more general framework. - -One way to view Newton's method is through the inverse of ``f`` (assuming it exists): if ``f(\alpha) = 0`` then ``\alpha = f^{-1}(0)``. - -If ``f`` has a simple zero at ``\alpha`` and is locally invertible (that is some ``f^{-1}`` exists) then the update step for Newton's method can be identified with: - -* fitting a polynomial to the local inverse function of ``f`` going through through the point ``(f(x_0),x_0)``, -* and matching the slope of ``f`` at the same point. - -That is, we can write ``g(y) = h_0 + h_1 (y-f(x_0))``. Then ``g(f(x_0)) = x_0 = h_0``, so ``h_0 = x_0``. From ``g'(f(x_0)) = 1/f'(x_0)``, we get ``h_1 = 1/f'(x_0)``. That is, ``g(y) = x_0 + (y-f(x_0))/f'(x_0)``. At ``y=0,`` we get the update step ``x_1 = g(0) = x_0 - f(x_0)/f'(x_0)``. - - -A similar viewpoint can be used to create derivative-free methods. - - -For example, the [secant method](https://en.wikipedia.org/wiki/Secant_method) can be seen as the result of fitting a degree-``1`` polynomial approximation for ``f^{-1}`` through two points ``(f(x_0),x_0)`` and ``(f(x_1), x_1)``. - - -Again, expressing this approximation as ``g(y) = h_0 + h_1(y-f(x_1))`` leads to ``g(f(x_1)) = x_1 = h_0``. -Substituting ``f(x_0)`` gives ``g(f(x_0)) = x_0 = x_1 + h_1(f(x_0)-f(x_1))``. Solving for ``h_1`` leads to ``h_1=(x_1-x_0)/(f(x_1)-f(x_0))``. Then ``x_2 = g(0) = x_1 + (x_1-x_0)/(f(x_1)-f(x_0)) \cdot f(x_1)``. This is the first step of the secant method: - -```math -x_{n+1} = x_n - f(x_n) \frac{x_n - x_{n-1}}{f(x_n) - f(x_{n-1})}. -``` - -That is, where the next step of Newton's method comes from the intersection of the tangent line at ``x_n`` with the ``x``-axis, the next step of the secant method comes from the intersection of the secant line defined by ``x_n`` and ``x_{n-1}`` with the ``x`` axis. That is, the secant method simply replaces ``f'(x_n)`` with the slope of the secant line between ``x_n`` and ``x_{n-1}``. - - -We code the update step as `λ2`: - -```julia; -λ2(f0,f1,x0,x1) = x1 - f1 * (x1-x0) / (f1-f0) -``` - -Then we can run a few steps to identify the zero of sine starting at ``3`` and ``4`` - -```julia; hold=true; term=true -x0,x1 = 4,3 -f0,f1 = sin.((x0,x1)) -@show x1,f1 - -x0,x1 = x1, λ2(f0,f1,x0,x1) -f0,f1 = f1, sin(x1) -@show x1,f1 - -x0,x1 = x1, λ2(f0,f1,x0,x1) -f0,f1 = f1, sin(x1) -@show x1,f1 - -x0,x1 = x1, λ2(f0,f1,x0,x1) -f0,f1 = f1, sin(x1) -@show x1,f1 - -x0,x1 = x1, λ2(f0,f1,x0,x1) -f0,f1 = f1, sin(x1) -x1,f1 -``` - - -Like Newton's method, the secant method coverges quickly for this problem (though its rate is less than the quadratic rate of Newton's method). - - -This method is included in `Roots` as `Secant()` (or `Order1()`): - -```julia; -find_zero(sin, (4,3), Secant()) -``` - - - -Though the derivative is related to the slope of the secant line, that is in the limit. The convergence of the secant method is not as fast as Newton's method, though at each step of the secant method, only one new function evaluation is needed, so it can be more efficient for functions that are expensive to compute or differentiate. - - -Let ``\epsilon_{n+1} = x_{n+1}-\alpha``, where ``\alpha`` is assumed to be the *simple* zero of ``f(x)`` that the secant method converges to. A [calculation](https://math.okstate.edu/people/binegar/4513-F98/4513-l08.pdf) shows that - -```math -\begin{align*} -\epsilon_{n+1} &\approx \frac{x_n-x_{n-1}}{f(x_n)-f(x_{n-1})} \frac{(1/2)f''(\alpha)(e_n-e_{n-1})}{x_n-x_{n-1}} \epsilon_n \epsilon_{n-1}\\ -& \approx \frac{f''(\alpha)}{2f'(\alpha)} \epsilon_n \epsilon_{n-1}\\ -&= C \epsilon_n \epsilon_{n-1}. -\end{align*} -``` - -The constant `C` is similar to that for Newton's method, and reveals potential troubles for the secant method similar to those of Newton's method: a poor initial guess (the initial error is too big), the second derivative is too large, the first derivative too flat near the answer. - -Assuming the error term has the form ``\epsilon_{n+1} = A|\epsilon_n|^\phi`` and substituting into the above leads to the equation - -```math -\frac{A^{1-1/\phi}}{C} = |\epsilon_n|^{1 - \phi +1/\phi}. -``` - -The left side being a constant suggests ``\phi`` solves: ``1 - \phi + 1/\phi = 0`` or ``\phi^2 -\phi - 1 = 0``. The solution is the golden ratio, ``(1 + \sqrt{5})/2 \approx 1.618\dots``. - - -### Steffensen's method - -Steffensen's method is a secant-like method that converges with ``|\epsilon_{n+1}| \approx C |\epsilon_n|^2``. The secant is taken between the points ``(x_n,f(x_n))`` and ``(x_n + f(x_n), f(x_n + f(x_n))``. Like Newton's method this requires ``2`` function evaluations per step. Steffensen's is implemented through `Roots.Steffensen()`. Steffensen's method is more sensitive to the initial guess than other methods, so in practice must be used with care, though it is a starting point for many higher-order derivative-free methods. - - - -## Inverse quadratic interpolation - -Inverse quadratic interpolation fits a quadratic polynomial through three points, not just two like the Secant method. The third being ``(f(x_2), x_2)``. - - -For example, here is the inverse quadratic function, ``g(y)``, going through three points marked with red dots. The blue dot is found from ``(g(0), 0)``. - -```julia; hold=true; echo=false - -a,b,c = 1,2,3 -fa,fb,fc = -1,1/4,1 -g(y) = (y-fb)*(y-fa)/(fc-fb)/(fc-fa)*c + (y-fc)*(y-fa)/(fb-fc)/(fb-fa)*b + (y-fc)*(y-fb)/(fa-fc)/(fa-fb)*a -ys = range(-2,2, length=100) -xs = g.(ys) -plot(xs, ys, legend=false) -scatter!([a,b,c],[fa,fb,fc], color=:red, markersize=5) -scatter!([g(0)],[0], color=:blue, markersize=5) -plot!(zero, color=:blue) -``` - - -Here we use `SymPy` to identify the degree-``2`` polynomial as a function of ``y``, then evaluate it at ``y=0`` to find the next step: - - -```julia -@syms y hs[0:2] xs[0:2] fs[0:2] -H(y) = sum(hᵢ*(y - fs[end])^i for (hᵢ,i) ∈ zip(hs, 0:2)) - -eqs = [H(fᵢ) ~ xᵢ for (xᵢ, fᵢ) ∈ zip(xs, fs)] -ϕ = solve(eqs, hs) -hy = subs(H(y), ϕ) -``` - -The value of `hy` at ``y=0`` yields the next guess based on the past three, and is given by: - -```julia; -q⁻¹ = hy(y => 0) -``` - - -Though the above can be simplified quite a bit when computed by hand, here we simply make this a function with `lambdify` which we will use below. - -```julia; -λ3 = lambdify(q⁻¹) # fs, then xs -``` - -(`SymPy`'s `lambdify` function, by default, picks the order of its argument lexicographically, in this case they will be the `f` values then the `x` values.) - -An inverse quadratic step is utilized by Brent's method, as possible, to yield a rapidly convergent bracketing algorithm implemented as a default zero finder in many software languages. `Julia`'s `Roots` package implements the method in `Roots.Brent()`. An inverse cubic interpolation is utilized by [Alefeld, Potra, and Shi](https://dl.acm.org/doi/10.1145/210089.210111) which gives an asymptotically even more rapidly convergent algorithm then Brent's (implemented in `Roots.AlefeldPotraShi()` and also `Roots.A42()`). This is used as a finishing step in many cases by the default hybrid `Order0()` method of `find_zero`. - -In a bracketing algorithm, the next step should reduce the size of the bracket, so the next iterate should be inside the current bracket. However, quadratic convergence does not guarantee this to happen. As such, sometimes a subsitute method must be chosen. - -[Chandrapatla's](https://www.google.com/books/edition/Computational_Physics/cC-8BAAAQBAJ?hl=en&gbpv=1&pg=PA95&printsec=frontcover) method, is a bracketing method utilizing an inverse quadratic step as the centerpiece. The key insight is the test to choose between this inverse quadratic step and a bisection step. This is done in the following based on values of ``\xi`` and ``\Phi`` defined within: - -```julia; -function chandrapatla(f, u, v, λ; verbose=false) - a,b = promote(float(u), float(v)) - fa,fb = f(a),f(b) - @assert fa * fb < 0 - - if abs(fa) < abs(fb) - a,b,fa,fb = b,a,fb,fa - end - - c, fc = a, fa - - maxsteps = 100 - for ns in 1:maxsteps - - Δ = abs(b-a) - m, fm = (abs(fa) < abs(fb)) ? (a, fa) : (b, fb) - ϵ = eps(m) - if Δ ≤ 2ϵ - return m - end - @show m,fm - iszero(fm) && return m - - ξ = (a-b)/(c-b) - Φ = (fa-fb)/(fc-fb) - - if Φ^2 < ξ < 1 - (1-Φ)^2 - xt = λ(fa,fc,fb, a,c,b) # inverse quadratic - else - xt = a + (b-a)/2 - end - - ft = f(xt) - - isnan(ft) && break - - if sign(fa) == sign(ft) - c,fc = a,fa - a,fa = xt,ft - else - c,b,a = b,a,xt - fc,fb,fa = fb,fa,ft - end - - verbose && @show ns, a, fa - - end - error("no convergence: [a,b] = $(sort([a,b]))") -end -``` - -Like bisection, this method ensures that ``a`` and ``b`` is a bracket, but it moves ``a`` to the newest estimate, so does not maintain that ``a < b`` throughout. - -We can see it in action on the sine function. Here we pass in ``\lambda``, but in a real implementation (as in `Roots.Chandrapatla()`) we would have programmed the algorithm to compute the inverse quadratic value. - -```julia; term=true -chandrapatla(sin, 3, 4, λ3, verbose=true) -``` - - - -The condition `Φ^2 < ξ < 1 - (1-Φ)^2` can be visualized. Assume `a,b=0,1`, `fa,fb=-1/2,1`, Then `c < a < b`, and `fc` has the same sign as `fa`, but what values of `fc` will satisfy the inequality? - -```julia; -ξ(c,fc) = (a-b)/(c-b) -Φ(c,fc) = (fa-fb)/(fc-fb) -Φl(c,fc) = Φ(c,fc)^2 -Φr(c,fc) = 1 - (1-Φ(c,fc))^2 -a,b = 0, 1 -fa,fb = -1/2, 1 -region = Lt(Φl, ξ) & Lt(ξ,Φr) -plot(region, xlims=(-2,a), ylims=(-3,0)) -``` - -When `(c,fc)` is in the shaded area, the inverse quadratic step is chosen. We can see that `fc < fa` is needed. - -For these values, this area is within the area where a implicit quadratic step will result in a value between `a` and `b`: - -```julia; -l(c,fc) = λ3(fa,fb,fc,a,b,c) -region₃ = ImplicitEquations.Lt(l,b) & ImplicitEquations.Gt(l,a) -plot(region₃, xlims=(-2,0), ylims=(-3,0)) -``` - -There are values in the parameter space where this does not occur. - -## Tolerances - -The `chandrapatla` algorithm typically waits until `abs(b-a) <= 2eps(m)` (where ``m`` is either ``b`` or ``a`` depending on the size of ``f(a)`` and ``f(b)``) is satisfied. Informally this means the algorithm stops when the two bracketing values are no more than a small amount apart. What is a "small amount?" - -To understand, we start with the fact that floating point numbers are an approximation to real numbers. - -Floating point numbers effectively represent a number in scientific -notation in terms of - -* a sign (plus or minus) , -* a *mantissa* (a number in ``[1,2)``, in binary ), and -* an exponent (to represent a power of ``2``). - -The mantissa is of the form `1.xxxxx...xxx` where there are ``m`` -different `x`s each possibly a `0` or `1`. The `i`th `x` indicates if the term `1/2^i` should be -included in the value. The mantissa is the sum of `1` plus the -indicated values of `1/2^i` for `i` in `1` to `m`. So the last `x` represents if `1/2^m` should be -included in the sum. As such, the -mantissa represents a discrete set of values, separated by `1/2^m`, as -that is the smallest difference possible. - -For example if `m=2` then the possible value for the mantissa are `11 => 1 + 1/2 + 1/4 = 7/4`, -`10 => 1 + 1/2 = 6/4`, `01 => 1 + 1/4 = 5/4`. and `00 => 1 = 4/4`, values separated by `1/4 = 1/2^m`. - -For ``64``-bit floating point numbers `m=52`, so the values in the mantissa differ by `1/2^52 = 2.220446049250313e-16`. This is the value of `eps()`. - -However, this "gap" between numbers is for values when the exponent is `0`. That is the numbers in `[1,2)`. For values in `[2,4)` the gap is twice, between `[1/2,1)` the gap is half. That is the gap depends on the size of the number. The gap between `x` and its next largest floating point number is given by `eps(x)` and that always satisfies `eps(x) <= eps() * abs(x)`. - -One way to think about this is the difference between `x` and the next largest floating point values is *basically* `x*(1+eps()) - x` or `x*eps()`. - -For the specific example, `abs(b-a) <= 2eps(m)` means that the gap between `a` and `b` is essentially 2 floating point values from the ``x`` value with the smallest ``f(x)`` value. - - -For bracketing methods that is about as good as you can get. However, once floating values are understood, the absolute best you can get for a bracketing interval would be -* along the way, a value `f(c)` is found which is *exactly* `0.0` -* the endpoints of the bracketing interval are *adjacent* floating point values, meaning the interval can not be bisected and `f` changes sign between the two values. - - -There can be problems when the stopping criteria is `abs(b-a) <= 2eps(m))` and the answer is `0.0` that require engineering around. For example, the algorithm above for the function `f(x) = -40*x*exp(-x)` does not converge when started with `[-9,1]`, even though `0.0` is an obvious zero. - - - -```julia; hold=true -fu(x) = -40*x*exp(-x) -chandrapatla(fu, -9, 1, λ3) -``` - -Here the issue is `abs(b-a)` is tiny (of the order `1e-119`) but `eps(m)` is even smaller. - - - - - -For non-bracketing methods, like Newton's method or the secant method, different criteria are useful. -There may not be a bracketing interval for `f` (for example `f(x) = (x-1)^2`) so the second criteria above might need to be restated in terms of the last two iterates, ``x_n`` and ``x_{n-1}``. Calling this difference ``\Delta = |x_n - x_{n-1}|``, we might stop if ``\Delta`` is small enough. As there are scenarios where this can happen, but the function is not at a zero, a check on the size of ``f`` is needed. - -However, there may be no floating point value where ``f`` is exactly `0.0` so checking the size of `f(x_n)` requires some agreement. - -First if `f(x_n)` is `0.0` then it makes sense to call `x_n` an *exact zero* of ``f``, even though this may hold even if `x_n`, a floating point value, is not mathematically an *exact* zero of ``f``. (Consider `f(x) = x^2 - 2x + 1`. Mathematically, this is identical to `g(x) = (x-1)^2`, but `f(1 + eps())` is zero, while `g(1+eps())` is `4.930380657631324e-32`. - -However, there may never be a value with `f(x_n)` exactly `0.0`. (The value of `sin(pi)` is not zero, for example, as `pi` is an approximation to ``\pi``, as well the `sin` of values adjacent to `float(pi)` do not produce `0.0` exactly.) - - -Suppose `x_n` is the closest floating number to ``\alpha``, the zero. Then the relative rounding error, ``(`` `x_n` ``- \alpha)/\alpha``, will be a value ``(1 + \delta)`` with ``\delta`` less than `eps()`. - -How far then can `f(x_n)` be from ``0 = f(\alpha)``? - -```math -f(x_n) = f(x_n - \alpha + \alpha) = f(\alpha + \alpha \cdot \delta) = f(\alpha \cdot (1 + \delta)), -``` - -Assuming ``f`` has a derivative, the linear approximation gives: - -```math -f(x_n) \approx f(\alpha) + f'(\alpha) \cdot (\alpha\delta) = f'(\alpha) \cdot \alpha \delta -``` - -So we should consider `f(x_n)` an *approximate zero* when it is on the scale of -``f'(\alpha) \cdot \alpha \delta``. - -That ``\alpha`` factor means we consider a *relative* tolerance for `f`. -Also important -- when `x_n` is close to `0`, -is the need for an *absolute* tolerance, one not dependent on the size of `x`. -So a good condition to check if `f(x_n)` is small is - -`abs(f(x_n)) <= abs(x_n) * rtol + atol`, or `abs(f(x_n)) <= max(abs(x_n) * rtol, atol)` - -where the relative tolerance, `rtol`, would absorb an estimate for ``f'(\alpha)``. - - -Now, in Newton's method the update step is ``f(x_n)/f'(x_n)``. Naturally when ``f(x_n)`` is close to ``0``, the update step is small and ``\Delta`` will be close to ``0``. *However*, should ``f'(x_n)`` be large, then ``\Delta`` can also be small and the algorithm will possibly stop, as ``x_{n+1} \approx x_n`` -- but not necessarily ``x_{n+1} \approx \alpha``. So termination on ``\Delta`` alone can be off. Checking if ``f(x_{n+1})`` is an approximate zero is also useful to include in a stopping criteria. - -One thing to keep in mind is that the right-hand side of the rule `abs(f(x_n)) <= abs(x_n) * rtol + atol`, as a function of `x_n`, goes to `Inf` as `x_n` increases. So if `f` has `0` as an asymptote (like `e^(-x)`) for large enough `x_n`, the rule will be `true` and `x_n` could be counted as an approximate zero, despite it not being one. - -So a modified criteria for convergence might look like: - -* stop if ``\Delta`` is small and `f` is an approximate zero with some tolerances -* stop if `f` is an approximate zero with some tolerances, but be mindful that this rule can identify mathematically erroneous answers. - -It is not uncommon to assign `rtol` to have a value like `sqrt(eps())` to account for accumulated floating point errors and the factor of ``f'(\alpha)``, though in the `Roots` package it is set smaller by default. - - -## Questions - -###### Question - -Let `f(x) = tanh(x)` (the hyperbolic tangent) and `fp(x) = sech(x)^2`, its derivative. - -Does *Newton's* method (using `Roots.Newton()`) converge starting at `1.0`? - -```julia; hold=true; echo=false -yesnoq("yes") -``` - -Does *Newton's* method (using `Roots.Newton()`) converge starting at `1.3`? - -```julia; hold=true; echo=false -yesnoq("no") -``` - -Does the secant method (using `Roots.Secant()`) converge starting at `1.3`? (a second starting value will automatically be chosen, if not directly passed in.) - -```julia; hold=true; echo=false -yesnoq("yes") -``` - -###### Question - -For the function `f(x) = x^5 - x - 1` both Newton's method and the secant method will converge to the one root when started from `1.0`. Using `verbose=true` as an argument to `find_zero`, (e.g., `find_zero(f, x0, Roots.Secant(), verbose=true)`) how many *more* steps does the secant method need to converge? - -```julia; hold=true; echo=false -numericq(2) -``` - -Do the two methods converge to the exact same value? - -```julia; hold=true; echo=false -yesnoq("yes") -``` - -###### Question - -Let `f(x) = exp(x) - x^4` and `x0=8.0`. How many steps (iterations) does it take for the secant method to converge using the default tolerances? - -```julia; hold=true; echo=false -numericq(10, 1) -``` - -###### Question - -Let `f(x) = exp(x) - x^4` and a starting bracket be `x0 = [8.9]`. Then calling `find_zero(f,x0, verbose=true)` will show that 49 steps are needed for exact bisection to converge. What about with the `Roots.Brent()` algorithm, which uses inverse quadratic steps when it can? - -It takes how many steps? - -```julia; hold=true; echo=false -numericq(36, 1) -``` - -The `Roots.A42()` method uses inverse cubic interpolation, as possible, how many steps does this method take to converge? - -```julia; hold=true; echo=false -numericq(3, 1) -``` - -The large difference is due to how the tolerances are set within `Roots`. The `Brent method gets pretty close in a few steps, but takes a much longer time to get close enough for the default tolerances - - -###### Question - -Consider this crazy function defined by: - -```julia; eval=false -f(x) = cos(100*x)-4*erf(30*x-10) -``` - -(The `erf` function is the (error function](https://en.wikipedia.org/wiki/Error_function) and is in the `SpecialFunctions` package loaded with `CalculusWithJulia`.) - -Make a plot over the interval $[-3,3]$ to see why it is called "crazy". - -Does `find_zero` find a zero to this function starting from $0$? - - -```julia; hold=true; echo=false -yesnoq("yes") -``` - -If so, what is the value? - -```julia; hold=true; echo=false -f(x) = cos(100*x)-4*erf(30*x-10) -val = find_zero(f, 0) -numericq(val) -``` - -If not, what is the reason? - -```julia; hold=true; echo=false -choices = [ -"The zero is a simple zero", -"The zero is not a simple zero", -"The function oscillates too much to rely on the tangent line approximation far from the zero", -"We can find an answer" -] -answ = 4 -radioq(choices, answ, keep_order=true) -``` - - -Does `find_zero` find a zero to this function starting from $1$? - - -```julia; hold=true; echo=false -yesnoq(false) -``` - -If so, what is the value? - -```julia; hold=true; echo=false -numericq(-999.999) -``` - -If not, what is the reason? - -```julia; hold=true; echo=false -choices = [ -"The zero is a simple zero", -"The zero is not a simple zero", -"The function oscillates too much to rely on the tangent line approximations far from the zero", -"We can find an answer" -] -answ = 3 -radioq(choices, answ, keep_order=true) -``` diff --git a/CwJ/derivatives/newtons-method.js b/CwJ/derivatives/newtons-method.js deleted file mode 100644 index a328710..0000000 --- a/CwJ/derivatives/newtons-method.js +++ /dev/null @@ -1,72 +0,0 @@ -// newton's method - -const b = JXG.JSXGraph.initBoard('jsxgraph', { - boundingbox: [-3,5,3,-5], axis:true -}); - - -var f = function(x) {return x*x*x*x*x - x - 1}; -var fp = function(x) { return 4*x*x*x*x - 1}; -var x0 = 0.85; - -var nm = function(x) { return x - f(x)/fp(x);}; - -var l = b.create('point', [-1.5,0], {name:'', size:0}); -var r = b.create('point', [1.5,0], {name:'', size:0}); -var xaxis = b.create('line', [l,r]) - - -var P0 = b.create('glider', [x0,0,xaxis], {name:'x0'}); -var P0a = b.create('point', [function() {return P0.X();}, - function() {return f(P0.X());}], {name:''}); - -var P1 = b.create('point', [function() {return nm(P0.X());}, - 0], {name:''}); -var P1a = b.create('point', [function() {return P1.X();}, - function() {return f(P1.X());}], {name:''}); - -var P2 = b.create('point', [function() {return nm(P1.X());}, - 0], {name:''}); -var P2a = b.create('point', [function() {return P2.X();}, - function() {return f(P2.X());}], {name:''}); - -var P3 = b.create('point', [function() {return nm(P2.X());}, - 0], {name:''}); -var P3a = b.create('point', [function() {return P3.X();}, - function() {return f(P3.X());}], {name:''}); - -var P4 = b.create('point', [function() {return nm(P3.X());}, - 0], {name:''}); -var P4a = b.create('point', [function() {return P4.X();}, - function() {return f(P4.X());}], {name:''}); -var P5 = b.create('point', [function() {return nm(P4.X());}, - 0], {name:'x5', strokeColor:'black'}); - - - - - -P0a.setAttribute({fixed:true}); -P1.setAttribute({fixed:true}); -P1a.setAttribute({fixed:true}); -P2.setAttribute({fixed:true}); -P2a.setAttribute({fixed:true}); -P3.setAttribute({fixed:true}); -P3a.setAttribute({fixed:true}); -P4.setAttribute({fixed:true}); -P4a.setAttribute({fixed:true}); -P5.setAttribute({fixed:true}); - -var sc = '#000000'; -b.create('segment', [P0,P0a], {strokeColor:sc, strokeWidth:1}); -b.create('segment', [P0a, P1], {strokeColor:sc, strokeWidth:1}); -b.create('segment', [P1,P1a], {strokeColor:sc, strokeWidth:1}); -b.create('segment', [P1a, P2], {strokeColor:sc, strokeWidth:1}); -b.create('segment', [P2,P2a], {strokeColor:sc, strokeWidth:1}); -b.create('segment', [P2a, P3], {strokeColor:sc, strokeWidth:1}); -b.create('segment', [P3,P3a], {strokeColor:sc, strokeWidth:1}); -b.create('segment', [P3a, P4], {strokeColor:sc, strokeWidth:1}); -b.create('segment', [P4,P4a], {strokeColor:sc, strokeWidth:1}); -b.create('segment', [P4a, P5], {strokeColor:sc, strokeWidth:1}); - -b.create('functiongraph', [f, -1.5, 1.5]) diff --git a/CwJ/derivatives/newtons_method.jmd b/CwJ/derivatives/newtons_method.jmd deleted file mode 100644 index f4ddcc2..0000000 --- a/CwJ/derivatives/newtons_method.jmd +++ /dev/null @@ -1,1432 +0,0 @@ -# Newton's method - - -This section uses these add-on packages: - -```julia -using CalculusWithJulia -using Plots -using SymPy -using Roots -``` - -```julia; echo=false; results="hidden" -using CalculusWithJulia.WeaveSupport -using ImplicitPlots - -fig_size = (800, 600) -const frontmatter = ( - title = "Newton's method", - description = "Calculus with Julia: Newton's method", - tags = ["CalculusWithJulia", "derivatives", "newton's method"], -); - -nothing -``` - ----- - -The Babylonian method is an algorithm to find an approximate value for $\sqrt{k}$. -It was described by the first-century Greek mathematician Hero of -[Alexandria](http://en.wikipedia.org/wiki/Babylonian_method). - -The method starts with some initial guess, called $x_0$. It then -applies a formula to produce an improved guess. This is repeated until -the improved guess is accurate enough or it is clear the algorithm -fails to work. - -For the Babylonian method, the next guess, $x_{i+1}$, is derived from the current guess, $x_i$. In mathematical notation, this is the updating step: - - -```math -x_{i+1} = \frac{1}{2}(x_i + \frac{k}{x_i}) -``` - - -We use this algorithm to approximate the square root of $2$, a value known to -the Babylonians. - -Start with $x$, then form $x/2 + 1/x$, from this again form $x/2 + 1/x$, repeat. - -We represent this step using a function - -```julia -babylon(x) = x/2 + 1/x -``` - -Let's look starting with $x = 2$ as a rational number: - -```julia; hold=true -x₁ = babylon(2//1) -x₁, x₁^2.0 -``` - -Our estimate improved from something which squared to $4$ down to something which squares to $2.25.$ A big improvement, but there is still more to come. Had we done one more step: - - -```julia; -x₂ = (babylon ∘ babylon)(2//1) -x₂, x₂^2.0 -``` - -We now see accuracy until the third decimal point. - -```julia; -x₃ = (babylon ∘ babylon ∘ babylon)(2//1) -x₃, x₃^2.0 -``` - -This is now accurate to the sixth decimal point. That is about as far -as we, or the Bablyonians, would want to go by hand. Using rational -numbers quickly grows out of hand. The next step shows the explosion. - -```julia; -reduce((x,step) -> babylon(x), 1:4, init=2//1) -``` - -(In the above, we used `reduce` to repeat a function call ``4`` times, as an alternative to the composition operation. In this section we show a few styles to do this repetition before introducing a packaged function.) - - -However, with the advent of floating point numbers, the method stays quite manageable: - - -```julia; hold=true; -xₙ = reduce((x, step) -> babylon(x), 1:6, init=2.0) -xₙ, xₙ^2 -``` - -We can see that the algorithm - to the precision offered by floating -point numbers - has resulted in an answer `1.414213562373095`. This -answer is an *approximation* to the actual answer. Approximation is necessary, -as $\sqrt{2}$ is an irrational number and so can never be exactly -represented in floating point. That being said, we can see that the value -of $f(x)$ is accurate to the last decimal place, so our approximation -is very close and is achieved in a few steps. - -## Newton's generalization - -Let $f(x) = x^3 - 2x -5$. The value of ``2`` is almost a zero, but not quite, as $f(2) = --1$. We can check that there are no *rational* roots. Though there is -a method to solve the cubic it may be difficult to compute and will -not be as generally applicable as some algorithm like the Babylonian -method to produce an approximate answer. - -Is there some generalization to the Babylonian method? - -We know that the tangent line is a good approximation to the function -at the point. Looking at this graph gives a hint as to an algorithm: - -```julia; hold=true; echo=false -f(x) = x^3 - 2x - 5 -fp(x) = 3x^2 - 2 -c = 2 -p = plot(f, 1.75, 2.25, legend=false) -plot!(x->f(2) + fp(2)*(x-2)) -plot!(zero) -scatter!(p, [c], [f(c)], color=:orange, markersize=3) -p -``` - -The tangent line and the function nearly agree near $2$. So much so, -that the intersection point of the tangent line with the $x$ axis -nearly hides the actual zero of $f(x)$ that is near $2.1$. - -That is, it seems that the intersection of the tangent line and the -$x$ axis should be an improved approximation for the zero of the -function. - -Let $x_0$ be $2$, and $x_1$ be the intersection point of the tangent line -at $(x_0, f(x_0))$ with the $x$ axis. Then by the definition of the -tangent line: - -```math -f'(x_0) = \frac{\Delta y }{\Delta x} = \frac{f(x_0)}{x_0 - x_1}. -``` - -This can be solved for $x_1$ to give $x_1 = x_0 - f(x_0)/f'(x_0)$. In general, if we had $x_i$ and used the intersection point of the tangent line to produce $x_{i+1}$ we would have Newton's method: - -```math -x_{i+1} = x_i - \frac{f(x_i)}{f'(x_i)}. -``` - - -Using automatic derivatives, as brought in with the `CalculusWithJulia` package, we can implement this algorithm. - - -The algorithm above starts at $2$ and then becomes: - -```julia; -f(x) = x^3 - 2x - 5 -x0 = 2.0 -x1 = x0 - f(x0) / f'(x0) -``` - -We can see we are closer to a zero: - -```julia; -f(x0), f(x1) -``` - -Trying again, we have - -```julia; -x2 = x1 - f(x1)/ f'(x1) -x2, f(x2), f(x1) -``` - -And again: - -```julia; -x3 = x2 - f(x2)/ f'(x2) -x3, f(x3), f(x2) -``` - - -```julia; -x4 = x3 - f(x3)/ f'(x3) -x4, f(x4), f(x3) -``` - -We see now that $f(x_4)$ is within machine tolerance of $0$, so we -call $x_4$ an *approximate zero* of $f(x)$. - - -> **Newton's method:** Let $x_0$ be an initial guess for a zero of -> $f(x)$. Iteratively define $x_{i+1}$ in terms of the just -> generated $x_i$ by: -> ```math -> x_{i+1} = x_i - f(x_i) / f'(x_i). -> ``` -> Then for -> reasonable functions and reasonable initial guesses, the sequence of -> points converges to a zero of $f$. - -On the computer, we know that actual convergence will likely never -occur, but accuracy to a certain tolerance can often be achieved. - - - -In the example above, we kept track of the previous values. This is -unnecessary if only the answer is sought. In that case, the update -step could use the same variable. Here we use `reduce`: - -```julia;hold=true; -xₙ = reduce((x, step) -> x - f(x)/f'(x), 1:4, init=2) -xₙ, f(xₙ) -``` - -In practice, the algorithm is implemented not by repeating the update step a fixed number of times, rather by repeating the step until either we -converge or it is clear we won't converge. For good guesses and most -functions, convergence happens quickly. - - - -!!! note - Newton looked at this same example in 1699 (B.T. Polyak, *Newton's - method and its use in optimization*, European Journal of Operational - Research. 02/2007; 181(3):1086-1096.) though his technique was - slightly different as he did not use the derivative, *per se*, but - rather an approximation based on the fact that his function was a - polynomial (though identical to the derivative). Raphson (1690) - proposed the general form, hence the usual name of the Newton-Raphson - method. - -#### Examples - -##### Example: visualizing convergence - -This graphic demonstrates the method and the rapid convergence: - -```julia; echo=false -function newtons_method_graph(n, f, a, b, c) - - xstars = [c] - xs = [c] - ys = [0.0] - - plt = plot(f, a, b, legend=false, size=fig_size) - plot!(plt, [a, b], [0,0], color=:black) - - - ts = range(a, stop=b, length=50) - for i in 1:n - x0 = xs[end] - x1 = x0 - f(x0)/D(f)(x0) - push!(xstars, x1) - append!(xs, [x0, x1]) - append!(ys, [f(x0), 0]) - end - plot!(plt, xs, ys, color=:orange) - scatter!(plt, xstars, 0*xstars, color=:orange, markersize=5) - plt -end -nothing -``` - -```julia; hold=true; echo=false; cache=true -### {{{newtons_method_example}}} - -caption = """ - -Illustration of Newton's Method converging to a zero of a function. - -""" -n = 6 - -fn, a, b, c = x->log(x), .15, 2, .2 - -anim = @animate for i=1:n - newtons_method_graph(i-1, fn, a, b, c) -end - -imgfile = tempname() * ".gif" -gif(anim, imgfile, fps = 1) - -ImageFile(imgfile, caption) -``` - ----- - -This interactive graphic (built using [JSXGraph](https://jsxgraph.uni-bayreuth.de/wp/index.html)) allows the adjustment of the point `x0`, initially at ``0.85``. Five iterations of Newton's method are illustrated. Different positions of `x0` clearly converge, others will not. - -```=html -
-``` - -```ojs -//| echo: false -//| output: false - -JXG = require("jsxgraph"); - -// newton's method - -b = JXG.JSXGraph.initBoard('jsxgraph', { - boundingbox: [-3,5,3,-5], axis:true -}); - - -f = function(x) {return x*x*x*x*x - x - 1}; -fp = function(x) { return 4*x*x*x*x - 1}; -x0 = 0.85; - -nm = function(x) { return x - f(x)/fp(x);}; - -l = b.create('point', [-1.5,0], {name:'', size:0}); -r = b.create('point', [1.5,0], {name:'', size:0}); -xaxis = b.create('line', [l,r]) - - -P0 = b.create('glider', [x0,0,xaxis], {name:'x0'}); -P0a = b.create('point', [function() {return P0.X();}, - function() {return f(P0.X());}], {name:''}); - -P1 = b.create('point', [function() {return nm(P0.X());}, - 0], {name:''}); -P1a = b.create('point', [function() {return P1.X();}, - function() {return f(P1.X());}], {name:''}); - -P2 = b.create('point', [function() {return nm(P1.X());}, - 0], {name:''}); -P2a = b.create('point', [function() {return P2.X();}, - function() {return f(P2.X());}], {name:''}); - -P3 = b.create('point', [function() {return nm(P2.X());}, - 0], {name:''}); -P3a = b.create('point', [function() {return P3.X();}, - function() {return f(P3.X());}], {name:''}); - -P4 = b.create('point', [function() {return nm(P3.X());}, - 0], {name:''}); -P4a = b.create('point', [function() {return P4.X();}, - function() {return f(P4.X());}], {name:''}); -P5 = b.create('point', [function() {return nm(P4.X());}, - 0], {name:'x5', strokeColor:'black'}); - - - - - -P0a.setAttribute({fixed:true}); -P1.setAttribute({fixed:true}); -P1a.setAttribute({fixed:true}); -P2.setAttribute({fixed:true}); -P2a.setAttribute({fixed:true}); -P3.setAttribute({fixed:true}); -P3a.setAttribute({fixed:true}); -P4.setAttribute({fixed:true}); -P4a.setAttribute({fixed:true}); -P5.setAttribute({fixed:true}); - -sc = '#000000'; -b.create('segment', [P0,P0a], {strokeColor:sc, strokeWidth:1}); -b.create('segment', [P0a, P1], {strokeColor:sc, strokeWidth:1}); -b.create('segment', [P1,P1a], {strokeColor:sc, strokeWidth:1}); -b.create('segment', [P1a, P2], {strokeColor:sc, strokeWidth:1}); -b.create('segment', [P2,P2a], {strokeColor:sc, strokeWidth:1}); -b.create('segment', [P2a, P3], {strokeColor:sc, strokeWidth:1}); -b.create('segment', [P3,P3a], {strokeColor:sc, strokeWidth:1}); -b.create('segment', [P3a, P4], {strokeColor:sc, strokeWidth:1}); -b.create('segment', [P4,P4a], {strokeColor:sc, strokeWidth:1}); -b.create('segment', [P4a, P5], {strokeColor:sc, strokeWidth:1}); - -b.create('functiongraph', [f, -1.5, 1.5]) - -``` - - - -##### Example: numeric not algebraic - -For the function $f(x) = \cos(x) - x$, we see that SymPy can not solve symbolically for a zero: - -```julia; -@syms x::real -solve(cos(x) - x, x) -``` - -We can find a numeric solution, even though there is no closed-form answer. Here we try Newton's method: - -```julia; hold=true -f(x) = cos(x) - x -x = .5 -x = x - f(x)/f'(x) # 0.7552224171056364 -x = x - f(x)/f'(x) # 0.7391416661498792 -x = x - f(x)/f'(x) # 0.7390851339208068 -x = x - f(x)/f'(x) # 0.7390851332151607 -x = x - f(x)/f'(x) -x, f(x) -``` - -To machine tolerance the answer is a zero, even though the exact answer is irrational and all finite floating point values can be represented as rational numbers. - -##### Example - -Use Newton's method to find the *largest* real solution to ``e^x = x^6``. - -A plot shows us roughly where the value lies: - -```julia; hold=true -f(x) = exp(x) -g(x) = x^6 -plot(f, 0, 25, label="f") -plot!(g, label="g") -``` - -Clearly by ``20`` the two paths diverge. We know exponentials eventually grow faster than powers, and this is seen in the graph. - -To use Newton's method to find the intersection point. Stop when the increment ``f(x)/f'(x)`` is smaller than `1e-4`. -We need to turn the solution to an equation into a value where a function is ``0``. Just moving the terms to one side of the equals sign gives ``e^x - x^6 = 0``, or the ``x`` we seek is a solution to ``h(x)=0`` with ``h(x) = e^x - x^6``. - - -```julia; hold=true; term=true -h(x) = exp(x) - x^6 -x = 20 -for step in 1:10 - delta = h(x)/h'(x) - x = x - delta - @show step, x, delta -end -``` - -So it takes ``8`` steps to get an increment that small and about `10` steps to get to full convergence. - -##### Example division as multiplication - -[Newton-Raphson Division](http://tinyurl.com/kjj9w92) is a means to divide by multiplying. - -Why would you want to do that? Well, even for computers division is -harder (read slower) than multiplying. The trick is that $p/q$ is -simply $p \cdot (1/q)$, so finding a means to compute a reciprocal by -multiplying will reduce division to multiplication. - - -Well suppose we have $q$, we could try to use Newton's method to find -$1/q$, as it is a solution to $f(x) = x - 1/q$. The Newton update step -simplifies to: - -```math -x - f(x) / f'(x) \quad\text{or}\quad x - (x - 1/q)/ 1 = 1/q -``` - -That doesn't really help, as Newton's method is just $x_{i+1} = 1/q$. -That is, it just jumps to the answer, the one we want to compute by some other means! - - -Trying again, we simplify the update step for a related function: -$f(x) = 1/x - q$ with $f'(x) = -1/x^2$ and then one step of the process is: - -```math -x_{i+1} = x_i - (1/x_i - q)/(-1/x_i^2) = -qx^2_i + 2x_i. -``` - -Now for $q$ in the interval $[1/2, 1]$ we want to get a *good* initial -guess. Here is a claim. We can use $x_0=48/17 - 32/17 \cdot q$. Let's check -graphically that this is a reasonable initial approximation to $1/q$: - -```julia; hold=true - -plot(q -> 1/q, 1/2, 1, label="1/q") -plot!(q -> 1/17 * (48 - 32q), label="linear approximation") -``` - - - -It can be shown that we have for any $q$ in $[1/2, 1]$ with initial guess $x_0 = -48/17 - 32/17\cdot q$ that Newton's method will converge to ``16`` digits in no more -than this many steps: - -```math -\log_2(\frac{53 + 1}{\log_2(17)}). -``` - - - -```julia; -a = log2((53 + 1)/log2(17)) -ceil(Integer, a) -``` - -That is ``4`` steps suffices. - -For $q = 0.80$, to find $1/q$ using the above we have - -```julia; hold=true -q = 0.80 -x = (48/17) - (32/17)*q -x = -q*x*x + 2*x -x = -q*x*x + 2*x -x = -q*x*x + 2*x -x = -q*x*x + 2*x -``` - -This method has basically $18$ multiplication and addition operations -for one division, so it naively would seem slower, but timing this -shows the method is competitive with a regular division. - -## Wrapping in a function - -In the previous examples, we saw fast convergence, guaranteed converge in ``4`` steps, and an example where ``8`` steps were needed to get the requested level of approximation. Newton's method usually converges quickly, but may converge slowly, and may not converge at all. Automating the task to avoid repeatedly running the update step is -a task best done by the computer. - -The `while` loop is a -good way to repeat commands until some condition is met. With this, we -present a simple function implementing Newton's method, we iterate -until the update step gets really small (the `atol`) or the -convergence takes more than ``50`` steps. (There are other, better choices that could be used to determine when the algorithm should stop, these are just easy to understand.) - -```julia; -function nm(f, fp, x0) - atol = 1e-14 - ctr = 0 - delta = Inf - while (abs(delta) > atol) && (ctr < 50) - delta = f(x0)/fp(x0) - x0 = x0 - delta - ctr = ctr + 1 - end - - ctr < 50 ? x0 : NaN -end -``` - - -##### Examples - - -- Find a zero of $\sin(x)$ starting at $x_0=3$: - -```julia; -nm(sin, cos, 3) -``` - -This is an approximation for $\pi$, that historically found use, as the convergence is fast. - -- Find a solution to $x^5 = 5^x$ near $2$: - -Writing a function to handle this, we have: - -```julia; -k(x) = x^5 - 5^x -``` - -We could find the derivative by hand, but use the automatic one instead: - -```julia; -alpha = nm(k, k', 2) -alpha, f(alpha) -``` - -### Functions in the Roots package - -Typing in the `nm` function might be okay once, but would be tedious -if it was needed each time. Besides, it isn't as robust to different inputs as possible. The `Roots` package provides a `Newton` -method for `find_zero`. - - -To use a different method with `find_zero`, the calling pattern is `find_zero(f, x, M)` where `f` represent the function(s), `x` the initial point(s), and `M` the method. Here we have: - -```julia -find_zero((sin, cos), 3, Roots.Newton()) -``` - -Or, if a derivative is not specified, one can be computed using automatic differentiation: - -```julia; hold=true -f(x) = sin(x) -find_zero((f, f'), 2, Roots.Newton()) -``` - -The argument `verbose=true` will force a print out of a message summarizing the convergence and showing each step. - -```julia; hold=true -f(x) = exp(x) - x^4 -find_zero((f,f'), 8, Roots.Newton(); verbose=true) -``` - - - -##### Example: intersection of two graphs - -Find the intersection point between $f(x) = \cos(x)$ and $g(x) = 5x$ near $0$. - -We have Newton's method to solve for zeros of $f(x)$, i.e. when $f(x) = -0$. Here we want to solve for $x$ with $f(x) = g(x)$. To do so, we -make a new function $h(x) = f(x) - g(x)$, that is $0$ when $f(x)$ -equals $g(x)$: - -```julia; hold=true -f(x) = cos(x) -g(x) = 5x -h(x) = f(x) - g(x) -x0 = find_zero((h,h'), 0, Roots.Newton()) -x0, h(x0), f(x0), g(x0) -``` - ----- - -We redo the above using a *parameter* for the ``5``, as there are some options on how it would be done. We let `f(x,p) = cos(x) - p*x`. Then we can use `Roots.Newton` by also defining a derivative: - -```julia; hold=true -f(x,p) = cos(x) - p*x -fp(x,p) = -sin(x) - p -xn = find_zero((f,fp), pi/4, Roots.Newton(); p=5) -xn, f(xn, 5) -``` - -To use automatic differentiation is not straightforward, as we must hold the `p` fixed. For this, we introduce a closure that fixes `p` and differentiates in the `x` variable (called `u` below): - -```julia; hold=true -f(x,p) = cos(x) - p*x -fp(x,p) = (u -> f(u,p))'(x) -xn = find_zero((f,fp), pi/4, Roots.Newton(); p=5) -``` - -##### Example: Finding $c$ in Rolle's Theorem - -The function $r(x) = \sqrt{1 - \cos(x^2)^2}$ has a zero at $0$ and one at ``a`` near ``1.77``. - -```julia; -r(x) = sqrt(1 - cos(x^2)^2) -plot(r, 0, 1.77) -``` - -As $f(x)$ is differentiable between $0$ and $a$, Rolle's theorem says -there will be value where the derivative is $0$. Find that value. - -This value will be a zero of the derivative. A graph shows it should be near $1.2$, so we use that as a starting value to get the answer: - -```julia; -find_zero((r',r''), 1.2, Roots.Newton()) -``` - - - - -## Convergence rates - -Newton's method is famously known to have "quadratic convergence." What -does this mean? Let the error in the $i$th step be called $e_i = x_i - -\alpha$. Then Newton's method satisfies a bound of the type: - -```math -\lvert e_{i+1} \rvert \leq M_i \cdot e_i^2. -``` - -If $M$ were just a constant and we suppose $e_0 = 10^{-1}$ then $e_1$ -would be less than $M 10^{-2}$ and $e_2$ less than $M^2 10^{-4}$, -$e_3$ less than $M^3 10^{-8}$ and $e_4$ less than $M^4 10^{-16}$ which -for $M=1$ is basically the machine precision when values are near -``1``. That is for some problems, with a good initial guess it will -take around ``4`` or so steps to converge. - -To identify ``M``, let ``\alpha`` be the zero of ``f`` to be approximated. Assume - -* The function ``f`` has at continuous second derivative in a neighborhood of ``\alpha``. -* The value ``f'(\alpha)`` is *non-zero* in the neighborhood of ``\alpha``. - -Then this linearization holds at each $x_i$ in the above neighborhood: - -```math -f(x) = f(x_i) + f'(x_i) \cdot (x - x_i) + \frac{1}{2} f''(\xi) \cdot (x-x_i)^2. -``` - -The value $\xi$ is from the mean value theorem and is between $x$ and $x_i$. - -Dividing by ``f'(x_i)`` and setting ``x=\alpha`` (as $f(\alpha)=0$) leaves - -```math -0 = \frac{f(x_i)}{f'(x_i)} + (\alpha-x_i) + \frac{1}{2}\cdot \frac{f''(\xi)}{f'(x_i)} \cdot (\alpha-x_i)^2. -``` - -For this value, we have - -```math -\begin{align*} -x_{i+1} - \alpha -&= \left(x_i - \frac{f(x_i)}{f'(x_i)}\right) - \alpha\\ -&= \left(x_i - \alpha \right) - \frac{f(x_i)}{f'(x_i)}\\ -&= (x_i - \alpha) + \left( -(\alpha - x_i) + \frac{1}{2}\frac{f''(\xi) \cdot(\alpha - x_i)^2}{f'(x_i)} -\right)\\ -&= \frac{1}{2}\frac{f''(\xi)}{f'(x_i)} \cdot(x_i - \alpha)^2. -\end{align*} -``` - -That is - -```math -e_{i+1} = \frac{1}{2}\frac{f''(\xi)}{f'(x_i)} e_i^2. -``` - - -This convergence to ``\alpha`` will be quadratic *if*: - -- The initial guess $x_0$ is not too far from $\alpha$, so $e_0$ is - managed. - -- The derivative at $\alpha$ is not too close to $0$, hence, by continuity ``f'(x_i)`` is not too close to ``0``. (As it appears in - the denominator). That is, the function can't be too flat, which - should make sense, as then the tangent line is nearly parallel to - the $x$ axis and would intersect far away. - -- The function ``f`` has a continuous second derivative at ``\alpha``. - -- The second derivative is not too big (in absolute value) near ``\alpha``. - A large second derivative means the function is very concave, - which means it is "turning" a lot. In this case, the function turns - away from the tangent line quickly, so the tangent line's zero is - not necessarily a good approximation to the actual zero, $\alpha$. - - -!!! note - The basic tradeoff: methods like Newton's are faster than the - bisection method in terms of function calls, but are not guaranteed to - converge, as the bisection method is. - - -What can go wrong when one of these isn't the case is illustrated next: - -### Poor initial step - -```julia; hold=true; echo=false; cache=true -### {{{newtons_method_poor_x0}}} -caption = """ - -Illustration of Newton's Method converging to a zero of a function, -but slowly as the initial guess, is very poor, and not close to the -zero. The algorithm does converge in this illustration, but not quickly and not to the nearest root from -the initial guess. - -""" - -fn, a, b, c = x -> sin(x) - x/4, -15, 20, 2pi - -n = 20 -anim = @animate for i=1:n - newtons_method_graph(i-1, fn, a, b, c) -end - -imgfile = tempname() * ".gif" -gif(anim, imgfile, fps = 2) - -ImageFile(imgfile, caption) -``` - - -```julia; hold=true; echo=false; cache=true -# {{{newtons_method_flat}}} -caption = L""" - -Illustration of Newton's method failing to coverge as for some $x_i$, -$f'(x_i)$ is too close to ``0``. In this instance after a few steps, the -algorithm just cycles around the local minimum near $0.66$. The values -of $x_i$ repeat in the pattern: $1.0002, 0.7503, -0.0833, 1.0002, -\dots$. This is also an illustration of a poor initial guess. If there -is a local minimum or maximum between the guess and the zero, such -cycles can occur. - -""" - -fn, a, b, c = x -> x^5 - x + 1, -1.5, 1.4, 0.0 - -n=7 -anim = @animate for i=1:n - newtons_method_graph(i-1, fn, a, b, c) -end -imgfile = tempname() * ".gif" -gif(anim, imgfile, fps = 1) - -ImageFile(imgfile, caption) -``` - -### The second derivative is too big - -```julia; hold=true; echo=false; cache=true -# {{{newtons_method_cycle}}} - -fn, a, b, c, = x -> abs(x)^(0.49), -2, 2, 1.0 -caption = L""" - -Illustration of Newton's Method not converging. Here the second -derivative is too big near the zero - it blows up near $0$ - and the -convergence does not occur. Rather the iterates increase in their -distance from the zero. - -""" - -n=10 -anim = @animate for i=1:n - newtons_method_graph(i-1, fn, a, b, c) -end - -imgfile = tempname() * ".gif" -gif(anim, imgfile, fps = 2) - -ImageFile(imgfile, caption) -``` -### The tangent line at some xᵢ is flat - -```julia; hold=true; echo=false; cache=true -# {{{newtons_method_wilkinson}}} - -caption = L""" - -The function $f(x) = x^{20} - 1$ has two bad behaviours for Newton's -method: for $x < 1$ the derivative is nearly $0$ and for $x>1$ the -second derivative is very big. In this illustration, we have an -initial guess of $x_0=8/9$. As the tangent line is fairly flat, the -next approximation is far away, $x_1 = 1.313\dots$. As this guess is -is much bigger than $1$, the ratio $f(x)/f'(x) \approx -x^{20}/(20x^{19}) = x/20$, so $x_i - x_{i-1} \approx (19/20)x_i$ -yielding slow, linear convergence until $f''(x_i)$ is moderate. For -this function, starting at $x_0=8/9$ takes 11 steps, at $x_0=7/8$ -takes 13 steps, at $x_0=3/4$ takes ``55`` steps, and at $x_0=1/2$ it takes -$204$ steps. - -""" - - -fn,a,b,c = x -> x^20 - 1, .7, 1.4, 8/9 -n = 10 - -anim = @animate for i=1:n - newtons_method_graph(i-1, fn, a, b, c) -end -imgfile = tempname() * ".gif" -gif(anim, imgfile, fps = 1) - -ImageFile(imgfile, caption) -``` - -###### Example - -Suppose $\alpha$ is a simple zero for $f(x)$. (The value $\alpha$ is -a zero of multiplicity $k$ if $f(x) = (x-\alpha)^kg(x)$ where -$g(\alpha)$ is not zero. A simple zero has multiplicity $1$. If -$f'(\alpha) \neq 0$ and the second derivative exists, then a zero -$\alpha$ will be simple.) Around $\alpha$, quadratic convergence should -apply. However, consider the function $g(x) = f(x)^k$ for some integer -$k \geq 2$. Then $\alpha$ is still a zero, but the derivative of $g$ -at $\alpha$ is zero, so the tangent line is basically flat. This will -slow the convergence up. We can see that the update step $g'(x)/g(x)$ -becomes $(1/k) f'(x)/f(x)$, so an extra factor is introduced. - -The calculation that produces the quadratic convergence now becomes: - -```math -x_{i+1} - \alpha = (x_i - \alpha) - \frac{1}{k}(x_i-\alpha + \frac{f''(\xi)}{2f'(x_i)}(x_i-\alpha)^2) = -\frac{k-1}{k} (x_i-\alpha) + \frac{f''(\xi)}{2kf'(x_i)}(x_i-\alpha)^2. -``` - -As $k > 1$, the $(x_i - \alpha)$ term dominates, and we see the -convergence is linear with $\lvert e_{i+1}\rvert \approx (k-1)/k -\lvert e_i\rvert$. - - - -## Questions - -###### Question - -Look at this graph with $x_0$ marked with a point: - -```julia; hold=true; echo=false -import SpecialFunctions: airyai -p = plot(airyai, -3.3, 0, legend=false); -plot!(p, zero, -3.3, 0); -scatter!(p, [-2.8], [0], color=:orange, markersize=5); -annotate!(p, [(-2.8, 0.2, "x₀")]) -p -``` - -If one step of Newton's method was used, what would be the value of $x_1$? - -```julia; hold=true; echo=false -choices = ["``-2.224``", "``-2.80``", "``-0.020``", "``0.355``"] -answ = 1 -radioq(choices, answ, keep_order=true) -``` - -###### Question - -Look at this graph of some increasing, concave up $f(x)$ with initial point $x_0$ marked. Let ``\alpha`` be the zero. - -```julia; hold=true; echo=false -p = plot(x -> x^2 - 2, .75, 2.2, legend=false); -plot!(p, zero, color=:green); -scatter!(p, [1],[0], color=:orange, markersize=5); -annotate!(p, [(1,.25, "x₀"), (sqrt(2), .2, "α")]); -p -``` - - -What can be said about $x_1$? - -```julia; hold=true; echo=false -choices = [ -L"It must be $x_1 > \alpha$", -L"It must be $x_1 < x_0$", -L"It must be $x_0 < x_1 < \alpha$" -] -answ = 1 -radioq(choices, answ) -``` - ----- - -Look at this graph of some increasing, concave up $f(x)$ with initial point $x_0$ marked. Let $\alpha$ be the zero. - -```julia; hold=true; echo=false -p = plot(x -> x^2 - 2, .75, 2.2, legend=false); -plot!(p, zero, .75, 2.2, color=:green); -scatter!(p, [2],[0], color=:orange, markersize=5); -annotate!(p, [(2,.25, "x₀"), (sqrt(2), .2, "α")]); -p -``` - - -What can be said about $x_1$? - -```julia; hold=true; echo=false -choices = [ -L"It must be $x_1 < \alpha$", -L"It must be $x_1 > x_0$", -L"It must be $\alpha < x_1 < x_0$" -] -answ = 3 -radioq(choices, answ) -``` - ----- - -Suppose $f(x)$ is increasing and concave up. From the tangent line representation: $f(x) = f(c) + f'(c)\cdot(x-c) + f''(\xi)/2 \cdot(x-c)^2$, explain why it must be that the graph of $f(x)$ lies on or *above* the tangent line. - -```julia; hold=true; echo=false -choices = [ -L"As $f''(\xi)/2 \cdot(x-c)^2$ is non-negative, we must have $f(x) - (f(c) + f'(c)\cdot(x-c)) \geq 0$.", -L"As $f''(\xi) < 0$ it must be that $f(x) - (f(c) + f'(c)\cdot(x-c)) \geq 0$.", -L"This isn't true. The function $f(x) = x^3$ at $x=0$ provides a counterexample" -] -answ = 1 -radioq(choices, answ) -``` - -This question can be used to give a proof for the previous two questions, which can be answered by considering the graphs alone. Combined, they say that if a function is increasing and concave up and ``\alpha`` is a zero, then if ``x_0 < \alpha`` it will be ``x_1 > \alpha``, and for any ``x_i > \alpha``, ``\alpha <= x_{i+1} <= x_\alpha``, so the sequence in Newton's method is decreasing and bounded below; conditions for which it is guaranteed mathematically there will be convergence. - - -###### Question - - -Let $f(x) = x^2 - 3^x$. This has derivative $2x - 3^x \cdot -\log(3)$. Starting with $x_0=0$, what does Newton's method converge on? - -```julia; hold=true; echo=false -f(x) = x^2 - 3^x; -fp(x) = 2x - 3^x*log(3); -val = Roots.newton(f, fp, 0); -numericq(val, 1e-14) -``` - -###### Question - - -Let $f(x) = \exp(x) - x^4$. There are 3 zeros for this function. Which one does Newton's method converge to when $x_0=2$? - - - -```julia; hold=true; echo=false -f(x) = exp(x) - x^4; -fp(x) = exp(x) - 4x^3; -xstar= Roots.newton(f, fp, 2); -numericq(xstar, 1e-1) -``` - -###### Question - - - -Let $f(x) = \exp(x) - x^4$. As mentioned, there are 3 zeros for this function. Which one does Newton's method converge to when $x_0=8$? - - - -```julia; hold=true; echo=false -f(x) = exp(x) - x^4; -fp(x) = exp(x) - 4x^3; -xstar = Roots.newton(f, fp, 8); -numericq(xstar, 1e-1) -``` - -###### Question - - -Let $f(x) = \sin(x) - \cos(4\cdot x)$. - -Starting at $\pi/8$, solve for the root returned by Newton's method - - -```julia; hold=true; echo=false -k1=4 -f(x) = sin(x) - cos(k1*x); -fp(x) = cos(x) + k1*sin(k1*x); -val = Roots.newton(f, fp, pi/(2k1)); -numericq(val) -``` - - -###### Question - -Using Newton's method find a root to $f(x) = \cos(x) - x^3$ starting at $x_0 = 1/2$. - -```julia; hold=true; echo=false -f(x) = cos(x) - x^3 -val = Roots.newton(f,f', 1/2) -numericq(val) -``` - - - -###### Question - -Use Newton's method to find a root of $f(x) = x^5 + x -1$. Make a quick graph to find a reasonable starting point. - -```julia; hold=true; echo=false -f(x) = x^5 + x - 1 -val = Roots.newton(f,f', -1) -numericq(val) -``` - -###### Question - - - -```julia; hold=true;echo=false -##Consider the following illustration of Newton's method: -caption = """ -Illustration of Newton's method. Moving the point ``x_0`` shows different behaviours of the algorithm. -""" -## JSXGraph(:derivatives, "newtons-method.js", caption) -nothing -``` - -For the following graph, graphically consider the algorithm for a few different starting points. - -```julia; hold=true; echo=false -# placeholder until CWJ bumps up a version? -plot(x -> x^5 - x - 1, -1, 2) -``` - -If ``x_0`` is ``1`` what occurs? - -```julia;echo=false -nm_choices = [ -"The algorithm converges very quickly. A good initial point was chosen.", -"The algorithm converges, but slowly. The initial point is close enough to the answer to ensure decreasing errors.", -"The algrithm fails to converge, as it cycles about" -] -radioq(nm_choices, 1, keep_order=true) -``` - - -When ``x_0 = 1.0`` the following values are true for ``f``: - -```julia; echo=false -ff(x) = x^5 - x - 1 -α = find_zero(ff, 1) -function error_terms(x) - (e₀=x-α, f₀′= f'(x), f̄₀′′=f''(α), ē₁ = 1/2*f''(α)/f'(x)*(x-α)^2) -end -error_terms(1.0) -``` - -Where the values `f̄₀′′` and `ē₁` are worst-case estimates when ``\xi`` is between ``x_0`` and the zero. - -Does the magnitude of the error increase or decrease in the first step? - -```julia; hold=true; echo=false -radioq(["Appears to increase", "It decreases"],2,keep_order=true) -``` - - -If ``x_0`` is set near ``0.40`` what happens? - - -```julia; hold=true;echo=false -radioq(nm_choices, 3, keep_order=true) -``` - -When ``x_0 = 0.4`` the following values are true for ``f``: - -```julia; hold=true; echo=false -error_terms(0.4) -``` - -Where the values `f̄₀′′` and `ē₁` are worst-case estimates when ``\xi`` is between ``x_0`` and the zero. - -Does the magnitude of the error increase or decrease in the first step? - -```julia; hold=true;echo=false -radioq(["Appears to increase", "It decreases"],1,keep_order=true) -``` - - - -If ``x_0`` is set near ``0.75`` what happens? - - -```julia; hold=true;echo=false -radioq(nm_choices, 3, keep_order=true) -``` - - - - - -###### Question - -Will Newton's method converge for the function $f(x) = x^5 - x + 1$ starting at $x=1$? - -```julia; hold=true; echo=false -choices = [ -"Yes", -"No. The initial guess is not close enough", -"No. The second derivative is too big", -L"No. The first derivative gets too close to $0$ for one of the $x_i$"] -answ = 2 -radioq(choices, answ, keep_order=true) -``` - - - -###### Question - -Will Newton's method converge for the function $f(x) = 4x^5 - x + 1$ starting at $x=1$? - - -```julia; hold=true; echo=false -choices = [ -"Yes", -"No. The initial guess is not close enough", -"No. The second derivative is too big, or does not exist", -L"No. The first derivative gets too close to $0$ for one of the $x_i$"] -answ = 2 -radioq(choices, answ, keep_order=true) -``` - -###### Question - -Will Newton's method converge for the function $f(x) = x^{10} - 2x^3 - x + 1$ starting from $0.25$? - - -```julia; hold=true; echo=false -choices = [ -"Yes", -"No. The initial guess is not close enough", -"No. The second derivative is too big, or does not exist", -L"No. The first derivative gets too close to $0$ for one of the $x_i$"] -answ = 1 -radioq(choices, answ, keep_order=true) -``` - -###### Question - -Will Newton's method converge for $f(x) = 20x/(100 x^2 + 1)$ starting at $0.1$? - - -```julia; hold=true; echo=false -choices = [ -"Yes", -"No. The initial guess is not close enough", -"No. The second derivative is too big, or does not exist", -L"No. The first derivative gets too close to $0$ for one of the $x_i$"] -answ = 4 -radioq(choices, answ, keep_order=true) -``` - - - -###### Question - -Will Newton's method converge to a zero for $f(x) = \sqrt{(1 - x^2)^2}$? - - -```julia; hold=true; echo=false -choices = [ -"Yes", -"No. The initial guess is not close enough", -"No. The second derivative is too big, or does not exist", -L"No. The first derivative gets too close to $0$ for one of the $x_i$"] -answ = 3 -radioq(choices, answ, keep_order=true) -``` - -###### Question - -Use Newton's method to find a root of $f(x) = 4x^4 - 5x^3 + 4x^2 -20x -6$ starting at $x_0 = 0$. - -```julia; hold=true; echo=false -f(x) = 4x^4 - 5x^3 + 4x^2 -20x -6 -val = find_zero((f,f') , 0, Roots.Newton()) -numericq(val) -``` - -###### Question - -Use Newton's method to find a zero of $f(x) = \sin(x) - x/2$ that is *bigger* than $0$. - -```julia; hold=true; echo=false -f(x) = sin(x) - x/2 -val = find_zero((f,f'), 2, Roots.Newton()) -numericq(val) -``` - - -###### Question - -The Newton baffler (defined below) is so named, as Newton's method will fail to find the root for most starting points. - -```julia; -function newton_baffler(x) - if ( x - 0.0 ) < -0.25 - 0.75 * ( x - 0 ) - 0.3125 - elseif ( x - 0 ) < 0.25 - 2.0 * ( x - 0 ) - else - 0.75 * ( x - 0 ) + 0.3125 - end -end -``` - - -Will Newton's method find the zero at $0.0$ starting at $1$? - - -```julia; hold=true; echo=false -yesnoq("no") -``` - -Considering this plot: - -```julia; hold=true; -plot(newton_baffler, -1.1, 1.1) -``` - -Starting with $x_0=1$, you can see why Newton's method will fail. Why? - -```julia; hold=true; echo=false -choices = [ -L"It doesn't fail, it converges to $0$", -L"The tangent lines for $|x| > 0.25$ intersect at $x$ values with $|x| > 0.25$", -L"The first derivative is $0$ at $1$" -] -answ = 2 -radioq(choices, answ) -``` - - -This function does not have a small first derivative; or a large second derivative; and the bump up can be made as close to the origin as desired, so the starting point can be very close to the zero. However, even though the conditions of the error term are satisfied, the error term does not apply, as ``f`` is not continuously differentiable. - - -###### Question - -Let $f(x) = \sin(x) - x/4$. Starting at $x_0 = 2\pi$ Newton's method will converge to a value, but it will take many steps. Using the argument `verbose=true` for `find_zero`, how many steps does it take: - -```julia; hold=true; echo=false -f(x) = sin(x) - x/4 -x₀ = 2π -tracks = Roots.Tracks() -find_zero((f,f'), x₀, Roots.Newton(); tracks=tracks) -val = tracks.steps -numericq(val, 2) -``` - -What is the zero that is found? - -```julia; hold=true; echo=false -val = Roots.newton(f,f', 2pi) -numericq(val) -``` - -Is this the closest zero to the starting point, $x_0$? - -```julia; hold=true; echo=false -yesnoq("no") -``` - -###### Question - -Quadratic convergence of Newton's method only applies to *simple* -roots. For example, we can see (using the `verbose=true` argument to -the `Roots` package's `newton` method, that it only takes $4$ steps to -find a zero to $f(x) = \cos(x) - x$ starting at $x_0 = 1$. But it takes -many more steps to find the same zero for $f(x) = (\cos(x) - x)^2$. - -How many? - -```julia; hold=true; echo=false -val = 24 -numericq(val, 2) -``` - -###### Question: Implicit equations - -The equation $x^2 + x\cdot y + y^2 = 1$ is a rotated ellipse. - -```julia; hold=true; echo=false - -f(x,y) = x^2 + x * y + y^2 - 1 -implicit_plot(f, xlims=(-2,2), ylims=(-2,2), legend=false) -``` - -Can we find which point on its graph has the largest $y$ value? - -This would be straightforward *if* we could write $y(x) = \dots$, for then we would simply find the critical points and investiate. But we can't so easily solve for $y$ interms of $x$. However, we can use Newton's method to do so: - -```julia; -function findy(x) - fn = y -> (x^2 + x*y + y^2) - 1 - fp = y -> (x + 2y) - find_zero((fn, fp), sqrt(1 - x^2), Roots.Newton()) -end -``` - -For a *fixed* x, this solves for $y$ in the equation: $F(y) = x^2 + x \cdot y + y^2 - 1 = 0$. It should be that $(x,y)$ is a solution: - -```julia; hold=true; -x = .75 -y = findy(x) -x^2 + x*y + y^2 ## is this 1? -``` - -So we have a means to find $y(x)$, but it is implicit. - -Using `find_zero`, find the value $x$ which maximizes `y` by finding a zero of `y'`. Use this to find the point $(x,y)$ with largest $y$ value. - -```julia; hold=true; echo=false -xstar = find_zero(findy', 0.5) -ystar = findy(xstar) -choices = ["``(-0.57735, 1.15470)``", - "``(0,0)``", - "``(0, -0.57735)``", - "``(0.57735, 0.57735)``"] -answ = 1 -radioq(choices, answ) -``` - -(Using automatic derivatives works for values identified with `find_zero` *as long as* the initial point has its type the same as that of `x`.) - -###### Question - -In the last problem we used an *approximate* derivative (forward difference) in place of the derivative. -This can introduce an error due to the approximation. Would Newton's method still converge if the derivative in the algorithm were replaced with an approximate derivative? In general, this can often be done *but* the convergence can be *slower* and the sensitivity to a poor initial guess even greater. - -Three common approximations are given by the -difference quotient for a fixed $h$: $f'(x_i) \approx (f(x_i+h)-f(x_i))/h$; -the secant line approximation: $f'(x_i) \approx (f(x_i) - f(x_{i-1})) / (x_i - x_{i-1})$; and the -Steffensen approximation $f'(x_i) \approx (f(x_i + f(x_i)) - f(x_i)) / f(x_i)$ (using $h=f(x_i)$). - - -Let's revisit the $4$-step convergence of Newton's method to the root of $f(x) = 1/x - q$ when $q=0.8$. Will these methods be as fast? - - -Let's define the above approximations for a given `f`: - -```julia; -q₀ = 0.8 -fq(x) = 1/x - q₀ -secant_approx(x0,x1) = (fq(x1) - f(x0)) / (x1 - x0) -diffq_approx(x0, h) = secant_approx(x0, x0+h) -steff_approx(x0) = diffq_approx(x0, fq(x0)) -``` - -Then using the difference quotient would look like: - -```julia; hold=true; -Δ = 1e-6 -x1 = 42/17 - 32/17 * q₀ -x1 = x1 - fq(x1) / diffq_approx(x1, Δ) # |x1 - xstar| = 0.06511395862036995 -x1 = x1 - fq(x1) / diffq_approx(x1, Δ) # |x1 - xstar| = 0.003391809999860218; etc -``` - -The Steffensen method would look like: - -```julia; hold=true; -x1 = 42/17 - 32/17 * q₀ -x1 = x1 - fq(x1) / steff_approx(x1) # |x1 - xstar| = 0.011117056291670258 -x1 = x1 - fq(x1) / steff_approx(x1) # |x1 - xstar| = 3.502579696146313e-5; etc. -``` - -And the secant method like: - -```julia; hold=true; -Δ = 1e-6 -x1 = 42/17 - 32/17 * q₀ -x0 = x1 - Δ # we need two initial values -x0, x1 = x1, x1 - fq(x1) / secant_approx(x0, x1) # |x1 - xstar| = 8.222358365284066e-6 -x0, x1 = x1, x1 - fq(x1) / secant_approx(x0, x1) # |x1 - xstar| = 1.8766323799379592e-6; etc. -``` - -Repeat each of the above algorithms until `abs(x1 - 1.25)` is `0` (which will happen for this problem, though not in general). Record the steps. - -* Does the difference quotient need *more* than $4$ steps? - -```julia; hold=true; echo=false -yesnoq(false) -``` - -* Does the secant method need *more* than $4$ steps? - -```julia; hold=true; echo=false -yesnoq(true) -``` - -* Does the Steffensen method need *more* than 4 steps? - -```julia; hold=true; echo=false -yesnoq(false) -``` - -All methods work quickly with this well-behaved problem. In general -the convergence rates are slightly different for each, with the -Steffensen method matching Newton's method and the difference quotient -method being slower in general. All can be more sensitive to the initial guess. diff --git a/CwJ/derivatives/numeric_derivatives.jmd b/CwJ/derivatives/numeric_derivatives.jmd deleted file mode 100644 index 65e4547..0000000 --- a/CwJ/derivatives/numeric_derivatives.jmd +++ /dev/null @@ -1,347 +0,0 @@ -# Numeric derivatives - -This section uses these add-on packages: - -```julia -using CalculusWithJulia -using Plots -using ForwardDiff -using SymPy -using Roots -``` - -```julia; echo=false; results="hidden" -using CalculusWithJulia.WeaveSupport - -const frontmatter = ( - title = "Numeric derivatives", - description = "Calculus with Julia: Numeric derivatives", - tags = ["CalculusWithJulia", "derivatives", "numeric derivatives"], -); -nothing -``` - ----- - -`SymPy` returns symbolic derivatives. Up to choices of simplification, these answers match those that would be derived by hand. This is useful when comparing with known answers and for seeing the structure of the answer. However, there are times we just want to work with the answer numerically. For that we have other options within `Julia`. We discuss approximate derivatives and automatic derivatives. The latter will find wide usage in these notes. - -### Approximate derivatives - -By approximating the limit of the secant line with a value for a small, but positive, $h$, we get an approximation to the derivative. That is - -```math -f'(x) \approx \frac{f(x+h) - f(x)}{h}. -``` - -This is the forward-difference approximation. The central difference approximation looks both ways: - -```math -f'(x) \approx \frac{f(x+h) - f(x-h)}{2h}. -``` - -Though in general they are different, they are both -approximations. The central difference is usually more accurate for the -same size $h$. However, both are susceptible to round-off errors. The -numerator is a subtraction of like-size numbers - a perfect -opportunity to lose precision. - -As such there is a balancing act: - -* if $h$ is too small the round-off errors are problematic, -* if $h$ is too big, the approximation to the limit is not good. - -For the forward -difference $h$ values around $10^{-8}$ are typically good, for the central -difference, values around $10^{-6}$ are typically good. - -##### Example - -Let's verify that the forward difference isn't too far off. - -```julia; -f(x) = exp(-x^2/2) -c = 1 -h = 1e-8 -fapprox = (f(c+h) - f(c)) / h -``` - -We can compare to the actual with: - -```julia; -@syms x -df = diff(f(x), x) -factual = N(df(c)) -abs(factual - fapprox) -``` - -The error is about ``1`` part in ``100`` million. - -The central difference is better here: - -```julia; hold=true -h = 1e-6 -cdapprox = (f(c+h) - f(c-h)) / (2h) -abs(factual - cdapprox) -``` - ----- - -The [FiniteDifferences](https://github.com/JuliaDiff/FiniteDifferences.jl) and [FiniteDiff](https://github.com/JuliaDiff/FiniteDiff.jl) packages provide performant interfaces for differentiation based on finite differences. - -### Automatic derivatives - -There are some other ways to compute derivatives numerically that give -much more accuracy at the expense of slightly increased computing -time. Automatic differentiation is the general name for a few -different approaches. These approaches promise less complexity - in -some cases - than symbolic derivatives and more accuracy than -approximate derivatives; the accuracy is on the order of -machine precision. - -The `ForwardDiff` package provides one of [several](https://juliadiff.org/) ways for `Julia` to compute automatic derivatives. `ForwardDiff` is well suited for functions encountered in these notes, which depend on at most a few variables and output no more than a few values at once. - - -The `ForwardDiff` package was loaded in this section; in general its features are available when the `CalculusWithJulia` package is loaded, as that package provides a more convenient interface. -The `derivative` function is not exported by `FiniteDiff`, so its usage requires qualification. To illustrate, to find the derivative of $f(x)$ at a *point* we have this syntax: - -```julia -ForwardDiff.derivative(f, c) # derivative is qualified by a module name -``` - -The `CalculusWithJulia` package defines an operator `D` which goes from finding a derivative at a point with `ForwardDiff.derivative` to defining a function which evaluates the derivative at each point. It is defined along the lines of `D(f) = x -> ForwardDiff.derivative(f,x)` in parallel to how the derivative operation for a function is defined mathematically from the definition for its value at a point. - - -Here we see the error in estimating $f'(1)$: - -```julia; -fauto = D(f)(c) # D(f) is a function, D(f)(c) is the function called on c -abs(factual - fauto) -``` - -In this case, it is exact. - - -The `D` operator is defined for most all functions in `Julia`, though, like the `diff` operator in `SymPy` there are some for which it won't work. - -##### Example - -For $f(x) = \sqrt{1 + \sin(\cos(x))}$ compare the difference between the forward derivative with $h=1e-8$ and that computed by `D` at $x=\pi/4$. - -The forward derivative is found with: - -```julia; -𝒇(x) = sqrt(1 + sin(cos(x))) -𝒄, 𝒉 = pi/4, 1e-8 -fwd = (𝒇(𝒄+𝒉) - 𝒇(𝒄))/𝒉 -``` - -That given by `D` is: - -```julia; -ds_value = D(𝒇)(𝒄) -ds_value, fwd, ds_value - fwd -``` - -Finally, `SymPy` gives an exact value we use to compare: - -```julia; -𝒇𝒑 = diff(𝒇(x), x) -``` - -```julia -actual = N(𝒇𝒑(PI/4)) -actual - ds_value, actual - fwd -``` - -#### Convenient notation - -`Julia` allows the possibility of extending functions to different -types. Out of the box, the `'` notation is not employed for functions, -but is used for matrices. It is used in postfix position, as with -`A'`. We can define it to do the same thing as `D` for functions and -then, we can evaluate derivatives with the familiar `f'(x)`. -This is done in `CalculusWithJulia` along the lines of `Base.adjoint(f::Function) = D(f)`. - - -Then, we have, for example: - -```julia; hold=true; -f(x) = sin(x) -f'(pi), f''(pi) -``` - - - - - -##### Example - -Suppose our task is to find a zero of the second derivative of $k(x) = -e^{-x^2/2}$ in $[0, 10]$, a known bracket. The `D` function takes a second argument to indicate the order of the derivative (e.g., `D(f,2)`), but we use the more familiar notation: - - -```julia; hold=true -k(x) = exp(-x^2/2) -find_zero(k'', 0..10) -``` - -We pass in the function object, `k''`, and not the evaluated function. - - -## Recap on derivatives in Julia - -A quick summary for finding derivatives in `Julia`, as there are $3$ different manners: - -* Symbolic derivatives are found using `diff` from `SymPy` -* Automatic derivatives are found using the notation `f'` using `ForwardDiff.derivative` -* approximate derivatives at a point, `c`, for a given `h` are found with `(f(c+h)-f(c))/h`. - - -For example, here all three are computed and compared: - -```julia; hold=true -f(x) = exp(-x)*sin(x) - -c = pi -h = 1e-8 - -fp = diff(f(x),x) - -fp, fp(c), f'(c), (f(c+h) - f(c))/h -``` - -!!! note - The use of `'` to find derivatives provided by `CalculusWithJulia` is convenient, and used extensively in these notes, but it needs to be noted that it does **not conform** with the generic meaning of `'` within `Julia`'s wider package ecosystem and may cause issue with linear algebra operations; the symbol is meant for the adjoint of a matrix. - - -## Questions - - -##### Question - -Find the derivative using a forward difference approximation of $f(x) = x^x$ at the point $x=2$ using `h=0.1`: - -```julia; hold=true; echo=false -f(x) = x^x -c, h = 2, 0.1 -val = (f(c+h) - f(c))/h -numericq(val) -``` - -Using `D` or `f'` find the value using automatic differentiation - -```julia; hold=true; echo=false -f(x) = x^x -c = 2 -val = f'(c) -numericq(val) -``` - - - -##### Question - -Mathematically, as the value of `h` in the forward difference gets -smaller the forward difference approximation gets better. On the -computer, this is thwarted by floating point representation issues (in -particular the error in subtracting two like-sized numbers in forming -$f(x+h)-f(x)$.) - -For `1e-16` what is the error (in absolute value) in finding the forward difference -approximation for the derivative of $\sin(x)$ at $x=0$? - -```julia; hold=true; echo=false -f(x) = sin(x) -h = 1e-16 -c = 0 -approx = (f(c+h)-f(c))/h -val = abs(cos(0) - approx) -numericq(val) -``` - -Repeat for $x=\pi/4$: - - -```julia; hold=true; echo=false -f(x) = sin(x) -h = 1e-16 -c = pi/4 -approx = (f(c+h)-f(c))/h -val = abs(cos(0) - approx) -numericq(val) -``` - - - - - -###### Question - -Let $f(x) = x^x$. Using `D`, find $f'(3)$. - -```julia; hold=true; echo=false -f(x) = x^x -val = D(f)(3) -numericq(val) -``` - -###### Question - - -Let $f(x) = \lvert 1 - \sqrt{1 + x}\rvert$. Using `D`, find $f'(3)$. - -```julia; hold=true; echo=false -f(x) = abs(1 - sqrt(1 + x)) -val = D(f)(3) -numericq(val) -``` - - -###### Question - - -Let $f(x) = e^{\sin(x)}$. Using `D`, find $f'(3)$. - -```julia; hold=true; echo=false -f(x) = exp(sin(x)) -val = D(f)(3) -numericq(val) -``` - - - - -###### Question - -For `Julia`'s -`airyai` function find a numeric derivative using the -forward difference. For $c=3$ and $h=10^{-8}$ find the forward -difference approximation to $f'(3)$ for the `airyai` function. - -```julia; hold=true; echo=false -h = 1e-8 -c = 3 -val = (airyai(c+h) - airyai(c))/h -numericq(val) -``` - - -###### Question - -Find the rate of change with respect to time of the function $f(t)= 64 - 16t^2$ at $t=1$. - -```julia; hold=true; echo=false -fp_(t) = -16*2*t -c = 1 -numericq(fp_(c)) -``` - -###### Question - -Find the rate of change with respect to height, $h$, of $f(h) = 32h^3 - 62 h + 12$ at $h=2$. - -```julia; hold=true; echo=false -fp_(h) = 3*32h^2 - 62 -c = 2 -numericq(fp_(2)) -``` diff --git a/CwJ/derivatives/optimization-trapezoid.js b/CwJ/derivatives/optimization-trapezoid.js deleted file mode 100644 index eae3b4e..0000000 --- a/CwJ/derivatives/optimization-trapezoid.js +++ /dev/null @@ -1,36 +0,0 @@ -// inscribe trapezoid -var R = 5; -var Delta = 0.5 -const b = JXG.JSXGraph.initBoard('jsxgraph', { - boundingbox: [-R-Delta,R+Delta,R+Delta,-1], axis:true -}); - -var xax = b.create("segment", [[0,0],[R,0]]); - -var P4 = b.create("glider", [R/2,0, xax], {name: "P_4=(r,0)"}); - - - -var CL = b.create('point', [function() {return -P4.X()},0], {name:''}); -var CR = b.create('point', [function() {return P4.X()},0], {name:''}); -var C = b.create('semicircle', [CL,CR]); - -var Crestricted = b.create("functiongraph", - [function(x) { - r = P4.X(); - y = Math.sqrt(r*r - x*x); - return y; - }, 0, function() {return P4.X()}]); - -var P3 = b.create("glider", [ - P4.X()/2, - Math.sqrt(P4.X()*P4.X()*(1 - 1/4)), - Crestricted], {name:"P_3=(x,y)"}); - -var P1 = b.create('point', [function() {return -Math.abs(P4.X());}, - function() {return P4.Y();}], {name:'P_1'}); -var P2 = b.create('point', [function() {return -P3.X();}, - function() {return P3.Y();}], {name:'P_2'}); - -var poly = b.create('polygon',[P1, P2, P3, P4], { borders:{strokeColor:'black'} }); -b.create('text',[-1.5,.25, function(){ return 'Area='+ poly.Area().toFixed(1); }]); diff --git a/CwJ/derivatives/optimization.jmd b/CwJ/derivatives/optimization.jmd deleted file mode 100644 index 8ba8471..0000000 --- a/CwJ/derivatives/optimization.jmd +++ /dev/null @@ -1,1402 +0,0 @@ -# Optimization - -This section uses these add-on packages: - -```julia -using CalculusWithJulia -using Plots -using Roots -using SymPy -``` - -```julia; echo=false; results="hidden" -using CalculusWithJulia.WeaveSupport - -fig_size = (800, 600) -frontmatter = ( - title = "Optimization", - description = "Calculus with Julia: Optimization", - tags = ["CalculusWithJulia", "derivatives", "optimization"], -); - -nothing -``` - ----- - -A basic application of calculus is to answer -questions which relate to the largest or smallest a function can be given -some constraints. - - -For example, - - -> Of all rectangles with perimeter ``20``, which has of the largest area? - - -The main tool is the extreme value theorem of Bolzano and Fermat's -theorem about critical points, which combined say: - -> If the function $f(x)$ is continuous on $[a,b]$ and differentiable -> on $(a,b)$, then the extrema exist and must -> occur at either an end point or a critical point. - - -Though not all of our problems lend themselves to a description of a -continuous function on a closed interval, if they do, we have an -algorithmic prescription to find the absolute extrema of a function: - -1) Find the critical points. For this we can use a root-finding algorithm like `find_zero`. - -2) Evaluate the function values at the critical points and at the end points. - -3) Identify the largest and smallest values. - -With the computer we can take some shortcuts, as we will be able to -graph our function to see where the extreme values will be, and in particular if they occur at end points or critical points. - -## Fixed perimeter and area - -The simplest way to investigate the maximum or minimum value of a -function over a closed interval is to just graph it and look. - -We began with the question of which rectangles of perimeter ``20`` have -the largest area? The figure shows a few different rectangles with -this perimeter and their respective areas. - -```julia; hold=true; echo=false; cache=true -### {{{perimeter_area_graphic}}} - - -function perimeter_area_graphic_graph(n) - h = 1 + 2n - w = 10-h - plt = plot([0,0,w,w,0], [0,h,h,0,0], legend=false, size=fig_size, - xlim=(0,10), ylim=(0,10)) - scatter!(plt, [w], [h], color=:orange, markersize=5) - annotate!(plt, [(w/2, h/2, "Area=$(round(w*h,digits=1))")]) - plt -end - -caption = """ - -Some possible rectangles that satisfy the constraint on the perimeter and their area. - -""" -n = 6 -anim = @animate for i=1:n - perimeter_area_graphic_graph(i-1) -end - -imgfile = tempname() * ".gif" -gif(anim, imgfile, fps = 1) - -ImageFile(imgfile, caption) -``` - - -The basic mathematical approach is to find a function of a single -variable to maximize or minimize. In this case we have two variables -describing a rectangle: a base $b$ and height $h$. Our formulas are the area of a rectangle: - -```math -A = bh, -``` - -and the formula for the perimeter of a rectangle: - -```math -P = 2b + 2h = 20. -``` - -From this last one, we see that $b$ can be no bigger than ``10`` and no -smaller than ``0`` from the restriction put in place through the -perimeter. Solving for $h$ in terms of $b$ then yields this -restatement of the problem: - -Maximize $A(b) = b \cdot (10 - b)$ over the interval $[0,10]$. - -This is exactly the form needed to apply our theorem about the -existence of extrema (a continuous function on a closed -interval). Rather than solve analytically by taking a derivative, we -simply graph to find the value: - -```julia; -Area(b) = b * (10 - b) -plot(Area, 0, 10) -``` - -You should see the maximum occurs at $b=5$ by symmetry, so $h=5$ as -well, and the maximum area is then $25$. This gives the satisfying -answer that among all rectangles of fixed perimeter, that with the -largest area is a square. As well, this indicates a common result: -there is often some underlying symmetry in the answer. - -### Exploiting polymorphism - -Before moving on, let's see a slightly different way to do this -problem with `Julia`, where we trade off some algebra for a bit of -abstraction. This technique was discussed in the section on -[functions](../precalc/functions.html). - -Let's first write area as a function of both base and height: - -```julia; -A(b, h) = b * h -``` - -From the constraint given by the perimeter being a fixed value we -can solve for `h` in terms of `b`. We write this as a function: - -```julia; -h(b) = (20 - 2b) / 2 -``` - -To get `A(b)` we simply need to substitute `h(b)` into our -formula for the area, `A`. However, instead of doing the substitution -ourselves using algebra we let `Julia` do it through composition of -functions: - -```julia; -A(b) = A(b, h(b)) -``` - -Now we can solve graphically as before, or numerically, such as here where -we search for zeros of the derivative: - -```julia; -find_zeros(A', 0, 10) # find_zeros in `Roots`, -``` - -(As a reminder, the notation `A'` is defined in `CalculusWithJulia` using the `derivative` function from the `ForwardDiff` package.) - - - -!!! note - Look at the last definition of `A`. The function `A` appears on both sides, though on the left side with one argument and on the right with two. These are two "methods" of a *generic* function, `A`. `Julia` allows multiple definitions for the same name as long as the arguments (their number and type) can disambiguate which to use. In this instance, when one argument is passed in then the last defintion is used (`A(b,h(b))`), whereas if two are passed in, then the method that multiplies both arguments is used. The advantage of multiple dispatch is illustrated: the same concept - area - has one function name, though there may be different ways to compute the area, so there is more than one implementation. - - -#### Example: Norman windows - -Here is a similar, though more complicated, example where the analytic -approach can be a bit more tedious, but the graphical one mostly -satisfying, though we do use a numerical algorithm to find -an exact final answer. - -Let a "[Norman](https://en.wikipedia.org/wiki/Norman_architecture)" -window consist of a rectangular window of top length $x$ and side -length $y$ and a half circle on top. The goal is to maximize the area -for a fixed value of the perimeter. Again, assume this perimeter is ``20`` -units. - -This figure shows two such windows, one with base length given by -$x=3$, the other with base length given by $x=4$. The one with base -length $4$ seems to have much bigger area, what value of $x$ will lead to the largest area? - -```julia; hold=true; echo=false -ts = range(0, stop=pi, length=50) -x1,y1 = 4, 4.85840 -x2,y2 = 3, 6.1438 -delta = 4 -p = plot(delta .+ x1*[0, 1,1,0], y1*[0,0,1,1], linetype=:polygon, fillcolor=:blue, legend=false) -plot!(p, x2*[0, 1,1,0], y2*[0,0,1,1], linetype=:polygon, fillcolor=:blue) - -plot!(p, delta .+ x1/2 .+ x1/2*cos.(ts), y1.+x1/2*sin.(ts), linetype=:polygon, fillcolor=:red) -plot!(p, x2/2 .+ x2/2*cos.(ts), y2 .+ x2/2*sin.(ts), linetype=:polygon, fillcolor=:red) -p -``` - -For this problem, we have two equations. - -The area is the area of the rectangle plus the area of the half circle ($\pi r^2/2$ with $r=x/2$). - -```math -A = xy + \pi(x/2)^2/2 -``` - -In `Julia` this is - -```julia; -Aᵣ(x, y) = x*y + pi*(x/2)^2 / 2 -``` - - -The perimeter consists of ``3`` sides of the rectangle and the perimeter -of half a circle ($\pi r$, with $r=x/2$): - -```math -P = 2y + x + \pi(x/2) = 20 -``` - -We solve for $y$ in the first with $y = (20 - x - \pi(x/2))/2$ so that in `Julia` we have: - -```julia; -y(x) = (20 - x - pi * x/2) / 2 -``` - -And then we substitute in `y(x)` for `y` in the area formula through: - -```julia; -Aᵣ(x) = Aᵣ(x, y(x)) -``` - -Of course both $x$ and $y$ are non-negative. The latter forces $x$ to -be no more than $x=20/(1+\pi/2)$. - -This leaves us the calculus problem of finding an absolute maximum of -a continuous function over the closed interval -$[0, 20/(1+\pi/2)]$. Our theorem tells us this maximum must occur, we -now proceed to find it. - -We begin by simply graphing and estimating the values of the maximum and -where it occurs. - -```julia; -plot(Aᵣ, 0, 20/(1+pi/2)) -``` - -The naked eye sees that maximum value is somewhere around $27$ and -occurs at $x\approx 5.6$. Clearly from the graph, we know the maximum -value happens at the critical point and there is only one such -critical point. - -As reading the maximum from the graph is more difficult than reading a -$0$ of a function, we plot the derivative using our approximate -derivative. - -```julia; -plot(Aᵣ', 5.5, 5.7) -``` - -We confirm that the critical point is around $5.6$. - -#### Using `find_zero` to locate critical points. - -Rather than zoom in graphically, we now use a root-finding algorithm, -to find a more precise value of the zero of ``A'``. We know that the -maximum will occur at a critical point, a zero of the derivative. The -`find_zero` function from the `Roots` package provides a non-linear -root-finding algorithm based on the bisection method. The only thing -to keep track of is that solving $f'(x) = 0$ means we use the -derivative and not the original function. - -We see from the graph that $[0, 20/(1+\pi/2)]$ will provide a bracket, as there is only one relative maximum: - -```julia; -x′ = find_zero(Aᵣ', (0, 20/(1+pi/2))) -``` - -This value is the lone critical point, and in this case gives -the position of the value that will maximize the function. The value and maximum area are then given by: - -```julia; -(x′, Aᵣ(x′)) -``` - -(Compare this answer to the previous, is the square the figure of -greatest area for a fixed perimeter, or just the figure amongst all -rectangles? See [Isoperimetric inequality](https://en.wikipedia.org/wiki/Isoperimetric_inequality) for an answer.) - - -### Using `argmax` to identify where a function is maximized - -This value that maximizes a function is sometimes referred to as the -*argmax*, or argument which maximizes the function. In `Julia` the -`argmax(f,domain)` function is defined to "Return a value ``x`` in the -domain of ``f`` for which ``f(x)`` is maximized. If there are multiple -maximal values for ``f(x)`` then the first one will be found." The -domain is some iterable collection. In the mathematical world this -would be an interval ``[a,b]``, but on the computer it is an -approximation, such as is returned by `range` below. Without out -having to take a derivative, as above, but sacrificing some accuracy, -the task of identifying `x` for where `A` is maximum, could be done -with - -```julia -argmax(Aᵣ, range(0, 20/(1+pi/2), length=10000)) -``` - - -#### A symbolic approach - -We could also do the above problem symbolically with the aid of `SymPy`. Here are the steps: - -```julia; -@syms 𝒘::real 𝒉::real - -𝑨₀ = 𝒘 * 𝒉 + pi * (𝒘/2)^2 / 2 -𝑷erim = 2*𝒉 + 𝒘 + pi * 𝒘/2 -𝒉₀ = solve(𝑷erim - 20, 𝒉)[1] -𝑨₁ = 𝑨₀(𝒉 => 𝒉₀) -𝒘₀ = solve(diff(𝑨₁,𝒘), 𝒘)[1] -``` - - -We know that `𝒘₀` is the maximum in this example from our previous -work. We shall see soon, that just knowing that the second derivative -is negative at `𝒘₀` would suffice to know this. Here we check that -condition: - -```julia; -diff(𝑨₁, 𝒘, 𝒘)(𝒘 => 𝒘₀) -``` - -As an aside, compare the steps involved above for a symbolic solution to those of previous work for a numeric solution: - -```julia; hold=true -Aᵣ(w, h) = w*h + pi*(w/2)^2 / 2 -h(w) = (20 - w - pi * w/2) / 2 -Aᵣ(w) = A(w, h(w)) -find_zero(A', (0, 20/(1+pi/2))) # 40 / (pi + 4) -``` - -They are similar, except we solved for `h0` symbolically, rather than by hand, when we solved for `h(w)`. - -##### Example - -```julia; hold=true; echo=false -caption = """ -The figure shows a trapezoid inscribed in a circle. By adjusting the point ``P_3 = (x,y)`` on the upper-half circle, the area of the trapezoid changes. What value of ``x`` will produce the maximum area for a given ``r`` (from ``P_4``, which can also be adjusted)? By playing around with different values of ``P_3`` and ``P_4`` the answer can be guessed. -""" - -#JSXGraph(:derivatives, "optimization-trapezoid.js", caption) -nothing -``` - -```julia; hold=true; echo=false -function trapezoid(r) - plot(x -> sqrt(1 - x^2), -1, 1, legend=false, aspect_ratio=:equal) - plot!([-1,1,r,-r,-1], [0,0,sqrt(1-r^2), sqrt(1-r^2), 0], lw=3, color=:red) -end -trapezoid(0.75) -``` - - -A trapezoid is *inscribed* in the upper-half circle of radius ``r``. The trapezoid is found be connecting the points ``(x,y)`` (in the first quadrant) with ``(r, 0)``, ``(-r,0)``, and ``(-x, y)``. Find the maximum area. (The above figure has ``x=0.75`` and ``r=1``.) - -Here the constraint is simply ``r^2 = x^2 + y^2`` with ``x`` and ``y`` being non-negative. The area is then found through the average of the two lengths times the height. Using `height` for `y`, we have: - -```julia -@syms x::positive r::positive -hₜ = sqrt(r^2 - x^2) -aₜ = (2x + 2r)/2 * hₜ -possible_sols = solve(diff(aₜ, x) ~ 0, x) # possibly many solutions -x0 = first(possible_sols) # only solution is also found from first or [1] indexing -``` - -The other values of interest can be found through substitution. For example: - -```julia -hₜ(x => x0) -``` - -## Trigonometry problems - -Many maximization and minimization problems involve triangles, which -in turn use trigonometry in their description. Here is an example, the -"ladder corner problem." (There are many other [ladder](http://www.mathematische-basteleien.de/ladder.htm) problems.) - - -A ladder is to be moved through a two-dimensional hallway which has a -bend and gets narrower after the bend. The hallway is ``8`` feet wide then -``5`` feet wide. What is the longest such ladder that can be navigated -around the corner? - -The figure shows a ladder of length $l_1 + l_2$ that got stuck - it -was too long. - - -```julia; hold=true; echo=false -p = plot([0, 0, 15], [15, 0, 0], color=:blue, legend=false) -plot!(p, [5, 5, 15], [15, 8, 8], color=:blue) -plot!(p, [0,14.53402874075368], [12.1954981558864, 0], linewidth=3) -plot!(p, [0,5], [8,8], color=:orange) -plot!(p, [5,5], [0,8], color=:orange) -annotate!(p, [(13, 1/2, "θ"), - (2.5, 11, "l₂"), (10, 5, "l₁"), (2.5, 7.0, "l₂ ⋅ cos(θ)"), - (5.1, 4, "l₁ ⋅ sin(θ)")]) -``` - - - -We approach this problem in reverse. It is easy to see when a ladder -is too long. It gets stuck at some angle $\theta$. So for each -$\theta$ we find that ladder length that is just too long. Then we -find the minimum length of all these ladders that are too long. If a -ladder is this length or more it will get stuck for some -angle. However, if it is less than this length it will not get stuck. -So to maximize a ladder length, we minimize a different -function. Neat. - -Now, to find the length $l = l_1 + l_2$ as a function of $\theta$. - -We need to brush off our trigonometry, in particular right triangle -trigonometry. We see from the figure that $l_1$ is the hypotenuse of a -right triangle with opposite side $8$ and $l_2$ is the hypotenuse of a -right triangle with adjacent side $5$. So, $8/l_1 = \sin\theta$ and -$5/l_2 = \cos\theta$. - - -That is, we have - -```julia; -l(l1, l2) = l1 + l2 -l1(t) = 8/sin(t) -l2(t) = 5/cos(t) - -l(t) = l(l1(t), l2(t)) # or simply l(t) = 8/sin(t) + 5/cos(t) -``` - -Our goal is to minimize this function for all angles between $0$ and $90$ degrees, or $0$ and $\pi/2$ radians. - -This is not a continuous function on a closed interval - it is -undefined at the endpoints. That being said, a quick plot will -convince us that the minimum occurs at a critical point and there is -only one critical point in $(0, \pi/2)$. - -```julia; -delta = 0.2 -plot(l, delta, pi/2 - delta) -``` - -The graph shows the minimum occurs between ``0.50`` and ``1.00`` radians, a bracket for the derivative. Here we find ``x`` and the minimum value: - -```julia; hold=true; -x = find_zero(l', (0.5, 1.0)) -x, l(x) -``` - -That is, any ladder less than this length can get around the hallway. - -## Rate times time problems - -Ethan Hunt, a top secret spy, has a mission to chase a bad guy. Here -is what we know: - -* Ethan likes to run. He can run at ``10`` miles per hour. -* He can drive a car - usually some concept car by BMW - at ``30`` miles per hour, but only on the road. - -For his mission, he needs to go ``10`` miles west and ``5`` `miles north. He -can do this by: - -* just driving ``8.310`` miles west then ``5`` miles north, or -* just running the diagonal distance, or -* driving $0 < x < 10$ miles west, then running on the diagonal - - -A quick analysis says: - -* It would take $(10+5)/30$ hours to just drive -* It would take $\sqrt{10^2 + 5^2}/10$ hours to just run - -Now, if he drives $x$ miles west ($0 < x < 10$) he would run an amount -given by the hypotenuse of a triangle with lengths $5$ and $10-x$. His -time driving would be $x/30$ and his time running would be -$\sqrt{5^2+(10-x)^2}/10$ for a total of: - -```math -T(x) = x/30 + \sqrt{5^2 + (10-x)^2}/10, \quad 0 < x < 10 -``` - -With the endpoints given by -$T(0) = \sqrt{10^2 + 5^2}/10$ and $T(10) = (10 + 5)/30$. - - -Let's plot $T(x)$ over the interval $(0,10)$ and look: - -```julia; -T(x) = x/30 + sqrt(5^2 + (10-x)^2)/10 -``` - -```julia; -plot(T, 0, 10) -``` - -The minimum happens way out near 8. We zoom in a bit: - -```julia; -plot(T, 7, 9) -``` - -It appears to be around ``8.3``. We now use `find_zero` to refine our -guess at the critical point using $[7,9]$: - -```julia; -α = find_zero(T', (7, 9)) -``` - -Okay, got it. Around``8.23``. So is our minimum time - -```julia; -T(α) -``` - -We know this is a relative minimum, but not that it is the global -minimum over the closed time interlal. For that we must also check the -endpoints: - -```julia; -sqrt(10^2 + 5^2)/10, T(α), (10+5)/30 -``` - -Ahh, we see that $T(x)$ is not continuous on $[0, 10]$, as it jumps at -$x=10$ down to an even smaller amount of $1/2$. It may not look as -impressive as a miles-long sprint, but Mr. Hunt is advised by Benji to drive -the whole way. - -### Rate times time ... the origin story - -```julia;hold=true; echo=false -### {{{lhopital_43}}} - -imgfile = "figures/fcarc-may2016-fig43-250.gif" -caption = L""" - -Image number $43$ from l'Hospital's calculus book (the first). A -traveler leaving location $C$ to go to location $F$ must cross two -regions separated by the straight line $AEB$. We suppose that in the -region on the side of $C$, he covers distance $a$ in time $c$, and -that on the other, on the side of $F$, distance $b$ in the same time -$c$. We ask through which point $E$ on the line $AEB$ he should pass, -so as to take the least possible time to get from $C$ to $F$? (From -http://www.ams.org/samplings/feature-column/fc-2016-05.) - - -""" -ImageFile(:derivatives, imgfile, caption) -``` - -The last example is a modern day illustration of a problem of calculus -dating back to l'Hospital. His parameterization is a bit -different. Let's change his by taking two points $(0, a)$ and -$(L,-b)$, with $a,b,L$ positive values. Above the $x$ axis travel -happens at rate $r_0$, and below, travel happens at rate $r_1$, again, -both positive. What value $x$ in $[0,L]$ will minimize the total travel time? - -We approach this symbolically with `SymPy`: - -```julia; -@syms x::positive a::positive b::positive L::positive r0::positive r1::positive - -d0 = sqrt(x^2 + a^2) -d1 = sqrt((L-x)^2 + b^2) - -t = d0/r0 + d1/r1 # time = distance/rate -dt = diff(t, x) # look for critical points -``` - -The answer will occur at a critical point or an endpoint, either $x=0$ or $x=L$. - -The structure of `dt` is too complicated for simply calling `solve` to find the critical points. Instead we help `SymPy` out a bit. We are solving an equation of the form $a/b + c/d = 0$. These solutions will also be solutions of $(a/b)^2 - (c/d)^2=0$ or even $a^2d^2 - c^2b^2 = 0$. This follows as solutions to $u+v=0$, also solve $(u+v)\cdot(u-v)=0$, or $u^2 - v^2=0$. Setting $u=a/b$ and $v=c/d$ completes the comparison. - -We can get these terms - $a$, $b$, $c$, and $d$ - as follows: - -```julia; -t1, t2 = dt.args # the `args` property returns the arguments to the outer function (+ in this case) -``` - -The equivalent of $a^2d^2 - c^2 b^2$ is found using the generic functions `numerator` and `denominator` to access the numerator and denominator of the fractions: - -```julia; -ex = numerator(t1^2)*denominator(t2^2) - denominator(t1^2)*numerator(t2^2) -``` - -This is a polynomial in the `x` variable of degree $4$, as seen here where the `sympy.Poly` function is used to identify the symbols of the polynomial from the parameters: - -```julia; -p = sympy.Poly(ex, x) # a0 + a1⋅x + a2⋅x^2 + a3⋅x^3 + a4⋅x^4 -p.coeffs() -``` - -Fourth degree polynomials can be solved. The critical points of the -original equation will be among the ``4`` solutions given. However, the result -is complicated. The -[article](http://www.ams.org/samplings/feature-column/fc-2016-05) -- from -which the figure came -- states that "In today's textbooks the problem, -usually involving a river, involves walking along one bank and then -swimming across; this corresponds to setting $g=0$ in l'Hospital's -example, and leads to a quadratic equation." Let's see that case, -which we can get in our notation by taking $b=0$: - -```julia; -q = ex(b=>0) -factor(q) -``` - -We see two terms: one with $x=L$ and another quadratic. For the simple -case $r_0=r_1$, a straight line is the best solution, and this -corresponds to $x=L$, which is clear from the formula above, as we -only have one solution to the following: - -```julia; -solve(q(r1=>r0), x) -``` - -Well, not so fast. We need to check the other endpoint, $x=0$: - -```julia; -ta = t(b=>0, r1=>r0) -ta(x=>0), ta(x=>L) -``` - -The value at $x=L$ is smaller, as $L^2 + a^2 \leq (L+a)^2$. (Well, that was a bit pedantic. The travel rates being identical means the fastest path will also be the shortest path and that is clearly ``x=L`` and not ``x=0``.) - - -Now, if, say, travel above the line is half as slow as travel along, then $2r_0 = r_1$, and the critical points will be: - -```julia; -out = solve(q(r1 => 2r0), x) -``` - -It is hard to tell which would minimize time without more work. To check a case ($a=1, L=2, r_0=1$) we might have - -```julia; -x_straight = t(r1 =>2r0, b=>0, x=>out[1], a=>1, L=>2, r0 => 1) # for x=L -``` - -Compared to the smaller ($x=\sqrt{3}a/3$): - -```julia; -x_angle = t(r1 =>2r0, b=>0, x=>out[2], a=>1, L=>2, r0 => 1) -``` - -What about $x=0$? - -```julia; -x_bent = t(r1 =>2r0, b=>0, x=>0, a=>1, L=>2, r0 => 1) -``` - -The value of $x=\sqrt{3}a/3$ minimizes time: - -```julia; -min(x_straight, x_angle, x_bent) -``` - -The traveler in this case is advised to head to the $x$ axis at $x=\sqrt{3}a/3$ and then travel along the $x$ axis. - - - -Will this approach always be true? Consider different parameters, say we -switch the values of $a$ and $L$ so $a > L$: - -```julia; -pts = [0, out...] -m,i = findmin([t(r1 =>2r0, b=>0, x=>u, a=>2, L=>1, r0 => 1) for u in pts]) # min, index -m, pts[i] -``` - -Here traveling directly to the point $(L,0)$ is fastest. Though travel -is slower, the route is more direct and there is no time saved by -taking the longer route with faster travel for part of it. - - - - -## Unbounded domains - -Maximize the function $xe^{-(1/2) x^2}$ over the interval $[0, \infty)$. - -Here the extreme value theorem doesn't technically apply, as we don't -have a closed interval. However, **if** we can eliminate the endpoints as -candidates, then we should be able to convince ourselves the maximum -must occur at a critical point of $f(x)$. (If not, then convince yourself for all sufficiently large $M$ the maximum over $[0,M]$ occurs at -a critical point, not an endpoint. Then let $M$ go to infinity. In general, for an optimization problem of a continuous function on the interval ``(a,b)`` if the right limit at ``a`` and left limit at ``b`` can be ruled out as candidates, the optimal value must occur at a critical point.) - -So to approach this problem we first graph it over a wide interval. - -```julia; -f(x) = x * exp(-x^2) -plot(f, 0, 100) -``` - -Clearly the action is nearer to ``1`` than ``100``. We try graphing the -derivative near that area: - -```julia; -plot(f', 0, 5) -``` - -This shows the value of interest near $0.7$ for a critical point. We use `find_zero` with $[0,1]$ as a bracket - -```julia; -c = find_zero(f', (0, 1)) -``` - -The maximum is then at - -```julia; -f(c) -``` - -##### Example: Minimize the surface area of a can - - -For a more applied problem of this type (infinite domain), consider a -can of some soft drink that is to contain ``355``ml which is ``355`` cubic -centimeters. We use metric units, as the relationship between volume -(cubic centimeters) and fluid amount (ml) is clear. A can to hold -this amount is produced in the shape of cylinder with radius $r$ and -height $h$. The materials involved give the surface area, which would -be: - -```math -SA = h \cdot 2\pi r + 2 \cdot \pi r^2. -``` - -The volume satisfies: - -```math -V = 355 = h \cdot \pi r^2. -``` - -Find the values of $r$ and $h$ which minimize the surface area. - - -First the surface area in both variables is given by - -```julia; -SA(h, r) = h * 2pi * r + 2pi * r^2 -``` - -Solving from the constraint on the volume for `h` in terms of `r` yields: - -```julia; -canheight(r) = 355 / (pi * r^2) -``` - -Composing gives a function of `r` alone: - -```julia; -SA(r) = SA(canheight(r), r) -``` - -This is minimized subject to the constraint that $r \geq 0$. A quick -glance shows that as $r$ gets close to $0$, the can must get -infinitely tall to contain that fixed volume, and would have infinite -surface area as the $1/r^2$ in the first term implies. On the other -hand, as $r$ goes to infinity, the height must go to ``0`` to make a -really flat can. Again, we would have infinite surface area, as the -$r^2$ term at the end indicates. With this observation, we can rule -out the endpoints as possible minima, so any minima must occur at a -critical point. - -We start by making a graph, making an educated guess that the answer -is somewhere near a real life answer, or around ``3``-``5`` cms in radius: - - -```julia; -plot(SA, 2, 10) -``` - -The minimum looks to be around $4$cm and is clearly between $2$cm and -$6$cm. We can use `find_zero` to zero in on the value of the critical point: - -```julia; -rₛₐ = find_zero(SA', (2, 6)) -``` - -Okay, $3.837...$ is our answer for $r$. Use this to get $h$: - -```julia; -canheight(rₛₐ) -``` - - - -This produces a can which is about square in profile. This is not how -most cans look though. Perhaps our model is too simple, or the cans -are optimized for some other purpose than minimizing materials. - - - -## Questions - -###### Question - -A geometric figure has area given in terms of two measurements by -$A=\pi a b$ and perimeter $P = \pi (a + b)$. If the perimeter is fixed -to be 20 units long, what is the maximal area the figure can be? - -```julia; hold=true; echo=false -A(a,b) = pi*a*b -P = 20 -b1(a) = 20/pi - a -A(a) = A(a, b1(a)) -x = find_zero(A', (0, 10)) -val = A(x) -numericq(val) -``` - -###### Question - -A geometric figure has area given in terms of two measurements by -$A=\pi a b$ and perimeter $P=\pi \cdot \sqrt{a^2 + b^2}/2$. If the -perimeter is 20 units long, what is the maximal area? - -```julia; hold=true; echo=false -A(a,b) = pi*a*b -P = 20 -b1(a) = sqrt((P*2/pi)^2 - a^2) -A(a) = A(a, b1(a)) -x = find_zero(A', (0, 10)) -val = A(x) -numericq(val) -``` - - -###### Question - -A rancher with ``10`` meters of fence wishes to make a pen adjacent to an -existing fence. The pen will be a rectangle with one edge using the -existing fence. Say that has length $x$, then $10 = 2y + x$, with $y$ -the other dimension of the pen. What is the maximum area that can be -made? - -```julia; hold=true; echo=false -p = plot(; legend=false, aspect_ratio=:equal, axis=nothing, border=:none) -plot!([0,10, 10, 0, 0], [0,0,10,10,0]; linewidth=3) -plot!(p, [10,14,14,10], [2, 2, 8,8]; linewidth = 1) -annotate!(p, [(15, 5, "x"), (12,1, "y")]) -p -``` - - - - -```julia; hold=true; echo=false -Ar(y) = (10-2y)*y; -val = Ar(find_zero(Ar', 5)) -numericq(val, 1e-3) -``` - -Is there "symmetry" in the answer between $x$ and $y$? - -```julia; hold=true; echo=false -yesnoq("no") -``` - -What is you were do do two pens like this back to back, then the -answer would involve a rectangle. Is there symmetry in the answer now? - -```julia; hold=true; echo=false -yesnoq("yes") -``` - - - -###### Question - -A rectangle of sides $w$ and $h$ has fixed area $20$. What is the *smallest* perimeter it can have? - -```julia; hold=true; echo=false -Prim(x,y) = 2x + 2y -Prim(x) = Prim(x, 20/x) -xstar = find_zero(Prim', 5) -val = Prim(xstar) -numericq(val) -``` - - -###### Question - -A rectangle of sides $w$ and $h$ has fixed area $20$. What is the *largest* perimeter it can have? - -```julia; hold=true; echo=false -choices = [ -"It can be infinite", -"It is also 20", -"``17.888``" -] -answ = 1 -radioq(choices, answ) -``` - -###### Question - -A cardboard box is to be designed with a square base and an open top holding a fixed volume ``V``. What dimensions yield the minimal surface area? - - - -If this problem were approached symbolically, we might see the following code. First: - -```julia;eval=false -@syms V::positive x::positive z::positive -SA = 1 * x * x + 4 * x * z -``` - -What does this express? - -```julia; hold=true; echo=false -radioq(( -"The box has a square base with open top, so `x*x` is the amount of material in the base; the 4 sides each have `x*z` area.", -"The volume is a fixed amount, so is `x*x*z`, with sides suitably labeled", -"The surface area of a box is `6x*x`, so this is wrong." -), 1) -``` - -What does this command express? - -```julia; eval=false -SA = subs(SA, z => V / x^2) -``` - -```julia; hold=true; echo=false -radioq(( -"This command replaces `z` with an expression in `x` using the constraint of fixed volume `V`", -"This command replaces `z`, reparameterizing in `V` instead.", -"This command is merely algebraic simplification" -), 1) -``` - -What does this command find? - -```julia; eval=false -solve(diff(SA, x) ~ 0, x) -``` - - -```julia; hold=true; echo=false -radioq(( -"This solves ``SA'=0``, that is it find critical points of a continuously differentiable function", -"This solves for ``V`` the fixed, but unknown volume", -"This checks the values of `SA` at the end points of the domain" -), 1) -``` - -What do these commands do? - -```julia; eval=false -cps = solve(diff(SA, x) ~ 0, x) -xx = filter(isreal, cps)[1] -diff(SA, x, x)(xx) > 0 -``` - -```julia; hold=true; echo=false -radioq(( -"This applies the second derivative test to the lone *real* critical point showing there is a local minimum at that point.", -"This applies the first derivative test to the lone *real* critical point showing there is a local minimum at that point.", -"This finds the ``4`th derivative of `SA`" -), 1) -``` - - - - - - -###### Question - -A rain gutter is constructed from a 30" wide sheet of tin by bending -it into thirds. If the sides are bent 90 degrees, then the -cross-sectional area would be $100 = 10^2$. This is not the largest -possible amount. For example, if the sides are bent by 45 degrees, the cross sectional area is: - -```julia; hold=true; echo=false -2 * (1/2 * 10*cos(pi/4) * 10 * sin(pi/4)) + 10*sin(pi/4) * 10 -``` - -Find a value in degrees that gives the maximum. (The first task is to -write the area in terms of $\theta$. - - -```julia; hold=true; echo=false -function Ar(t) - opp = 10 * sin(t) - adj = 10 * cos(t) - 2 * opp * adj/2 + opp * 10 -end -t = find_zero(Ar', pi/4); ## Has issues with order=8 algorithm, tol > 1e-14 is needed -val = t * 180/pi; -numericq(val, 1e-3) -``` - -###### Question Non-Norman windows - -Suppose our new "Norman" window has half circular tops at the top and bottom? If the perimeter is fixed at $20$ and the dimensions of the rectangle are $x$ for the width and $y$ for the height. - -What is the value of $y$ that maximizes the area? - -```julia; hold=true; echo=false -P = 20 -A(x,y) = x*y + pi * (x/2)^2 -y(x) = (P - pi*x)/2 # P = 2y + 2pi*x/2 -A(x) = A(x,y(x)) -x0 = find_zero(D(A), (0, 10)) -val = y(x0) -numericq(val) # 0 -``` - - -###### Question (Thanks https://www.math.ucdavis.edu/~kouba) - -A movie screen projects on a wall 20 feet high beginning 10 feet above -the floor. This figure shows $\theta$ for $x=30$: - -```julia; hold=true; echo=false -p = plot([0, 30,30], [0,0,10], xlim=(0, 32), color=:blue, legend=false) -plot!(p, [30, 30], [10, 30], color=:blue, linewidth=4) -plot!(p, [0, 30,30,0], [0,10,30,0], color=:orange) -annotate!(p, [(x,y,l) for (x,y,l) in zip([15, 5, 31, 31], [1.5, 3.5, 5, 20], ["x=30", "θ", "10", "20"])]) -``` - -What value of $x$ gives the largest angle $\theta$? (In degrees.) - -```julia; hold=true; echo=false -theta(x) = atan(30/x) - atan(10/x) -val = find_zero(D(theta), 20); ## careful where one starts -val = theta(val) * 180/pi -numericq(val, 1e-1) -``` - - -###### Question - -A maximum likelihood estimator is a value derived by maximizing a function. For example, if - -```julia; -Likhood(t) = t^3 * exp(-3t) * exp(-2t) * exp(-4t) ## 0 <= t <= 10 -``` - -Then `Likhood(t)` is continuous and has single peak, so the maximum occurs -at the lone critical point. It turns out that this problem is bit sensitive to an initial condition, so we bracket - -```julia -find_zero(Likhood', (0.1, 0.5)) -``` - -Now if $Likhood(t) = \exp(-3t) \cdot \exp(-2t) \cdot \exp(-4t), \quad 0 \leq t \leq 10$, by graphing, explain why the same approach won't work: - -```julia; hold=true; echo=false -choices=["It does work and the answer is x = 2.27...", - L" $Likhood(t)$ is not continuous on $0$ to $10$", - L" $Likhood(t)$ takes its maximum at a boundary point - not a critical point"]; -answ = 3; -radioq(choices, answ) -``` - -##### Question - -Let $x_1$, $x_2$, $x_n$ be a set of unspecified numbers in a data -set. Form the expression $s(x) = (x-x_1)^2 + \cdots (x-x_n)^2$. What -is the smallest this can be (in $x$)? - -We approach this using `SymPy` and $n=10$ - -```julia; hold=true; eval=false -@syms s xs[1:10] -s(x) = sum((x-xi)^2 for xi in xs) -cps = solve(diff(s(x), x), x) -``` - -Run the above code. Baseed on the critical points found, what do you guess will be the -minimum value in terms of the values $x_1$, $x_2, \dots$? - -```julia; hold=true; echo=false -choices=[ -"The mean, or average, of the values", -"The median, or middle number, of the values", -L"The square roots of the values squared, $(x_1^2 + \cdots x_n^2)^2$" -] -answ = 1 -radioq(choices, answ) -``` - -###### Question - -Minimize the function $f(x) = 2x + 3/x$ over $(0, \infty)$. - -```julia; hold=true; echo=false -f(x) = 2x + 3/x; -val = find_zero(f', 1); -numericq(val, 1e-3) -``` - - -###### Question - -Of all rectangles of area 4, find the one with smallest perimeter. What is the perimeter? - -```julia; hold=true; echo=false -# 4 = xy -Prim(x) = 2x + 2*(4/x); -val = find_zero(D(Prim), 1); -numericq(Prim(val), 1e-3) ## a square! -``` - - -###### Question - -A running track is in the shape of two straight aways and two half -circles. The total distance (perimeter) is 400 meters. Suppose $w$ is -the width (twice the radius of the circles) and $h$ is the -height. What dimensions minimize the sum $w + h$? - -You have $P(w, h) = 2\pi \cdot (w/2) + 2\cdot(h-w)$. - - -```julia; hold=true; echo=false -Ar(w,h) = w + h -h(w) = (400 - 2pi*w/2 + 2w) / 2 -Ar(w) = Ar(w, h(w)) ## linear -val = Ar(0) -numericq(val) -``` - -###### Question - -A cell phone manufacturer wishes to make a rectangular phone with -total surface area of 12,000 $mm^2$ and maximal screen area. The -screen is surrounded by bezels with sizes of 8$mm$ on the long sides -and 32$mm$ on the short sides. (So, for example, the screen width is -shorter by $2\cdot 8$ mm than the phone width.) - -What are the dimensions (width and -height) that allow the maximum screen area? - -The width is: - -```julia; hold=true; echo=false -#A = w*h = 12000 -w(h) = 12_000 / h -S(w, h) = (w- 2*8) * (h - 2*32) -S(h) = S(w(h), h) -hstar =find_zero(D(S), 500) -wstar = w(hstar) -numericq(wstar) -``` - -The height is? - -```julia; hold=true; echo=false -w(h) = 12_000 / h -S(w, h) = (w- 2*8) * (h - 2*32) -S(h) = S(w(h), h) -hstar =find_zero(D(S), 500) -numericq(hstar) -``` - -###### Question - -Find the value $x > 0$ which minimizes the distance from the graph of -$f(x) = \log_e(x) - x$ to the origin $(0,0)$. - -```julia; hold=true; echo=false -f(x) = log(x) - x -p = plot(f, 0.2, 2, ylim=(-2,0.25), legend=false, linewidth=3) -plot!(p, [0,0], [-2, 0.25], color=:blue) -plot!(p, [0,2],[0,0], color=:blue) -xs = [0,1]; ys = [0, f(1)] -scatter!(p, xs,ys, color=:orange) -plot!(p, xs, ys, color=:orange, linewidth=3) -annotate!(p, [(.75, f(.5)/2, "d = $(round(sqrt(.5^2 + f(.5)^2), digits=2))")]) -p -``` - -```julia; hold=true; echo=false -d2(x) = sqrt((0-x)^2 + (0 - f(x))^2) -xstar = find_zero(D(d2), 1) -val = d2(xstar) -numericq(val) -``` - - -###### Question - -```julia; hold=true; echo=false -### {{{lhopital_40}}} -imgfile ="figures/fcarc-may2016-fig40-300.gif" -caption = L""" - -Image number $40$ from l'Hospital's calculus book (the first calculus book). Among all the cones that can be inscribed in a sphere, determine which one has the largest lateral area. (From http://www.ams.org/samplings/feature-column/fc-2016-05) - -""" -ImageFile(:derivatives, imgfile, caption) -``` - -The figure above poses a problem about cones in spheres, which can be reduced to a two-dimensional problem. Take a sphere of radius $r=1$, and imagine a secant line of length $l$ connecting $(-r, 0)$ to another point $(x,y)$ with $y>0$. Rotating that line around the $x$ axis produces a cone and its lateral surface is given by $SA=\pi \cdot y \cdot l$. Write $SA$ as a function of $x$ and solve. - -The largest lateral surface area is: - -```julia; hold=true; echo=false -r = 1 -SA(r,l) = pi * r * l -y(x) = sqrt(1 - x^2) -l(x) = sqrt((x-(-1))^2 + y(x)^2) -SA(x) = SA(y(x), l(x)) -cp = find_zero(SA', (-1/2, 1/2)) -val = SA(cp) -numericq(val) -``` - -The surface area of a sphere of radius $1$ is $4\pi r^2 = 4 \pi$. This is how many times greater than that of the largest cone? - -```julia; hold=true; echo=false -choices = ["exactly four times", -L"exactly $\pi$ times", -L"about $2.6$ times as big", -"about the same"] -answ = 1 -radioq(choices, answ) -``` - -###### Question - -In the examples the functions `argmax(f, itr)` and `findmin(collection)` were used. These have mathematical analogs. What is `argmax(f,itr)` in terms of math notation, where ``vs`` is the iterable collection of values: - -```julia; hold=true; echo=false -choices = [ - raw"``\{v \mid v \text{ in } vs, f(v) = \max(f(vs))\}``", - raw"``\{f(v) \mid v \text{ in } vs, f(v) = \max(f(vs))\}``", - raw"``\{i \mid v_i \text{ in } vs, f(v_i) = \max(f(vs))\}``" - ] -radioq(choices, 1) -``` - -The functions are related: `findmax` returns the maximum value and an index in the collection for which the value will be largest; `argmax` returns an element of the set for which the function is largest, so `argmax(identify, itr)` should correspond to the index found by `findmax` (through `itr[findmax(itr)[2]`) - -###### Question - -Let ``f(x) = (a/x)^x`` for ``a,x > 0``. When is this maximized? The following might be helpful - -```julia; hold=true -@syms x::positive a::postive -diff((a/x)^x, x) -``` - -This can be `solve`d to discover the answer. - -```julia; hold=true; echo=false -choices = [ -"``e``", -"``a/e``", -"``e/a``", -"``a \\cdot e``" -] -answ=2 -radioq(choices, answ) -``` - -###### Question - -The ladder problem has an trigonometry-free solution. We show one attributed to [Asma](http://www.mathematische-basteleien.de/ladder.htm). - -```julia; hold=true; echo=false -plt = plot(; axis=nothing, border=:none, legend=false, aspect_ratio=:equal) -a,b = 1, 2 -p = 1/2 -x = a/p -plot!(plt, [0, b*(1+p), 0, 0], [0, 0, a+x, 0]) -plot!(plt, [b,b,0,0],[0,a,a,0]) -annotate!(plt, [(b/2,0, "b"), (0,a/2,"a"), (0,a+x/2,"x"), (b+b*p/2,0,"bp")]) -plt -``` - -Introducing a variable ``p``, we get, following the above figure, the ladder of length ``c`` touching the wall at ``b+bp`` and ``a + x``. - -Using similar triangles, we have: - -```julia; hold=true -@syms a::positive b::positive p::positive x::positive -solve(x/b ~ (x+a)/(b + b*p), x) -``` - -With ``x = a/p`` we get by Pythagorean's theorem that - -```math -\begin{align*} -c^2 &= (a + a/p)^2 + (b + bp)^2 \\ - &= a^2(1 + \frac{1}{p})^2 + b^2(1+p)^2. -\end{align*} -``` - -The ladder problem minimizes ``c`` or equivalently ``c^2``. - -Why is the following set of commands useful in this task: - -```julia; eval=false -c2 = a^2*(1 + 1/p)^2 + b^2*(1+p)^2 -c2p = diff(c2, p) -eq = numer(together(c2p)) -solve(eq ~ 0, p) -``` - -```julia; hold=true; echo=false -choices = ["It finds the critical points", - "It finds the minimal value of `c`", - "It finds the minimal value of `p`"] -radioq(choices, 1) -``` - -The polynomial `nu` is what degree in `p`? - -```julia; hold=true; echo=false -numericq(4) -``` - -The only positive real solution for ``p`` from ``nu`` is - -```julia; hold=true;echo=false -choices = [ -raw" ``(a/b)^{2/3}``", -raw" ``1``", -raw" ``\sqrt{3}/2 \cdot (a/b)^{2/3}``" -] -radioq(choices, 1) -``` - -###### Question - -In [Hall](https://www.maa.org/sites/default/files/hall03010308158.pdf) we can find a dozen optimization problems related to the following figure of the parabola ``y=x^2`` a point ``P=(a,a^2)``, ``a > 0``, and its normal line. We will do two. - - -```julia; hold=true, echo=false -p = plot(; legend=false, aspect_ratio=:equal, axis=nothing, border=:none) - b = 2. - plot!(p, x -> x^2, -b, b) - plot!(p, [-b,b], [0,0]) - plot!(p, [0,0], [0, b^2]) - a = 1 - scatter!(p, [a],[a^2]) - m = 2a - plot!(p, x -> a^2 + m*(x-a), 1/2, b) - mₚ = -1/m - plot!(p, x -> a^2 + mₚ*(x-a)) - scatter!(p, [-3/2], [(3/2)^2]) - annotate!(p, [(1+1/4, 1+1/8, "P"), (-3/2-1/4, (-3/2)^2-1/4, "Q")]) -p -``` - -What do these commands do? - -```julia; hold=true; -@syms x::real, a::real -mₚ = - 1 / diff(x^2, x)(a) -solve(x^2 - (a^2 + mₚ*(x-a)) ~ 0, x) -``` - -```julia; hold=true; echo=false -radioq(( -"It finds the ``x`` value of the intersection points of the normal line and the parabola", -"It finds the tangent line", -"It finds the point ``P``" -), 1) -``` - -Numerically, find the value of ``a`` that minimizes the ``y`` coordinate of ``Q``. - -```julia; hold=true; echo=false -y(a) = (-a - 1/(2a))^2 -a = find_zero(y', 1) -numericq(a) -``` - - -Numerically find the value of ``a`` that minimizes the length of the line seqment ``PQ``. - -```juila; hold=true; echo=false -x(a) = -a - 1/(2a) -d(a) = (a-x(a))^2 + (a^2 - x(a)^2)^2 -a = find_zero(d', 1) -numericq(a) -``` diff --git a/CwJ/derivatives/process.jl b/CwJ/derivatives/process.jl deleted file mode 100644 index 3ef5ee5..0000000 --- a/CwJ/derivatives/process.jl +++ /dev/null @@ -1,44 +0,0 @@ -using WeavePynb - -using CwJWeaveTpl - - - -fnames = [ - "derivatives", ## more questions - "numeric_derivatives", - - "mean_value_theorem", - "optimization", - "curve_sketching", - - "linearization", - "newtons_method", - "lhopitals_rule", ## Okay - -but could beef up questions.. - - - "implicit_differentiation", ## add more questions? - "related_rates", - "taylor_series_polynomials" -] - - -process_file(nm; cache=:off) = CwJWeaveTpl.mmd(nm * ".jmd", cache=cache) - -function process_files(;cache=:user) - for f in fnames - @show f - process_file(f, cache=cache) - end -end - - - - -""" -## TODO derivatives - -tangent lines intersect at avearge for a parabola - -Should we have derivative results: inverse functions, logarithmic differentiation... -""" diff --git a/CwJ/derivatives/related_rates.jmd b/CwJ/derivatives/related_rates.jmd deleted file mode 100644 index ea96a82..0000000 --- a/CwJ/derivatives/related_rates.jmd +++ /dev/null @@ -1,781 +0,0 @@ -# Related rates - -This section uses these add-on packaages: - -```julia -using CalculusWithJulia -using Plots -using Roots -using SymPy -``` - -```julia; echo=false; results="hidden" -using CalculusWithJulia.WeaveSupport - -fig_size=(800, 600) -const frontmatter = ( - title = "Related rates", - description = "Calculus with Julia: Related rates", - tags = ["CalculusWithJulia", "derivatives", "related rates"], -); -nothing -``` - ----- - -Related rates problems involve two (or more) unknown quantities that -are related through an equation. As the two variables depend on each -other, also so do their rates - change with respect to some variable -which is often time, though exactly how remains to be -discovered. Hence the name "related rates." - -#### Examples - -The following is a typical "book" problem: - -> A screen saver displays the outline of a ``3`` cm by ``2`` cm rectangle and -> then expands the rectangle in such a way that the ``2`` cm side is -> expanding at the rate of ``4`` cm/sec and the proportions of the -> rectangle never change. How fast is the area of the rectangle -> increasing when its dimensions are ``12`` cm by ``8`` cm? -> [Source.](http://oregonstate.edu/instruct/mth251/cq/Stage9/Practice/ratesProblems.html) - -```julia; hold=true; echo=false; cache=true -### {{{growing_rects}}} -## Secant line approaches tangent line... -function growing_rects_graph(n) - w = (t) -> 2 + 4t - h = (t) -> 3/2 * w(t) - t = n - 1 - - w_2 = w(t)/2 - h_2 = h(t)/2 - - w_n = w(5)/2 - h_n = h(5)/2 - - plt = plot(w_2 * [-1, -1, 1, 1, -1], h_2 * [-1, 1, 1, -1, -1], xlim=(-17,17), ylim=(-17,17), - legend=false, size=fig_size) - annotate!(plt, [(-1.5, 1, "Area = $(round(Int, 4*w_2*h_2))")]) - plt - - -end -caption = L""" - -As $t$ increases, the size of the rectangle grows. The ratio of width to height is fixed. If we know the rate of change in time for the width ($dw/dt$) and the height ($dh/dt$) can we tell the rate of change of *area* with respect to time ($dA/dt$)? - -""" -n=6 - -anim = @animate for i=1:n - growing_rects_graph(i) -end - -imgfile = tempname() * ".gif" -gif(anim, imgfile, fps = 1) - -ImageFile(imgfile, caption) -``` - -Here we know $A = w \cdot h$ and we know some things about how $w$ and -$h$ are related *and* about the rate of how both $w$ and $h$ grow in -time $t$. That means that we could express this growth in terms of -some functions $w(t)$ and $h(t)$, then we can figure out that the area - as a function of $t$ - will be expressed as: - -```math -A(t) = w(t) \cdot h(t). -``` - -We would get by the product rule that the *rate of change* of area with respect to time, $A'(t)$ is just: - -```math -A'(t) = w'(t) h(t) + w(t) h'(t). -``` - -As an aside, it is fairly conventional to suppress the $(t)$ part of -the notation $A=wh$ and to use the Leibniz notation for derivatives: - -```math -\frac{dA}{dt} = \frac{dw}{dt} h + w \frac{dh}{dt}. -``` - -This relationship is true for all $t$, but the problem discusses a -certain value of $t$ - when $w(t)=8$ and $h(t) = 12$. At this same -value of $t$, we have $w'(t) = 4$ and so $h'(t) = 6$. Substituting these 4 values into the 4 unknowns in the formula for $A'(t)$ gives: - -```math -A'(t) = 4 \cdot 12 + 8 \cdot 6 = 96. -``` - -Summarizing, from the relationship between $A$, $w$ and $t$, there is -a relationship between their rates of growth with respect to $t$, a -time variable. Using this and known values, we can compute. In this -case, $A'$ at the specific $t$. - - -We could also have done this differently. We would recognize the following: - -- The area of a rectangle is just: - -```julia; -A(w,h) = w * h -``` - -- The width - expanding at a rate of $4t$ from a starting value of $2$ - must satisfy: - -```julia; -w(t) = 2 + 4*t -``` - -- The height is a constant proportion of the width: - -```julia; -h(t) = 3/2 * w(t) -``` - -This means again that area depends on $t$ through this formula: - -```julia; -A(t) = A(w(t), h(t)) -``` - - -This is why the rates of change are related: as $w$ and $h$ change in -time, the functional relationship with $A$ means $A$ also changes in -time. - - - -Now to answer the question, when the width is 8, we must have that $t$ is: - -```julia; -tstar = find_zero(x -> w(x) - 8, [0, 4]) # or solve by hand to get 3/2 -``` - -The question is to find the rate the area is increasing at the given -time $t$, which is $A'(t)$ or $dA/dt$. We get this by performing the -differentiation, then substituting in the value. - -Here we do so with the aid of `Julia`, though this problem could readily be done "by hand." - -We have expressed $A$ as a function of $t$ by composition, so can differentiate that: - -```julia; -A'(tstar) -``` - - ----- - -Now what? Why is ``96`` of any interest? It is if the value at a specific -time is needed. But in general, a better question might be to -understand if there is some pattern to the numbers in the figure, -these being $6, 54, 150, 294, 486, 726$. Their differences are the -*average* rate of change: - -```julia; -xs = [6, 54, 150, 294, 486, 726] -ds = diff(xs) -``` - -Those seem to be increasing by a fixed amount each time, which we can see by one more application of `diff`: - -```julia; -diff(ds) -``` - -How can this relationship be summarized? Well, let's go back to what we know, though this time using symbolic math: - -```julia; -@syms t -diff(A(t), t) -``` - -This should be clear: the rate of change, $dA/dt$, is increasing -linearly, hence the second derivative, $dA^2/dt^2$ would be constant, -just as we saw for the average rate of change. - -So, for this problem, a constant rate of change in width and height -leads to a linear rate of change in area, put otherwise, linear growth -in both width and height leads to quadratic growth in area. - -##### Example - -A ladder, with length $l$, is leaning against a wall. We parameterize -this problem so that the top of the ladder is at $(0,h)$ and the -bottom at $(b, 0)$. Then $l^2 = h^2 + b^2$ is a constant. - -If the ladder starts to slip away at the base, but remains in contact -with the wall, express the rate of change of $h$ with respect to $t$ -in terms of $db/dt$. - -We have from implicitly differentiating in $t$ the equation $l^2 = h^2 + b^2$, noting that $l$ is a constant, that: - -```math -0 = 2h \frac{dh}{dt} + 2b \frac{db}{dt}. -``` - - -Solving, yields: - -```math -\frac{dh}{dt} = -\frac{b}{h} \cdot \frac{db}{dt}. -``` - - -* If when $l = 12$ it is known that $db/dt = 2$ when $b=4$, find $dh/dt$. - -We just need to find $h$ for this value of $b$, as the other two quantities in the last equation are known. - -But $h = \sqrt{l^2 - b^2}$, so the answer is: - - -```julia; -length, bottom, dbdt = 12, 4, 2 -height = sqrt(length^2 - bottom^2) --bottom/height * dbdt -``` - -* What happens to the rate as $b$ goes to $l$? - -As $b$ goes to $l$, $h$ goes to ``0``, so $b/h$ blows up. Unless $db/dt$ -goes to $0$, the expression will become $-\infty$. - -!!! note - Often, this problem is presented with ``db/dt`` having a constant rate. In this case, the ladder problem defies physics, as ``dh/dt`` eventually is faster than the speed of light as ``h \rightarrow 0+``. In practice, were ``db/dt`` kept at a constant, the ladder would necessarily come away from the wall. The trajectory would follow that of a tractrix were there no gravity to account for. - - -##### Example - -```julia; hold=true; echo=false -caption = "A man and woman walk towards the light." - -imgfile = "figures/long-shadow-noir.png" -ImageFile(:derivatives, imgfile, caption) -``` - -Shadows are a staple of film noir. In the photo, suppose a man and a woman walk towards a street light. As they approach the light the length of their shadow changes. - -Suppose, we focus on the ``5`` foot tall woman. Her shadow comes from a streetlight ``15`` feet high. She is walking at ``3`` feet per second towards the light. What is the rate of change of her shadow? - -The setup for this problem involves drawing a right triangle with height ``12`` and base given by the distance ``x`` from the light the woman is *plus* the length ``l`` of the shadow. There is a similar triangle formed by the woman's height with length ``l``. Equating the ratios of the sided gives: - -```math -\frac{5}{l} = \frac{12}{x + l} -``` - -As we need to take derivatives, we work with the reciprocal relationship: - -```math -\frac{l}{5} = \frac{x + l}{12} -``` - -Differentiating in ``t`` gives: - -```math -\frac{l'}{5} = \frac{x' + l'}{12} -``` - -Or - -```math -l' \cdot (\frac{1}{5} - \frac{1}{12}) = \frac{x'}{12} -``` - -Solving for ``l'`` gives an answer in terms of ``x'`` the rate the woman is walking. In this description ``x`` is getting shorter, so ``x'`` would be ``-3`` feet per second and the shadow length would be decreasing at a rate proportional to the walking speed. - -##### Example - -```julia; hold=true; echo=false -p = plot(; axis=nothing, border=:none, legend=false, aspect_ratio=:equal) -scatter!(p, [0],[50], color=:yellow, markersize=50) -plot!(p, [0, 50], [0,0], linestyle=:dash) -plot!(p, [0,50], [50,0], linestyle=:dot) -plot!(p, [25,25],[25,0], linewidth=5, color=:black) -plot!(p, [25,50], [0,0], linewidth=2, color=:black) -``` - -The sun is setting at the rate of ``1/20`` radian/min, and appears to be dropping perpendicular to the horizon, as depicted in the figure. How fast is the shadow of a ``25`` meter wall lengthening at the moment when the shadow is ``25`` meters long? - -Let the shadow length be labeled ``x``, as it appears on the ``x`` axis above. Then we have by right-angle trigonometry: - -```math -\tan(\theta) = \frac{25}{x} -``` - -of ``x\tan(\theta) = 25``. - -As ``t`` evolves, we know ``d\theta/dt`` but what is ``dx/dt``? Using implicit differentiation yields: - -```math -\frac{dx}{dt} \cdot \tan(\theta) + x \cdot (\sec^2(\theta)\cdot \frac{d\theta}{dt}) = 0 -``` - -Substituting known values and identifying ``\theta=\pi/4`` when the shadow length, ``x``, is ``25`` gives: - -```math -\frac{dx}{dt} \cdot \tan(\pi/4) + 25 \cdot((4/2) \cdot \frac{-1}{20} = 0 -``` - -This can be solved for the unknown: ``dx/dt = 50/20``. - - - -##### Example - -A batter hits a ball toward third base at ``75`` ft/sec and runs toward first base at a rate of ``24`` ft/sec. At what rate does the distance between the ball and the batter change when ``2`` seconds have passed? - - -We will answer this with `SymPy`. First we create some symbols for the movement of the ball towardsthird base, `b(t)`, the runner toward first base, `r(t)`, and the two velocities. We use symbolic functions for the movements, as we will be differentiating them in time: - -```julia -@syms b() r() v_b v_r -d = sqrt(b(t)^2 + r(t)^2) -``` - -The distance formula applies to give ``d``. As the ball and runner are moving in a perpendicular direction, the formula is easy to apply. - -We can differentiate `d` in terms of `t` and in process we also find the derivatives of `b` and `r`: - -```julia -db, dr = diff(b(t),t), diff(r(t),t) # b(t), r(t) -- symbolic functions -dd = diff(d,t) # d -- not d(t) -- an expression -``` - -The slight difference in the commands is due to `b` and `r` being symbolic functions, whereas `d` is a symbolic expression. Now we begin substituting. First, from the problem `db` is just the velocity in the ball's direction, or `v_b`. Similarly for `v_r`: - -```julia -ddt = subs(dd, db => v_b, dr => v_r) -``` - -Now, we can substitute in for `b(t)`, as it is `v_b*t`, etc.: - -```julia -ddt₁ = subs(ddt, b(t) => v_b * t, r(t) => v_r * t) -``` - -This finds the rate of change of time for any `t` with symbolic values of the velocities. (And shows how the answer doesn't actually depend on ``t``.) The problem's answer comes from a last substitution: - -```julia -ddt₁(t => 2, v_b => 75, v_r => 24) -``` - -Were this done by "hand," it would be better to work with distance squared to avoid the expansion of complexity from the square root. That is, using implicit differentiation: - -```math -\begin{align*} -d^2 &= b^2 + r^2\\ -2d\cdot d' &= 2b\cdot b' + 2r\cdot r'\\ -d' &= (b\cdot b' + r \cdot r')/d\\ -d' &= (tb'\cdot b' + tr' \cdot r')/d\\ -d' &= \left((b')^2 + (r')^2\right) \cdot \frac{t}{d}. -\end{align*} -``` - -##### Example - -```julia; hold=true; echo=false; cache=true -###{{{baseball_been_berry_good}}} -## Secant line approaches tangent line... -function baseball_been_berry_good_graph(n) - - v0 = 15 - x = (t) -> 50t - y = (t) -> v0*t - 5 * t^2 - - - ns = range(.25, stop=3, length=8) - - t = ns[n] - ts = range(0, stop=t, length=50) - xs = map(x, ts) - ys = map(y, ts) - - degrees = atand(y(t)/(100-x(t))) - degrees = degrees < 0 ? 180 + degrees : degrees - - plt = plot(xs, ys, legend=false, size=fig_size, xlim=(0,150), ylim=(0,15)) - plot!(plt, [x(t), 100], [y(t), 0.0], color=:orange) - annotate!(plt, [(55, 4,"θ = $(round(Int, degrees)) degrees"), - (x(t), y(t), "($(round(Int, x(t))), $(round(Int, y(t))))")]) - -end -caption = L""" - -The flight of the ball as being tracked by a stationary outfielder. This ball will go over the head of the player. What can the player tell from the quantity $d\theta/dt$? - -""" -n = 8 - - -anim = @animate for i=1:n - baseball_been_berry_good_graph(i) -end - - -imgfile = tempname() * ".gif" -gif(anim, imgfile, fps = 1) - -ImageFile(imgfile, caption) -``` - - -A baseball player stands ``100`` meters from home base. A batter hits the -ball directly at the player so that the distance from home plate is -$x(t)$ and the height is $y(t)$. - -The player tracks the flight of the ball in terms of the angle -$\theta$ made between the ball and the player. This will satisfy: - -```math -\tan(\theta) = \frac{y(t)}{100 - x(t)}. -``` - -What is the rate of change of $\theta$ with respect to $t$ in terms of that of $x$ and $y$? - -We have by the chain rule and quotient rule: - -```math -\sec^2(\theta) \theta'(t) = \frac{y'(t) \cdot (100 - x(t)) - y(t) \cdot (-x'(t))}{(100 - x(t))^2}. -``` - -If we have $x(t) = 50t$ and $y(t)=v_{0y} t - 5 t^2$ when is the rate of change of the angle happening most quickly? - - -The formula for $\theta'(t)$ is - -```math -\theta'(t) = \cos^2(\theta) \cdot \frac{y'(t) \cdot (100 - x(t)) - y(t) \cdot (-x'(t))}{(100 - x(t))^2}. -``` - - -This question requires us to differentiate *again* in $t$. Since we -have fairly explicit function for $x$ and $y$, we will use `SymPy` to -do this. - -```julia; -@syms theta() - -v0 = 5 -x(t) = 50t -y(t) = v0*t - 5 * t^2 -eqn = tan(theta(t)) - y(t) / (100 - x(t)) -``` - -```julia; -thetap = diff(theta(t),t) -dtheta = solve(diff(eqn, t), thetap)[1] -``` - -We could proceed directly by evaluating: - -```julia; -d2theta = diff(dtheta, t)(thetap => dtheta) -``` - -That is not so tractable, however. - -It helps to simplify -$\cos^2(\theta(t))$ using basic right-triangle trigonometry. Recall, $\theta$ comes from a right triangle with -height $y(t)$ and length $(100 - x(t))$. The cosine of this angle will -be $100 - x(t)$ divided by the length of the hypotenuse. So we can -substitute: - -```julia; -dtheta₁ = dtheta(cos(theta(t))^2 => (100 -x(t))^2/(y(t)^2 + (100-x(t))^2)) -``` - - -Plotting reveals some interesting things. For $v_{0y} < 10$ we have graphs that look like: - -```julia; -plot(dtheta₁, 0, v0/5) -``` - -The ball will drop in front of the player, and the change in $d\theta/dt$ is monotonic. - - - -But let's rerun the code with $v_{0y} > 10$: - -```julia; hold=true -v0 = 15 -x(t) = 50t -y(t) = v0*t - 5 * t^2 -eqn = tan(theta(t)) - y(t) / (100 - x(t)) -thetap = diff(theta(t),t) -dtheta = solve(diff(eqn, t), thetap)[1] -dtheta₁ = subs(dtheta, cos(theta(t))^2, (100 - x(t))^2/(y(t)^2 + (100 - x(t))^2)) -plot(dtheta₁, 0, v0/5) -``` - - -In the second case we have a different shape. The graph is not -monotonic, and before the peak there is an inflection point. Without -thinking too hard, we can see that the greatest change in the angle is -when it is just above the head ($t=2$ has $x(t)=100$). - -That these two graphs differ so, means that the player may be able to -read if the ball is going to go over his or her head by paying -attention to the how the ball is being tracked. - -##### Example - -Hipster pour-over coffee is made with a conical coffee filter. The -cone is actually a [frustum](http://en.wikipedia.org/wiki/Frustum) of -a cone with small diameter, say $r_0$, chopped off. We will parameterize -our cone by a value $h \geq 0$ on the $y$ axis and an angle $\theta$ -formed by a side and the $y$ axis. Then the coffee filter is the part -of the cone between some $h_0$ (related $r_0=h_0 \tan(\theta)$) and $h$. - -The volume of a cone of height $h$ is $V(h) = \pi/3 h \cdot -R^2$. From the geometry, $R = h\tan(\theta)$. The volume of the -filter then is: - -```math -V = V(h) - V(h_0). -``` - -What is $dV/dh$ in terms of $dR/dh$? - -Differentiating implicitly gives: - - -```math -\frac{dV}{dh} = \frac{\pi}{3} ( R(h)^2 + h \cdot 2 R \frac{dR}{dh}). -``` - -We see that it depends on $R$ and the change in $R$ with respect to $h$. However, we visualize $h$ - the height - so it is better to re-express. Clearly, $dR/dh = \tan\theta$ and using $R(h) = h \tan(\theta)$ we get: - -```math -\frac{dV}{dh} = \pi h^2 \tan^2(\theta). -``` - -The rate of change goes down as $h$ gets smaller ($h \geq h_0$) and gets bigger for bigger $\theta$. - -How do the quantities vary in time? - -For an incompressible fluid, by balancing the volume leaving with how -it leaves we will have $dh/dt$ is the ratio of the cross-sectional -area at bottom over that at the height of the fluid $(\pi \cdot (h_0\tan(\theta))^2) / -(\pi \cdot ((h\tan\theta))^2)$ times the outward velocity of the fluid. - -That is $dh/dt = (h_0/h)^2 \cdot v$. Which makes sense - larger openings -($h_0$) mean more fluid lost per unit time so the height change -follows, higher levels ($h$) means the change in height is slower, as -the cross-sections have more volume. - - -By [Torricelli's](http://en.wikipedia.org/wiki/Torricelli's_law) law, -the out velocity follows the law $v = \sqrt{2g(h-h_0)}$. This gives: - -```math -\frac{dh}{dt} = \frac{h_0^2}{h^2} \cdot v = \frac{h_0^2}{h^2} \sqrt{2g(h-h_0)}. -``` - -If $h >> h_0$, then $\sqrt{h-h_0} = \sqrt{h}\sqrt(1 - h_0/h) \approx \sqrt{h}(1 - (1/2)(h_0/h)) \approx \sqrt{h}$. So the rate of change of height in time is like $1/h^{3/2}$. - - -Now, by the chain rule, we have then the rate of change of volume with respect to time, $dV/dt$, is: - -```math -\begin{align*} -\frac{dV}{dt} &= -\frac{dV}{dh} \cdot \frac{dh}{dt}\\ -&= \pi h^2 \tan^2(\theta) \cdot \frac{h_0^2}{h^2} \sqrt{2g(h-h_0)} \\ -&= \pi \sqrt{2g} \cdot (r_0)^2 \cdot \sqrt{h-h_0} \\ -&\approx \pi \sqrt{2g} \cdot r_0^2 \cdot \sqrt{h}. -\end{align*} -``` - - - -This rate depends on the square of the size of the -opening ($r_0^2$) and the square root of the height ($h$), but not the -angle of the cone. - - -## Questions - -###### Question - -Supply and demand. Suppose demand for product $XYZ$ is $d(x)$ and supply -is $s(x)$. The excess demand is $d(x) - s(x)$. Suppose this is positive. How does this influence -price? Guess the "law" of economics that applies: - -```julia; hold=true; echo=false -choices = [ -"The rate of change of price will be ``0``", -"The rate of change of price will increase", -"The rate of change of price will be positive and will depend on the rate of change of excess demand." -] -answ = 3 -radioq(choices, answ, keep_order=true) -``` - -(Theoretically, when demand exceeds supply, prices increase.) - -###### Question - -Which makes more sense from an economic viewpoint? - -```julia; hold=true; echo=false -choices = [ -"If the rate of change of unemployment is negative, the rate of change of wages will be negative.", -"If the rate of change of unemployment is negative, the rate of change of wages will be positive." -] -answ = 2 -radioq(choices, answ, keep_order=true) -``` - -(Colloquially, "the rate of change of unemployment is negative" means the unemployment rate is going down, so there are fewer workers available to fill new jobs.) - -###### Question - -In chemistry there is a fundamental relationship between pressure -($P$), temperature ($T)$ and volume ($V$) given by $PV=cT$ where $c$ -is a constant. Which of the following would be true with respect to time? - -```julia; hold=true; echo=false -choices = [ -L"The rate of change of pressure is always increasing by $c$", -"If volume is constant, the rate of change of pressure is proportional to the temperature", -"If volume is constant, the rate of change of pressure is proportional to the rate of change of temperature", -"If pressure is held constant, the rate of change of pressure is proportional to the rate of change of temperature"] -answ = 3 -radioq(choices, answ, keep_order=true) -``` - -###### Question - -A pebble is thrown into a lake causing ripples to form expanding -circles. Suppose one of the circles expands at a rate of ``1`` foot per second and -the radius of the circle is ``10`` feet, what is the rate of change of -the area enclosed by the circle? - -```julia; hold=true; echo=false -# a = pi*r^2 -# da/dt = pi * 2r * drdt -r = 10; drdt = 1 -val = pi * 2r * drdt -numericq(val, units=L"feet$^2$/second") -``` - -###### Question - -A pizza maker tosses some dough in the air. The dough is formed in a -circle with radius ``10``. As it rotates, its area increases at a rate of -``1`` inch$^2$ per second. What is the rate of change of the radius? - -```julia; hold=true; echo=false -# a = pi*r^2 -# da/dt = pi * 2r * drdt -r = 10; dadt = 1 -val = dadt /( pi * 2r) -numericq(val, units="inches/second") -``` - -###### Question - - -An FBI agent with a powerful spyglass is located in a boat anchored -400 meters offshore. A gangster under surveillance is driving along -the shore. Assume the shoreline is straight and that the gangster is 1 -km from the point on the shore nearest to the boat. If the spyglasses -must rotate at a rate of $\pi/4$ radians per minute to track -the gangster, how fast is the gangster moving? (In kilometers per minute.) -[Source.](http://oregonstate.edu/instruct/mth251/cq/Stage9/Practice/ratesProblems.html) - - -```julia; hold=true; echo=false -## tan(theta) = x/y -## sec^2(theta) dtheta/dt = 1/y dx/dt (y is constant) -## dxdt = y sec^2(theta) dtheta/dt -dthetadt = pi/4 -y0 = .4; x0 = 1.0 -theta = atan(x0/y0) -val = y0 * sec(theta)^2 * dthetadt -numericq(val, units="kilometers/minute") -``` - - -###### Question - -A flood lamp is installed on the ground 200 feet from a vertical -wall. A six foot tall man is walking towards the wall at the rate of -4 feet per second. How fast is the tip of his shadow moving down the -wall when he is 50 feet from the wall? -[Source.](http://oregonstate.edu/instruct/mth251/cq/Stage9/Practice/ratesProblems.html) -(As the question is written the answer should be positive.) - -```julia; hold=true; echo=false -## y/200 = 6/x -## dydt = 200 * 6 * -1/x^2 dxdt -x0 = 200 - 50 -dxdt = 4 -val = 200 * 6 * (1/x0^2) * dxdt -numericq(val, units="feet/second") -``` - - -###### Question - - -Consider the hyperbola $y = 1/x$ and think of it as a slide. A -particle slides along the hyperbola so that its x-coordinate is -increasing at a rate of $f(x)$ units/sec. If its $y$-coordinate is -decreasing at a constant rate of $1$ unit/sec, what is $f(x)$? -[Source.](http://oregonstate.edu/instruct/mth251/cq/Stage9/Practice/ratesProblems.html) - -```julia; hold=true; echo=false -choices = [ -"``f(x) = 1/x``", -"``f(x) = x^0``", -"``f(x) = x``", -"``f(x) = x^2``" -] -answ = 4 -radioq(choices, answ, keep_order=true) -``` - -###### Question - -A balloon is in the shape of a sphere, fortunately, as this gives -a known formula, $V=4/3 \pi r^3$, for the volume. If the balloon is being filled with a rate of -change of volume per unit time is $2$ and the radius is $3$, what is -rate of change of radius per unit time? - -```julia; hold=true; echo=false -r, dVdt = 3, 2 -drdt = dVdt / (4 * pi * r^2) -numericq(drdt, units="units per unit time") -``` - -###### Question - -Consider the curve $f(x) = x^2 - \log(x)$. For a given $x$, the tangent line intersects the $y$ axis. Where? - -```julia; hold=true; echo=false -choices = [ -"``y = 1 - x^2 - \\log(x)``", -"``y = 1 - x^2``", -"``y = 1 - \\log(x)``", -"``y = x(2x - 1/x)``" -] -answ = 1 -radioq(choices, answ) -``` - -If $dx/dt = -1$, what is $dy/dt$? - -```julia; hold=true; echo=false -choices = [ -"``dy/dt = 2x + 1/x``", -"``dy/dt = 1 - x^2 - \\log(x)``", -"``dy/dt = -2x - 1/x``", -"``dy/dt = 1``" -] -answ=1 -radioq(choices, answ) -``` diff --git a/CwJ/derivatives/symbolic_derivatives.jmd b/CwJ/derivatives/symbolic_derivatives.jmd deleted file mode 100644 index f579056..0000000 --- a/CwJ/derivatives/symbolic_derivatives.jmd +++ /dev/null @@ -1,218 +0,0 @@ -# Symbolic derivatives - -This section uses this add-on package: - -```julia -using TermInterface -``` - -```julia; echo=false -const frontmatter = ( - title = "Symbolic derivatives", - description = "Calculus with Julia: Symbolic derivatives", - tags = ["CalculusWithJulia", "derivatives", "symbolic derivatives"], -); -``` - ----- - -The ability to breakdown an expression into operations and their -arguments is necessary when trying to apply the differentiation -rules. Such rules are applied from the outside in. Identifying -the proper "outside" function is usually most of the battle when finding derivatives. - -In the following example, we provide a sketch of a framework to differentiate expressions by a chosen symbol to illustrate how the outer function drives the task of differentiation. - -The `Symbolics` package provides native symbolic manipulation abilities for `Julia`, similar to `SymPy`, though without the dependence on `Python`. The `TermInterface` package, used by `Symbolics`, provides a generic interface for expression manipulation for this package that *also* is implemented for `Julia`'s expressions and symbols. - -An expression is an unevaluated portion of code that for our purposes -below contains other expressions, symbols, and numeric literals. They -are held in the `Expr` type. A symbol, such as `:x`, is distinct from -a string (e.g. `"x"`) and is useful to the programmer to distinguish -between the contents a variable points to from the name of the -variable. Symbols are fundamental to metaprogramming in `Julia`. An -expression is a specification of some set of statements to execute. A -numeric literal is just a number. - -The three main functions from `TermInterface` we leverage are `istree`, `operation`, and `arguments`. The `operation` function returns the "outside" function of an expression. For example: - -```julia -operation(:(sin(x))) -``` - -We see the `sin` function, referred to by a symbol (`:sin`). -The `:(...)` above *quotes* the argument, and does not evaluate it, hence `x` need not be defined above. (The `:` notation is used to create both symbols and expressions.) - - -The arguments are the terms that the outside function is called on. For our purposes there may be ``1`` (*unary*), ``2`` (*binary*), or more than ``2`` (*nary*) arguments. (We ignore zero-argument functions.) For example: - -```julia -arguments(:(-x)), arguments(:(pi^2)), arguments(:(1 + x + x^2)) -``` - -(The last one may be surprising, but all three arguments are passed to the `+` function.) - - -Here we define a function to decide the *arity* of an expression based on the number of arguments it is called with: - -```julia -function arity(ex) - n = length(arguments(ex)) - n == 1 ? Val(:unary) : - n == 2 ? Val(:binary) : Val(:nary) -end -``` - - -Differentiation must distinguish between expressions, variables, and -numbers. Mathematically expressions have an "outer" function, whereas variables and numbers can be directly differentiated. The `istree` -function in `TermInterface` returns `true` when passed an expression, -and `false` when passed a symbol or numeric literal. The latter two -may be distinguished by `isa(..., Symbol)`. - -Here we create a function, `D`, that when it encounters an expression it *dispatches* to a specific method of `D` based on the outer operation and arity, otherwise if it encounters a symbol or a numeric literal it does the differentiation: - -```julia -function D(ex, var=:x) - if istree(ex) - op, args = operation(ex), arguments(ex) - D(Val(op), arity(ex), args, var) - elseif isa(ex, Symbol) && ex == :x - 1 - else - 0 - end -end -``` - -Now to develop methods for `D` for different "outside" functions and arities. - -Addition can be unary (`:(+x)` is a valid quoting, even if it might simplify to the symbol `:x` when evaluated), *binary*, or *nary*. Here we implement the *sum rule*: - -```julia -D(::Val{:+}, ::Val{:unary}, args, var) = D(first(args), var) - -function D(::Val{:+}, ::Val{:binary}, args, var) - a′, b′ = D.(args, var) - :($a′ + $b′) -end - -function D(::Val{:+}, ::Val{:nary}, args, var) - a′s = D.(args, var) - :(+($a′s...)) -end -``` - -The `args` are always held in a container, so the unary method must pull out the first one. The binary case should read as: apply `D` to each of the two arguments, and then create a quoted expression containing the sum of the results. The dollar signs interpolate into the quoting. (The "primes" are unicode notation achieved through `\prime[tab]` and not operations.) The *nary* case does something similar, only uses splatting to produce the sum. - -Subtraction must also be implemented in a similar manner, but not for the *nary* case: - -```julia -function D(::Val{:-}, ::Val{:unary}, args, var) - a′ = D(first(args), var) - :(-$a′) -end -function D(::Val{:-}, ::Val{:binary}, args, var) - a′, b′ = D.(args, var) - :($a′ - $b′) -end -``` - -The *product rule* is similar to addition, in that ``3`` cases are considered: - -```julia -D(op::Val{:*}, ::Val{:unary}, args, var) = D(first(args), var) - -function D(::Val{:*}, ::Val{:binary}, args, var) - a, b = args - a′, b′ = D.(args, var) - :($a′ * $b + $a * $b′) -end - -function D(op::Val{:*}, ::Val{:nary}, args, var) - a, bs... = args - b = :(*($(bs...))) - a′ = D(a, var) - b′ = D(b, var) - :($a′ * $b + $a * $b′) -end -``` - -The *nary* case above just peels off the first factor and then uses the binary product rule. - -Division is only a binary operation, so here we have the *quotient rule*: - -```julia -function D(::Val{:/}, ::Val{:binary}, args, var) - u,v = args - u′, v′ = D(u, var), D(v, var) - :( ($u′*$v - $u*$v′)/$v^2 ) -end -``` - -Powers are handled a bit differently. The power rule would require checking if the exponent does not contain the variable of differentiation, exponential derivatives would require checking the base does not contain the variable of differentation. Trying to implement both would be tedious, so we use the fact that ``x = \exp(\log(x))`` (for `x` in the domain of `log`, more care is necessary if `x` is negative) to differentiate: - -```julia -function D(::Val{:^}, ::Val{:binary}, args, var) - a, b = args - D(:(exp($b*log($a))), var) # a > 0 assumed here -end -``` - - -That leaves the task of defining a rule to differentiate both `exp` and `log`. -We do so with *unary* definitions. In the following we also implement `sin` and `cos` rules: - -```julia -function D(::Val{:exp}, ::Val{:unary}, args, var) - a = first(args) - a′ = D(a, var) - :(exp($a) * $a′) -end - -function D(::Val{:log}, ::Val{:unary}, args, var) - a = first(args) - a′ = D(a, var) - :(1/$a * $a′) -end - -function D(::Val{:sin}, ::Val{:unary}, args, var) - a = first(args) - a′ = D(a, var) - :(cos($a) * $a′) -end - -function D(::Val{:cos}, ::Val{:unary}, args, var) - a = first(args) - a′ = D(a, var) - :(-sin($a) * $a′) -end -``` - -The pattern is similar for each. The `$a′` factor is needed due to the *chain rule*. The above illustrates the simple pattern necessary to add a derivative rule for a function. More could be, but for this example the above will suffice, as now the system is ready to be put to work. - - -```julia -ex₁ = :(x + 2/x) -D(ex₁, :x) -``` - -The output does not simplify, so some work is needed to identify `1 - 2/x^2` as the answer. - - -```julia -ex₂ = :( (x + sin(x))/sin(x)) -D(ex₂, :x) -``` - -Again, simplification is not performed. - -Finally, we have a second derivative taken below: - -```julia -ex₃ = :(sin(x) - x - x^3/6) -D(D(ex₃, :x), :x) -``` - - -The length of the expression should lead to further appreciation for simplification steps taken when doing such a computation by hand. diff --git a/CwJ/derivatives/taylor_series_polynomials.jmd b/CwJ/derivatives/taylor_series_polynomials.jmd deleted file mode 100644 index 4258ab8..0000000 --- a/CwJ/derivatives/taylor_series_polynomials.jmd +++ /dev/null @@ -1,1217 +0,0 @@ -# Taylor Polynomials and other Approximating Polynomials - -This section uses these add-on packages: - -```julia -using CalculusWithJulia -using Plots -using SymPy -using Unitful -``` - - -```julia; echo=false; results="hidden" -using CalculusWithJulia.WeaveSupport -using Roots - -fig_size = (800, 600) -const frontmatter = ( - title = "Taylor Polynomials and other Approximating Polynomials", - description = "Calculus with Julia: Taylor Polynomials and other Approximating Polynomials", - tags = ["CalculusWithJulia", "derivatives", "taylor polynomials and other approximating polynomials"], -); -nothing -``` - -The tangent line was seen to be the "best" linear approximation to a -function at a point $c$. Approximating a function by a linear function -gives an easier to use approximation at the expense of accuracy. It -suggests a tradeoff between ease and accuracy. Is there a way to gain more accuracy at the expense of ease? - -Quadratic functions are still fairly easy to work with. Is it possible to find the best "quadratic" -approximation to a function at a point $c$. - -More generally, for a given $n$, what would be the best polynomial of -degree $n$ to approximate $f(x)$ at $c$? - -We will see in this section how the Taylor polynomial answers these -questions, and is the appropriate generalization of the tangent -line approximation. - - -```julia; hold=true; echo=false; cache=true -###{{{taylor_animation}}} -taylor(f, x, c, n) = series(f, x, c, n+1).removeO() -function make_taylor_plot(u, a, b, k) - k = 2k - plot(u, a, b, title="plot of T_$k", linewidth=5, legend=false, size=fig_size, ylim=(-2,2.5)) - if k == 1 - plot!(zero, range(a, stop=b, length=100)) - else - plot!(taylor(u, x, 0, k), range(a, stop=b, length=100)) - end -end - - - -@syms x -u = 1 - cos(x) -a, b = -2pi, 2pi -n = 8 -anim = @animate for i=1:n - make_taylor_plot(u, a, b, i) -end - -imgfile = tempname() * ".gif" -gif(anim, imgfile, fps = 1) - - -caption = L""" - -Illustration of the Taylor polynomial of degree $k$, $T_k(x)$, at $c=0$ and its graph overlayed on that of the function $1 - \cos(x)$. - -""" - -ImageFile(imgfile, caption) -``` - -## The secant line and the tangent line - -We approach this general problem **much** more indirectly than is needed. We introducing notations that are attributed to Newton and proceed from there. By leveraging `SymPy` we avoid tedious computations and *hopefully* gain some insight. - -Suppose ``f(x)`` is a function which is defined in a neighborhood of -$c$ and has as many continuous derivatives as we care to take at $c$. - - -We have two related formulas: - -* The *secant line* connecting $(c, f(c))$ and $(c+h, f(c+h))$ for a - value of $h>0$ is given in point-slope form by - -```math -sl(x) = f(c) + \frac{(f(c+h) - f(c))}{h} \cdot (x-c). -``` - -The slope is the familiar approximation to the derivative: $(f(c+h)-f(c))/h$. - -* The *tangent line* to the graph of $f(x)$ at $x=c$ is described by - the function - -```math -tl(x) = f(c) + f'(c) \cdot(x - c). -``` - -The key is the term multiplying ``(x-c)`` for the secant line is an approximation to the related term for the tangent line. -That is, the secant line approximates the tangent -line, which is the linear function that -best approximates the function at the point $(c, f(c))$. -This is -quantified by the *mean value theorem* which states under our -assumptions on ``f(x)`` that there exists some $\xi$ between $x$ and -$c$ for which: - -```math -f(x) - tl(x) = \frac{f''(\xi)}{2} \cdot (x-c)^2. -``` - - -The term "best" is deserved, as any other straight line will differ -at least in an $(x-c)$ term, which in general is larger than an -$(x-c)^2$ term for $x$ "near" $c$. - - -(This is a consequence of Cauchy's mean value theorem with ``F(c) = f(c) - f'(c)\cdot(c-x)`` and ``G(c) = (c-x)^2`` - -```math -\begin{align*} -\frac{F'(\xi)}{G'(\xi)} &= -\frac{f'(\xi) - f''(\xi)(\xi-x) - f(\xi)\cdot 1}{2(\xi-x)} \\ -&= -f''(\xi)/2\\ -&= \frac{F(c) - F(x)}{G(c) - G(x)}\\ -&= \frac{f(c) - f'(c)(c-x) - (f(x) - f'(x)(x-x))}{(c-x)^2 - (x-x)^2} \\ -&= \frac{f(c) + f'(c)(x-c) - f(x)}{(x-c)^2} -\end{align*} -``` - -That is, ``f(x) = f(c) + f'(c)(x-c) + f''(\xi)/2\cdot(x-c)^2``, or ``f(x)-tl(x)`` is as described.) - - -The secant line also has an interpretation that will generalize - it is the smallest order polynomial that goes through, or *interpolates*, the points $(c,f(c))$ and $(c+h, f(c+h))$. This is obvious from the construction - as this is how the slope is derived - but from the formula itself requires showing $tl(c) = f(c)$ and $tl(c+h) = f(c+h)$. The former is straightforward, as $(c-c) = 0$, so clearly $tl(c) = f(c)$. The latter requires a bit of algebra. - - -We have: - -> The best *linear* approximation at a point ``c`` is related to the *linear* polynomial interpolating the points ``c`` and ``c+h`` as ``h`` goes to ``0``. - -This is the relationship we seek to generalize through our round about approach below: - -> The best approximation at a point ``c`` by a polynomial of degree ``n`` or less is related to the polynomial interpolating through the points ``c, c+h, \dots, c+nh`` as ``h`` goes to ``0``. - -As in the linear case, there is flexibility in the exact points chosen for the interpolation. - ----- - -Now, we take a small detour to define some notation. Instead of -writing our two points as $c$ and $c+h,$ we use $x_0$ and -$x_1$. For any set of points $x_0, x_1, \dots, x_n$, -define the **divided differences** of $f$ inductively, as follows: - -```math -\begin{align} -f[x_0] &= f(x_0) \\ -f[x_0, x_1] &= \frac{f[x_1] - f[x_0]}{x_1 - x_0}\\ -\cdots &\\ -f[x_0, x_1, x_2, \dots, x_n] &= \frac{f[x_1, \dots, x_n] - f[x_0, x_1, x_2, \dots, x_{n-1}]}{x_n - x_0}. -\end{align} -``` - - -We see the first two values look familiar, and to generate more we just take certain ratios akin to those formed when finding a secant line. - - -With this notation the secant line can be re-expressed as: - -```math -sl(x) = f[c] + f[c, c+h] \cdot (x-c). -``` - -If we think of $f[c, c+h]$ as an approximate *first* derivative, we -have an even stronger parallel between a secant line $x=c$ and the -tangent line at $x=c$: ``tl(x) = f(c) + f'(c)\cdot (x-c)``. - -We use `SymPy` to investigate. First we create a *recursive* function to compute the divided differences: - -```julia; -divided_differences(f, x) = f(x) - -function divided_differences(f, x, xs...) - xs = sort(vcat(x, xs...)) - (divided_differences(f, xs[2:end]...) - divided_differences(f, xs[1:end-1]...)) / (xs[end] - xs[1]) -end -``` - -In the following, by adding a `getindex` method, we enable the `[]` notation of Newton to work with symbolic functions, like `u()` defined below, which is used in place of ``f``: - -```julia; -Base.getindex(u::SymFunction, xs...) = divided_differences(u, xs...) - -@syms x::real c::real h::positive u() -ex = u[c, c+h] -``` - -We can take a limit and see the familiar (yet differently represented) value of $u'(c)$: - -```julia; -limit(ex, h => 0) -``` - -The choice of points is flexible. Here we use ``c-h`` and ``c+h``: - -```julia -limit(u[c-h, c+h], h=>0) -``` - - -Now, let's look at: - -```julia; -ex₂ = u[c, c+h, c+2h] -simplify(ex₂) -``` - -Not so bad after simplification. The limit shows this to be an approximation to the second derivative divided by $2$: - -```julia; -limit(ex₂, h => 0) -``` - -(The expression is, up to a divisor of $2$, the second order forward -[difference equation](http://tinyurl.com/n4235xy), a well-known -approximation to $f''$.) - - -This relationship between higher-order divided differences and higher-order derivatives generalizes. This is expressed in this -[theorem](http://tinyurl.com/zjogv83): - -> Suppose $m=x_0 < x_1 < x_2 < \dots < x_n=M$ are distinct points. If $f$ has $n$ -> continuous derivatives then there exists a value $\xi$, where $m < \xi < M$, satisfying: - -```math -f[x_0, x_1, \dots, x_n] = \frac{1}{n!} \cdot f^{(n)}(\xi). -``` - -This immediately applies to the above, where we parameterized by $h$: -$x_0=c, x_1=c+h, x_2 = c+2h$. For then, as $h$ goes to $0$, it must be that $m, M -\rightarrow c$, and so the limit of the divided differences must -converge to $(1/2!) \cdot f^{(2)}(c)$, as $f^{(2)}(\xi)$ converges to $f^{(2)}(c)$. - -A proof based on Rolle's theorem appears in the appendix. - - -## Quadratic approximations; interpolating polynomials - -Why the fuss? The answer comes from a result of Newton on -*interpolating* polynomials. Consider a function $f$ and $n+1$ points -$x_0$, $x_1, \dots, x_n$. Then an interpolating polynomial is a -polynomial of least degree that goes through each point $(x_i, -f(x_i))$. The [Newton form](https://en.wikipedia.org/wiki/Newton_polynomial) of such a -polynomial can be written as: - -```math -\begin{align*} -f[x_0] &+ f[x_0,x_1] \cdot (x-x_0) + f[x_0, x_1, x_2] \cdot (x-x_0) \cdot (x-x_1) + \\ -& \cdots + f[x_0, x_1, \dots, x_n] \cdot (x-x_0)\cdot \cdots \cdot (x-x_{n-1}). -\end{align*} -``` - -The case $n=0$ gives the value $f[x_0] = f(c)$, which can be interpreted as the slope-$0$ line that goes through the point $(c,f(c))$. - -We are familiar with the case $n=1$, with $x_0=c$ and $x_1=c+h$, this becomes our secant-line formula: - -```math -f[c] + f[c, c+h](x-c). -``` - -As mentioned, we can verify directly that it -interpolates the points $(c,f(c))$ and $(c+h, f(c+h))$. He we let `SymPy` do the algebra: - -```julia; -p₁ = u[c] + u[c, c+h] * (x-c) -p₁(x => c) - u(c), p₁(x => c+h) - u(c+h) -``` - - -Now for something new. Take the $n=2$ case with -$x_0 = c$, $x_1 = c + h$, and $x_2 = c+2h$. Then the interpolating polynomial is: - -```math -f[c] + f[c, c+h](x-c) + f[c, c+h, c+2h](x-c)(x-(c+h)). -``` - -We add the next term to our previous polynomial and simplify - -```julia; -p₂ = p₁ + u[c, c+h, c+2h] * (x-c) * (x-(c+h)) -simplify(p₂) -``` - -We can check that this interpolates the three points. Notice that at -$x_0=c$ and $x_1=c+h$, the last term, $f[x_0, x_1, -x_2]\cdot(x-x_0)(x-x_1)$, vanishes, so we already have the polynomial -interpolating there. Only the -value $x_2=c+2h$ remains to be checked: - -```julia; -p₂(x => c+2h) - u(c+2h) -``` - -Hmm, doesn't seem correct - that was supposed to be $0$. The issue isn't the math, it is that SymPy needs to be encouraged to simplify: - -```julia; -simplify(p₂(x => c+2h) - u(c+2h)) -``` - -By contrast, at the point $x=c+3h$ we have no guarantee of interpolation, and indeed don't, as this expression is non always zero: - -```julia; -simplify(p₂(x => c+3h) - u(c+3h)) -``` - -Interpolating polynomials are of interest in their own right, but for now we want to use them as motivation for the best polynomial approximation of a certain degree for a function. Motivated by how the secant line leads to the tangent line, we note that coefficients of the quadratic interpolating polynomial above have limits as $h$ goes to $0$, leaving this polynomial: - -```math -f(c) + f'(c) \cdot (x-c) + \frac{1}{2!} \cdot f''(c) (x-c)^2. -``` - -This is clearly related to the tangent line approximation of $f(x)$ at -$x=c$, but carrying an extra quadratic term. - -Here we visualize the approximations with -the function $f(x) = \cos(x)$ at $c=0$. - -```julia; hold=true -f(x) = cos(x) -a, b = -pi/2, pi/2 -c = 0 -h = 1/4 - -fp = -sin(c) # by hand, or use diff(f), ... -fpp = -cos(c) - - -p = plot(f, a, b, linewidth=5, legend=false, color=:blue) -plot!(p, x->f(c) + fp*(x-c), a, b; color=:green, alpha=0.25, linewidth=5) # tangent line is flat -plot!(p, x->f(c) + fp*(x-c) + (1/2)*fpp*(x-c)^2, a, b; color=:green, alpha=0.25, linewidth=5) # a parabola -p -``` - -This graph illustrates that the extra quadratic term can track the -curvature of the function, whereas the tangent line itself can't. So, -we have a polynomial which is a "better" approximation, is it the best -approximation? - - -The Cauchy mean value theorem, as in the case of the tangent line, will guarantee the existence of $\xi$ between $c$ and $x$, for which - -```math -f(x) - \left(f(c) + f'(c) \cdot(x-c) + \frac{1}{2}\cdot f''(c) \cdot (x-c)^2 \right) = -\frac{1}{3!}f'''(\xi) \cdot (x-c)^3. -``` - -In this sense, the above quadratic polynomial, called the Taylor Polynomial of degree 2, is the best *quadratic* approximation to $f$, as the difference goes to $0$ at a rate of ``(x-c)^3``. - - -The graphs of the secant line and approximating parabola for $h=1/4$ are similar: - - -```julia; hold=true -f(x) = cos(x) -a, b = -pi/2, pi/2 -c = 0 -h = 1/4 - -x0, x1, x2 = c-h, c, c+h - -f0 = divided_differences(f, x0) -fd = divided_differences(f, x0, x1) -fdd = divided_differences(f, x0, x1, x2) - -p = plot(f, a, b, color=:blue, linewidth=5, legend=false) -plot!(p, x -> f0 + fd*(x-x0), a, b, color=:green, alpha=0.25, linewidth=5); -plot!(p, x -> f0 + fd*(x-x0) + fdd * (x-x0)*(x-x1), a,b, color=:green, alpha=0.25, linewidth=5); -p -``` - -Though similar, the graphs are **not** identical, as the interpolating - polynomials aren't the best approximations. For example, in the -tangent-line graph the parabola only intersects the cosine graph at -$x=0$, whereas for the secant-line graph - by definition - the -parabola intersects the graph at least $2$ times and the -interpolating polynomial $3$ times (at $x_0$, $x_1$, and $x_2$). - - - - -##### Example - -Consider the function $f(t) = \log(1 + t)$. We have mentioned that for $t$ small, the value $t$ is a good approximation. A better one becomes: - -```math -f(0) + f'(0) \cdot t + \frac{1}{2} \cdot f''(0) \cdot t^2 = 0 + 1t - \frac{t^2}{2} -``` - -A graph shows the difference: - -```julia; hold=true -f(t) = log(1 + t) -a, b = -1/2, 1 -plot(f, a, b, legend=false, linewidth=5) -plot!(t -> t, a, b) -plot!(t -> t - t^2/2, a, b) -``` - -Though we can see that the tangent line is a good approximation, the -quadratic polynomial tracks the logarithm better farther from $c=0$. - -##### Example - -A wire is bent in the form of a half circle with radius $R$ centered -at $(0,R)$, so the bottom of the wire is at the origin. A bead is -released on the wire at angle $\theta$. As time evolves, the bead will -slide back and forth. How? (Ignoring friction.) - - -Let $U$ be the potential energy, $U=mgh = mgR \cdot (1 - -\cos(\theta))$. The velocity of the object will depend on $\theta$ - -it will be $0$ at the high point, and largest in magnitude at the -bottom - and is given by $v(\theta) = R \cdot d\theta/ dt$. (The bead -moves along the wire so its distance traveled is $R\cdot \Delta -\theta$, this, then, is just the time derivative of distance.) - -By ignoring friction, the total energy is conserved giving: - -```math -K = \frac{1}{2}m v^2 + mgR \cdot (1 - \cos(\theta) = -\frac{1}{2} m R^2 (\frac{d\theta}{dt})^2 + mgR \cdot (1 - \cos(\theta)). -``` - -The value of $1-\cos(\theta)$ inhibits further work which would be possible were there an easier formula there. In fact, we could try the excellent approximation $1 - \theta^2/2$ from the quadratic approximation. Then we have: - -```math -K \approx \frac{1}{2} m R^2 (\frac{d\theta}{dt})^2 + mgR \cdot (1 - \theta^2/2). -``` - -Assuming equality and differentiating in $t$ gives by the chain rule: - -```math -0 = \frac{1}{2} m R^2 2\frac{d\theta}{dt} \cdot \frac{d^2\theta}{dt^2} - mgR \theta\cdot \frac{d\theta}{dt}. -``` - -This can be solved to give this relationship: - -```math -\frac{d^2\theta}{dt^2} = - \frac{g}{R}\theta. -``` - -The solution to this "equation" can be written (in some -parameterization) as $\theta(t)=A\cos \left(\omega t+\phi -\right)$. This motion is the well-studied simple [harmonic -oscillator](https://en.wikipedia.org/wiki/Harmonic_oscillator), a -model for a simple pendulum. - -#### Example: optimization - -Consider the following approach to finding the minimum or maximum of a function: - -* At ``x_k`` fit a quadratic polynomial to ``f(x)`` matching the derivative and second derivative of ``f``. -* Let ``x_{k+1}`` be at the vertex of this fitted quadratic polynomial -* Iterate to convergence - - -The polynomial in question will be the Taylor polynomial of degree ``2``: - -```math -T_2(x_k) = f(x_k) + f'(x_k)(x-x_k) + \frac{f''(x_k)}{2}(x - x_k)^2 -``` - -The vertex of this quadratic polynomial will be when its derivative is ``0`` which can be solved for ``x_{k+1}`` giving: - -```math -x_{k+1} = x_k - \frac{f'(x_k)}{f''(x_k)}. -``` - -This assumes ``f''(x_k)`` is non-zero. - -On inspection, it is seen that this is Newton's method applied to -``f'(x)``. This method, when convergent, finds a zero of ``f'(x)``. We -know that should the algorithm converge, it will have found a critical -point, not necessarily a value for a local extrema. - -## The Taylor polynomial of degree ``n`` - - -Starting with the Newton form of the interpolating polynomial of smallest degree: - -```math -\begin{align*} -f[x_0] &+ f[x_0,x_1] \cdot (x - x_0) + f[x_0, x_1, x_2] \cdot (x - x_0)\cdot(x-x_1) + \\ -& \cdots + f[x_0, x_1, \dots, x_n] \cdot (x-x_0) \cdot \cdots \cdot (x-x_{n-1}). -\end{align*} -``` - -and taking $x_i = c + i\cdot h$, for a given $n$, we have in the limit as $h > 0$ goes to zero that coefficients of this polynomial converge to the coefficients of the *Taylor Polynomial of degree n*: - -```math -f(c) + f'(c)\cdot(x-c) + \frac{f''(c)}{2!}(x-c)^2 + \cdots + \frac{f^{(n)}(c)}{n!} (x-c)^n. -``` - - - -This polynomial will be the best approximation of degree ``n`` or less -to the function $f$, near $c$. The error will be given - again by an -application of the Cauchy mean value theorem: - -```math -\frac{1}{(n+1)!} \cdot f^{(n+1)}(\xi) \cdot (x-c)^n -``` - -for some $\xi$ between $c$ and $x$. - - - -The Taylor polynomial for $f$ about $c$ of degree $n$ can be computed -by taking $n$ derivatives. For such a task, the computer is very -helpful. In `SymPy` the `series` function will compute the Taylor -polynomial for a given $n$. For example, here is the series expansion -to 10 terms of the function $\log(1+x)$ about $c=0$: - - -```julia; hold=true -c, n = 0, 10 -l = series(log(1 + x), x, c, n+1) -``` - -A pattern can be observed. - - - -Using `series`, we can see Taylor polynomials for several familiar functions: - - -```julia; -series(1/(1-x), x, 0, 10) # sum x^i for i in 0:n -``` - -```julia; -series(exp(x), x, 0, 10) # sum x^i/i! for i in 0:n -``` - -```julia; -series(sin(x), x, 0, 10) # sum (-1)^i * x^(2i+1) / (2i+1)! for i in 0:n -``` - - -```julia; -series(cos(x), x, 0, 10) # sum (-1)^i * x^(2i) / (2i)! for i in 0:n -``` - -Each of these last three have a pattern that can be expressed quite succinctly if the denominator is recognized as $n!$. - - -The output of `series` includes a big "Oh" term, which identifies the -scale of the error term, but also gets in the way of using the -output. `SymPy` provides the `removeO` method to strip this. (It is called as `object.removeO()`, as it is a method of an object in SymPy.) - - - -!!! note - A Taylor polynomial of degree ``n`` consists of ``n+1`` terms and an error term. The "Taylor series" is an *infinite* collection of terms, the first ``n+1`` matching the Taylor polynomial of degree ``n``. The fact that series are *infinite* means care must be taken when even talking about their existence, unlike a Tyalor polynomial, which is just a polynomial and exists as long as a sufficient number of derivatives are available. - - - -We define a function to compute Taylor polynomials from a function. The following returns a function, not a symbolic object, using `D`, from `CalculusWithJulia`, which is based on `ForwardDiff.derivative`, to find higher-order derivatives: - -```julia; -function taylor_poly(f, c=0, n=2) - x -> f(c) + sum(D(f, i)(c) * (x-c)^i / factorial(i) for i in 1:n) -end -``` - -With a function, we can compare values. -For example, here we see the difference between the Taylor polynomial and the answer for a small value of $x$: - -```julia; hold=true -a = .1 -f(x) = log(1+x) -Tn = taylor_poly(f, 0, 5) -Tn(a) - f(a) -``` - - -### Plotting - -Let's now visualize a function and the two approximations - the Taylor -polynomial and the interpolating polynomial. We use this function to -generate the interpolating polynomial as a function: - -```julia; -function newton_form(f, xs) - x -> begin - tot = divided_differences(f, xs[1]) - for i in 2:length(xs) - tot += divided_differences(f, xs[1:i]...) * prod([x-xs[j] for j in 1:(i-1)]) - end - tot - end -end -``` - -To see a plot, we have - -```julia; -𝒇(x) = sin(x) -𝒄, 𝒉, 𝒏 = 0, 1/4, 4 -int_poly = newton_form(𝒇, [𝒄 + i*𝒉 for i in 0:𝒏]) -tp = taylor_poly(𝒇, 𝒄, 𝒏) -𝒂, 𝒃 = -pi, pi -plot(𝒇, 𝒂, 𝒃; linewidth=5, label="f") -plot!(int_poly; color=:green, label="interpolating") -plot!(tp; color=:red, label="Taylor") -``` - -To get a better sense, we plot the residual differences here: - -```julia -d1(x) = 𝒇(x) - int_poly(x) -d2(x) = 𝒇(x) - tp(x) -plot(d1, 𝒂, 𝒃; color=:blue, label="interpolating") -plot!(d2; color=:green, label="Taylor") -``` - -The graph should be $0$ at each of the the points in `xs`, which we -can verify in the graph above. Plotting over a wider region shows a -common phenomenon that these polynomials approximate the function near -the values, but quickly deviate away: - - -In this graph we make a plot of the Taylor polynomial for different sizes of $n$ for the function ``f(x) = 1 - \cos(x)``: - - -```julia; hold=true -f(x) = 1 - cos(x) -a, b = -pi, pi -plot(f, a, b, linewidth=5, label="f") -plot!(taylor_poly(f, 0, 2), label="T₂") -plot!(taylor_poly(f, 0, 4), label="T₄") -plot!(taylor_poly(f, 0, 6), label="T₆") -``` - -Though all are good approximations near $c=0$, as more terms are -included, the Taylor polynomial becomes a better approximation over a wider -range of values. - - -##### Example: period of an orbiting satellite - -Kepler's third [law](http://tinyurl.com/y7oa4x2g) of planetary motion states: - -> The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. - -In formulas, $P^2 = a^3 \cdot (4\pi^2) / (G\cdot(M + m))$, where $M$ and $m$ are the respective masses. Suppose a satellite is in low earth orbit with a constant height, $a$. Use a Taylor polynomial to approximate the period using Kepler's third law to relate the quantities. - -Suppose $R$ is the radius of the earth and $h$ the height above the earth assuming $h$ is much smaller than $R$. The mass $m$ of a satellite is negligible to that of the earth, so $M+m=M$ for this purpose. We have: - -```math -P = \frac{2\pi}{\sqrt{G\cdot M}} \cdot (h+R)^{3/2} = \frac{2\pi}{\sqrt{G\cdot M}} \cdot R^{3/2} \cdot (1 + h/R)^{3/2} = P_0 \cdot (1 + h/R)^{3/2}, -``` - -where $P_0$ collects terms that involve the constants. - -We can expand $(1+x)^{3/2}$ to fifth order, to get: - -```math -(1+x)^{3/2} \approx 1 + \frac{3x}{2} + \frac{3x^2}{8} - \frac{1x^3}{16} + \frac{3x^4}{128} -\frac{3x^5}{256} -``` - -Our approximation becomes: - -```math -P \approx P_0 \cdot (1 + \frac{3(h/R)}{2} + \frac{3(h/R)^2}{8} - \frac{(h/R)^3}{16} + \frac{3(h/R)^4}{128} - \frac{3(h/R)^5}{256}). -``` - -Typically, if $h$ is much smaller than $R$ the first term is enough giving a formula like $P \approx P_0 \cdot(1 + \frac{3h}{2R})$. - - -A satellite phone utilizes low orbit satellites to relay phone communications. The [Iridium](http://www.kddi.com/english/business/cloud-network-voice/satellite/iridium/mobile/) system uses satellites with an elevation ``h=780km``. The radius of the earth is $3,959 miles$, the mass of the earth is $5.972 × 10^{24} kg$, and the gravitational [constant](https://en.wikipedia.org/wiki/Gravitational_constant), $G$ is $6.67408 \cdot 10^{-11}$ $m^3/(kg \cdot s^2)$. - -Compare the approximate value with ``1`` term to the exact value. - -```julia; -G = 6.67408e-11 -H = 780 * 1000 -R = 3959 * 1609.34 # 1609 meters per mile -M = 5.972e24 -P0, HR = (2pi)/sqrt(G*M) * R^(3/2), H/R - -Preal = P0 * (1 + HR)^(3/2) -P1 = P0 * (1 + 3*HR/2) -Preal, P1 -``` - -With terms out to the fifth power, we get a better approximation: - -```julia; -P5 = P0 * (1 + 3*HR/2 + 3*HR^2/8 - HR^3/16 + 3*HR^4/128 - 3*HR^5/256) -``` - -The units of the period above are in seconds. That answer here is about ``100`` minutes: - -```julia; -Preal/60 -``` - -When $H$ is much smaller than $R$ the approximation with ``5``th order is -really good, and serviceable with just ``1`` term. Next we check if this -is the same when $H$ is larger than $R$. - ----- - -The height of a [GPS satellite](http://www.gps.gov/systems/gps/space/) is about $12,550$ miles. Compute the period of a circular orbit and compare with the estimates. - -```julia; -Hₛ = 12250 * 1609.34 # 1609 meters per mile -HRₛ = Hₛ/R - -Prealₛ = P0 * (1 + HRₛ)^(3/2) -P1ₛ = P0 * (1 + 3*HRₛ/2) -P5ₛ = P0 * (1 + 3*HRₛ/2 + 3*HRₛ^2/8 - HRₛ^3/16 + 3*HRₛ^4/128 - 3*HRₛ^5/256) - -Prealₛ, P1ₛ, P5ₛ -``` - -We see the Taylor polynomial underestimates badly in this case. A reminder -that these approximations are locally good, but may not be good on all -scales. Here $h \approx 3R$. We can see from this graph -of $(1+x)^{3/2}$ and its ``5``th degree Taylor polynomial $T_5$ that it is a bad approximation when $x > 2$. - -```julia; echo=false -f1(x) = (1+x)^(3/2) -p2(x) = 1 + 3x/2 + 3x^2/8 - x^3/16 + 3x^4/128 - 3x^5/256 -plot(f1, -1, 3, linewidth=4, legend=false) -plot!(p2, -1, 3) -``` - ----- - -Finally, we show how to use the `Unitful` package. This package allows us to define different units, carry these -units through computations, and convert between similar units with -`uconvert`. In this example, we define several units, then show how -they can then be used as constants. - -```julia; hold=true -m, mi, kg, s, hr = u"m", u"mi", u"kg", u"s", u"hr" - -G = 6.67408e-11 * m^3 / kg / s^2 -H = uconvert(m, 12250 * mi) # unit convert miles to meter -R = uconvert(m, 3959 * mi) -M = 5.972e24 * kg - -P0, HR = (2pi)/sqrt(G*M) * R^(3/2), H/R -Preal = P0 * (1 + HR)^(3/2) # in seconds -Preal, uconvert(hr, Preal) # ≈ 11.65 hours -``` - -We see `Preal` has the right units - the units of mass and distance cancel leaving a measure of time - but it is hard to sense how long this is. Converting to hours, helps us see the satellite orbits about twice per day. - - - -##### Example: computing $\log(x)$ - -Where exactly does the value assigned to $\log(5)$ come from? The -value needs to be computed. At some level, many questions resolve down -to the basic operations of addition, subtraction, multiplication, and -division. Preferably not the latter, as division is slow. Polynomials -then should be fast to compute, and so computing logarithms using a -polynomial becomes desirable. - -But how? One can see details of a possible -way -[here](https://github.com/musm/Amal.jl/blob/master/src/log.jl). - -First, there is usually a reduction stage. In this phase, the problem -is transformed in a manner to one involving only a fixed interval of values. For this, -function values of $k$ and $m$ are found so that $x = 2^k \cdot (1+m)$ -*and* $\sqrt{2}/2 < 1+m < \sqrt{2}$. If these are found, then $\log(x)$ can be computed with -$k \cdot \log(2) + \log(1+m)$. The first value - a multiplication - can easily be -computed using pre-computed value of $\log(2)$, the second then *reduces* the problem to an interval. - - -Now, for this problem a further -trick is utilized, writing $s= f/(2+f)$ so that -$\log(1+m)=\log(1+s)-\log(1-s)$ for some small range of $s$ values. These combined make it possible to compute $\log(x)$ for any real $x$. - -To compute $\log(1\pm s)$, we can find a Taylor polynomial. Let's go out to degree $19$ and use `SymPy` to do the work: - -```julia; -@syms s -aₗ = series(log(1 + s), s, 0, 19) -bₗ = series(log(1 - s), s, 0, 19) -a_b = (aₗ - bₗ).removeO() # remove"Oh" not remove"zero" -``` - -This is re-expressed as $2s + s \cdot p$ with $p$ given by: - -```julia; -cancel(a_b - 2s/s) -``` - -Now, $2s = m - s\cdot m$, so the above can be reworked to be $\log(1+m) = m - s\cdot(m-p)$. - - -(For larger values of $m$, a similar, but different approximation, can be used to minimize floating point errors.) - - -How big can the error be between this *approximations* and $\log(1+m)$? We plot to see how big $s$ can be: - -```julia; -@syms v -plot(v/(2+v), sqrt(2)/2 - 1, sqrt(2)-1) -``` - -This shows, $s$ is as big as - -```julia; -Max = (v/(2+v))(v => sqrt(2) - 1) -``` - -The error term is like $2/19 \cdot \xi^{19}$ which is largest at this value of $M$. Large is relative - it is really small: - -```julia; -(2/19)*Max^19 -``` - -Basically that is machine precision. Which means, that as far as can be told on the computer, the value produced by $2s + s \cdot p$ is about as accurate as can be done. - -To try this out to compute $\log(5)$. We have $5 = 2^2(1+0.25)$, so $k=2$ and $m=0.25$. - -```julia -k, m = 2, 0.25 -𝒔 = m / (2+m) -pₗ = 2 * sum(𝒔^(2i)/(2i+1) for i in 1:8) # where the polynomial approximates the logarithm... - -log(1 + m), m - 𝒔*(m-pₗ), log(1 + m) - ( m - 𝒔*(m-pₗ)) - -``` - -The two values differ by less than $10^{-16}$, as advertised. Re-assembling then, we compare the computed values: - -```julia; -Δ = k * log(2) + (m - 𝒔*(m-pₗ)) - log(5) -``` - - -The actual code is different, as the Taylor polynomial isn't -used. The Taylor polynomial is a great approximation near a point, but -there might be better polynomial approximations for all values in an interval. -In this case there is, and that polynomial is used in the production -setting. This makes things a bit more efficient, but the basic idea -remains - for a prescribed accuracy, a polynomial approximation can -be found over a given interval, which can be cleverly utilized to -solve for all applicable values. - - -##### Example: higher order derivatives of the inverse function - -For notational purposes, let ``g(x)`` be the inverse function for ``f(x)``. Assume *both* functions have a Taylor polynomial expansion: - -```math -\begin{align*} -f(x_0 + \Delta_x) &= f(x_0) + a_1 \Delta_x + a_2 (\Delta_x)^2 + \cdots a_n + (\Delta_x)^n + \dots\\ -g(y_0 + \Delta_y) &= g(y_0) + b_1 \Delta_y + b_2 (\Delta_y)^2 + \cdots b_n + (\Delta_y)^n + \dots -\end{align*} -``` - -Then using ``x = g(f(x))``, we have expanding the terms and using ``\approx`` to drop the ``\dots``: - -```math -\begin{align*} -x_0 + \Delta_x &= g(f(x_0 + \Delta_x)) \\ -&\approx g(f(x_0) + \sum_{j=1}^n a_j (\Delta_x)^j) \\ -&\approx g(f(x_0)) + \sum_{i=1}^n b_i \left(\sum_{j=1}^n a_j (\Delta_x)^j \right)^i \\ -&\approx x_0 + \sum_{i=1}^{n-1} b_i \left(\sum_{j=1}^n a_j (\Delta_x)^j\right)^i + b_n \left(\sum_{j=1}^n a_j (\Delta_x)^j\right)^n -\end{align*} -``` - -That is: - -```math -b_n \left(\sum_{j=1}^n a_j (\Delta_x)^j \right)^n = -(x_0 + \Delta_x) - \left( x_0 + \sum_{i=1}^{n-1} b_i \left(\sum_{j=1}^n a_j (\Delta_x)^j \right)^i \right) -``` - -Solving for ``b_n = g^{(n)}(y_0) / n!`` gives the formal expression: - -```math -g^{(n)}(y_0) = n! \cdot \lim_{\Delta_x \rightarrow 0} -\frac{\Delta_x - \sum_{i=1}^{n-1} b_i \left(\sum_{j=1}^n a_j (\Delta_x)^j \right)^i}{ -\left(\sum_{j=1}^n a_j \left(\Delta_x^j\right)^i\right)^n} -``` - -(This is following [Liptaj](https://vixra.org/pdf/1703.0295v1.pdf)). - -We will use `SymPy` to take this limit for the first `4` derivatives. Here is some code that expands ``x + \Delta_x = g(f(x_0 + \Delta_x))`` and then uses `SymPy` to solve: - -```julia; -@syms x₀ Δₓ f′[1:4] g′[1:4] - -as(i) = f′[i]/factorial(i) -bs(i) = g′[i]/factorial(i) - -gᵏs = Any[] -eqns = Any[] -for n ∈ 1:4 - Δy = sum(as(j) * Δₓ^j for j ∈ 1:n) - left = x₀ + Δₓ - right = x₀ + sum(bs(i)*Δy^i for i ∈ 1:n) - - eqn = left ~ right - push!(eqns, eqn) - - gⁿ = g′[n] - ϕ = solve(eqn, gⁿ)[1] - - # replace g′ᵢs in terms of computed f′ᵢs - for j ∈ 1:n-1 - ϕ = subs(ϕ, g′[j] => gᵏs[j]) - end - - L = limit(ϕ, Δₓ => 0) - push!(gᵏs, L) - -end -gᵏs -``` - -We can see the expected `g' = 1/f'` (where the point of evalution is ``g(y) = 1/f'(f^{-1}(y))`` is not written). In addition, we get 3 more formulas, hinting that the answers grow rapidly in terms of their complexity. - -In the above, for each `n`, the code above sets up the two sides, `left` and `right`, of an equation involving the higher-order derivatives of ``g``. For example, when `n=2` we have: - -```julia; -eqns[2] -``` - - -The `solve` function is used to identify ``g^{(n)}`` represented in terms of lower-order derivatives of ``g``. These values have been computed and stored and are then substituted into `ϕ`. Afterwards a limit is taken and the answer recorded. - - - - - -## Questions - -###### Question - -Compute the Taylor polynomial of degree ``10`` for $\sin(x)$ about $c=0$ using `SymPy`. Based on the form, which formula seems appropriate: - -```julia; hold=true; echo=false -choices = [ -"``\\sum_{k=0}^{10} x^k``", -"``\\sum_{k=1}^{10} (-1)^{n+1} x^n/n``", -"``\\sum_{k=0}^{4} (-1)^k/(2k+1)! \\cdot x^{2k+1}``", -"``\\sum_{k=0}^{10} x^n/n!``" -] -answ = 3 -radioq(choices, answ) -``` - -###### Question - -Compute the Taylor polynomial of degree ``10`` for $e^x$ about $c=0$ using `SymPy`. Based on the form, which formula seems appropriate: - -```julia; hold=true; echo=false -choices = [ -"``\\sum_{k=0}^{10} x^k``", -"``\\sum_{k=1}^{10} (-1)^{n+1} x^n/n``", -"``\\sum_{k=0}^{4} (-1)^k/(2k+1)! \\cdot x^{2k+1}``", -"``\\sum_{k=0}^{10} x^n/n!``" -] -answ = 4 -radioq(choices, answ) -``` - - - -###### Question - -Compute the Taylor polynomial of degree ``10`` for $1/(1-x)$ about $c=0$ using `SymPy`. Based on the form, which formula seems appropriate: - -```julia; hold=true; echo=false -choices = [ -"``\\sum_{k=0}^{10} x^k``", -"``\\sum_{k=1}^{10} (-1)^{n+1} x^n/n``", -"``\\sum_{k=0}^{4} (-1)^k/(2k+1)! \\cdot x^{2k+1}``", -"``\\sum_{k=0}^{10} x^n/n!``" -] -answ = 1 -radioq(choices, answ) -``` - -###### Question - -Let $T_5(x)$ be the Taylor polynomial of degree ``5`` for the function $\sqrt{1+x}$ about $x=0$. What is the coefficient of the $x^5$ term? - -```julia; hold=true; echo=false -choices = [ -"``7/256``", -"``-5/128``", -"``1/5!``", -"``2/15``" -] -answ = 1 -radioq(choices, answ) -``` - -###### Question - -The ``5``th order Taylor polynomial for $\sin(x)$ about $c=0$ is: $x - x^3/3! + x^5/5!$. Use this to find the first ``3`` terms of the Taylor polynomial of $\sin(x^2)$ about $c=0$. - -They are: - -```julia; hold=true; echo=false -choices = [ -"``x^2 - x^6/3! + x^{10}/5!``", -"``x^2``", -"``x^2 \\cdot (x - x^3/3! + x^5/5!)``" -] -answ = 1 -radioq(choices, answ) -``` - - - -###### Question - -A more direct derivation of the form of the Taylor polynomial (here taken about $c=0$) is to *assume* a polynomial form that matches $f$: - -```math -f(x) = a + bx + cx^2 + dx^3 + ex^4 + \cdots -``` - -If this is true, then formally evaluating at $x=0$ gives $f(0) = a$, so $a$ is determined. Similarly, formally differentiating and evaluating at $0$ gives $f'(0) = b$. What is the result of formally differentiating $4$ times and evaluating at $0$: - -```julia; hold=true; echo=false -choices = ["``f''''(0) = e``", -"``f''''(0) = 4 \\cdot 3 \\cdot 2 e = 4! e``", -"``f''''(0) = 0``"] -answ = 2 -radioq(choices, answ) -``` - -###### Question - -How big an error is there in approximating $e^x$ by its ``5``th degree Taylor polynomial about $c=0$, $1 + x + x^2/2! + x^3/3! + x^4/4! + x^5/5!$, over $[-1,1]$? - -The error is known to be $( f^{(6)}(\xi)/6!) \cdot x^6$ for some $\xi$ in $[-1,1]$. - - -* The ``6``th derivative of $e^x$ is still $e^x$: - -```julia; hold=true; echo=false -yesnoq(true) -``` - -* Which is true about the function $e^x$: - -```julia; hold=true; echo=false -choices =["It is increasing", "It is decreasing", "It both increases and decreases"] -answ = 1 -radioq(choices, answ) -``` - - -* The maximum value of $e^x$ over $[-1,1]$ occurs at - -```julia; hold=true; echo=false -choices=["A critical point", "An end point"] -answ = 2 -radioq(choices, answ) -``` - -* Which theorem tells you that for a *continuous* function over *closed* interval, a maximum value will exist? - -```julia; hold=true; echo=false -choices = [ -"The intermediate value theorem", -"The mean value theorem", -"The extreme value theorem"] -answ = 3 -radioq(choices, answ) -``` - -* What is the *largest* possible value of the error: - -```julia; hold=true; echo=false -choices = [ -"``1/6!\\cdot e^1 \\cdot 1^6``", -"``1^6 \\cdot 1 \\cdot 1^6``"] -answ = 1 -radioq(choices,answ) -``` - -###### Question - -The error in using $T_k(x)$ to approximate $e^x$ over the interval $[-1/2, 1/2]$ is $(1/(k+1)!) e^\xi x^{k+1}$, for some $\xi$ in the interval. This is *less* than $1/((k+1)!) e^{1/2} (1/2)^{k+1}$. - -* Why? - -```julia; hold=true; echo=false -choices = [ -L"The function $e^x$ is increasing, so takes on its largest value at the endpoint and the function $|x^n| \leq |x|^n \leq (1/2)^n$", -L"The function has a critical point at $x=1/2$", -L"The function is monotonic in $k$, so achieves its maximum at $k+1$" -] -answ = 1 -radioq(choices, answ) -``` - -Assuming the above is right, find the smallest value $k$ guaranteeing a error no more than $10^{-16}$. - -```julia; hold=true; echo=false -f(k) = 1/factorial(k+1) * exp(1/2) * (1/2)^(k+1) -(f(13) > 1e-16 && f(14) < 1e-16) && numericq(14) -``` - -* The function $f(x) = (1 - x + x^2) \cdot e^x$ has a Taylor polynomial about ``0`` such that all coefficients are rational numbers. Is it true that the numerators are all either ``1`` or prime? (From the 2014 [Putnam](http://kskedlaya.org/putnam-archive/2014.pdf) exam.) - -Here is one way to get all the values bigger than 1: - -```julia; hold=true; -ex = (1 - x + x^2)*exp(x) -Tn = series(ex, x, 0, 100).removeO() -ps = sympy.Poly(Tn, x).coeffs() -qs = numer.(ps) -qs[qs .> 1] |> Tuple # format better for output -``` - -Verify by hand that each of the remaining values is a prime number to answer the question (Or you can use `sympy.isprime.(qs)`). - -Are they all prime or $1$? - -```julia; hold=true; echo=false -yesnoq(true) -``` - -## Appendix - -We mentioned two facts that could use a proof: the Newton form of the interpolating polynomial and the mean value theorem for divided differences. Our explanation tries to emphasize a parallel with the secant line's relationship with the tangent line. The standard way to discuss Taylor polynomials is different (also more direct) and so these two proofs are not in most calculus texts. - -A [proof](https://www.math.uh.edu/~jingqiu/math4364/interpolation.pdf) of the Newton form can be done knowing that the interpolating polynomial is unique and can be expressed either as - -```math -g(x)=a_0 + a_1 (x-x_0) + \cdots + a_n (x-x_0)\cdot\cdots\cdot(x-x_{n-1}) -``` - -*or* in this reversed form - -```math -h(x)=b_0 + b_1 (x-x_n) + b_2(x-x_n)(x-x_{n-1}) + \cdots + b_n (x-x_n)(x-x_{n-1})\cdot\cdots\cdot(x-x_1). -``` - -These two polynomials are of degree $n$ or less and have $u(x) = h(x)-g(x)=0$, by uniqueness. So the coefficients of $u(x)$ are $0$. We have that the coefficient of $x^n$ must be $a_n-b_n$ so $a_n=b_n$. Our goal is to express $a_n$ in terms of $a_{n-1}$ and $b_{n-1}$. Focusing on the $x^{n-1}$ term, we have: - -```math -\begin{align*} -b_n(x-x_n)(x-x_{n-1})\cdot\cdots\cdot(x-x_1) -&- a_n\cdot(x-x_0)\cdot\cdots\cdot(x-x_{n-1}) \\ -&= -a_n [(x-x_1)\cdot\cdots\cdot(x-x_{n-1})] [(x- x_n)-(x-x_0)] \\ -&= -a_n \cdot(x_n - x_0) x^{n-1} + p_{n-2}, -\end{align*} -``` - -where $p_{n-2}$ is a polynomial of at most degree $n-2$. (The expansion of $(x-x_1)\cdot\cdots\cdot(x-x_{n-1}))$ leaves $x^{n-1}$ plus some lower degree polynomial.) Similarly, we have -$a_{n-1}(x-x_0)\cdot\cdots\cdot(x-x_{n-2}) = a_{n-1}x^{n-1} + q_{n-2}$ and -$b_{n-1}(x-x_n)\cdot\cdots\cdot(x-x_2) = b_{n-1}x^{n-1}+r_{n-2}$. Combining, we get that the $x^{n-1}$ term of $u(x)$ is - -```math -(b_{n-1}-a_{n-1}) - a_n(x_n-x_0) = 0. -``` - -On rearranging, this yields $a_n = (b_{n-1}-a_{n-1}) / (x_n - x_0)$. By *induction* - that $a_i=f[x_0, x_1, \dots, x_i]$ and $b_i = f[x_n, x_{n-1}, \dots, x_{n-i}]$ (which has trivial base case) - this is $(f[x_1, \dots, x_n] - f[x_0,\dots x_{n-1}])/(x_n-x_0)$. - -Now, assuming the Newton form is correct, a -[proof](http://tinyurl.com/zjogv83) of the mean value theorem for -divided differences comes down to Rolle's theorem. Starting from the -Newton form of the polynomial and expanding in terms of -$1, x, \dots, x^n$ we see that -$g(x) = p_{n-1}(x) + f[x_0, x_1, \dots,x_n]\cdot x^n$, -where now $p_{n-1}(x)$ is a -polynomial of degree at most $n-1$. That is, the coefficient of -$x^n$ is $f[x_0, x_1, \dots, x_n]$. Consider the function $h(x)=f(x) - g(x)$. -It has zeros $x_0, x_1, \dots, x_n$. - -By Rolle's theorem, between any two such zeros $x_i, x_{i+1}$, $0 \leq i < n$ there must be a zero of the derivative of $h(x)$, say $\xi^1_i$. So $h'(x)$ has zeros $\xi^1_0 < \xi^1_1 < \dots < \xi^1_{n-1}$. - - -We visualize this with $f(x) = \sin(x)$ and $x_i = i$ for $i=0, 1, 2, 3$, The $x_i$ values are indicated with circles, the $\xi^1_i$ values indicated with squares: - -```julia; hold=true; echo=false -f(x) = sin(x) -xs = 0:3 -dd = divided_differences -g(x) = dd(f,0) + dd(f, 0,1)*x + dd(f, 0,1,2)*x*(x-1) + dd(f, 0,1,2,3)*x*(x-1)*(x-2) -h1(x) = f(x) - g(x) -cps = find_zeros(D(h1), -1, 4) -plot(h1, -1/4, 3.25, linewidth=3, legend=false) -scatter!(xs, h1.(xs), markersize=5) -scatter!(cps, h1.(cps), markersize=5, marker=:square) -``` - - - -Again by Rolle's theorem, between any pair of adjacent zeros $\xi^1_i, \xi^1_{i+1}$ there must be a zero $\xi^2_i$ of $h''(x)$. So there are $n-1$ zeros of $h''(x)$. Continuing, we see that there will be -$n+1-3$ zeros of $h^{(3)}(x)$, -$n+1-4$ zeros of $h^{4}(x)$, $\dots$, -$n+1-(n-1)$ zeros of $h^{n-1}(x)$, and finally -$n+1-n$ ($1$) zeros of $h^{(n)}(x)$. Call this last zero $\xi$. It satisfies $x_0 \leq \xi \leq x_n$. Further, -$0 = h^{(n)}(\xi) = f^{(n)}(\xi) - g^{(n)}(\xi)$. But $g$ is a degree $n$ polynomial, so the $n$th derivative is the coefficient of $x^n$ times $n!$. In this case we have $0 = f^{(n)}(\xi) - f[x_0, \dots, x_n] n!$. Rearranging yields the result. diff --git a/CwJ/differentiable_vector_calculus/Project.toml b/CwJ/differentiable_vector_calculus/Project.toml deleted file mode 100644 index c6d07d2..0000000 --- a/CwJ/differentiable_vector_calculus/Project.toml +++ /dev/null @@ -1,15 +0,0 @@ -[deps] -CSV = "336ed68f-0bac-5ca0-87d4-7b16caf5d00b" -Contour = "d38c429a-6771-53c6-b99e-75d170b6e991" -DataFrames = "a93c6f00-e57d-5684-b7b6-d8193f3e46c0" -DifferentialEquations = "0c46a032-eb83-5123-abaf-570d42b7fbaa" -ForwardDiff = "f6369f11-7733-5829-9624-2563aa707210" -IntervalSets = "8197267c-284f-5f27-9208-e0e47529a953" -JSON = "682c06a0-de6a-54ab-a142-c8b1cf79cde6" -LinearAlgebra = "37e2e46d-f89d-539d-b4ee-838fcccc9c8e" -MDBM = "dd61e66b-39ce-57b0-8813-509f78be4b4d" -Plots = "91a5bcdd-55d7-5caf-9e0b-520d859cae80" -PyPlot = "d330b81b-6aea-500a-939a-2ce795aea3ee" -QuadGK = "1fd47b50-473d-5c70-9696-f719f8f3bcdc" -Roots = "f2b01f46-fcfa-551c-844a-d8ac1e96c665" -SymPy = "24249f21-da20-56a4-8eb1-6a02cf4ae2e6" 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related things in `Julia`. This directory holds alternatives to the some choices made within these notes: - -## Symbolics - -* needs writing - -## Makie - -* needs updating diff --git a/CwJ/alternatives/SciML.jmd b/CwJ/alternatives/SciML.jmd deleted file mode 100644 index e497340..0000000 --- a/CwJ/alternatives/SciML.jmd +++ /dev/null @@ -1,625 +0,0 @@ -# The SciML suite of packages - - -The `Julia` ecosystem advances rapidly. For much of it, the driving force is the [SciML](https://github.com/SciML) organization (Scientific Machine Learning). - -In this section we describe some packages provided by this organization that could be used as alternatives to the ones utilized in these notes. Members of this organization created many packages for solving different types of differential equations, and have branched out from there. Many newer efforts of this organization have been to write uniform interfaces to other packages in the ecosystem, some of which are discussed below. We don't discuss the promise of SCIML: "Performance is considered a priority, and performance issues are considered bugs," as we don't pursue features like in-place modification, sparsity, etc. Interested readers should consult the relevant packages documentation. - -The basic structure to use these packages is the "problem-algorithm-solve" interface described in [The problem-algorithm-solve interface](../ODEs/solve.html). We also discussed this interface a bit in [ODEs](../ODEs/differential_equations.html). - -!!! note - These packages are in a process of rapid development and change to them is expected. These notes were written using the following versions: - -```julia -pkgs = ["Symbolics", "NonlinearSolve", "Optimization", "Integrals"] -import Pkg; Pkg.status(pkgs) -``` - -## Symbolic math (`Symbolics`) - -The `Symbolics`, `SymbolicUtils`, and `ModelingToolkit` packages are provided by this organization. These can be viewed as an alternative to `SymPy`, which is used throughout this set of notes. See the section on [Symbolics](./symbolics.html) for some additional details, the package [documentation](https://symbolics.juliasymbolics.org/stable/), or the documentation for [SymbolicsUtils](https://github.com/JuliaSymbolics/SymbolicUtils.jl). - - -## Solving equations - -Solving one or more equations (simultaneously) is different in the linear case (where solutions are readily found -- though performance can distinguish approaches -- and the nonlinear case -- where for most situations, numeric approaches are required. - -### `LinearSolve` - -The `LinearSolve` package aims to generalize the solving of linear equations. For many cases these are simply represented as matrix equations of the form `Ax=b`, from which `Julia` (borrowing from MATLAB) offers the interface `A \ b` to yield `x`. There are scenarios that don't naturally fit this structure and perhaps problems where different tolerances need to be specified, and the `LinearSolve` package aims to provide a common interface to handle these scenarios. As this set of notes doesn't bump into such, this package is not described here. In the symbolic case, the `Symbolics.solve_for` function was described in [Symbolics](./symbolics.html). - -### `NonlinearSolve` -The `NonlinearSolve` package can be seen as an alternative to the use of the `Roots` package in this set of notes. The package presents itself as "Fast implementations of root finding algorithms in Julia that satisfy the SciML common interface." - -The package is loaded through the following command: - -```julia -using NonlinearSolve -``` - -Unlike `Roots`, the package handles problems beyond the univariate case, as such the simplest problems have a little extra setup required. - -For example, suppose we want to use this package to solve for zeros of ``f(x) = x^5 - x - 1``. We could do so a few different ways. - -First, we need to define a `Julia` function representing `f`. We do so with: - -```julia -f(u, p) = @. (u^5 - u - 1) -``` - -The function definition expects a container for the "`x`" variables and allows the passing of a container to hold parameters. We could have used the dotted operations for the power and each subtraction to allow vectorization of these basic math operations, as `u` is a container of values. The `@.` macro makes adding the "dots" quite easy, as illustrated above. It converts "every function call or operator in expr into a `dot call`." - -A problem is set up with this function and an initial guess. The `@SVector` specification for the guess is for performance purposes and is provided by the `StaticArrays` package. - -```julia -using StaticArrays -u0 = @SVector[1.0] -prob = NonlinearProblem(f, u0) -``` - -The problem is solved by calling `solve` with an appropriate method specified. Here we use Newton's method. The derivative of `f` is computed automatically. - -```julia -soln = solve(prob, NewtonRaphson()) -``` - -The basic interface for retrieving the solution from the solution object is to use indexing: - -```julia -soln[] -``` - ----- - -!!! note - This interface is more performant than `Roots`, though it isn't an apples to oranges comparison as different stopping criteria are used by the two. In order to be so, we need to help out the call to `NonlinearProblem` to indicate the problem is non-mutating by adding a "`false`", as follows: - -```julia -using BenchmarkTools -@btime solve(NonlinearProblem{false}(f, @SVector[1.0]), NewtonRaphson()) -``` - -As compared to: - -```julia -import Roots -import ForwardDiff -g(x) = x^5 - x - 1 -gp(x) = ForwardDiff.derivative(g, x) -@btime solve(Roots.ZeroProblem((g, gp), 1.0), Roots.Newton()) -``` ----- - -This problem can also be solved using a bracketing method. The package has both `Bisection` and `Falsi` as possible methods. To use a bracketing method, the initial bracket must be specified. - -```julia -u0 = (1.0, 2.0) -prob = NonlinearProblem(f, u0) -``` - -And - -```julia -solve(prob, Bisection()) -``` - ----- - -Incorporating parameters is readily done. For example to solve ``f(x) = \cos(x) - x/p`` for different values of ``p`` we might have: - - -```julia -f(x, p) = @. cos(x) - x/p -u0 = (0, pi/2) -p = 2 -prob = NonlinearProblem(f, u0, p) -solve(prob, Bisection()) -``` - -!!! note - The *insignificant* difference in stopping criteria used by `NonlinearSolve` and `Roots` is illustrated in this example, where the value returned by `NonlinearSolve` differs by one floating point value: - -```julia -an = solve(NonlinearProblem{false}(f, u0, p), Bisection()) -ar = solve(Roots.ZeroProblem(f, u0), Roots.Bisection(); p=p) -nextfloat(an[]) == ar, f(an[], p), f(ar, p) -``` - - ----- - -We can solve for several parameters at once, by using an equal number of initial positions as follows: - -```julia -ps = [1, 2, 3, 4] -u0 = @SVector[1, 1, 1, 1] -prob = NonlinearProblem(f, u0, ps) -solve(prob, NewtonRaphson()) -``` - - -### Higher dimensions - - -We solve now for a point on the surface of the following `peaks` function where the gradient is ``0``. (The gradient here will be a vector-valued function from ``R^2 `` to ``R^2.``) First we define the function: - -```julia -function _peaks(x, y) - p = 3 * (1 - x)^2 * exp(-x^2 - (y + 1)^2) - p -= 10 * (x / 5 - x^3 - y^5) * exp(-x^2 - y^2) - p -= 1/3 * exp(-(x + 1)^2 - y^2) - p -end -peaks(u) = _peaks(u[1], u[2]) # pass container, take first two components -``` - -The gradient can be computed different ways within `Julia`, but here we use the fact that the `ForwardDiff` package is loaded by `NonlinearSolve`. Once the function is defined, the pattern is similar to above. We provide a starting point, create a problem, then solve: - -```julia -∇peaks(x, p=nothing) = NonlinearSolve.ForwardDiff.gradient(peaks, x) -u0 = @SVector[1.0, 1.0] -prob = NonlinearProblem(∇peaks, u0) -u = solve(prob, NewtonRaphson()) -``` - -We can see that this identified value is a "zero" through: - -```julia; error=true -∇peaks(u.u) -``` - -### Using Modeling toolkit to model the non-linear problem - -Nonlinear problems can also be approached symbolically using the `ModelingToolkit` package. There is one additional step necessary. - -As an example, we look to solve numerically for the zeros of ``x^5-x-\alpha`` for a parameter ``\alpha``. We can describe this equation as follows: - -```julia -using ModelingToolkit - -@variables x -@parameters α - -eq = x^5 - x - α ~ 0 -``` - -The extra step is to specify a "`NonlinearSystem`." It is a system, as in practice one or more equations can be considered. The `NonlinearSystem`constructor handles the details where the equation, the variable, and the parameter are specified. Below this is done using vectors with just one element: - -```julia -ns = NonlinearSystem([eq], [x], [α], name=:ns) -``` - -The `name` argument is special. The name of the object (`ns`) is assigned through `=`, but the system must also know this same name. However, the name on the left is not known when the name on the right is needed, so it is up to the user to keep them synchronized. The `@named` macro handles this behind the scenes by simply rewriting the syntax of the assignment: - -```julia -@named ns = NonlinearSystem([eq], [x], [α]) -``` - -With the system defined, we can pass this to `NonlinearProblem`, as was done with a function. The parameter is specified here, and in this case is `α => 1.0`. The initial guess is `[1.0]`: - -```julia -prob = NonlinearProblem(ns, [1.0], [α => 1.0]) -``` - -The problem is solved as before: - -```julia -solve(prob, NewtonRaphson()) -``` - -## Optimization (`Optimization.jl`) - -We describe briefly the `Optimization` package which provides a common interface to *numerous* optimization packages in the `Julia` ecosystem. We discuss only the interface for `Optim.jl` defined in `OptimizationOptimJL`. - -We begin with a simple example from first semester calculus: - -> Among all rectangles of fixed perimeter, find the one with the *maximum* area. - -If the perimeter is taken to be ``25``, the mathematical setup has a -constraint (``P=25=2x+2y``) and an objective (``A=xy``) to -maximize. In this case, the function to *maximize* is ``A(x) = x \cdot -(25-2x)/2``. This is easily done different ways, such as finding the -one critical point and identifying this as the point of maximum. - -To do this last step using `Optimization` we would have. - -```julia -height(x) = @. (25 - 2x)/2 -A(x, p=nothing) = @.(- x * height(x)) -``` - -The minus sign is needed here as optimization routines find *minimums*, not maximums. - -To use `Optimization` we must load the package **and** the underlying backend glue code we intend to use: - -```julia -using Optimization -using OptimizationOptimJL -``` - - -Next, we define an optimization function with information on how its derivatives will be taken. The following uses `ForwardDiff`, which is a good choice in the typical calculus setting, where there are a small number of inputs (just ``1`` here.) - -```julia -F = OptimizationFunction(A, Optimization.AutoForwardDiff()) -x0 = [4.0] -prob = OptimizationProblem(F, x0) -``` - -The problem is solved through the common interface with a specified method, in this case `Newton`: - -```julia -soln = solve(prob, Newton()) -``` - -!!! note - We use `Newton` not `NewtonRaphson` as above. Both methods are similar, but they come from different uses -- for latter for solving non-linear equation(s), the former for solving optimization problems. - -The solution is an object containing the identified answer and more. To get the value, use index notation: - -```julia -soln[] -``` - -The corresponding ``y`` value and area are found by: - -```julia -xstar = soln[] -height(xstar), A(xstar) -``` - -The `minimum` property also holds the identified minimum: - -```julia -soln.minimum # compare with A(soln[], nothing) -``` - -The package is a wrapper around other packages. The output of the underlying package is presented in the `original` property: - -```julia -soln.original -``` - ----- - - -This problem can also be approached symbolically, using `ModelingToolkit`. - -For example, we set up the problem with: - -```julia -using ModelingToolkit -@parameters P -@variables x -y = (P - 2x)/2 -Area = - x*y -``` - -The above should be self explanatory. To put into a form to pass to `solve` we define a "system" by specifying our objective function, the variables, and the parameters. - -```julia -@named sys = OptimizationSystem(Area, [x], [P]) -``` - -(This step is different, as before an `OptimizationFunction` was defined; we use `@named`, as above, to ensure the system has the same name as the identifier, `sys`.) - - -This system is passed to `OptimizationProblem` along with a specification of the initial condition (``x=4``) and the perimeter (``P=25``). A vector of pairs is used below: - -```julia -prob = OptimizationProblem(sys, [x => 4.0], [P => 25.0]; grad=true, hess=true) -``` - -The keywords `grad=true` and `hess=true` instruct for automatic derivatives to be taken as needed. These are needed in the choice of method, `Newton`, below. - -Solving this problem then follows the same pattern as before, again with `Newton` we have: - -```julia -solve(prob, Newton()) -``` - -(A derivative-free method like `NelderMead()` could be used and then the `grad` and `hess` keywords above would be unnecessary, though not harmful.) - ----- - -The related calculus problem: - -> Among all rectangles with a fixed area, find the one with *minimum* perimeter - - -could be similarly approached: - -```julia -@parameters Area -@variables x -y = Area/x # from A = xy -P = 2x + 2y -@named sys = OptimizationSystem(P, [x], [Area]) - -u0 = [x => 4.0] -p = [Area => 25.0] - -prob = OptimizationProblem(sys, u0, p; grad=true, hess=true) -soln = solve(prob, LBFGS()) -``` - -We used an initial guess of ``x=4`` above. The `LBFGS` method is -a computationally efficient modification of the -Broyden-Fletcher-Goldfarb-Shanno algorithm ... It is a quasi-Newton -method that updates an approximation to the Hessian using past -approximations as well as the gradient." On this problem it performs similarly to `Newton`, though in general may be preferable for higher-dimensional problems. - -### Two dimensional - -Scalar functions of two input variables can have their minimum value identified in the same manner using `Optimization.jl`. - -For example, consider the function - -```math -f(x,y) = (x + 2y - 7)^2 + (2x + y - 5)^2 -``` - -We wish to minimize this function. - - -We begin by defining a function in `Julia`: - -```julia -function f(u, p) - x, y = u - (x + 2y - 7)^2 + (2x + y - 5)^2 -end -``` - -We turn this into an optimization function by specifying how derivatives will be taken, as we will the `LBFGS` algorithm below: - -```julia -ff = OptimizationFunction(f, Optimization.AutoForwardDiff()) -``` - -We will begin our search at the origin. We have to be mindful to use floating point numbers here: - -```julia -u0 = [0.0, 0.0] # or zeros(2) -``` - -```julia -prob = OptimizationProblem(ff, u0) -``` - -Finally, we solve the values: - -```julia -solve(prob, LBFGS()) -``` - -The value of ``(1, 3)`` agrees with the contour graph, as it is a point in the interior of the contour for the smallest values displayed. - -```julia -using Plots - -xs = range(0, 2, length=100) -ys = range(2, 4, length=100) -contour(xs, ys, (x,y) -> f((x,y), nothing)) -``` - - -We could also use a *derivative-free* method, and skip a step: - -```julia -prob = OptimizationProblem(f, u0) # skip making an OptimizationFunction -solve(prob, NelderMead()) -``` - -## Integration (`Integrals.jl`) - -The `Integrals` package provides a common interface to different numeric integration packages in the `Julia` ecosystem. For example, `QuadGK` and `HCubature`. The value of this interface, over those two packages, is its non-differentiated access to other packages, which for some uses may be more performant. - -The package follows the same `problem-algorithm-solve` interface, as already seen. - -The interface is designed for ``1``-and-higher dimensional integrals. - -The package is loaded with - -```julia -using Integrals -``` - - -For a simple definite integral, such as ``\int_0^\pi \sin(x)dx``, we have: - -```julia -f(x, p) = sin(x) -prob = IntegralProblem(f, 0.0, pi) -soln = solve(prob, QuadGKJL()) -``` - -To get access to the answer, we can use indexing notation: - -```julia -soln[] -``` - -Comparing to just using `QuadGK`, the same definite integral would be -estimated with: - -```julia -using QuadGK -quadgk(sin, 0, pi) -``` - -The estimated upper bound on the error from `QuadGK`, is available through the `resid` property on the `Integrals` output: - -```julia -soln.resid -``` - - -The `Integrals` solution is a bit more verbose, but it is more flexible. For example, the `HCubature` package provides a similar means to compute ``n``- dimensional integrals. For this problem, the modifications would be: - -```julia -f(x, p) = sin.(x) -prob = IntegralProblem(f, [0.0], [pi]) -soln = solve(prob, HCubatureJL()) -``` - -```julia -soln[] -``` - -The estimated maximum error is also given by `resid`: - -```julia -soln.resid -``` - ----- - -As well, suppose we wanted to parameterize our function and then differentiate. - -Consider ``d/dp \int_0^\pi \sin(px) dx``. We can do this integral directly to get - -```math -\begin{align*} -\frac{d}{dp} \int_0^\pi \sin(px)dx -&= \frac{d}{dp}\left( \frac{-1}{p} \cos(px)\Big\rvert_0^\pi\right)\\ -&= \frac{d}{dp}\left( -\frac{\cos(p\cdot\pi)-1}{p}\right)\\ -&= \frac{\cos(p\cdot \pi) - 1)}{p^2} + \frac{\pi\cdot\sin(p\cdot\pi)}{p} -\end{align*} -``` - -Using `Integrals` with `QuadGK` we have: - -```julia -f(x, p) = sin(p*x) -function ∫sinpx(p) - prob = IntegralProblem(f, 0.0, pi, p) - solve(prob, QuadGKJL()) -end -``` - -We can compute values at both ``p=1`` and ``p=2``: - -```julia -∫sinpx(1), ∫sinpx(2) -``` - -To find the derivative in ``p`` , we have: - -```julia -ForwardDiff.derivative(∫sinpx, 1), ForwardDiff.derivative(∫sinpx, 2) -``` - - -(In `QuadGK`, the following can be differentiated `∫sinpx(p) = quadgk(x -> sin(p*x), 0, pi)[1]` as well. `Integrals` gives a consistent interface. - - -### Higher dimension integrals - -The power of a common interface is the ability to swap backends and the uniformity for different dimensions. Here we discuss integrals of scalar-valued and vector-valued functions. - -#### ``f: R^n \rightarrow R`` - -The area under a surface generated by ``z=f(x,y)`` over a rectangular region ``[a,b]\times[c,d]`` can be readily computed. The two coding implementations require ``f`` to be expressed as a function of a vector--*and* a parameter--and the interval to be expressed using two vectors, one for the left endpoints (`[a,c]`) and on for the right endpoints (`[b,d]`). - -For example, the area under the function ``f(x,y) = 1 + x^2 + 2y^2`` over ``[-1/2, 1/2] \times [-1,1]`` is computed by: - -```julia -f(x, y) = 1 + x^2 + 2y^2 # match math -fxp(x, p) = f(x[1], x[2]) # prepare for IntegralProblem -ls = [-1/2, -1] # left endpoints -rs = [1/2, 1] # right endpoints -prob = IntegralProblem(fxp, ls, rs) -soln = solve(prob, HCubatureJL()) -``` - -Of course, we could have directly defined the function (`fxp`) using indexing of the `x` variable. - ----- - -For non-rectangular domains a change of variable is required. - -For example, an integral to assist in finding the volume of a sphere might be - -```math -V = 2 \iint_R \sqrt{\rho^2 - x^2 - y^2} dx dy -``` - -where ``R`` is the disc of radius ``\rho`` in the ``x-y`` plane. - -The usual approach is to change to polar-coordinates and write this integral as - -```math -V = \int_0^{2\pi}\int_0^\rho \sqrt{\rho^2 - r^2} r dr d\theta -``` - -the latter being an integral over a rectangular domain. - -To compute this transformed integral, we might have: - -```julia -function vol_sphere(ρ) - f(rθ, p) = sqrt(ρ^2 - rθ[1]^2) * rθ[1] - ls = [0,0] - rs = [ρ, 2pi] - prob = IntegralProblem(f, ls, rs) - solve(prob, HCubatureJL()) -end - -vol_sphere(2) -``` - -If it is possible to express the region to integrate as ``G(R)`` where ``R`` is a rectangular region, then the change of variables formula, - -```math -\iint_{G(R)} f(x) dA = \iint_R (f\circ G)(u) |det(J_G(u)| dU -``` - -turns the integral into the non-rectangular domain ``G(R)`` into one over the rectangular domain ``R``. The key is to *identify* ``G`` and to compute the Jacobian. The latter is simply accomplished with `ForwardDiff.jacobian`. - -For an example, we find the moment of inertia about the axis of the unit square tilted counter-clockwise an angle ``0 \leq \alpha \leq \pi/2``. - -The counter clockwise rotation of a unit square by angle ``\alpha`` is described by - -```math -G(u, v) = \langle \cos(\alpha)\cdot u - \sin(\alpha)\cdot v, \sin(\alpha)\cdot u, +\cos(\alpha)\cdot v \rangle -``` - -So we have ``\iint_{G(R)} x^2 dA`` is computed by the following with ``\alpha=\pi/4``: - -```julia -import LinearAlgebra: det - - -𝑓(uv) = uv[1]^2 - -function G(uv) - - α = pi/4 # could be made a parameter - - u,v = uv - [cos(α)*u - sin(α)*v, sin(α)*u + cos(α)*v] -end - -f(u, p) = (𝑓∘G)(u) * det(ForwardDiff.jacobian(G, u)) - -prob = IntegralProblem(f, [0,0], [1,1]) -solve(prob, HCubatureJL()) -``` - -#### ``f: R^n \rightarrow R^m`` - -The `Integrals` package provides an interface for vector-valued functions. By default, the number of dimensions in the output is assumed to be ``1``, but the `nout` argument can adjust that. - - -Let ``f`` be vector valued with components ``f_1, f_2, \dots, f_m``, then the output below is the vector with components ``\iint_R f_1 dV, \iint_R f_2 dV, \dots, \iint_R f_m dV``. - - - -For a trivial example, we have: - -```julia -f(x, p) = [x[1], x[2]^2] -prob = IntegralProblem(f, [0,0],[3,4], nout=2) -solve(prob, HCubatureJL()) -``` diff --git a/CwJ/alternatives/interval_arithmetic.__jmd__ b/CwJ/alternatives/interval_arithmetic.__jmd__ deleted file mode 100644 index 78fd860..0000000 --- a/CwJ/alternatives/interval_arithmetic.__jmd__ +++ /dev/null @@ -1,107 +0,0 @@ -# Using interval arithemetic - -Highlighted here is the use of interval arithmetic for calculus problems. - -Unlike floating point math, where floating point values are an *approximation* to real numbers, interval arithmetic uses *interval* which are **guaranteed** to contain the given value. We use the `IntervalArithmetic` package and friends to work below, but note there is nothing magic about the concept. - -## Basic XXX - - - -## Using `IntervalRootFinding` to identify zeros of a function - -The `IntervalRootFinding` package provides a more *rigorous* alternative to `find_zeros`. This packages leverages the interval arithmetic features of `IntervalArithmetic`. -The `IntervalRootFinding` package provides a function `roots`, with usage similar to `find_zeros`. Intervals are specified with the notation `a..b`. In the following, we *qualify* `roots` to not conflict with the `roots` function from `SymPy`, which has already been loaded: - -```julia -import IntervalArithmetic -import IntervalRootFinding -``` - -```julia -u(x) = sin(x) - 0.1*x^2 + 1 -𝑱 = IntervalArithmetic.Interval(-10, 10) # cumbersome -10..10; needed here: .. means something in CalculusWithJulia -rts = IntervalRootFinding.roots(u, 𝑱) -``` - -The "zeros" are returned with an enclosing interval and a flag, which for the above indicates a unique zero in the interval. - -The intervals with a unique answer can be filtered and refined with a construct like the following: - -```julia -[find_zero(u, (IntervalArithmetic.interval(I).lo, IntervalArithmetic.interval(I).hi)) for I in rts if I.status == :unique] -``` - -The midpoint of the returned interval can be found by composing the `mid` function with the `interval` function of the package: - -```julia -[(IntervalArithmetic.mid ∘ IntervalArithmetic.interval)(I) for I in rts if I.status == :unique] -``` - - - -For some problems, `find_zeros` is more direct, as with this one: - - -```julia -find_zeros(u, (-10, 10)) -``` - -Which can be useful if there is some prior understanding of the zeros expected to be found. -However, `IntervalRootFinding` is more efficient computationally and *offers a guarantee* on the values found. - - - -For functions where roots are not "unique" a different output may appear: - -```julia; hold=true; -f(x) = x*(x-1)^2 -rts = IntervalRootFinding.roots(f, 𝑱) -``` - -The interval labeled `:unknown` contains a `0`, but it can't be proved by `roots`. - - -Interval arithmetic finds **rigorous** **bounds** on the range of `f` values over a closed interval `a..b` (the range is `f(a..b)`). "Rigorous" means the bounds are truthful and account for possible floating point issues. "Bounds" means the answer lies within, but the bound need not be the answer. - -This allows one -- for some functions -- to answer affirmatively questions like: - -* Is the function *always* positive on `a..b`? Negative? This can be done by checking if `0` is in the bound given by `f(a..b)`. If it isn't then one of the two characterizations is true. - -* Is the function *strictly increasing* on `a..b`? Strictly decreasing? These questions can be answered using the (upcoming) [derivative](../derivatives/derivatives.html). If the derivative is positive on `a..b` then `f` is strictly increasing, if negative on `a..b` then `f` is strictly decreasing. Finding the derivative can be done within the `IntervalArithmetic` framework using [automatic differentiation](../derivatives/numeric_derivatives.html), a blackbox operation denoted `f'` below. - -Combined, for some functions and some intervals these two questions can be answered affirmatively: - -* the interval does not contain a zero (`0 !in f(a..b)`) -* over the interval, the function crosses the `x` axis *once* (`f(a..a)` and `f(b..b)` are one positive and one negative *and* `f` is strictly monotone, or `0 !in f'(a..b)`) - -This allows the following (simplified) bisection-like algorithm to be used: - -* consider an interval `a..b` -* if the function is *always* positive or negative, it can be discarded as no zero can be in the interval -* if the function crosses the `x` axis *once* over this interval **then** there is a "unique" zero in the interval and the interval can be marked so and set aside -* if neither of the above *and* `a..b` is not too small already, then *sub-divide* the interval and repeat the above with *both* smaller intervals -* if `a..b` is too small, stop and mark it as "unknown" - -When terminated there will be intervals with unique zeros flagged and smaller intervals with an unknown status. - -Compared to the *bisection* algorithm -- which only knows for some intervals if that interval has one or more crossings -- this algorithm gives a more rigorous means to get all the zeros in `a..b`. - - - - - - -For a last example of the value of this package, this function, which appeared in our discussion on limits, is *positive* for **every** floating point number, but has two zeros snuck in at values within the floating point neighbors of $15/11$ - -```julia -t(x) = x^2 + 1 +log(abs( 11*x-15 ))/99 -``` - -The `find_zeros` function will fail on identifying any potential zeros of this function. Even the basic call of `roots` will fail, due to it relying on some smoothness of `f`. However, explicitly asking for `Bisection` shows the *potential* for one or more zeros near $15/11$: - -```julia -IntervalRootFinding.roots(t, 𝑱, IntervalRootFinding.Bisection) -``` - -(The basic algorithm above can be sped up using a variant of [Newton's](../derivatives/newton_method.html) method, this variant assumes some "smoothness" in the function `f`, whereas this `f` is not continuous at the point ``x=15/11``.) diff --git a/CwJ/alternatives/makie_plotting.jmd b/CwJ/alternatives/makie_plotting.jmd deleted file mode 100644 index 97b6605..0000000 --- a/CwJ/alternatives/makie_plotting.jmd +++ /dev/null @@ -1,1046 +0,0 @@ -# Calculus plots with Makie - - -The [Makie.jl webpage](https://github.com/JuliaPlots/Makie.jl) says - -> From the Japanese word Maki-e, which is a technique to sprinkle lacquer with gold and silver powder. Data is basically the gold and silver of our age, so let's spread it out beautifully on the screen! - -`Makie` itself is a metapackage for a rich ecosystem. We show how to -use the interface provided by the `GLMakie` backend to produce the -familiar graphics of calculus. - - -!!! note "Examples and tutorials" - `Makie` is a sophisticated plotting package, and capable of an enormous range of plots (cf. [examples](https://makie.juliaplots.org/stable/examples/plotting_functions/).) `Makie` also has numerous [tutorials](https://makie.juliaplots.org/stable/tutorials/) to learn from. These are far more extensive that what is described herein, as this section focuses just on the graphics from calculus. - -## Figures - -Makie draws graphics onto a canvas termed a "scene" in the Makie documentation. -A scene is an implementation detail, the basic (non-mutating) plotting commands described below return a `FigureAxisPlot` object, a compound object that combines a figure, an axes, and a plot object. The `show` method for these objects display the figure. - -For `Makie` there are the `GLMakie`, `WGLMakie`, and `CairoMakie` backends for different types of canvases. In the following, we have used `GLMakie`. `WGLMakie` is useful for incorporating `Makie` plots into web-based technologies. - -We begin by loading the main package and the `norm` function from the standard `LinearAlgebra` package: - -```julia -using GLMakie -import LinearAlgebra: norm - -``` - -The `Makie` developers have workarounds for the delayed time to first plot, but without utilizing these the time to load the package is lengthy. - - -## Points (`scatter`) - -The task of plotting the points, say ``(1,2)``, ``(2,3)``, ``(3,2)`` can be done different ways. Most plotting packages, and `Makie` is no exception, allow the following: form vectors of the ``x`` and ``y`` values then plot those with `scatter`: - -```julia -xs = [1,2,3] -ys = [2,3,2] -scatter(xs, ys) -``` - -The `scatter` function creates and returns an object, which when displayed shows the plot. - - -### `Point2`, `Point3` - -When learning about points on the Cartesian plane, a "`t`"-chart is often produced: - -``` -x | y ------ -1 | 2 -2 | 3 -3 | 2 -``` - -The `scatter` usage above used the columns. The rows are associated with the points, and these too can be used to produce the same graphic. -Rather than make vectors of ``x`` and ``y`` (and optionally ``z``) coordinates, it is more idiomatic to create a vector of "points." `Makie` utilizes a `Point` type to store a 2 or 3 dimensional point. The `Point2` and `Point3` constructors will be utilized. - -`Makie` uses a GPU, when present, to accelerate the graphic rendering. GPUs employ 32-bit numbers. Julia uses an `f0` to indicate 32-bit floating points. Hence the alternate types `Point2f0` to store 2D points as 32-bit numbers and `Points3f0` to store 3D points as 32-bit numbers are seen in the documentation for Makie. - - -We can plot a vector of points in as direct manner as vectors of their coordinates: - -```julia -pts = [Point2(1,2), Point2(2,3), Point2(3,2)] -scatter(pts) -``` - -A typical usage is to generate points from some vector-valued -function. Say we have a parameterized function `r` taking ``R`` into -``R^2`` defined by: - -```julia -r(t) = [sin(t), cos(t)] -``` - - -Then broadcasting values gives a vector of vectors, each identified with a point: - -```julia -ts = [1,2,3] -r.(ts) -``` - -We can broadcast `Point2` over this to create a vector of `Point` objects: - -```julia -pts = Point2.(r.(ts)) -``` - -These then can be plotted directly: - -```julia -scatter(pts) -``` - - -The ploting of points in three dimesions is essentially the same, save the use of `Point3` instead of `Point2`. - -```julia -r(t) = [sin(t), cos(t), t] -ts = range(0, 4pi, length=100) -pts = Point3.(r.(ts)) -scatter(pts; markersize=5) -``` - - ----- - -To plot points generated in terms of vectors of coordinates, the -component vectors must be created. The "`t`"-table shows how, simply -loop over each column and add the corresponding ``x`` or ``y`` (or ``z``) -value. This utility function does exactly that, returning the vectors -in a tuple. - -```julia -unzip(vs) = Tuple([vs[j][i] for j in eachindex(vs)] for i in eachindex(vs[1])) -``` - -!!! note - In the `CalculusWithJulia` package, `unzip` is implemented using `SplitApplyCombine.invert`. - -We might have then: - -```julia -scatter(unzip(r.(ts))...; markersize=5) -``` - -where splatting is used to specify the `xs`, `ys`, and `zs` to `scatter`. - -(Compare to `scatter(Point3.(r.(ts)))` or `scatter(Point3∘r).(ts))`.) - -### Attributes - -A point is drawn with a "marker" with a certain size and color. These attributes can be adjusted, as in the following: - -```julia -scatter(xs, ys; - marker=[:x,:cross, :circle], markersize=25, - color=:blue) -``` - -Marker attributes include - -* `marker` a symbol, shape. -* `marker_offset` offset coordinates -* `markersize` size (radius pixels) of marker - -A single value will be repeated. A vector of values of a matching size will specify the attribute on a per point basis. - -## Curves - -The curves of calculus are lines. The `lines` command of `Makie` will render a curve by connecting a series of points with straight-line segments. By taking a sufficient number of points the connect-the-dot figure can appear curved. - - -### Plots of univariate functions - -The basic plot of univariate calculus is the graph of a function ``f`` over an interval ``[a,b]``. This is implemented using a familiar strategy: produce a series of representative values between ``a`` and ``b``; produce the corresponding ``f(x)`` values; plot these as points and connect the points with straight lines. - - -To create regular values between `a` and `b` typically the `range` function or the range operator (`a:h:b`) are employed. The the related `LinRange` function is also an option. - - -For example: - -```julia -f(x) = sin(x) -a, b = 0, 2pi -xs = range(a, b, length=250) -lines(xs, f.(xs)) -``` - -`Makie` also will read the interval notation of `IntervalSets` and select its own set of intermediate points: - -```julia -lines(a..b, f) -``` - - -As with `scatter`, `lines` returns an object that produces a graphic when displayed. - -As with `scatter`, `lines` can can also be drawn using a vector of points: - -```julia -pts = [Point2(x, f(x)) for x ∈ xs] -lines(pts) -``` - -(Though the advantage isn't clear here, this will be useful when the points are generated in different manners.) - -When a `y` value is `NaN` or infinite, the connecting lines are not drawn: - -```julia -xs = 1:5 -ys = [1,2,NaN, 4, 5] -lines(xs, ys) -``` - -As with other plotting packages, this is useful to represent discontinuous functions, such as what occurs at a vertical asymptote or a step function. - - -#### Adding to a figure (`lines!`, `scatter!`, ...) - -To *add* or *modify* a scene can be done using a mutating version of a plotting primitive, such as `lines!` or `scatter!`. The names follow `Julia`'s convention of using an `!` to indicate that a function modifies an argument, in this case the underlying figure. - -Here is one way to show two plots at once: - -```julia -xs = range(0, 2pi, length=100) -lines(xs, sin.(xs)) -lines!(xs, cos.(xs)) -current_figure() -``` - -!!! note "Current figure" - The `current_figure` call is needed to have the figure display, as the returned value of `lines!` is not a figure object. (Figure objects display when shown as the output of a cell.) - - -We will see soon how to modify the line attributes so that the curves can be distinguished. - -The following shows the construction details in the graphic: - -```julia -xs = range(0, 2pi, length=10) -lines(xs, sin.(xs)) -scatter!(xs, sin.(xs); - markersize=10) -current_figure() -``` - - -As an example, this shows how to add the tangent line to a graph. The slope of the tangent line being computed by `ForwardDiff.derivative`. - -```julia -import ForwardDiff -f(x) = x^x -a, b= 0, 2 -c = 0.5 -xs = range(a, b, length=200) - -tl(x) = f(c) + ForwardDiff.derivative(f, c) * (x-c) - -lines(xs, f.(xs)) -lines!(xs, tl.(xs), color=:blue) -current_figure() -``` - -This example, modified from a [discourse](https://discourse.julialang.org/t/how-to-plot-step-functions-x-correctly-in-julia/84087/5) post by user `@rafael.guerra`, shows how to plot a step function (`floor`) using `NaN`s to create line breaks. The marker colors set for `scatter!` use `:white` to match the background color. - -```julia -x = -5:5 -δ = 5eps() # for rounding purposes; our interval is [i,i+1) ≈ [i, i+1-δ] -xx = Float64[] -for i ∈ x[1:end-1] - append!(xx, (i, i+1 - δ, NaN)) -end -yy = floor.(xx) - -lines(xx, yy) -scatter!(xx, yy, color=repeat([:black, :white, :white], length(xx)÷3)) - -current_figure() -``` - - - -### Text (`annotations`) - -Text can be placed at a point, as a marker is. To place text, the desired text and a position need to be specified along with any adjustments to the default attributes. - -For example: - -```julia -xs = 1:5 -pts = Point2.(xs, xs) -scatter(pts) -annotations!("Point " .* string.(xs), pts; - textsize = 50 .- 2*xs, - rotation = 2pi ./ xs) - -current_figure() -``` - -The graphic shows that `textsize` adjusts the displayed size and `rotation` adjusts the orientation. (The graphic also shows a need to manually override the limits of the `y` axis, as the `Point 5` is chopped off; the `ylims!` function to do so will be shown later.) - -Attributes for `text`, among many others, include: - -* `align` Specify the text alignment through `(:pos, :pos)`, where `:pos` can be `:left`, `:center`, or `:right`. -* `rotation` to indicate how the text is to be rotated -* `textsize` the font point size for the text -* `font` to indicate the desired font - - - -#### Line attributes - -In a previous example, we added the argument `color=:blue` to the `lines!` call. This was to set an attribute for the line being drawn. Lines have other attributes that allow different ones to be distinguished, as above where colors indicate the different graphs. - -Other attributes can be seen from the help page for `lines`, and include: - -* `color` set with a symbol, as above, or a string -* `label` a label for the line to display in a legend -* `linestyle` available styles are set by a symbol, one of `:dash`, `:dot`, `:dashdot`, or `:dashdotdot`. -* `linewidth` width of line -* `transparency` the `alpha` value, a number between ``0`` and ``1``, smaller numbers for more transparent. - - -#### Simple legends - -A simple legend displaying labels given to each curve can be produced by `axislegend`. For example: - -```julia -xs = 0..pi -lines(xs, x -> sin(x^2), label="sin(x^2)") -lines!(xs, x -> sin(x)^2, label = "sin(x)^2") -axislegend() - -current_figure() -``` - -Later, we will see how to control the placement of a legend within a figure. - -#### Titles, axis labels, axis ticks - -The basic plots we have seen are of type `FigureAxisPlot`. The "axis" part controls attributes of the plot such as titles, labels, tick positions, etc. These values can be set in different manners. On construction we can pass values to a named argument `axis` using a named tuple. - -For example: - -```julia -xs = 0..2pi -lines(xs, sin; - axis=(title="Plot of sin(x)", xlabel="x", ylabel="sin(x)") - ) -``` - -To access the `axis` element of a plot **after** the plot is constructed, values can be assigned to the `axis` property of the `FigureAxisPlot` object. For example: - -```julia -xs = 0..2pi -p = lines(xs, sin; - axis=(title="Plot of sin(x)", xlabel="x", ylabel="sin(x)") - ) -p.axis.xticks = MultiplesTicks(5, pi, "π") # label 5 times using `pi` - -current_figure() -``` - -The ticks are most easily set as a collection of values. Above, the `MultiplesTicks` function was used to label with multiples of ``\pi``. - -Later we will discuss how `Makie` allows for subsequent modification of several parts of the plot (not just the ticks) including the data. - -#### Figure resolution, ``x`` and ``y`` limits - -As just mentioned, the basic plots we have seen are of type `FigureAxisPlot`. The "figure" part can be used to adjust the background color or the resolution. As with attributes for the axis, these too can be passed to a simple constructor: - -```julia -lines(xs, sin; - axis=(title="Plot of sin(x)", xlabel="x", ylabel="sin(x)"), - figure=(;resolution=(300, 300)) - ) -``` - -The `;` in the tuple passed to `figure` is one way to create a *named* tuple with a single element. Alternatively, `(resolution=(300,300), )` -- with a trailing comma -- could have been used. - - - -To set the limits of the graph there are shorthand functions `xlims!`, `ylims!`, and `zlims!`. This might prove useful if vertical asymptotes are encountered, as in this example: - -```julia -f(x) = 1/x -a,b = -1, 1 -xs = range(-1, 1, length=200) -lines(xs, f.(xs)) -ylims!(-10, 10) - -current_figure() -``` - -This still leaves the artifact due to the vertical asymptote at ``0`` having different values from the left and the right. - - -### Plots of parametric functions - -A space curve is a plot of a function ``f:R^2 \rightarrow R`` or ``f:R^3 \rightarrow R``. - -To construct a curve from a set of points, we have a similar pattern in both ``2`` and ``3`` dimensions: - -```julia -r(t) = [sin(2t), cos(3t)] -ts = range(0, 2pi, length=200) -pts = Point2.(r.(ts)) # or (Point2∘r).(ts) -lines(pts) -``` - -Or - -```julia -r(t) = [sin(2t), cos(3t), t] -ts = range(0, 2pi, length=200) -pts = Point3.(r.(ts)) -lines(pts) -``` - - -Alternatively, vectors of the ``x``, ``y``, and ``z`` components can be produced and then plotted using the pattern `lines(xs, ys)` or `lines(xs, ys, zs)`. For example, using `unzip`, as above, we might have done the prior example with: - -```julia -xs, ys, zs = unzip(r.(ts)) -lines(xs, ys, zs) -``` - - -#### Aspect ratio - -A simple plot of a parametrically defined circle will show an ellipse, as the aspect ratio of the ``x`` and ``y`` axis is not ``1``. To enforce this, we can pass a value of `aspect=1` to the underlying "Axis" object. For example: - -```julia -ts = range(0, 2pi, length=100) -xs, ys = sin.(ts), cos.(ts) -lines(xs, ys; axis=(; aspect = 1)) -``` - -#### Tangent vectors (`arrows`) - -A tangent vector along a curve can be drawn quite easily using the `arrows` function. There are different interfaces for `arrows`, but we show the one which uses a vector of positions and a vector of "vectors". For the latter, we utilize the `derivative` function from `ForwardDiff`: - -```julia -r(t) = [sin(t), cos(t)] # vector, not tuple -ts = range(0, 4pi, length=200) -lines(Point2.(r.(ts))) - -nts = 0:pi/4:2pi -us = r.(nts) -dus = ForwardDiff.derivative.(r, nts) - -arrows!(Point2.(us), Point2.(dus)) - -current_figure() -``` - - -In 3 dimensions the differences are minor: - -```julia -r(t) = [sin(t), cos(t), t] # vector, not tuple -ts = range(0, 4pi, length=200) -lines(Point3.(r.(ts))) - -nts = 0:pi/2:(4pi-pi/2) -us = r.(nts) -dus = ForwardDiff.derivative.(r, nts) - -arrows!(Point3.(us), Point3.(dus)) - -current_figure() -``` - - -#### Arrow attributes - -Attributes for `arrows` include - -* `arrowsize` to adjust the size -* `lengthscale` to scale the size -* `arrowcolor` to set the color -* `arrowhead` to adjust the head -* `arrowtail` to adjust the tail - - - -## Surfaces - -Plots of surfaces in ``3`` dimensions are useful to help understand the behavior of multivariate functions. - -#### Surfaces defined through ``z=f(x,y)`` - -The "`peaks`" function defined below has a few prominent peaks: - -```julia -function peaks(x, y) - p = 3*(1-x)^2*exp(-x^2 - (y+1)^2) - p -= 10(x/5-x^3-y^5)*exp(-x^2-y^2) - p -= 1/3*exp(-(x+1)^2-y^2) - p -end -``` - -Here we see how `peaks` can be visualized over the region ``[-5,5]\times[-5,5]``: - -```julia -xs = ys = range(-5, 5, length=25) -surface(xs, ys, peaks) -``` - -The calling pattern `surface(xs, ys, f)` implies a rectangular grid over the ``x``-``y`` plane defined by `xs` and `ys` with ``z`` values given by ``f(x,y)``. - - -Alternatively a "matrix" of ``z`` values can be specified. For a function `f`, this is conveniently generated by the pattern `f.(xs, ys')`, the `'` being important to get a matrix of all ``x``-``y`` pairs through `Julia`'s broadcasting syntax. - -```julia -zs = peaks.(xs, ys') -surface(xs, ys, zs) -``` - - -To see how this graph is constructed, the points ``(x,y,f(x,y))`` are plotted over the grid and displayed. - -Here we downsample to illustrate: - -```julia -xs = ys = range(-5, 5, length=5) -pts = [Point3(x, y, peaks(x,y)) for x in xs for y in ys] -scatter(pts, markersize=25) -``` - - -These points are then connected. The `wireframe` function illustrates -just the frame: - -```julia -wireframe(xs, ys, peaks.(xs, ys'); linewidth=5) -``` - -The `surface` call triangulates the frame and fills in the shading: - -```julia -surface!(xs, ys, peaks.(xs, ys')) -current_figure() -``` - - - - -#### Parametrically defined surfaces - -A surface may be parametrically defined through a function ``r(u,v) = (x(u,v), y(u,v), z(u,v))``. For example, the surface generated by ``z=f(x,y)`` is of the form with ``r(u,v) = (u,v,f(u,v))``. - -The `surface` function and the `wireframe` function can be used to display such surfaces. In previous usages, the `x` and `y` values were vectors from which a 2-dimensional grid is formed. For parametric surfaces, a grid for the `x` and `y` values must be generated. This function will do so: - -```julia -function parametric_grid(us, vs, r) - n,m = length(us), length(vs) - xs, ys, zs = zeros(n,m), zeros(n,m), zeros(n,m) - for (i, uᵢ) in pairs(us) - for (j, vⱼ) in pairs(vs) - x,y,z = r(uᵢ, vⱼ) - xs[i,j] = x - ys[i,j] = y - zs[i,j] = z - end - end - (xs, ys, zs) -end -``` - -With the data suitably massaged, we can directly plot either a `surface` or `wireframe` plot. - ----- - -As an aside, The above can be done more campactly with nested list comprehensions: - -``` -xs, ys, zs = [[pt[i] for pt in r.(us, vs')] for i in 1:3] -``` - -Or using the `unzip` function directly after broadcasting: - -``` -xs, ys, zs = unzip(r.(us, vs')) -``` - ----- - -For example, a sphere can be parameterized by ``r(u,v) = (\sin(u)\cos(v), \sin(u)\sin(v), \cos(u))`` and visualized through: - -```julia -r(u,v) = [sin(u)*cos(v), sin(u)*sin(v), cos(u)] -us = range(0, pi, length=25) -vs = range(0, pi/2, length=25) -xs, ys, zs = parametric_grid(us, vs, r) - -surface(xs, ys, zs) -wireframe!(xs, ys, zs) -current_figure() -``` - -A surface of revolution for ``g(u)`` revolved about the ``z`` axis can be visualized through: - -```julia -g(u) = u^2 * exp(-u) -r(u,v) = (g(u)*sin(v), g(u)*cos(v), u) -us = range(0, 3, length=10) -vs = range(0, 2pi, length=10) -xs, ys, zs = parametric_grid(us, vs, r) - -surface(xs, ys, zs) -wireframe!(xs, ys, zs) -current_figure() -``` - - -A torus with big radius ``2`` and inner radius ``1/2`` can be visualized as follows - -```julia -r1, r2 = 2, 1/2 -r(u,v) = ((r1 + r2*cos(v))*cos(u), (r1 + r2*cos(v))*sin(u), r2*sin(v)) -us = vs = range(0, 2pi, length=25) -xs, ys, zs = parametric_grid(us, vs, r) - -surface(xs, ys, zs) -wireframe!(xs, ys, zs) -current_figure() -``` - - -A Möbius strip can be produced with: - -```julia -ws = range(-1/4, 1/4, length=8) -thetas = range(0, 2pi, length=30) -r(w, θ) = ((1+w*cos(θ/2))*cos(θ), (1+w*cos(θ/2))*sin(θ), w*sin(θ/2)) -xs, ys, zs = parametric_grid(ws, thetas, r) - -surface(xs, ys, zs) -wireframe!(xs, ys, zs) -current_figure() -``` - -## Contour plots (`contour`, `contourf`, `heatmap`) - -For a function ``z = f(x,y)`` an alternative to a surface plot, is a contour plot. That is, for different values of ``c`` the level curves ``f(x,y)=c`` are drawn. - -For a function ``f(x,y)``, the syntax for generating a contour plot follows that for `surface`. - -For example, using the `peaks` function, previously defined, we have a contour plot over the region ``[-5,5]\times[-5,5]`` is generated through: - -```julia -xs = ys = range(-5, 5, length=100) -contour(xs, ys, peaks) -``` - -The default of ``5`` levels can be adjusted using the `levels` keyword: - -```julia -contour(xs, ys, peaks; levels = 20) -``` - -The `levels` argument can also specify precisely what levels are to be drawn. - -The contour graph makes identification of peaks and valleys easy as the limits of patterns of nested contour lines. - - -A *filled* contour plot is produced by `contourf`: - -```julia -contourf(xs, ys, peaks) -``` - -A related, but alternative visualization, using color to represent magnitude is a heatmap, produced by the `heatmap` function. The calling syntax is similar to `contour` and `surface`: - - -```julia -heatmap(xs, ys, peaks) -``` - -This graph shows peaks and valleys through "hotspots" on the graph. - - -The `MakieGallery` package includes an example of a surface plot with both a wireframe and 2D contour graph added. It is replicated here using the `peaks` function scaled by ``5``. - -The function and domain to plot are described by: - -```julia -xs = ys = range(-5, 5, length=51) -zs = peaks.(xs, ys') / 5; -``` - -The `zs` were generated, as `wireframe` does not provide the interface for passing a function. - -The `surface` and `wireframe` are produced as follows. Here we manually create the figure and axis object so that we can set the viewing angle through the `elevation` argument to the axis object: - -```julia -fig = Figure() -ax3 = Axis3(fig[1,1]; - elevation=pi/9, azimuth=pi/16) -surface!(ax3, xs, ys, zs) -wireframe!(ax3, xs, ys, zs; - overdraw = true, transparency = true, - color = (:black, 0.1)) -current_figure() -``` - -To add the contour, a simple call via `contour!(scene, xs, ys, zs)` will place the contour at the ``z=0`` level which will make it hard to read. Rather, placing at the "bottom" of the figure is desirable. To identify that the minimum value, is identified (and rounded) and the argument `transformation = (:xy, zmin)` is passed to `contour!`: - -```julia -ezs = extrema(zs) -zmin, zmax = floor(first(ezs)), ceil(last(ezs)) -contour!(ax3, xs, ys, zs; - levels = 15, linewidth = 2, - transformation = (:xy, zmin)) -zlims!(zmin, zmax) -current_figure() -``` - -The `transformation` plot attribute sets the "plane" (one of `:xy`, `:yz`, or `:xz`) at a location, in this example `zmin`. - - -The manual construction of a figure and an axis object will be further discussed later. - - -### Three dimensional contour plots - -The `contour` function can also plot ``3``-dimensional contour plots. Concentric spheres, contours of ``x^2 + y^2 + z^2 = c`` for ``c > 0`` are presented by the following: - -```julia -f(x,y,z) = x^2 + y^2 + z^2 -xs = ys = zs = range(-3, 3, length=100) - -contour(xs, ys, zs, f) -``` - - -### Implicitly defined curves and surfaces - -Suppose ``f`` is a scalar-valued function. If `f` takes two variables for its input, then the equation ``f(x,y) = 0`` implicitly defines ``y`` as a function of ``x``; ``y`` can be visualized *locally* with a curve. If ``f`` takes three variables for its input, then the equation ``f(x,y,z)=0`` implicitly defines ``z`` as a function of ``x`` and ``y``; ``z`` can be visualized *locally* with a surface. - -#### Implicitly defined curves - -The graph of an equation is the collection of all ``(x,y)`` values -satisfying the equation. This is more general than the graph of a -function, which can be viewed as the graph of the equation -``y=f(x)``. An equation in ``x``-``y`` can be graphed if the set of -solutions to a related equation ``f(x,y)=0`` can be identified, as one -can move all terms to one side of an equation and define ``f`` as the -rule of the side with the terms. The implicit function theorem ensures that under some conditions, *locally* near a point ``(x, y)``, the value ``y`` can be represented as a function of ``x``. So, the graph of the equation ``f(x,y)=0`` can be produced by stitching together these local function representations. - -The contour graph can produce these graphs by setting the `levels` argument to `[0]`. - -```julia -f(x,y) = x^3 + x^2 + x + 1 - x*y # solve x^3 + x^2 + x + 1 = x*y -xs = range(-5, 5, length=100) -ys = range(-10, 10, length=100) - -contour(xs, ys, f.(xs, ys'); levels=[0]) -``` - -The `implicitPlots.jl` function uses the `Contour` package along with a `Plots` recipe to plot such graphs. Here we see how to use `Makie` in a similar manner: - -```julia; eval=false -import Contour - -function implicit_plot(xs, ys, f; kwargs...) - fig = Figure() - ax = Axis(fig[1,1]) - implicit_plot!(ax, xs, ys, f; kwargs...) - fig -end - -function implicit_plot!(ax, xs, ys, f; kwargs...) - z = [f(x, y) for x in xs, y in ys] - cs = Contour.contour(collect(xs), collect(ys), z, 0.0) - ls = Contour.lines(cs) - - isempty(ls) && error("empty") - - for l ∈ ls - us, vs = Contour.coordinates(l) - lines!(ax, us, vs; kwargs...) - end - -end -``` - - - - -#### Implicitly defined surfaces, ``F(x,y,z)=0`` - -To plot the equation ``F(x,y,z)=0``, for ``F`` a scalar-valued -function, again the implicit function theorem says that, under -conditions, near any solution ``(x,y,z)``, ``z`` can be represented as -a function of ``x`` and ``y``, so the graph will look likes surfaces -stitched together. The `Implicit3DPlotting` package takes an approach like -`ImplicitPlots` to represent these surfaces. It replaces the `Contour` -package computation with a ``3``-dimensional alternative provided -through the `Meshing` and `GeometryBasics` packages. - - -The `Implicit3DPlotting` package needs some maintenance, so we borrow the main functionality and wrap it into a function: - -```julia -import Meshing -import GeometryBasics - -function make_mesh(xlims, ylims, zlims, f, - M = Meshing.MarchingCubes(); # or Meshing.MarchingTetrahedra() - samples=(35, 35, 35), - ) - - lims = extrema.((xlims, ylims, zlims)) - Δ = xs -> last(xs) - first(xs) - xs = Vec(first.(lims)) - Δxs = Vec(Δ.(lims)) - - GeometryBasics.Mesh(f, Rect(xs, Δxs), M; samples = samples) -end -``` - -The `make_mesh` function creates a mesh that can be visualized with the `wireframe` or `mesh` plotting functions. - - -This example, plotting an implicitly defined sphere, comes from the -documentation of `Implicit3DPlotting`. The `f` in `make_mesh` is a -scalar-valued function of a vector: - -```julia -f(x) = sum(x.^2) - 1 -xs = ys = zs = (-5, 5) -m = make_mesh(xs, ys, zs, f) -wireframe(m) -``` - -Here we visualize an intersection of a sphere with another figure: - -```julia -r₂(x) = sum(x.^2) - 5/4 # a sphere -r₄(x) = sum(x.^4) - 1 -xs = ys = zs = -2:2 -m2,m4 = make_mesh(xs, ys, zs, r₂), make_mesh(xs, ys, zs, r₄) - -wireframe(m4, color=:yellow) -wireframe!(m2, color=:red) -current_figure() -``` - - -This example comes from [Wikipedia](https://en.wikipedia.org/wiki/Implicit_surface) showing an implicit surface of genus ``2``: - -```julia -f(x,y,z) = 2y*(y^2 -3x^2)*(1-z^2) + (x^2 +y^2)^2 - (9z^2-1)*(1-z^2) -zs = ys = xs = range(-5/2, 5/2, length=100) -m = make_mesh(xs, ys, zs, x -> f(x...)) -wireframe(m) -``` - - -(This figure does not render well through `contour(xs, ys, zs, f, levels=[0])`, as the hole is not shown.) - - -For one last example from Wikipedia, we have the Cassini oval which "can be defined as the point set for which the *product* of the distances to ``n`` given points is constant." That is: - -```julia -function cassini(λ, ps = ((1,0,0), (-1, 0, 0))) - n = length(ps) - x -> prod(norm(x .- p) for p ∈ ps) - λ^n -end -xs = ys = zs = range(-2, 2, length=100) -m = make_mesh(xs, ys, zs, cassini(1.05)) -wireframe(m) -``` - -## Vector fields. Visualizations of ``f:R^2 \rightarrow R^2`` - -The vector field ``f(x,y) = \langle y, -x \rangle`` can be visualized as a set of vectors, ``f(x,y)``, positioned at a grid. These arrows can be visualized with the `arrows` function. The `arrows` function is passed a vector of points for the anchors and a vector of points representing the vectors. - -We can generate these on a regular grid through: - -```julia -f(x, y) = [y, -x] -xs = ys = -5:5 -pts = vec(Point2.(xs, ys')) -dus = vec(Point2.(f.(xs, ys'))); -first(pts), first(dus) # show an example -``` - -Broadcasting over `(xs, ys')` ensures each pair of possible values is encountered. The `vec` call reshapes an array into a vector. - -Calling `arrows` on the prepared data produces the graphic: - -```julia -arrows(pts, dus) -``` - -The grid seems rotated at first glance; but is also confusing. This is due to the length of the vectors as the ``(x,y)`` values get farther from the origin. Plotting the *normalized* values (each will have length ``1``) can be done easily using `norm` (which is found in the standard `LinearAlgebra` library): - -```julia -dvs = dus ./ norm.(dus) -arrows(pts, dvs) -``` - -The rotational pattern becomes much clearer now. - -The `streamplot` function also illustrates this phenomenon. This implements an "algorithm [that] puts an arrow somewhere and extends the streamline in both directions from there. Then, it chooses a new position (from the remaining ones), repeating the the exercise until the streamline gets blocked, from which on a new starting point, the process repeats." - -The `streamplot` function expects a `Point` not a pair of values, so we adjust `f` slightly and call the function using the pattern `streamplot(g, xs, ys)`: - -```julia -f(x, y) = [y, -x] -g(xs) = Point2(f(xs...)) - -streamplot(g, -5..5, -5..5) -``` - -(We used interval notation to set the viewing range, a range could also be used.) - -!!! note - The calling pattern of `streamplot` is different than other functions, such as `surface`, in that the function comes first. - - -## Layoutables and Observables - -### Layoutables - -`Makie` makes it really easy to piece together figures from individual plots. To illustrate, we create a graphic consisting of a plot of a function, its derivative, and its second derivative. In our graphic, we also leave space for a label. - -!!! note - The Layout [Tutorial](https://makie.juliaplots.org/stable/tutorials/layout-tutorial/) has *much* more detail on this subject. - -The basic plotting commands, like `lines`, return a `FigureAxisPlot` object. For laying out our own graphic, we manage the figure and axes manually. The commands below create a figure, then assign axes to portions of the figure: - -```julia -F = Figure() -af = F[2,1:2] = Axis(F) -afp = F[3,1:end] = Axis(F) -afpp = F[4,:] = Axis(F) -``` - -The axes are named `af`, `afp` and `afpp`, as they will hold the respective graphs. The key here is the use of matrix notation to layout the graphic in a grid. The first one is row 2 and columns 1 through 2; the second row 3 and again all columns, the third is row 4 and all columns. - -In this figure, we want the ``x``-axis for each of the three graphics to be linked. This command ensures that: - -```julia -linkxaxes!(af, afp, afpp); -``` - -By linking axes, if one is updated, say through `xlims!`, the others will be as well. - -We now plot our functions. The key here is the mutating form of `lines!` takes an axis object to mutate as its first argument: - -```julia -f(x) = 8x^4 - 8x^2 + 1 -fp(x) = 32x^3 - 16x -fpp(x) = 96x^2 - 16 - -xs = -1..1 -lines!(af, xs, f) -lines!(afp, xs, fp) -lines!(afp, xs, zero, color=:blue) -lines!(afpp, xs, fpp) -lines!(afpp, xs, zero, color=:blue); -``` - -We can give title information to each axis: - -```julia -af.title = "f" -afp.title = "fp" -afpp.title = "fpp"; -``` - -Finally, we add a label in the first row, but for illustration purposes, only use the first column. - -```julia -Label(F[1,1], """ -Plots of f and its first and second derivatives. -When the first derivative is zero, the function -f has relative extrema. When the second derivative -is zero, the function f has an inflection point. -"""); -``` - -Finally we display the figure: - -```julia -F -``` - -### Observables - -The basic components of a plot in `Makie` can be updated [interactively](https://makie.juliaplots.org/stable/documentation/nodes/index.html#observables_interaction). `Makie` uses the `Observables` package which allows complicated interactions to be modeled quite naturally. In the following we give a simple example. - - - -In Makie, an `Observable` is a structure that allows its value to be -updated, similar to an array. When changed, observables can trigger -an event. Observables can rely on other observables, so events can be -cascaded. - -This simple example shows how an observable `h` can be used to create a collection of points representing a secant line. The figure shows the value for `h=3/2`. - -```julia -f(x) = sqrt(x) -c = 1 -xs = 0..3 -h = Observable(3/2) - -points = lift(h) do h - xs = [0,c,c+h,3] - tl = x -> f(c) + (f(c+h)-f(c))/h * (x-c) - [Point2(x, tl(x)) for x ∈ xs] -end - -lines(xs, f) -lines!(points) -current_figure() -``` - -We can update the value of `h` using `setindex!` notation (square brackets). For example, to see that the secant line is a good approximation to the tangent line as ``h \rightarrow 0`` we can set `h` to be `1/4` and replot: - -```julia -h[] = 1/4 -current_figure() -``` - -The line `h[] = 1/4` updated `h` which then updated `points` (a points is lifted up from `h`) which updated the graphic. (In these notes, we replot to see the change, but in an interactive session, the current *displayed* figure would be updated; no replotting would be necessary.) - - -Finally, this example shows how to add a slider to adjust the value of `h` with a mouse. The slider object is positioned along with a label using the grid reference, as before. - - -```julia -f(x) = sqrt(x) -c = 1 -xs = 0..3 - -F = Figure() -ax = Axis(F[1,1:2]) -h = Slider(F[2,2], range = 0.01:0.01:1.5, startvalue = 1.5) -Label(F[2,1], "Adjust slider to change `h`"; - justification = :left) - -points = lift(h.value) do h - xs = [0,c,c+h,3] - tl = x-> f(c) + (f(c+h)-f(c))/h * (x-c) - [Point2(x, tl(x)) for x ∈ xs] -end - -lines!(ax, xs, f) -lines!(ax, points) -current_figure() -``` - -The slider value is "lifted" by its `value` component, as shown. Otherwise, the above is fairly similar to just using an observable for `h`. diff --git a/CwJ/alternatives/plotly_plotting.jmd b/CwJ/alternatives/plotly_plotting.jmd deleted file mode 100644 index 84eaae5..0000000 --- a/CwJ/alternatives/plotly_plotting.jmd +++ /dev/null @@ -1,667 +0,0 @@ -# JavaScript based plotting libraries - -!!! alert "Not working with quarto" - Currently, the plots generated here are not rendering within quarto. - - - -This section uses this add-on package: - -```julia -using PlotlyLight -``` - -To avoid a dependence on the `CalculusWithJulia` package, we load two utility packages: - -```julia -using PlotUtils -using SplitApplyCombine -``` - ----- - -`Julia` has different interfaces to a few JavaScript plotting libraries, notably the [vega](https://vega.github.io/vega/) and [vega-lite](https://vega.github.io/vega-lite/) through the [VegaLite.jl](https://github.com/queryverse/VegaLite.jl) package, and [plotly](https://plotly.com/javascript/) through several interfaces: `Plots.jl`, `PlotlyJS.jl`, and `PlotlyLight.jl`. These all make web-based graphics, for display through a web browser. - -The `Plots.jl` interface is a backend for the familiar `Plots` package, making the calling syntax familiar, as is used throughout these notes. The `plotly()` command, from `Plots`, switches to this backend. - -The `PlotlyJS.jl` interface offers direct translation from `Julia` structures to the underlying `JSON` structures needed by plotly, and has mechanisms to call back into `Julia` from `JavaScript`. This allows complicated interfaces to be produced. - -Here we discuss `PlotlyLight` which conveniently provides the translation from `Julia` structures to the -`JSON` structures needed in a light-weight package, which plots quickly, without the delays due to compilation of the more complicated interfaces. Minor modifications would be needed to adjust the examples to work with `PlotlyJS` or `PlotlyBase`. The documentation for the `JavaScript` [library](https://plotly.com/javascript/) provides numerous examples which can easily be translated. The [one-page-reference](https://plotly.com/javascript/reference/) gives specific details, and is quoted from below, at times. - - -This discussion covers the basic of graphing for calculus purposes. It does not cover, for example, the faceting common in statistical usages, or the chart types common in business and statistics uses. The `plotly` library is much more extensive than what is reviewed below. - -## Julia dictionaries to JSON - -`PlotlyLight` uses the `JavaScript` interface for the `plotly` libraries. Unlike more developed interfaces, like the one for `Python`, `PlotlyLight` only manages the translation from `Julia` structures to `JavaScript` structures and the display of the results. - -The key to translation is the mapping for `Julia`'s dictionaries to -the nested `JSON` structures needed by the `JavaScript` library. - - -For example, an introductory [example](https://plotly.com/javascript/line-and-scatter/) for a scatter plot includes this `JSON` structure: - -```julia; eval=false -var trace1 = { - x: [1, 2, 3, 4], - y: [10, 15, 13, 17], - mode: 'markers', - type: 'scatter' -}; -``` - -The `{}` create a list, the `[]` an Array (or vector, as it does with `Julia`), the `name:` are keys. The above is simply translated via: - -```julia; -Config(x = [1,2,3,4], - y = [10, 15, 13, 17], - mode = "markers", - type = "scatter" - ) -``` - -The `Config` constructor (from the `EasyConfig` package loaded with `PlotlyLight`) is an interface for a dictionary whose keys are symbols, which are produced by the named arguments passed to `Config`. By nesting `Config` statements, nested `JavaScript` structures can be built up. As well, these can be built on the fly using `.` notation, as in: - -```julia; hold=true -cfg = Config() -cfg.key1.key2.key3 = "value" -cfg -``` - -To produce a figure with `PlotlyLight` then is fairly straightforward: data and, optionally, a layout are created using `Config`, then passed along to the `Plot` command producing a `Plot` object which has `display` methods defined for it. This will be illustrated through the examples. - - -## Scatter plot - -A basic scatter plot of points ``(x,y)`` is created as follows: - -```julia; hold=true -xs = 1:5 -ys = rand(5) -data = Config(x = xs, - y = ys, - type="scatter", - mode="markers" - ) -Plot(data) -``` - -The symbols `x` and `y` (and later `z`) specify the data to `plotly`. Here the `mode` is specified to show markers. - - -The `type` key specifies the chart or trace type. The `mode` specification sets the drawing mode for the trace. Above it is "markers". It can be any combination of "lines", "markers", or "text" joined with a "+" if more than one is desired. - -## Line plot - -A line plot is very similar, save for a different `mode` specification: - -```julia; hold=true -xs = 1:5 -ys = rand(5) -data = Config(x = xs, - y = ys, - type="scatter", - mode="lines" - ) -Plot(data) -``` - -The difference is solely the specification of the `mode` value, for a line plot it is "lines," for a scatter plot it is "markers" The `mode` "lines+markers" will plot both. The default for the "scatter" types is to use "lines+markers" for small data sets, and "lines" for others, so for this example, `mode` could be left off. - - -### Nothing - -The line graph plays connect-the-dots with the points specified by paired `x` and `y` values. *Typically*, when and `x` value is `NaN` that "dot" (or point) is skipped. However, `NaN` doesn't pass through the JSON conversion -- `nothing` can be used. - -```julia; hold=true -data = Config( - x=[0,1,nothing,3,4,5], - y = [0,1,2,3,4,5], - type="scatter", mode="markers+lines") -Plot(data) -``` - -## Multiple plots - -More than one graph or layer can appear on a plot. The `data` argument can be a vector of `Config` values, each describing a plot. For example, here we make a scatter plot and a line plot: - -```julia; hold=true -data = [Config(x = 1:5, - y = rand(5), - type = "scatter", - mode = "markers", - name = "scatter plot"), - Config(x = 1:5, - y = rand(5), - type = "scatter", - mode = "lines", - name = "line plot") - ] -Plot(data) -``` - -The `name` argument adjusts the name in the legend referencing the plot. This is produced by default. - - -### Adding a layer - -In `PlotlyLight`, the `Plot` object has a field `data` for storing a vector of configurations, as above. After a plot is made, this field can have values pushed onto it and the corresponding layers will be rendered when the plot is redisplayed. - -For example, here we plot the graphs of both the ``\sin(x)`` and ``\cos(x)`` over ``[0,2\pi]``. We used the utility `PlotUtils.adapted_grid` to select the points to use for the graph. - -```julia; hold=true -a, b = 0, 2pi - -xs, ys = PlotUtils.adapted_grid(sin, (a,b)) -p = Plot(Config(x=xs, y=ys, name="sin")) - -xs, ys = PlotUtils.adapted_grid(cos, (a,b)) -push!(p.data, Config(x=xs, y=ys, name="cos")) - -p # to display the plot -``` - -The values for `a` and `b` are used to generate the ``x``- and ``y``-values. These can also be gathered from the existing plot object. Here is one way, where for each trace with an `x` key, the extrema are consulted to update a list of left and right ranges. - -```julia; hold=true -xs, ys = PlotUtils.adapted_grid(x -> x^5 - x - 1, (0, 2)) # answer is (0,2) -p = Plot([Config(x=xs, y=ys, name="Polynomial"), - Config(x=xs, y=0 .* ys, name="x-axis", mode="lines", line=Config(width=5))] - ) -ds = filter(d -> !isnothing(get(d, :x, nothing)), p.data) -a=reduce(min, [minimum(d.x) for d ∈ ds]; init=Inf) -b=reduce(max, [maximum(d.x) for d ∈ ds]; init=-Inf) -(a, b) -``` - - - -## Interactivity - -`JavaScript` allows interaction with a plot as it is presented within a browser. (Not the `Julia` process which produced the data or the plot. For that interaction, `PlotlyJS` may be used.) The basic *default* features are: - -* The data producing a graphic are displayed on hover using flags. -* The legend may be clicked to toggle whether the corresponding graph is displayed. -* The viewing region can be narrowed using the mouse for selection. -* The toolbar has several features for panning and zooming, as well as adjusting the information shown on hover. - -Later we will see that ``3``-dimensional surfaces can be rotated interactively. - - -## Plot attributes - -Attributes of the markers and lines may be adjusted when the data configuration is specified. A selection is shown below. Consult the reference for the extensive list. - -### Marker attributes - -A marker's attributes can be adjusted by values passed to the `marker` key. Labels for each marker can be assigned through a `text` key and adding `text` to the `mode` key. For example: - -```julia; hold=true -data = Config(x = 1:5, - y = rand(5), - mode="markers+text", - type="scatter", - name="scatter plot", - text = ["marker $i" for i in 1:5], - textposition = "top center", - marker = Config(size=12, color=:blue) - ) -Plot(data) -``` - -The `text` mode specification is necessary to have text be displayed -on the chart, and not just appear on hover. The `size` and `color` -attributes are recycled; they can be specified using a vector for -per-marker styling. Here the symbol `:blue` is used to specify a -color, which could also be a name, such as `"blue"`. - -#### RGB Colors - -The `ColorTypes` package is the standard `Julia` package providing an -`RGB` type (among others) for specifying red-green-blue colors. To -make this work with `Config` and `JSON3` requires some type-piracy -(modifying `Base.string` for the `RGB` type) to get, say, `RGB(0.5, -0.5, 0.5)` to output as `"rgb(0.5, 0.5, 0.5)"`. (RGB values in -JavaScript are integers between ``0`` and ``255`` or floating point -values between ``0`` and ``1``.) A string with this content can be -specified. Otherwise, something like the following can be used to -avoid the type piracy: - -```julia; -struct rgb - r - g - b -end -PlotlyLight.JSON3.StructTypes.StructType(::Type{rgb}) = PlotlyLight.JSON3.StructTypes.StringType() -Base.string(x::rgb) = "rgb($(x.r), $(x.g), $(x.b))" -``` - - -With these defined, red-green-blue values can be used for colors. For example to give a range of colors, we might have: - -```julia; hold=true -cols = [rgb(i,i,i) for i in range(10, 245, length=5)] -sizes = [12, 16, 20, 24, 28] -data = Config(x = 1:5, - y = rand(5), - mode="markers+text", - type="scatter", - name="scatter plot", - text = ["marker $i" for i in 1:5], - textposition = "top center", - marker = Config(size=sizes, color=cols) - ) -Plot(data) -``` - -The `opacity` key can be used to control the transparency, with a value between ``0`` and ``1``. - -#### Marker symbols - -The `marker_symbol` key can be used to set a marker shape, with the basic values being: `circle`, `square`, `diamond`, `cross`, `x`, `triangle`, `pentagon`, `hexagram`, `star`, `diamond`, `hourglass`, `bowtie`, `asterisk`, `hash`, `y`, and `line`. Add `-open` or `-open-dot` modifies the basic shape. - -```julia; hold=true -markers = ["circle", "square", "diamond", "cross", "x", "triangle", "pentagon", - "hexagram", "star", "diamond", "hourglass", "bowtie", "asterisk", - "hash", "y", "line"] -n = length(markers) -data = [Config(x=1:n, y=1:n, mode="markers", - marker = Config(symbol=markers, size=10)), - Config(x=1:n, y=2 .+ (1:n), mode="markers", - marker = Config(symbol=markers .* "-open", size=10)), - Config(x=1:n, y=4 .+ (1:n), mode="markers", - marker = Config(symbol=markers .* "-open-dot", size=10)) - ] -Plot(data) -``` - - -### Line attributes - -The `line` key can be used to specify line attributes, such as `width` (pixel width), `color`, or `dash`. - -The `width` key specifies the line width in pixels. - -The `color` key specifies the color of the line drawn. - -The `dash` key specifies the style for the drawn line. Values can be set by string from "solid", "dot", "dash", "longdash", "dashdot", or "longdashdot" or set by specifying a pattern in pixels, e.g. "5px,10px,2px,2px". - -The `shape` attribute determine how the points are connected. The default is `linear`, but other possibilities are `hv`, `vh`, `hvh`, `vhv`, `spline` for various patterns of connectivity. The following example, from the plotly documentation, shows the differences: - - -```julia; hold=true -shapes = ["linear", "hv", "vh", "hvh", "vhv", "spline"] -data = [Config(x = 1:5, y = 5*(i-1) .+ [1,3,2,3,1], mode="lines+markers", type="scatter", - name=shape, - line=Config(shape=shape) - ) for (i, shape) ∈ enumerate(shapes)] -Plot(data) -``` - -### Text - -The text associated with each point can be drawn on the chart, when "text" is included in the `mode` or shown on hover. - - -The onscreen text is passed to the `text` attribute. The [`texttemplate`](https://plotly.com/javascript/reference/scatter/#scatter-texttemplate) key can be used to format the text with details in the accompanying link. - -Similarly, the `hovertext` key specifies the text shown on hover, with [`hovertemplate`](https://plotly.com/javascript/reference/scatter/#scatter-hovertemplate) used to format the displayed text. - - - - -### Filled regions - -The `fill` key for a chart of mode `line` specifies how the area -around a chart should be colored, or filled. The specification are -declarative, with values in "none", "tozeroy", "tozerox", "tonexty", -"tonextx", "toself", and "tonext". The value of "none" is the default, unless stacked traces are used. - -In the following, to highlight the difference between ``f(x) = \cos(x)`` and ``p(x) = 1 - x^2/2`` the area from ``f`` to the next ``y`` is declared; for ``p``, the area to ``0`` is declared. - -```julia; hold=true -xs = range(-1, 1, 100) -data = [ - Config( - x=xs, y=cos.(xs), - fill = "tonexty", - fillcolor = "rgba(0,0,255,0.25)", # to get transparency - line = Config(color=:blue) - ), - Config( - x=xs, y=[1 - x^2/2 for x ∈ xs ], - fill = "tozeroy", - fillcolor = "rgba(255,0,0,0.25)", # to get transparency - line = Config(color=:red) - ) -] -Plot(data) -``` - -The `toself` declaration is used below to fill in a polygon: - -```julia; hold=true -data = Config( - x=[-1,1,1,-1,-1], y = [-1,1,-1,1,-1], - fill="toself", - type="scatter") -Plot(data) -``` - -## Layout attributes - -The `title` key sets the main title; the `title` key in the `xaxis` configuration sets the ``x``-axis title (similarly for the ``y`` axis). - - -The legend is shown when ``2`` or more charts or specified, by default. This can be adjusted with the `showlegend` key, as below. The legend shows the corresponding `name` for each chart. - -```julia; hold=true -data = Config(x=1:5, y=rand(5), type="scatter", mode="markers", name="legend label") -lyt = Config(title = "Main chart title", - xaxis = Config(title="x-axis label"), - yaxis = Config(title="y-axis label"), - showlegend=true - ) -Plot(data, lyt) -``` - -The `xaxis` and `yaxis` keys have many customizations. For example: `nticks` specifies the maximum number of ticks; `range` to set the range of the axis; `type` to specify the axis type from "linear", "log", "date", "category", or "multicategory;" and `visible` - -The aspect ratio of the chart can be set to be equal through the `scaleanchor` key, which specifies another axis to take a value from. For example, here is a parametric plot of a circle: - -```julia; hold=true -ts = range(0, 2pi, length=100) -data = Config(x = sin.(ts), y = cos.(ts), mode="lines", type="scatter") -lyt = Config(title = "A circle", - xaxis = Config(title = "x"), - yaxis = Config(title = "y", - scaleanchor = "x") - ) -Plot(data, lyt) -``` - - -#### Annotations - -Text annotations may be specified as part of the layout object. Annotations may or may not show an arrow. Here is a simple example using a vector of annotations. - -```julia; hold=true -data = Config(x = [0, 1], y = [0, 1], mode="markers", type="scatter") -layout = Config(title = "Annotations", - xaxis = Config(title="x", - range = (-0.5, 1.5)), - yaxis = Config(title="y", - range = (-0.5, 1.5)), - annotations = [ - Config(x=0, y=0, text = "(0,0)"), - Config(x=1, y=1.2, text = "(1,1)", showarrow=false) - ] - ) -Plot(data, layout) -``` - - -The following example is more complicated use of the elements previously described. It mimics an image from [Wikipedia](https://en.wikipedia.org/wiki/List_of_trigonometric_identities) for trigonometric identities. The use of ``\LaTeX`` does not seem to be supported through the `JavaScript` interface; unicode symbols are used instead. The `xanchor` and `yanchor` keys are used to position annotations away from the default. The `textangle` key is used to rotate text, as desired. - -```julia, hold=true -alpha = pi/6 -beta = pi/5 -xₘ = cos(alpha)*cos(beta) -yₘ = sin(alpha+beta) -r₀ = 0.1 - -data = [ - Config( - x = [0,xₘ, xₘ, 0, 0], - y = [0, 0, yₘ, yₘ, 0], - type="scatter", mode="line" - ), - Config( - x = [0, xₘ], - y = [0, sin(alpha)*cos(beta)], - fill = "tozeroy", - fillcolor = "rgba(100, 100, 100, 0.5)" - ), - Config( - x = [0, cos(alpha+beta), xₘ], - y = [0, yₘ, sin(alpha)*cos(beta)], - fill = "tonexty", - fillcolor = "rgba(200, 0, 100, 0.5)", - ), - Config( - x = [0, cos(alpha+beta)], - y = [0, yₘ], - line = Config(width=5, color=:black) - ) -] - -lyt = Config( - height=450, - showlegend=false, - xaxis=Config(visible=false), - yaxis = Config(visible=false, scaleanchor="x"), - annotations = [ - - Config(x = r₀*cos(alpha/2), y = r₀*sin(alpha/2), - text="α", showarrow=false), - Config(x = r₀*cos(alpha+beta/2), y = r₀*sin(alpha+beta/2), - text="β", showarrow=false), - Config(x = cos(alpha+beta) + r₀*cos(pi+(alpha+beta)/2), - y = yₘ + r₀*sin(pi+(alpha+beta)/2), - xanchor="center", yanchor="center", - text="α+β", showarrow=false), - Config(x = xₘ + r₀*cos(pi/2+alpha/2), - y = sin(alpha)*cos(beta) + r₀ * sin(pi/2 + alpha/2), - text="α", showarrow=false), - Config(x = 1/2 * cos(alpha+beta), - y = 1/2 * sin(alpha+beta), - text = "1"), - Config(x = xₘ/2*cos(alpha), y = xₘ/2*sin(alpha), - xanchor="center", yanchor="bottom", - text = "cos(β)", - textangle=-rad2deg(alpha), - showarrow=false), - Config(x = xₘ + sin(beta)/2*cos(pi/2 + alpha), - y = sin(alpha)*cos(beta) + sin(beta)/2*sin(pi/2 + alpha), - xanchor="center", yanchor="top", - text = "sin(β)", - textangle = rad2deg(pi/2-alpha), - showarrow=false), - - Config(x = xₘ/2, - y = 0, - xanchor="center", yanchor="top", - text = "cos(α)⋅cos(β)", showarrow=false), - Config(x = 0, - y = yₘ/2, - xanchor="right", yanchor="center", - text = "sin(α+β)", - textangle=-90, - showarrow=false), - Config(x = cos(alpha+beta)/2, - y = yₘ, - xanchor="center", yanchor="bottom", - text = "cos(α+β)", showarrow=false), - Config(x = cos(alpha+beta) + (xₘ - cos(alpha+beta))/2, - y = yₘ, - xanchor="center", yanchor="bottom", - text = "sin(α)⋅sin(β)", showarrow=false), - Config(x = xₘ, y=sin(alpha)*cos(beta) + (yₘ - sin(alpha)*cos(beta))/2, - xanchor="left", yanchor="center", - text = "cos(α)⋅sin(β)", - textangle=90, - showarrow=false), - Config(x = xₘ, - y = sin(alpha)*cos(beta)/2, - xanchor="left", yanchor="center", - text = "sin(α)⋅cos(β)", - textangle=90, - showarrow=false) - ] -) - -Plot(data, lyt) -``` - -## Parameterized curves - -In ``2``-dimensions, the plotting of a parameterized curve is similar to that of plotting a function. In ``3``-dimensions, an extra ``z``-coordinate is included. - -To help, we define an `unzip` function as an interface to `SplitApplyCombine`'s `invert` function: - -```julia -unzip(v) = SplitApplyCombine.invert(v) -``` - -Earlier, we plotted a two dimensional circle, here we plot the related helix. - -```julia; hold=true -helix(t) = [cos(t), sin(t), t] - -ts = range(0, 4pi, length=200) - -xs, ys, zs = unzip(helix.(ts)) - -data = Config(x=xs, y=ys, z=zs, - type = "scatter3d", # <<- note the 3d - mode = "lines", - line=(width=2, - color=:red) - ) - -Plot(data) -``` - -The main difference is the chart type, as this is a ``3``-dimensional plot, "scatter3d" is used. - -### Quiver plots - -There is no `quiver` plot for `plotly` using JavaScript. In ``2``-dimensions a text-less annotation could be employed. In ``3``-dimensions, the following (from [stackoverflow.com](https://stackoverflow.com/questions/43164909/plotlypython-how-to-plot-arrows-in-3d)) is a possible workaround where a line segment is drawn and capped with a small cone. Somewhat opaquely, we use `NamedTuple` for an iterator to create the keys for the data below: - - -```julia; hold=true -helix(t) = [cos(t), sin(t), t] -helix′(t) = [-sin(t), cos(t), 1] -ts = range(0, 4pi, length=200) -xs, ys, zs = unzip(helix.(ts)) -helix_trace = Config(; NamedTuple(zip((:x,:y,:z), unzip(helix.(ts))))..., - type = "scatter3d", # <<- note the 3d - mode = "lines", - line=(width=2, - color=:red) - ) - -tss = pi/2:pi/2:7pi/2 -rs, r′s = helix.(tss), helix′.(tss) - -arrows = [ - Config(x = [x[1], x[1]+x′[1]], - y = [x[2], x[2]+x′[2]], - z = [x[3], x[3]+x′[3]], - mode="lines", type="scatter3d") - for (x, x′) ∈ zip(rs, r′s) -] - -tips = rs .+ r′s -lengths = 0.1 * r′s - -caps = Config(; - NamedTuple(zip([:x,:y,:z], unzip(tips)))..., - NamedTuple(zip([:u,:v,:w], unzip(lengths)))..., - type="cone", anchor="tail") - -data = vcat(helix_trace, arrows, caps) - -Plot(data) -``` - -If several arrows are to be drawn, it might be more efficient to pass multiple values in for the `x`, `y`, ... values. They expect a vector. In the above, we create ``1``-element vectors. - -## Contour plots - -A contour plot is created by the "contour" trace type. The data is prepared as a vector of vectors, not a matrix. The following has the interior vector corresponding to slices ranging over ``x`` for a fixed ``y``. With this, the construction is straightforward using a comprehension: - -```julia; hold=true -f(x,y) = x^2 - 2y^2 - -xs = range(0,2,length=25) -ys = range(0,2, length=50) -zs = [[f(x,y) for x in xs] for y in ys] - -data = Config( - x=xs, y=ys, z=zs, - type="contour" -) - -Plot(data) -``` - - -The same `zs` data can be achieved by broadcasting and then collecting as follows: - -```julia; hold=true -f(x,y) = x^2 - 2y^2 - -xs = range(0,2,length=25) -ys = range(0,2, length=50) -zs = collect(eachrow(f.(xs', ys))) - -data = Config( - x=xs, y=ys, z=zs, - type="contour" -) - -Plot(data) -``` - -The use of just `f.(xs', ys)` or `f.(xs, ys')`, as with other plotting packages, is not effective, as `JSON3` writes matrices as vectors (with linear indexing). - -## Surface plots - -The chart type "surface" allows surfaces in ``3`` dimensions to be plotted. - -### Surfaces defined by ``z = f(x,y)`` - -Surfaces defined through a scalar-valued function are drawn quite naturally, save for needing to express the height data (``z`` axis) using a vector of vectors, and not a matrix. - -```julia; hold=true -peaks(x,y) = 3 * (1-x)^2 * exp(-(x^2) - (y+1)^2) - - 10*(x/5 - x^3 - y^5) * exp(-x^2-y^2) - 1/3 * exp(-(x+1)^2 - y^2) - -xs = range(-3,3, length=50) -ys = range(-3,3, length=50) -zs = [[peaks(x,y) for x in xs] for y in ys] - -data = Config(x=xs, y=ys, z=zs, - type="surface") - -Plot(data) -``` - -### Parametrically defined surfaces - -For parametrically defined surfaces, the ``x`` and ``y`` values also correspond to matrices. Her we see a pattern to plot a torus. The [`aspectmode`](https://plotly.com/javascript/reference/layout/scene/#layout-scene-aspectmode) instructs the scene's axes to be drawn in proportion with the axes' ranges. - -```julia; hold=true -r, R = 1, 5 -X(theta,phi) = [(r*cos(theta)+R)*cos(phi), (r*cos(theta)+R)*sin(phi), r*sin(theta)] - -us = range(0, 2pi, length=25) -vs = range(0, pi, length=25) - -xs = [[X(u,v)[1] for u in us] for v in vs] -ys = [[X(u,v)[2] for u in us] for v in vs] -zs = [[X(u,v)[3] for u in us] for v in vs] - -data = Config( - x = xs, y = ys, z = zs, - type="surface", - mode="scatter3d" -) - -lyt = Config(scene=Config(aspectmode="data")) - -Plot(data, lyt) -``` diff --git a/CwJ/alternatives/symbolics.jmd b/CwJ/alternatives/symbolics.jmd deleted file mode 100644 index 5a43022..0000000 --- a/CwJ/alternatives/symbolics.jmd +++ /dev/null @@ -1,806 +0,0 @@ -# Symbolics.jl - -There are a few options in `Julia` for symbolic math, for example, the `SymPy` package which wraps a Python library. This section describes a collection of native `Julia` packages providing many features of symbolic math. - -## About - -The `Symbolics` package bills itself as a "fast and modern Computer Algebra System (CAS) for a fast and modern programming language." This package relies on the `SymbolicUtils` package and is built upon by the `ModelingToolkit` package, which is only briefly touched on here. - - -We begin by loading the `Symbolics` package which when loaded re-exports the `SymbolicUtils` package. - -```julia -using Symbolics -``` - -## Symbolic variables - -Symbolic math at its core involves symbolic variables, which essentially defer evaluation until requested. The creation of symbolic variables differs between the two packages discussed here. - -`SymbolicUtils` creates variables which carry `Julia` type information (e.g. `Int`, `Float64`, ...). This type information carries through operations involving these variables. Symbolic variables can be created with the `@syms` macro. For example - -```julia -@syms x y::Int f(x::Real)::Real -``` - -This creates `x` a symbolic value with symbolic type `Number`, `y` a symbolic variable holding integer values, and `f` a symbolic function of a single real variable outputting a real variable. - -The non-exported `symtype` function reveals the underlying type: - -```julia -import Symbolics.SymbolicUtils: symtype - -symtype(x), symtype(y) -``` - -For `y`, the symbolic type being real does not imply the type of `y` is a subtype of `Real`: - -```julia -isa(y, Real) -``` - - -We see that the function `f` when called with `y` would return a value of (symbolic) type `Real`: - -```julia -f(y) |> symtype -``` - -As the symbolic type of `x` is `Number` -- which is not a subtype of `Real` -- the following will error: - -```julia; error=true -f(x) -``` - - - -!!! note - The `SymPy` package also has an `@syms` macro to create variables. Though their names agree, they do different things. Using both packages together would require qualifying many shared method names. For `SymbolicUtils`, the `@syms` macro uses `Julia` types to parameterize the variables. In `SymPy` it is possible to specify *assumptions* on the variables, but that is different and not useful for dispatch without some extra effort. - -### Variables in Symbolics - -For `Symbolics`, symbolic variables are created using a wrapper around an underlying `SymbolicUtils` object. This wrapper, `Num`, is a subtype of `Real` (the underlying `SymbolicUtils` object may have symbolic type `Real`, but it won't be a subtype of `Real`.) - -Symbolic values are created with the `@variables` macro. For example: - -```julia -@variables x y::Int z[1:3]::Int f(..)::Int -``` - -This creates -* a symbolic value `x` of `symtype` `Real` -* a symbolic value `y` of `symtype` `Int` -* a vector of symbolic values each of `symtype` `Int` -* a symbolic function `f` returning an object of `symtype` `Int` - -The symbolic type reflects that of the underlying object behind the `Num` wrapper: - -```julia -typeof(x), symtype(x), typeof(Symbolics.value(x)) -``` - (The `value` method unwraps the `Num` wrapper.) - -### Variables in ModelingToolkit - -The `ModelingToolkit` package has a slightly different declaration for variables, and is described next. First the package is loaded - -```julia -using ModelingToolkit -``` - -`ModelingToolkit` re-exports all of the `Symbolics` package when loaded. - -The role of `ModelingToolkit` is that "it allows for users to give a high-level description of a model for symbolic preprocessing to analyze and enhance the model." This symbolic description allows for variables to be identified as "parameters" or "variables". For example, to parameterize a quadratic equation: - -```julia -@parameters a b c -@variables x -y = a*x^2 + b*x + c -``` - -The numeric solution of the quadratic equation (solving for ``y=0``) would involved specifying values for the parameters and then numerically solving. This separation of parameters and variables is similar to the `f(x, p)` pattern of function definition. - -The typical usage is multi-variable. This example is from the package's documentationto describe a differential equation: - -```julia -@parameters t σ ρ β -@variables x(t) y(t) z(t) -D = Differential(t) -``` - -The `D` will be described a bit later, but it formally specifies a derivative in the `t` variable. The `x(t)`, `y(t)`, `z(t)` are symbolic functions of `t`, so expressions like `D(x)` below mean the time derivative of an unknown function `x`. Here are how the Lorenz equation equations are specified: - -```julia -eqs = [D(x) ~ σ * (y - x), - D(y) ~ x * (ρ - z) - y, - D(z) ~ x * y - β * z] -``` - -## Symbolic expressions - -Symbolic expressions are built up from symbolic variables through natural `Julia` idioms. `SymbolicUtils` privileges a few key operations: `Add`, `Mul`, `Pow`, and `Div`. For examples: - -```julia -@syms x y -typeof(x + y) # `Add` -``` - -```julia -typeof(x * y) # `Mul` -``` - -Whereas, applying a function leaves a different type: - -```julia -typeof(sin(x)) -``` - -The `Term` wrapper just represents the effect of calling a function (in this case `sin`) on its arguments (in this case `x`). - -This happens in the background with symbolic variables in `Symbolics`: - -```julia -@variables x -typeof(sin(x)), typeof(Symbolics.value(sin(x))) -``` - -### Tree structure to expressions - -The `TermInterface` package is used by `SymbolicUtils` to explore the tree structure of an expression. The main methods are (cf. [SymbolicUtils.jl](https://symbolicutils.juliasymbolics.org/#expression_interface)): - -* `istree(ex)`: `true` if `ex` is not a *leaf* node (like a symbol or numeric literal) -* `operation(ex)`: the function being called (if `istree` returns `true`) -* `arguments(ex)`: the arguments to the function begin called -* `symtype(ex)`: the inferred type of the expression - -In addition, the `issym` function, to determine if `x` is of type `Sym`, is useful to distinguish *leaf* nodes, as will be illustrated below. - -These methods can be used to "walk" the tree: - -```julia -@syms x y -ex = 1 + x^2 + y -operation(ex) # the outer function is `+` -``` - - -```julia -arguments(ex) # `+` is n-ary, in this case with 3 arguments -``` - -```julia -ex1 = arguments(ex)[3] # terms have been reordered -operation(ex1) # operation for `x^2` is `^` -``` - -```julia -a, b = arguments(ex1) -``` - -```julia -istree(ex1), istree(a) -``` - -Here `a` is not a "tree", as it has no operation or arguments, it is just a variable (the `x` variable). - -The value of `symtype` is the *inferred* type of an expression, which may not match the actual type. For example, - -```julia -@variables x::Int -symtype(x), symtype(sin(x)), symtype(x/x), symtype(x / x^2) -``` - -The last one, is not likely to be an integer, but that is the inferred type in this case. - -##### Example - -As an example, we write a function to find the free symbols in a symbolic expression comprised of `SymbolicUtils` variables. (The `Symbolics.get_variables` also does this task.) To find the symbols involves walking the expression tree until a leaf node is found and then adding that to our collection if it matches `issym`. - -```julia -import Symbolics.SymbolicUtils: issym -free_symbols(ex) = (s=Set(); free_symbols!(s, ex); s) -function free_symbols!(s, ex) - if istree(ex) - for a ∈ arguments(ex) - free_symbols!(s, a) - end - else - issym(ex) && push!(s, ex) # push new symbol onto set - end -end -``` - -```julia -@syms x y z -ex = sin(x + 1)*cos(z) -free_symbols(ex) -``` - -## Expression manipulation - -### Substitute - -The `substitute` command is used to replace values with other values. For example: - -```julia -@variables x y z -ex = 1 + x + x^2/2 + x^3/6 -substitute(ex, x=>1) -``` - -This defines a symbolic expression, then substitutes the value `1` in for `x`. The `Pair` notation is useful for a *single* substitution. When there is more than one substitution, a dictionary is used: - -```julia -w = x^3 + y^3 - 2z^3 -substitute(w, Dict(x=>2, y=>3)) -``` - -The `fold` argument can be passed `false` to inhibit evaluation of values. Compare: - -```julia -ex = 1 + sqrt(x) -substitute(ex, x=>2), substitute(ex, x=>2, fold=false) -``` - -Or - -```julia -ex = sin(x) -substitute(ex, x=>π), substitute(ex, x=>π, fold=false) -``` - -### Simplify - -Algebraic operations with symbolic values can involve an exponentially increasing number of terms. As such, some simplification rules are applied after an operation to reduce the complexity of the computed value. - -For example, `0+x` should simplify to `x`, as well `1*x`, `x^0`, or `x^1` should each simplify, to some natural answer. - -`SymbolicUtils` also [simplifies](https://symbolicutils.juliasymbolics.org/#simplification) several other expressions, including: - -* `-x` becomes `(-1)*x` -* `x * x` becomes `x^2` (and `x^n` if more terms). Meaning this expression is represented as a power, not a product -* `x + x` becomes `2*x` (and `n*x` if more terms). Similarly, this represented as a product, not a sum. -* `p/q * x` becomes `(p*x)/q)`, similarly `p/q * x/y` becomes `(p*x)/(q*y)`. (Division wraps multiplication.) - -In `SymbolicUtils`, this *rewriting* is accomplished by means of *rewrite rules*. The package makes it easy to apply user-written rewrite rules. - -### Rewriting - -Many algebraic simplifications are done by the `simplify` command. For example, the basic trigonometric identities are applied: - -```julia -@variables x -ex = sin(x)^2 + cos(x)^2 -ex, simplify(ex) -``` - -The `simplify` function applies a series of rewriting rule until the expression stabilizes. The rewrite rules can be user generated, if desired. For example, the Pythagorean identity of trigonometry, just used, can be implement with this rule: - -```julia -r = @acrule(sin(~x)^2 + cos(~x)^2 => one(~x)) -ex |> Symbolics.value |> r |> Num -``` - -The rewrite rule, `r`, is defined by the `@acrule` macro. The `a` is for associative, the `c` for commutative, assumptions made by the macro. (The `c` means `cos(x)^2 + sin(x)^2` will also simplify.) Rewrite rules are called on the underlying `SymbolicUtils` expression, so we first unwrap, then after re-wrap. - -The above expression for `r` is fairly easy to appreciate. The value `~x` matches the same variable or expression. So the above rule will also simplify more complicated expressions: - -```julia -@variables y z -ex1 = substitute(ex, x => sin(x + y + z)) -ex1 |> Symbolics.value |> r |> Num -``` - -Rewrite rules when applied return the rewritten expression, if there is a match, or `nothing`. - -Rules involving two values are also easily created. This one, again, comes from the set of simplifications defined for trigonometry and exponential simplifications: - -```julia -r = @rule(exp(~x)^(~y) => exp(~x * ~y)) # (e^x)^y -> e^(x*y) -ex = exp(-x+z)^y -ex, ex |> Symbolics.value |> r |> Num -``` - -This rule is not commutative or associative, as `x^y` is not the same as `y^x` and `(x^y)^z` is not `x^(y^z)` in general. - - -The application of rules can be filtered through qualifying predicates. This artificial example uses `iseven` which returns `true` for even numbers. Here we subtract `1` when a number is not even, and otherwise leave the number alone. We do this with two rules: - -```julia -reven = @rule ~x::iseven => ~x -rodd = @rule ~x::(!iseven) => ~x - 1 -r = SymbolicUtils.Chain([rodd, reven]) -r(2), r(3) -``` - -The `Chain` function conveniently allows the sequential application of rewrite rules. - - -The notation `~x` is called a "slot variable" in the [documentation](https://symbolicutils.juliasymbolics.org/rewrite/) for `SymbolicUtils`. It matches a single expression. To match more than one expression, a "segment variable", denoted with two `~`s is used. - -### Creating functions - -By utilizing the tree-like nature of a symbolic expression, a `Julia` expression can be built from an symbolic expression easily enough. The `Symbolics.toexpr` function does this: - -```julia -ex = exp(-x + z)^y -Symbolics.toexpr(ex) -``` - -This output shows an internal representation of the steps for computing the value `ex` given different inputs. (The number `(-1)` multiplies `x`, this is added to `z` and the result passed to `exp`. That values is then used as the base for `^` with exponent `y`.) - -Such `Julia` expressions are one step away from building `Julia` functions for evaluating symbolic expressions fast (though with some technical details about "world age" to be reckoned with). The `build_function` function with the argument `expression=Val(false)` will compile a `Julia` function: - -```julia -h = build_function(ex, x, y, z; expression=Val(false)) -h(1, 2, 3) -``` - -The above is *similar* to substitution: - -```julia -substitute(ex, Dict(x=>1, y=>2, z=>3)) -``` - -However, `build_function` will be **significantly** more performant, which when many function calls are used -- such as with plotting -- is a big advantage. - -!!! note - The documentation colorfully says "`build_function` is kind of like if `lambdify` (from `SymPy`) ate its spinach." - -The above, through passing ``3`` variables after the expression, creates a function of ``3`` variables. Functions of a vector of inputs can also be created, just by expressing the variables in that manner: - -```julia -h1 = build_function(ex, [x, y, z]; expression=Val(false)) -h1([1, 2, 3]) # not h1(1,2,3) -``` - -##### Example - -As an example, here we use the `Roots` package to find a zero of a function defined symbolically: - -```julia -import Roots -@variables x -ex = x^5 - x - 1 -λ = build_function(ex, x; expression=Val(false)) -Roots.find_zero(λ, (1, 2)) -``` - -### Plotting - -Using `Plots`, the plotting of symbolic expressions is similar to the plotting of a function, as there is a plot recipe that converts the expression into a function via `build_function`. - -For example, - -```julia -using Plots -@variables x -plot(x^x^x, 0, 2) -``` - -A parametric plot is easily defined: - -```julia -plot(sin(x), cos(x), 0, pi/4) -``` - -Expressions to be plotted can represent multivariate functions. - -```julia -@variables x y -ex = 3*(1-x)^2*exp(-x^2 - (y+1)^2) - 10(x/5-x^3-y^5)*exp(-x^2-y^2) - 1/3*exp(-(x+1)^2-y^2) -xs = ys = range(-5, 5, length=100) -surface(xs, ys, ex) -``` - -The ordering of the variables is determined by `Symbolics.get_variables`: - -```julia -Symbolics.get_variables(ex) -``` - - - -### Polynomial manipulations - -There are some facilities for manipulating polynomial expressions in `Symbolics`. A polynomial, mathematically, is an expression involving one or more symbols with coefficients from a collection that has, at a minimum, addition and multiplication defined. The basic building blocks of polynomials are *monomials*, which are comprised of products of powers of the symbols. Mathematically, monomials are often allowed to have a multiplying coefficient and may be just a coefficient (if each symbol is taken to the power ``0``), but here we consider just expressions of the type ``x_1^{a_1} \cdot x_2^{a_2} \cdots x_k^{a_k}`` with the ``a_i > 0`` as monomials. - -With this understanding, then an expression can be broken up into monomials with a possible leading coefficient (possibly ``1``) *and* terms which are not monomials (such as a constant or a more complicated function of the symbols). This is what is returned by the `polynomial_coeffs` function. - -For example - -```julia -@variables a b c x -d, r = polynomial_coeffs(a*x^2 + b*x + c, (x,)) -``` - -The first term output is dictionary with keys which are the monomials and with values which are the coefficients. The second term, the residual, is all the remaining parts of the expression, in this case just the constant `c`. - -The expression can then be reconstructed through - -```julia -r + sum(v*k for (k,v) ∈ d) -``` - -The above has `a,b,c` as parameters and `x` as the symbol. This separation is designated by passing the desired polynomial symbols to `polynomial_coeff` as an iterable. (Above as a ``1``-element tuple.) - -More complicated polynomials can be similarly decomposed: - -```julia -@variables a b c x y z -ex = a*x^2*y*z + b*x*y^2*z + c*x*y*z^2 -d, r = polynomial_coeffs(ex, (x, y, z)) -``` - -The (sparse) decomposition of the polynomial is returned through `d`. The same pattern as above can be used to reconstruct the expression. -To extract the coefficient for a monomial term, indexing can be used. Of note, is an expression like `x^2*y*z` could *possibly* not equal the algebraically equal `x*y*z*x`, as they are only equal after some simplification, but the keys are in a canonical form, so this is not a concern: - -```julia -d[x*y*z*x], d[z*y*x^2] -``` - -The residual term will capture any non-polynomial terms: - -```julia -ex = sin(x) - x + x^3/6 -d, r = polynomial_coeffs(ex, (x,)) -r -``` - -To find the degree of a monomial expression, the `degree` function is available. Here it is applied to each monomial in `d`: - -```julia -[degree(k) for (k,v) ∈ d] -``` - -The `degree` function will also identify the degree of more complicated terms: - -```julia -degree(1 + x + x^2) -``` - -Mathematically the degree of the ``0`` polynomial may be ``-1`` or undefined, but here it is ``0``: - -```julia -degree(0), degree(1), degree(x), degree(x^a) -``` - -The coefficients are returned as *values* of a dictionary, and dictionaries are unsorted. To have a natural map between polynomials of a single symbol in the standard basis and a vector, we could use a pattern like this: - -```julia -@variables x a0 as[1:10] -p = a0 + sum(as[i]*x^i for i ∈ eachindex(collect(as))) -d, r = polynomial_coeffs(p, (x,)) -d -``` - -To sort the values we can use a pattern like the following: - -```julia -vcat(r, [d[k] for k ∈ sort(collect(keys(d)), by=degree)]) -``` - ----- - -Rational expressions can be decomposed into a numerator and denominator using the following idiom, which ensures the outer operation is division (a binary operation): - -```julia -@variables x -ex = (1 + x + x^2) / (1 + x + x^2 + x^3) -function nd(ex) - ex1 = Symbolics.value(ex) - (operation(ex1) == /) || return (ex, one(ex)) - Num.(arguments(ex1)) -end -nd(ex) -``` - -With this, the study of asymptotic behaviour of a univariate rational expression would involve an investigation like the following: - -```julia -m,n = degree.(nd(ex)) -m > n ? "limit is infinite" : m < n ? "limit is 0" : "limit is a constant" -``` - -### Vectors and matrices - -Symbolic vectors and matrices can be created with a specified size: - -```julia -@variables v[1:3] M[1:2, 1:3] N[1:3, 1:3] -``` - -Computations, like finding the determinant below, are lazy unless the values are `collect`ed: - -```julia -using LinearAlgebra -det(N) -``` - -```julia -det(collect(N)) -``` - -Similarly, with `norm`: - -```julia -norm(v) -``` - -and - -```julia -norm(collect(v)) -``` - -Matrix multiplication is also deferred, but the size compatability of the matrices and vectors is considered early: - -```julia -M*N, N*N, M*v -``` - -This errors, as the matrix dimensions are not compatible for multiplication: - -```julia; error=true -N*M -``` - -Similarly, linear solutions can be symbolically specified: - -```julia -@variables R[1:2, 1:2] b[1:2] -R \ b -``` - -```julia -collect(R \ b) -``` - - - -### Algebraically solving equations - -The `~` operator creates a symbolic equation. For example - -```julia -@variables x y -x^5 - x ~ 1 -``` - -or - -```julia -eqs = [5x + 2y, 6x + 3y] .~ [1, 2] -``` - -The `Symbolics.solve_for` function can solve *linear* equations. For example, - -```julia -Symbolics.solve_for(eqs, [x, y]) -``` - -The coefficients can be symbolic. Two examples could be: - -```julia -@variables m b x y -eq = y ~ m*x + b -Symbolics.solve_for(eq, x) -``` - - -```julia -@variables a11 a12 a22 x y b1 b2 -R,X,b = [a11 a12; 0 a22], [x; y], [b1, b2] -eqs = R*X .~ b -``` - -```julia -Symbolics.solve_for(eqs, [x,y]) -``` - -### Limits - -As of writing, there is no extra functionality provided by `Symbolics` for computing limits. - -### Derivatives - -`Symbolics` provides the `derivative` function to compute the derivative of a function with respect to a variable: - -```julia -@variables a b c x -y = a*x^2 + b*x + c -yp = Symbolics.derivative(y, x) -``` - -Or to find a critical point: - -```julia -Symbolics.solve_for(yp ~ 0, x) # linear equation to solve -``` - - -The derivative computation can also be broken up into an expression indicating the derivative and then a function to apply the derivative rules: - -```julia -D = Differential(x) -D(y) -``` - -and then - -```julia -expand_derivatives(D(y)) -``` - - -Using `Differential`, differential equations can be specified. An example was given in [ODEs](../ODEs/differential_equations.html), using `ModelingToolkit`. - -Higher order derivatives can be done through composition: - -```julia -D(D(y)) |> expand_derivatives -``` - - -Differentials can also be multiplied to create operators for taking higher-order derivatives: - -```julia -@variables x y -ex = (x - y^2)/(x^2 + y^2) -Dx, Dy = Differential(x), Differential(y) -Dxx, Dxy, Dyy = Dx*Dx, Dx*Dy, Dy*Dy -[Dxx(ex) Dxy(ex); Dxy(ex) Dyy(ex)] .|> expand_derivatives -``` - -In addition to `Symbolics.derivative` there are also the helper functions, such as `hessian` which performs the above - -```julia -Symbolics.hessian(ex, [x,y]) -``` - -The `gradient` function is also defined - -```julia -@variables x y z -ex = x^2 - 2x*y + z*y -Symbolics.gradient(ex, [x, y, z]) -``` - -The `jacobian` function takes an array of expressions: - -```julia -@variables x y -eqs = [ x^2 - y^2, 2x*y] -Symbolics.jacobian(eqs, [x,y]) -``` - - -### Integration - -The `SymbolicNumericIntegration` package provides a means to integrate *univariate* expressions through its `integrate` function. - - -Symbolic integration can be approached in different ways. SymPy implements part of the Risch algorithm in addition to other algorithms. Rules-based algorithms could also be implemented. - -For a trivial example, here is a rule that could be used to integrate a single integral - -```julia -@syms x ∫(x) - -is_var(x) = (xs = Symbolics.get_variables(x); length(xs) == 1 && xs[1] === x) -r = @rule ∫(~x::is_var) => x^2/2 - -r(∫(x)) -``` - - -The `SymbolicNumericIntegration` package includes many more predicates for doing rules-based integration, but it primarily approaches the task in a different manner. - -If ``f(x)`` is to be integrated, a set of *candidate* answers is generated. The following is **proposed** as an answer: ``\sum q_i \Theta_i(x)``. Differentiating the proposed answer leads to a *linear system of equations* that can be solved. - -The example in the [paper](https://arxiv.org/pdf/2201.12468v2.pdf) describing the method is with ``f(x) = x \sin(x)`` and the candidate thetas are ``{x, \sin(x), \cos(x), x\sin(x), x\cos(x)}`` so that the propose answer is: - -```math -\int f(x) dx = q_1 x + q_2 \sin(x) + q_3 \cos(x) + q_4 x \sin(x) + q_4 x \cos(x) -``` - -We differentiate the right hand side: - -```julia -@variables q[1:5] x -ΣqᵢΘᵢ = dot(collect(q), (x, sin(x), cos(x), x*sin(x), x*cos(x))) -simplify(Symbolics.derivative(ΣqᵢΘᵢ, x)) -``` - -This must match ``x\sin(x)`` so we have by -equating coefficients of the respective terms: - -```math -q_2 + q_5 = 0, \quad q_4 = 0, \quad q_1 = 0, \quad q_3 = 0, \quad q_5 = -1 -``` - -That is ``q_2=1``, ``q_5=-1``, and the other coefficients are ``0``, giving -an answer computed with: - -```julia -d = Dict(q[i] => v for (i,v) ∈ enumerate((0,1,0,0,-1))) -substitute(ΣqᵢΘᵢ, d) -``` - -The package provides an algorithm for the creation of candidates and the means to solve when possible. The `integrate` function is the main entry point. It returns three values: `solved`, `unsolved`, and `err`. The `unsolved` is the part of the integrand which can not be solved through this package. It is `0` for a given problem when `integrate` is successful in identifying an antiderivative, in which case `solved` is the answer. The value of `err` is a bound on the numerical error introduced by the algorithm. - -To see, we have: - -```julia -using SymbolicNumericIntegration -@variables x - -integrate(x * sin(x)) -``` - -The second term is `0`, as this integrand has an identified antiderivative. - -```julia -integrate(exp(x^2) + sin(x)) -``` - -This returns `exp(x^2)` for the unsolved part, as this function has no simple antiderivative. - -Powers of trig functions have antiderivatives, as can be deduced using integration by parts. When the fifth power is used, there is a numeric aspect to the algorithm that is seen: - -```julia -u,v,w = integrate(sin(x)^5) -``` - -The derivative of `u` matches up to some numeric tolerance: - -```julia -Symbolics.derivative(u, x) - sin(x)^5 -``` - ----- - -The integration of rational functions (ratios of polynomials) can be done algorithmically, provided the underlying factorizations can be identified. The `SymbolicNumericIntegration` package has a function `factor_rational` that can identify factorizations. - -```julia -import SymbolicNumericIntegration: factor_rational -@variables x -u = (1 + x + x^2)/ (x^2 -2x + 1) -v = factor_rational(u) -``` - -The summands in `v` are each integrable. We can see that `v` is a reexpression through - -```julia -simplify(u - v) -``` - -The algorithm is numeric, not symbolic. This can be seen in these two factorizations: - -```julia -u = 1 / expand((x^2-1)*(x-2)^2) -v = factor_rational(u) -``` - -or - -```julia -u = 1 / expand((x^2+1)*(x-2)^2) -v = factor_rational(u) -``` - -As such, the integrals have numeric differences from their mathematical counterparts: - -```julia -a,b,c = integrate(u) -``` - -We can see a bit of how much through the following, which needs a tolerance set to identify the rational numbers of the mathematical factorization correctly: - -```julia -cs = [first(arguments(term)) for term ∈ arguments(a)] # pick off coefficients -``` - -```julia -rationalize.(cs; tol=1e-8) -``` diff --git a/CwJ/derivatives/Project.toml b/CwJ/derivatives/Project.toml deleted file mode 100644 index f6a9189..0000000 --- a/CwJ/derivatives/Project.toml +++ /dev/null @@ -1,17 +0,0 @@ -[deps] -DataFrames = "a93c6f00-e57d-5684-b7b6-d8193f3e46c0" -EllipsisNotation = "da5c29d0-fa7d-589e-88eb-ea29b0a81949" -ForwardDiff = "f6369f11-7733-5829-9624-2563aa707210" -ImplicitPlots = "55ecb840-b828-11e9-1645-43f4a9f9ace7" -IntervalArithmetic = "d1acc4aa-44c8-5952-acd4-ba5d80a2a253" -IntervalConstraintProgramming = "138f1668-1576-5ad7-91b9-7425abbf3153" -LaTeXStrings = "b964fa9f-0449-5b57-a5c2-d3ea65f4040f" -MDBM = "dd61e66b-39ce-57b0-8813-509f78be4b4d" -Plots = "91a5bcdd-55d7-5caf-9e0b-520d859cae80" -Printf = "de0858da-6303-5e67-8744-51eddeeeb8d7" -QuadGK = "1fd47b50-473d-5c70-9696-f719f8f3bcdc" -Roots = "f2b01f46-fcfa-551c-844a-d8ac1e96c665" -SymPy = "24249f21-da20-56a4-8eb1-6a02cf4ae2e6" -TaylorSeries = "6aa5eb33-94cf-58f4-a9d0-e4b2c4fc25ea" -TermInterface = "8ea1fca8-c5ef-4a55-8b96-4e9afe9c9a3c" -Unitful = "1986cc42-f94f-5a68-af5c-568840ba703d" diff --git a/CwJ/derivatives/curve_sketching.jmd b/CwJ/derivatives/curve_sketching.jmd deleted file mode 100644 index b41065a..0000000 --- a/CwJ/derivatives/curve_sketching.jmd +++ /dev/null @@ -1,543 +0,0 @@ -# Curve Sketching - -This section uses the following add-on packages: - -```julia -using CalculusWithJulia -using Plots -using SymPy -using Roots -using Polynomials # some name clash with SymPy -``` - - - -```julia; echo=false; results="hidden" -using CalculusWithJulia.WeaveSupport - -fig_size=(800, 600) -const frontmatter = ( - title = "Curve Sketching", - description = "Calculus with Julia: Curve Sketching", - tags = ["CalculusWithJulia", "derivatives", "curve sketching"], -); -nothing -``` - ----- - -The figure illustrates a means to *sketch* a sine curve - identify as -many of the following values as you can: - -* asymptotic behaviour (as ``x \rightarrow \pm \infty``), -* periodic behaviour, -* vertical asymptotes, -* the $y$ intercept, -* any $x$ intercept(s), -* local peaks and valleys (relative extrema). -* concavity - -With these, a sketch fills in between the -points/lines associated with these values. - - -```julia; hold=true; echo=false; cache=true -### {{{ sketch_sin_plot }}} - - -function sketch_sin_plot_graph(i) - f(x) = 10*sin(pi/2*x) # [0,4] - deltax = 1/10 - deltay = 5/10 - - zs = find_zeros(f, 0-deltax, 4+deltax) - cps = find_zeros(D(f), 0-deltax, 4+deltax) - xs = range(0, stop=4*(i-2)/6, length=50) - if i == 1 - ## plot zeros - title = "Plot the zeros" - p = scatter(zs, 0*zs, title=title, xlim=(-deltax,4+deltax), ylim=(-10-deltay,10+deltay), legend=false) - elseif i == 2 - ## plot extrema - title = "Plot the local extrema" - p = scatter(zs, 0*zs, title=title, xlim=(-deltax,4+deltax), ylim=(-10-deltay,10+deltay), legend=false) - scatter!(p, cps, f.(cps)) - else - ## sketch graph - title = "sketch the graph" - p = scatter(zs, 0*zs, title=title, xlim=(-deltax,4+deltax), ylim=(-10-deltay,10+deltay), legend=false) - scatter!(p, cps, f.(cps)) - plot!(p, xs, f.(xs)) - end - p -end - - -caption = L""" - -After identifying asymptotic behaviours, -a curve sketch involves identifying the $y$ intercept, if applicable; the $x$ intercepts, if possible; the local extrema; and changes in concavity. From there a sketch fills in between the points. In this example, the periodic function $f(x) = 10\cdot\sin(\pi/2\cdot x)$ is sketched over $[0,4]$. - -""" - - - -n = 8 -anim = @animate for i=1:n - sketch_sin_plot_graph(i) -end - -imgfile = tempname() * ".gif" -gif(anim, imgfile, fps = 1) - -ImageFile(imgfile, caption) -``` - -Though this approach is most useful for hand-sketches, the underlying -concepts are important for properly framing graphs made with the -computer. - -We can easily make a graph of a function over a specified -interval. What is not always so easy is to pick an interval that shows -off the features of interest. In the section on -[rational](../precalc/rational_functions.html) functions there was a -discussion about how to draw graphs for rational functions so that -horizontal and vertical asymptotes can be seen. These are properties -of the "large." In this section, we build on this, but concentrate now -on more local properties of a function. - -##### Example - -Produce a graph of the function $f(x) = x^4 -13x^3 + 56x^2-92x + 48$. - -We identify this as a fourth-degree polynomial with postive leading -coefficient. Hence it will eventually look $U$-shaped. If we graph -over a too-wide interval, that is all we will see. Rather, we do some -work to produce a graph that shows the zeros, peaks, and valleys of -$f(x)$. To do so, we need to know the extent of the zeros. We can try -some theory, but instead we just guess and if that fails, will work harder: - -```julia; -f(x) = x^4 - 13x^3 + 56x^2 -92x + 48 -rts = find_zeros(f, -10, 10) -``` - -As we found $4$ roots, we know by the fundamental theorem of algebra we have them all. This means, our graph need not focus on values much larger than $6$ or much smaller than $1$. - -To know where the peaks and valleys are, we look for the critical points: - -```julia; -cps = find_zeros(f', 1, 6) -``` - -Because we have the $4$ distinct zeros, we must have the peaks and -valleys appear in an interleaving manner, so a search over $[1,6]$ -finds all three critical points and without checking, they must -correspond to relative extrema. - -Next we identify the *inflection points* which are among the zeros of the second derivative (when defined): - -```julia -ips = find_zeros(f'', 1, 6) -``` - -If there is no sign change for either ``f'`` or ``f''`` over ``[a,b]`` then the sketch of ``f`` on this interval must be one of: - -* increasing and concave up (if ``f' > 0`` and ``f'' > 0``) -* increasing and concave down (if ``f' > 0`` and ``f'' < 0``) -* decreasing and concave up (if ``f' < 0`` and ``f'' > 0``) -* decreasing and concave down (if ``f' < 0`` and ``f'' < 0``) - -This aids in sketching the graph between the critical points and inflection points. - - -We finally check that if we were to just use $[0,7]$ as a domain to -plot over that the function doesn't get too large to mask the -oscillations. This could happen if the $y$ values at the end points -are too much larger than the $y$ values at the peaks and valleys, as -only so many pixels can be used within a graph. For this we have: - -```julia; -f.([0, cps..., 7]) -``` - -The values at $0$ and at $7$ are a bit large, as compared to the -relative extrema, and since we know the graph is eventually -$U$-shaped, this offers no insight. So we narrow the range a bit for -the graph: - -```julia; -plot(f, 0.5, 6.5) -``` - - ----- - -This sort of analysis can be automated. The plot "recipe" for polynomials from the `Polynomials` package does similar considerations to choose a viewing window: - -```julia -xₚ = variable(Polynomial) -plot(f(xₚ)) # f(xₚ) of Polynomial type -``` - - - - -##### Example - -Graph the function - -```math -f(x) = \frac{(x-1)\cdot(x-3)^2}{x \cdot (x-2)}. -``` - -Not much to do here if you are satisfied with a graph that only gives insight into the asymptotes of this rational function: - -```julia; -𝒇(x) = ( (x-1)*(x-3)^2 ) / (x * (x-2) ) -plot(𝒇, -50, 50) -``` - -We can see the slant asymptote and hints of vertical asymptotes, but, -we'd like to see more of the basic features of the graph. - -Previously, we have discussed rational functions and their -asymptotes. This function has numerator of degree ``3`` and denominator of -degree ``2``, so will have a slant asymptote. As well, the zeros of the -denominator, $0$ and $-2$, will lead to vertical asymptotes. - -To identify how wide a viewing window should be, for the rational -function the asymptotic behaviour is determined after the concavity is -done changing and we are past all relative extrema, so we should take -an interval that includes all potential inflection points and critical -points: - -```julia; -𝒇cps = find_zeros(𝒇', -10, 10) -poss_ips = find_zero(𝒇'', (-10, 10)) -extrema(union(𝒇cps, poss_ips)) -``` - -So a range over $[-5,5]$ should display the key features including the slant asymptote. - -Previously we used the `rangeclamp` function defined in `CalculusWithJulia` to avoid the distortion that vertical asymptotes can have: - -```julia; -plot(rangeclamp(𝒇), -5, 5) -``` - -With this graphic, we can now clearly see in the graph the two zeros at $x=1$ and $x=3$, the vertical asymptotes at $x=0$ and $x=2$, and the slant asymptote. - ----- - -Again, this sort of analysis can be systematized. The rational function type in the `Polynomials` package takes a stab at that, but isn't quite so good at capturing the slant asymptote: - -```julia -xᵣ = variable(RationalFunction) -plot(𝒇(xᵣ)) # f(x) of RationalFunction type -``` - - -##### Example - -Consider the function ``V(t) = 170 \sin(2\pi\cdot 60 \cdot t)``, a model for the alternating current waveform for an outlet in the United States. Create a graph. - -Blindly trying to graph this, we will see immediate issues: - -```julia -V(t) = 170 * sin(2*pi*60*t) -plot(V, -2pi, 2pi) -``` - -Ahh, this periodic function is *too* rapidly oscillating to be plotted without care. We recognize this as being of the form ``V(t) = a\cdot\sin(c\cdot t)``, so where the sine function has a period of ``2\pi``, this will have a period of ``2\pi/c``, or ``1/60``. So instead of using ``(-2\pi, 2\pi)`` as the interval to plot over, we need something much smaller: - - -```julia -plot(V, -1/60, 1/60) -``` - - -##### Example - -Plot the function ``f(x) = \ln(x/100)/x``. - -We guess that this function has a *vertical* asymptote at ``x=0+`` and a horizontal asymptote as ``x \rightarrow \infty``, we verify through: - -```julia -@syms x -ex = log(x/100)/x -limit(ex, x=>0, dir="+"), limit(ex, x=>oo) -``` - -The ``\ln(x/100)`` part of ``f`` goes ``-\infty`` as ``x \rightarrow 0+``; yet ``f(x)`` is eventually positive as ``x \rightarrow 0``. So a graph should - -* not show too much of the vertical asymptote -* capture the point where ``f(x)`` must cross ``0`` -* capture the point where ``f(x)`` has a relative maximum -* show enough past this maximum to indicate to the reader the eventual horizontal asyptote. - -For that, we need to get the ``x`` intercepts and the critical points. The ``x/100`` means this graph has some scaling to it, so we first look between ``0`` and ``200``: - -```julia -find_zeros(ex, 0, 200) # domain is (0, oo) -``` - -Trying the same for the critical points comes up empty. We know there is one, but it is past ``200``. Scanning wider, we see: - -```julia -find_zeros(diff(ex,x), 0, 500) -``` - - -So maybe graphing over ``[50, 300]`` will be a good start: - -```julia -plot(ex, 50, 300) -``` - -But it isn't! The function takes its time getting back towards ``0``. We know that there must be a change of concavity as ``x \rightarrow \infty``, as there is a horizontal asymptote. We looks for the anticipated inflection point to ensure our graph includes that: - -```julia -find_zeros(diff(ex, x, x), 1, 5000) -``` - -So a better plot is found by going well beyond that inflection point: - -```julia -plot(ex, 75, 1500) -``` - - - -## Questions - -###### Question - -Consider this graph - -```julia; hold=true; echo=false -f(x) = (x-2)* (x-2.5)*(x-3) / ((x-1)*(x+1)) -p = plot(f, -20, -1-.3, legend=false, xlim=(-15, 15), color=:blue) -plot!(p, f, -1 + .2, 1 - .02, color=:blue) -plot!(p, f, 1 + .05, 20, color=:blue) -``` - -What kind of *asymptotes* does it appear to have? - -```julia; hold=true; echo=false -choices = [ -L"Just a horizontal asymptote, $y=0$", -L"Just vertical asymptotes at $x=-1$ and $x=1$", -L"Vertical asymptotes at $x=-1$ and $x=1$ and a horizontal asymptote $y=1$", -L"Vertical asymptotes at $x=-1$ and $x=1$ and a slant asymptote" -] -answ = 4 -radioq(choices, answ) -``` - -###### Question - -Consider the function ``p(x) = x + 2x^3 + 3x^3 + 4x^4 + 5x^5 +6x^6``. Which interval shows more than a ``U``-shaped graph that dominates for large ``x`` due to the leading term being ``6x^6``? - -(Find an interval that contains the zeros, critical points, and inflection points.) - - -```julia; hold=true; echo=false -choices = ["``(-5,5)``, the default bounds of a calculator", -"``(-3.5, 3.5)``, the bounds given by Cauchy for the real roots of ``p``", -"``(-1, 1)``, as many special polynomials have their roots in this interval", -"``(-1.1, .25)``, as this constains all the roots, the critical points, and inflection points and just a bit more" -] -radioq(choices, 4, keep_order=true) -``` -###### Question - -Let ``f(x) = x^3/(9-x^2)``. - -What points are *not* in the domain of ``f``? - -```julia; echo=false -qchoices = [ - "The values of `find_zeros(f, -10, 10)`: `[-3, 0, 3]`", - "The values of `find_zeros(f', -10, 10)`: `[-5.19615, 0, 5.19615]`", - "The values of `find_zeros(f'', -10, 10)`: `[-3, 0, 3]`", - "The zeros of the numerator: `[0]`", - "The zeros of the denominator: `[-3, 3]`", - "The value of `f(0)`: `0`", - "None of these choices" -] -radioq(qchoices, 5, keep_order=true) -``` - -The ``x``-intercepts are: - -```julia; hold=true; echo=false -radioq(qchoices, 4, keep_order=true) -``` - -The ``y``-intercept is: -```julia; hold=true; echo=false -radioq(qchoices, 6, keep_order=true) -``` - -There are *vertical asymptotes* at ``x=\dots``? - -```julia; hold=true; echo=false -radioq(qchoices, 5) -``` - -The *slant* asymptote has slope? - -```julia; hold=true; echo=false -numericq(1) -``` - -The function has critical points at - -```julia; hold=true,echo=false -radioq(qchoices, 2, keep_order=true) -``` - -The function has relative extrema at - -```julia; hold=true;echo=false -radioq(qchoices, 7, keep_order=true) -``` - -The function has inflection points at - -```julia; hold=true;echo=false -radioq(qchoices, 7, keep_order=true) -``` - - - -###### Question - -A function ``f`` has - -* zeros of ``\{-0.7548\dots, 2.0\}``, -* critical points at ``\{-0.17539\dots, 1.0, 1.42539\dots\}``, -* inflection points at ``\{0.2712\dots,1.2287\}``. - -Is this a possible graph of ``f``? - -```julia; hold=true;echo=false -f(x) = x^4 - 3x^3 + 2x^2 + x - 2 -plot(f, -1, 2.5, legend=false) -``` - -```julia; hold=true;echo=false -yesnoq("yes") -``` - -###### Question - -Two models for population growth are *exponential* growth: $P(t) = P_0 a^t$ and -[logistic growth](https://en.wikipedia.org/wiki/Logistic_function#In_ecology:_modeling_population_growth): $P(t) = K P_0 a^t / (K + P_0(a^t - 1))$. The exponential growth model has growth rate proportional to the current population. The logistic model has growth rate depending on the current population *and* the available resources (which can limit growth). - - -Letting $K=10$, $P_0=5$, and $a= e^{1/4}$. A plot over $[0,5]$ shows somewhat similar behaviour: - -```julia; -K, P0, a = 50, 5, exp(1/4) -exponential_growth(t) = P0 * a^t -logistic_growth(t) = K * P0 * a^t / (K + P0*(a^t-1)) - -plot(exponential_growth, 0, 5) -plot!(logistic_growth) -``` - -Does a plot over $[0,50]$ show qualitatively similar behaviour? - -```julia; hold=true; echo=false -yesnoq(true) -``` - -Exponential growth has $P''(t) = P_0 a^t \log(a)^2 > 0$, so has no inflection point. By plotting over a sufficiently wide interval, can you answer: does the logistic growth model have an inflection point? - -```julia; hold=true; echo=false -yesnoq(true) -``` - -If yes, find it numerically: - -```julia; hold=true; echo=false -val = find_zero(D(logistic_growth,2), (0, 20)) -numericq(val) -``` - -The available resources are quantified by $K$. As $K \rightarrow \infty$ what is the limit of the logistic growth model: - -```julia; hold=true; echo=false -choices = [ -"The exponential growth model", -"The limit does not exist", -"The limit is ``P_0``"] -answ = 1 -radioq(choices, answ) -``` - -##### Question - -The plotting algorithm for plotting functions starts with a small -initial set of points over the specified interval ($21$) and then -refines those sub-intervals where the second derivative is determined -to be large. - -Why are sub-intervals where the second derivative is large different than those where the second derivative is small? - -```julia; hold=true; echo=false -choices = [ -"The function will increase (or decrease) rapidly when the second derivative is large, so there needs to be more points to capture the shape", -"The function will have more curvature when the second derivative is large, so there needs to be more points to capture the shape", -"The function will be much larger (in absolute value) when the second derivative is large, so there needs to be more points to capture the shape", -] -answ = 2 -radioq(choices, answ) -``` - -##### Question - -Is there a nice algorithm to identify what domain a function should be -plotted over to produce an informative graph? -[Wilkinson](https://www.cs.uic.edu/~wilkinson/Publications/plotfunc.pdf) -has some suggestions. (Wilkinson is well known to the `R` community as -the specifier of the grammar of graphics.) It is mentioned that -"finding an informative domain for a given function depends on at least -three features: periodicity, asymptotics, and monotonicity." - -Why would periodicity matter? - -```julia; hold=true; echo=false -choices = [ -"An informative graph only needs to show one or two periods, as others can be inferred.", -"An informative graph need only show a part of the period, as the rest can be inferred.", -L"An informative graph needs to show several periods, as that will allow proper computation for the $y$ axis range."] -answ = 1 -radioq(choices, answ) -``` - -Why should asymptotics matter? - -```julia; hold=true; echo=false -choices = [ -L"A vertical asymptote can distory the $y$ range, so it is important to avoid too-large values", -L"A horizontal asymptote must be plotted from $-\infty$ to $\infty$", -"A slant asymptote must be plotted over a very wide domain so that it can be identified." -] -answ = 1 -radioq(choices, answ) -``` - -Monotonicity means increasing or decreasing. This is important for what reason? - -```julia; hold=true; echo=false -choices = [ -"For monotonic regions, a large slope or very concave function might require more care to plot", -"For monotonic regions, a function is basically a straight line", -"For monotonic regions, the function will have a vertical asymptote, so the region should not be plotted" -] -answ = 1 -radioq(choices, answ) -``` diff --git a/CwJ/derivatives/derivatives.jmd b/CwJ/derivatives/derivatives.jmd deleted file mode 100644 index d59aaff..0000000 --- a/CwJ/derivatives/derivatives.jmd +++ /dev/null @@ -1,1527 +0,0 @@ -# Derivatives - -This section uses these add-on packages: - -```julia -using CalculusWithJulia -using Plots -using SymPy -``` - - -```julia; echo=false; results="hidden" -using CalculusWithJulia.WeaveSupport -using DataFrames - -const frontmatter = ( - title = "Derivatives", - description = "Calculus with Julia: Derivatives", - tags = ["CalculusWithJulia", "derivatives", "derivatives"], -); - -fig_size=(800, 600) - -nothing -``` - ----- - -Before defining the derivative of a function, let's begin with two -motivating examples. - -##### Example: Driving - -Imagine motoring along down highway ``61`` leaving Minnesota on the way to -New Orleans; though lost in listening to music, still mindful of the -speedometer and odometer, both prominently placed on the dashboard of -the car. - -The speedometer reads ``60`` miles per hour, what is the odometer doing? -Besides recording total distance traveled, it is incrementing -dutifully every hour by ``60`` miles. Why? Well, the well-known formula relating distance, time and rate of travel is - -```math -\text{distance} = \text{ rate } \times \text{ time.} -``` - -If the rate is a constant ``60`` miles/hour, then in one hour the distance traveled is ``60`` miles. - -Of course, the odometer isn't just incrementing once per hour, it is incrementing once every ``1/10``th of a mile. How much time does that take? Well, we would need to solve $1/10=60 \cdot t$ which means $t=1/600$ hours, better known as once every ``6`` seconds. - -Using some mathematical notation, would give $x(t) = v\cdot t$, where -$x$ is position at time $t$, $v$ is the *constant* velocity and $t$ the time -traveled in hours. A simple graph of the first three hours of travel would show: - -```julia; hold=true; -position(t) = 60 * t -plot(position, 0, 3) -``` - -Oh no, we hit traffic. In the next ``30`` minutes we only traveled -``15`` miles. We were so busy looking out for traffic, the speedometer was -not checked. What would the average speed have been? Though in the ``30`` -minutes of stop-and-go traffic, the displayed speed may have varied, the *average speed* -would simply be the change in distance over the change in time, or -$\Delta x / \Delta t$. That is - -```julia -15/(1/2) -``` - - -Now suppose that after $6$ hours of travel the GPS in the car gives us a readout of distance traveled -as a function of time. The graph looks like this: - -```julia; hold=true; echo=false -function position(t) - t <= 3 ? 60*t : - t <= 3.5 ? position(3) + 30*(t-3) : - t <= 4 ? position(3.5) + 75 * (t-3.5) : - t <= 4.5 ? position(4) : position(4.5) + 60*(t-4.5) -end -plot(position, 0, 6) -``` - -We can see with some effort that the slope is steady for the first three hours, is slightly less between $3$ and -$3.5$ hours, then is a bit steeper for the next half hour. After that, it is flat for the -about half an hour, then the slope continues on with same value as in the first -``3`` hours. What does that say about our speed during our trip? - -Based on the graph, what was the average speed over the first three hours? Well, we traveled ``180`` miles, and took ``3`` hours: - -```julia -180/3 -``` - -What about the next half hour? Squinting shows the amount traveled was ``15`` miles (``195 - 180``) and it took ``1/2`` an hour: - -```julia -15/(1/2) -``` - -And the half hour after that? The average speed is found from the distance traveled, ``37.5`` miles, divided by the time, ``1/2`` hour: - -```julia -37.5 / (1/2) -``` - -Okay, so there was some speeding involved. - -The next half hour the car did not move. What was the average speed? Well the change in position was ``0``, but the time was ``1/2`` hour, so the average was ``0``. - -Perhaps a graph of the speed is a bit more clear. We can do this based on the above: - -```julia -function speed(t) - 0 < t <= 3 ? 60 : - t <= 3.5 ? 30 : - t <= 4 ? 75 : - t <= 4.5 ? 0 : 60 -end -plot(speed, 0, 6) -``` - -The jumps, as discussed before, are artifacts of the graphing -algorithm. What is interesting, is we could have derived the graph of -`speed` from that of `x` by just finding the slopes of the line -segments, and we could have derived the graph of `x` from that of -`speed`, just using the simple formula relating distance, rate, and -time. - -!!! note - We were pretty loose with some key terms. There is a - distinction between "speed" and "velocity", this being the speed - is the absolute value of velocity. Velocity incorporates a - direction as well as a magnitude. Similarly, distance traveled and - change in position are not the same thing when there is back - tracking involved. The total distance traveled is computed with - the speed, the change in position is computed with the - velocity. When there is no change of sign, it is a bit more - natural, perhaps, to use the language of speed and distance. - -##### Example: Galileo's ball and ramp experiment - -One of history's most famous experiments was performed by -[Galileo](http://en.wikipedia.org/wiki/History_of_experiments) where -he rolled balls down inclined ramps, making note of distance traveled -with respect to time. As Galileo had no ultra-accurate measuring device, -he needed to slow movement down by controlling the angle of the -ramp. With this, he could measure units of distance per units of time. -(Click through to *Galileo and Perspective* [Dauben](http://www.mcm.edu/academic/galileo/ars/arshtml/mathofmotion1.html).) - - -Suppose that no matter what the incline was, Galileo observed that in -units of the distance traveled in the first second that the distance -traveled between subsequent seconds was $3$ times, then $5$ times, then -$7$ times, ... This table summarizes. - -```julia; hold=true; echo=false -ts = [0,1,2,3,4,5] -dxs = [0,1,3, 5, 7, 9] -ds = [0,1,4,9,16,25] -d = DataFrame(t=ts, delta=dxs, distance=ds) -table(d) -``` - -A graph of distance versus time could be found by interpolating between the measured points: - -```julia; -ts = [0,1,2,3,4, 5] -xs = [0,1,4,9,16,25] -plot(ts, xs) -``` - -The graph looks almost quadratic. What would the following questions have yielded? - -* What is the average speed between $0$ and $3$? - -```julia -(9-0) / (3-0) # (xs[4] - xs[1]) / (ts[4] - ts[1]) -``` - -* What is the average speed between $2$ and $3$? - -```julia -(9-4) / (3-2) # (xs[4] - xs[3]) / (ts[4] - ts[3]) -``` - -From the graph, we can tell that the slope of the line connecting -$(2,4)$ and $(3,9)$ will be greater than that connecting $(0,0)$ and -$(3,9)$. In fact, given the shape of the graph (concave up), the line -connecting $(0,0)$ with any point will have a slope less than or equal -to any of the line segments. - -The average speed between $k$ and $k+1$ for this graph is: - -```julia -xs[2]-xs[1], xs[3] - xs[2], xs[4] - xs[3], xs[5] - xs[4] -``` - -We see it increments by $2$. The acceleration is the rate of change of -speed. We see the rate of change of speed is constant, as the speed -increments by ``2`` each time unit. - -Based on this - and given Galileo's insight - it appears the -acceleration for a falling body subject to gravity will be -**constant** and the position as a function of time will be quadratic. - -## The slope of the secant line - -In the above examples, we see that the average speed is computed using -the slope formula. This can be generalized for any univariate function -$f(x)$: - -> The average rate of change between $a$ and $b$ is $(f(b) - f(a)) / -> (b - a)$. It is typical to express this as $\Delta y/ \Delta x$, -> where $\Delta$ means "change". - -Geometrically, this is the slope of the line connecting the points -$(a, f(a))$ and $(b, f(b))$. This line is called a -[secant](http://en.wikipedia.org/wiki/Secant_line) line, which is just -a line intersecting two specified points on a curve. - - -Rather than parameterize this problem using $a$ and $b$, we let $c$ and $c+h$ represent the two values for $x$, then the secant-line-slope formula becomes - -```math -m = \frac{f(c+h) - f(c)}{h}. -``` - -## The slope of the tangent line - -The slope of the secant line represents the average rate of change -over a given period, $h$. What if this rate is so variable, that it -makes sense to take smaller and smaller periods $h$? In fact, what if -$h$ goes to $0$? - -```julia; hold=true; echo=false; cache=true -function secant_line_tangent_line_graph(n) - f(x) = sin(x) - c = pi/3 - h = 2.0^(-n) * pi/4 - m = (f(c+h) - f(c))/h - - xs = range(0, stop=pi, length=50) - plt = plot(f, 0, pi, legend=false, size=fig_size) - plot!(plt, xs, f(c) .+ cos(c)*(xs .- c), color=:orange) - plot!(plt, xs, f(c) .+ m*(xs .- c), color=:black) - scatter!(plt, [c,c+h], [f(c), f(c+h)], color=:orange, markersize=5) - - plot!(plt, [c, c+h, c+h], [f(c), f(c), f(c+h)], color=:gray30) - annotate!(plt, [(c+h/2, f(c), text("h", :top)), (c + h + .05, (f(c) + f(c + h))/2, text("f(c+h) - f(c)", :left))]) - - plt -end -caption = L""" - -The slope of each secant line represents the *average* rate of change between $c$ and $c+h$. As $h$ goes towards $0$, we recover the slope of the tangent line, which represents the *instantatneous* rate of change. - -""" - - - -n = 5 -anim = @animate for i=0:n - secant_line_tangent_line_graph(i) -end - -imgfile = tempname() * ".gif" -gif(anim, imgfile, fps = 1) - -ImageFile(imgfile, caption) -``` - - - -The graphic suggests that the slopes of the secant line converge to -the slope of a "tangent" line. That is, for a given $c$, this -limit exists: - -```math -\lim_{h \rightarrow 0} \frac{f(c+h) - f(c)}{h}. -``` - -We will define the tangent line at $(c, f(c))$ to be the line through -the point with the slope from the limit above - provided that limit -exists. Informally, the tangent line is the line through the point -that best approximates the function. - -```julia; hold=true; echo=false; cache=true -function line_approx_fn_graph(n) - f(x) = sin(x) - c = pi/3 - h = round(2.0^(-n) * pi/2, digits=2) - m = cos(c) - - Delta = max(f(c) - f(c-h), f(min(c+h, pi/2)) - f(c)) - - p = plot(f, c-h, c+h, legend=false, xlims=(c-h,c+h), ylims=(f(c)-Delta,f(c)+Delta )) - plot!(p, x -> f(c) + m*(x-c)) - scatter!(p, [c], [f(c)]) - p -end -caption = L""" - -The tangent line is the best linear approximation to the function at the point $(c, f(c))$. As the viewing window zooms in on $(c,f(c))$ we - can see how the graph and its tangent line get more similar. - -""" - -n = 6 -anim = @animate for i=1:n - line_approx_fn_graph(i) -end - -imgfile = tempname() * ".gif" -gif(anim, imgfile, fps = 1) - -ImageFile(imgfile, caption) -``` - -The tangent line is not just a line that intersects the graph in one -point, nor does it need only intersect the line in just one point. - -!!! note - This last point was certainly not obvious at - first. [Barrow](http://www.maa.org/sites/default/files/0746834234133.di020795.02p0640b.pdf), - who had Newton as a pupil, and was the first to sketch a proof of - part of the Fundamental Theorem of Calculus, understood a tangent - line to be a line that intersects a curve at only one point. - - -##### Example - -What is the slope of the tangent line to $f(x) = \sin(x)$ at $c=0$? - -We need to compute the limit $(\sin(c+h) - \sin(c))/h$ which is the -limit as $h$ goes to $0$ of $\sin(h)/h.$ We know this to be ``1.`` - -```julia; hold=true -f(x) = sin(x) -c = 0 -tl(x) = f(c) + 1 * (x - c) -plot(f, -pi/2, pi/2) -plot!(tl, -pi/2, pi/2) -``` - -## The derivative - -The limit of the slope of the secant line gives an operation: for each -$c$ in the domain of $f$ there is a number (the slope of the tangent -line) or it does not exist. That is, there is a derived function from -$f$. Call this function the *derivative* of $f$. - - -There are many notations for the derivative, mostly we use the "prime" notation: - -```math -f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}. -``` - -The limit above is identical, only it uses $x$ instead of $c$ to -emphasize that we are thinking of a function now, and not just a value -at a point. - - -The derivative is related to a function, but at times it is more convenient to write only the expression defining the rule of the function. In that case, we use this notation for the derivative ``[\text{expression}]'``. - -### Some basic derivatives - -- **The power rule**. What is the derivative of the monomial $f(x) = x^n$? We need to look - at $(x+h)^n - x^n$ for positive, integer-value $n$. Let's look at a case, $n=5$ - -```julia -@syms x::real h::real -n = 5 -ex = expand((x+h)^n - x^n) -``` - -All terms have an `h` in them, so we cancel it out: - -```julia -cancel(ex/h, h) -``` - -We see the lone term `5x^4` without an $h$, so as we let $h$ go to $0$, this will be the limit. That is, $f'(x) = 5x^4$. - - -For integer-valued, positive, $n$, the binomial theorem gives an -expansion $(x+h)^n = x^n + nx^{n-1}\cdot h^1 + n\cdot(n-1)x^{n-2}\cdot h^2 + \cdots$. Subtracting $x^n$ -then dividing by $h$ leaves just the term $nx^{n-1}$ without a power -of $h$, so the limit, in general, is just this term. That is: - -```math -[x^n]' = nx^{n-1}. -``` - - -It isn't a special case, but when $n=0$, we also have the above -formula applies, as $x^0$ is the constant $1$, and all constant -functions will have a derivative of $0$ at all $x$. We will see that in -general, the power rule applies for any $n$ where $x^n$ is defined. - -- What is the derivative of $f(x) = \sin(x)$? We know that $f'(0)= 1$ - by the earlier example with ``(\sin(0+h)-\sin(0))/h = \sin(h)/h``, here we solve in general. - -We need to consider the difference $\sin(x+h) - \sin(x)$: - -```julia -sympy.expand_trig(sin(x+h) - sin(x)) # expand_trig is not exposed in `SymPy` -``` - -That used the formula $\sin(x+h) = \sin(x)\cos(h) + \sin(h)\cos(x)$. - -We could then rearrange the secant line slope formula to become: - -```math -\cos(x) \cdot \frac{\sin(h)}{h} + \sin(x) \cdot \frac{\cos(h) - 1}{h} -``` - -and take a limit. If the answer isn't clear, we can let `SymPy` do this work: - -```julia -limit((sin(x+h) - sin(x))/ h, h => 0) -``` - -From the formula ``[\sin(x)]' = \cos(x)`` we can easily get the *slope* of the tangent line to ``f(x) = \sin(x)`` at ``x=0`` by simply evaluating ``\cos(0) = 1``. - -- Let's see what the derivative of $\ln(x) = \log(x)$ is (using base ``e`` for ``\log`` unless otherwise indicated). We have - -```math -\frac{\log(x+h) - \log(x)}{h} = \frac{1}{h}\log(\frac{x+h}{x}) = \log((1+h/x)^{1/h}). -``` - -As noted earlier, Cauchy saw the limit as $u$ goes to $0$ of $f(u) = (1 + -u)^{1/u}$ is $e$. Re-expressing the above we can get $1/h \cdot -\log(f(h/x))$. The limit as $h$ goes to $0$ of this is found from -the composition rules for limits: as $\lim_{h \rightarrow 0} f(h/x) = -e^{1/x}$, and since $\log(x)$ is continuous at $e^{1/x}$ we get this expression has a limit of $1/x$. - -We verify through: - -```julia -limit((log(x+h) - log(x))/h, h => 0) -``` - -- The derivative of ``f(x) = e^x`` can also be done from a limit. We have - -```math -\frac{e^{x+h} - e^x}{h} = \frac{e^x \cdot(e^h -1)}{h}. -``` - -Earlier, we saw that $\lim_{h \rightarrow 0}(e^h - 1)/h = 1$. With this, we get -$[e^x]' = e^x$, that is it is a function satisfying $f'=f$. - - ----- - -There are several different -[notations](http://en.wikipedia.org/wiki/Notation_for_differentiation) -for derivatives. Some are historical, some just add -flexibility. We use the prime notation of Lagrange: ``f'(x)``, ``u'`` and ``[\text{expr}]'``, -where the first emphasizes that the derivative is a function with a -value at ``x``, the second emphasizes the derivative operates on -functions, the last emphasizes that we are taking the derivative of -some expression. - - -There are many other notations: - -- The Leibniz notation uses the infinitesimals: ``dy/dx`` to relate to - ``\Delta y/\Delta x``. This notation is very common, and especially - useful when more than one variable is involved. `SymPy` uses - Leibniz notation in some of its output, expressing somethings such - as: - -```math -f'(x) = \frac{d}{d\xi}(f(\xi)) \big|_{\xi=x}. -``` - - The notation - ``\big|`` - on the right-hand side separates the tasks of finding the - derivative and evaluating the derivative at a specific value. - -- Euler used `D` for the operator `D(f)`. This was initially used by - [Argobast](http://jeff560.tripod.com/calculus.html). The notation `D(f)(c)` would be needed to evaluate the derivative at a point. - -- Newton used a "dot" above the variable, ``\dot{x}(t)``, which is still widely used in physics to indicate a derivative in time. This inidicates take the derivative and then plug in ``t``. - -- The notation ``[expr]'(c)`` or ``[expr]'\big|_{x=c}``would similarly mean, take the derivative of the expression and **then** evaluate at ``c``. - - - - -## Rules of derivatives - -We could proceed in a similar manner -- using limits to find other -derivatives, but let's not. If we have a function $f(x) = x^5 -\sin(x)$, it would be nice to leverage our previous work on the -derivatives of $f(x) =x^5$ and $g(x) = \sin(x)$, rather than derive an -answer from scratch. - - -As with limits and continuity, it proves very useful to consider rules -that make the process of finding derivatives of combinations of -functions a matter of combining derivatives of the individual functions in some manner. - -We already have one such rule: - -### Power rule - -We have seen for integer ``n \geq 0`` the formula: - -```math -[x^n]' = n x^{n-1}. -``` - -This will be shown true for all real exponents. - -### Sum rule - -Let's consider $k(x) = a\cdot f(x) + b\cdot g(x)$, what is its derivative? That is, in terms of $f$, $g$ and their derivatives, can we express $k'(x)$? - -We can rearrange $(k(x+h) - k(x))$ as follows: - -```math -(a\cdot f(x+h) + b\cdot g(x+h)) - (a\cdot f(x) + b \cdot g(x)) = -a\cdot (f(x+h) - f(x)) + b \cdot (g(x+h) - g(x)). -``` - -Dividing by $h$, we see that this becomes - -```math -a\cdot \frac{f(x+h) - f(x)}{h} + b \cdot \frac{g(x+h) - g(x)}{h} \rightarrow a\cdot f'(x) + b\cdot g'(x). -``` - -That is ``[a\cdot f(x) + b \cdot g(x)]' = a\cdot f'(x) + b\cdot g'(x)``. - - -This holds two rules: the derivative of a constant times a function is -the constant times the derivative of the function; and the derivative -of a sum of functions is the sum of the derivative of the functions. - -This example shows a useful template: - -```math -\begin{align*} -[2x^2 - \frac{x}{3} + 3e^x]' & = 2[\square]' - \frac{[\square]]}{3} + 3[\square]'\\ -&= 2[x^2]' - \frac{[x]'}{3} + 3[e^x]'\\ -&= 2(2x) - \frac{1}{3} + 3e^x\\ -&= 4x - \frac{1}{3} + 3e^x -\end{align*} -``` - -### Product rule - -Other rules can be similarly derived. `SymPy` can give us them as -well. Here we define two symbolic functions `u` and `v` and let `SymPy` -derive a formula for the derivative of a product of functions: - -```julia; hold=true; -@syms u() v() -f(x) = u(x) * v(x) -limit((f(x+h) - f(x))/h, h => 0) -``` - -The output uses the Leibniz notation to represent that the derivative -of $u(x) \cdot v(x)$ is the $u$ times the derivative of $v$ evaluated -at ``x`` plus $v$ times the derivative of $u$ evaluated at ``x``. A -common shorthand is $[uv]' = u'v + uv'$. - -This example shows a useful template for the product rule: - -```math -\begin{align*} -[(x^2+1)\cdot e^x]' &= [\square]' \cdot (\square) + (\square) \cdot [\square]'\\ -&= [x^2 + 1]' \cdot (e^x) + (x^2+1) \cdot [e^x]'\\ -&= (2x)\cdot e^x + (x^2+1)\cdot e^x -\end{align*} -``` - - - -### Quotient rule - -The derivative of $f(x) = u(x)/v(x)$ - a ratio of functions - can be -similarly computed. The result will be $[u/v]' = (u'v - uv')/u^2$: - -```julia; hold=true; -@syms u() v() -f(x) = u(x) / v(x) -limit((f(x+h) - f(x))/h, h => 0) -``` - -This example shows a useful template for the quotient rule: - -```math -\begin{align*} -[\frac{x^2+1}{e^x}]' &= \frac{[\square]' \cdot (\square) - (\square) \cdot [\square]'}{(\square)^2}\\ -&= \frac{[x^2 + 1]' \cdot (e^x) - (x^2+1) \cdot [e^x]'}{(e^x)^2}\\ -&= \frac{(2x)\cdot e^x - (x^2+1)\cdot e^x}{e^{2x}} -\end{align*} -``` - -##### Examples - -Compute the derivative of ``f(x) = (1 + \sin(x)) + (1 + x^2)``. - -As written we can identify ``f(x) = u(x) + v(x)`` with -``u=(1 + \sin(x))``, ``v=(1 + x^2)``. The sum rule immediately applies to give: - -```math -f'(x) = (\cos(x)) + (2x). -``` - ----- - -Compute the derivative of ``f(x) = (1 + \sin(x)) \cdot (1 + x^2)``. - -The same ``u`` and ``v`` my be identified. The product rule readily applies to yield: - -```math -f'(x) = u'v + uv' = \cos(x) \cdot (1 + x^2) + (1 + \sin(x)) \cdot (2x). -``` - ----- - -Compute the derivative of ``f(x) = (1 + \sin(x)) / (1 + x^2)``. - -The same ``u`` and ``v`` my be identified. The quotient rule readily applies to yield: - -```math -f'(x) = u'v - uv' = \frac{\cos(x) \cdot (1 + x^2) - (1 + \sin(x)) \cdot (2x)}{(1+x^2)^2}. -``` - ----- - -Compute the derivative of ``f(x) = (x-1) \cdot (x-2)``. - -This can be done using the product rule *or* by expanding the polynomial and using the power and sum rule. As this polynomial is easy to expand, we do both and compare: - -```math -[(x-1)(x-2)]' = [x^2 - 3x + 2]' = 2x -3. -``` - -Whereas the product rule gives: - -```math -[(x-1)(x-2)]' = 1\cdot (x-2) + (x-1)\cdot 1 = 2x - 3. -``` - ----- - -Find the derivative of $f(x) = (x-1)(x-2)(x-3)(x-4)(x-5)$. - -We could expand this, as above, but without computer assistance the potential for error is high. Instead we will use the product rule on the product of ``5`` terms. - -Let's first treat the case of $3$ products: - -```math -[u\cdot v\cdot w]' =[ u \cdot (vw)]' = u' (vw) + u [vw]' = u'(vw) + u[v' w + v w'] = -u' vw + u v' w + uvw'. -``` - -This pattern generalizes, clearly, to: -```math -[f_1\cdot f_2 \cdots f_n]' = f_1' f_2 \cdots f_n + f_1 \cdot f_2' \cdot (f_3 \cdots f_n) + \dots + -f_1 \cdots f_{n-1} \cdot f_n'. -``` - -There are $n$ terms, each where one of the $f_i$s have a derivative. Were we to multiply top and bottom by $f_i$, we would get each term looks like: $f \cdot f_i'/f_i$. - -With this, we can proceed. Each term $x-i$ has derivative $1$, so the answer to $f'(x)$, with $f$ as above, is $f'(x) = f(x)/(x-1) + f(x)/(x-2) + f(x)/(x-3) + f(x)/(x-4) + f(x)/(x-5)$, that is: - -```math -f'(x) = (x-2)(x-3)(x-4)(x-5) + (x-1)(x-3)(x-4)(x-5) + (x-1)(x-2)(x-4)(x-5) + (x-1)(x-2)(x-3)(x-5) + (x-1)(x-2)(x-3)(x-4). -``` - ----- - -Find the derivative of ``x\sin(x)`` evaluated at ``\pi``. - -```math -[x\sin(x)]'\big|_{x=\pi} = (1\sin(x) + x\cos(x))\big|_{x=\pi} = (\sin(\pi) + \pi \cdot \cos(\pi)) = -\pi. -``` - -### Chain rule - -Finally, the derivative of a composition of functions can be computed -using pieces of each function. This gives a rule called the *chain -rule*. Before deriving, let's give a slight motivation. - - -Consider the output of a factory for some widget. It depends on two steps: -an initial manufacturing step and a finishing step. The number of -employees is important in how much is initially manufactured. Suppose -$x$ is the number of employees and $g(x)$ is the amount initially -manufactured. Adding more employees increases the amount made by the -made-up rule $g(x) = \sqrt{x}$. The finishing step depends on how much -is made by the employees. If $y$ is the amount made, then $f(y)$ is -the number of widgets finished. Suppose for some reason that $f(y) = -y^2.$ - -How many widgets are made as a function of employees? The composition -$u(x) = f(g(x))$ would provide that. Changes in the initial manufacturing step lead to changes in how much is initially made; changes in the initial amount made leads to changes in the finished products. Each change contributes to the overall change. - -What is the effect of adding employees on the rate of output of widgets? -In this specific case we know the answer, as $(f \circ g)(x) = x$, so -the answer is just the rate is $1$. - -In general, we want to express $\Delta f / \Delta x$ in a form so that we can take a limit. - -But what do we know? We know $\Delta g / \Delta x$ and $\Delta f/\Delta y$. Using $y=g(x)$, this suggests that we might have luck with the right side of this equation: - -```math -\frac{\Delta f}{\Delta x} = \frac{\Delta f}{\Delta y} \cdot \frac{\Delta y}{\Delta x}. -``` - - -Interpreting this, we get the *average* rate of change in the -composition can be thought of as a product: The *average* rate of -change of the initial step ($\Delta y/ \Delta x$) times the *average* -rate of the change of the second step evaluated not at $x$, but at -$y$, $\Delta f/ \Delta y$. - - -Re-expressing using derivative notation with $h$ would be: - - -```math -\frac{f(g(x+h)) - f(g(x))}{h} = \frac{f(g(x+h)) - f(g(x))}{g(x+h) - g(x)} \cdot \frac{g(x+h) - g(x)}{h}. -``` - -The left hand side will converge to the derivative of $u(x)$ or $[f(g(x))]'$. - -The right most part of the right side would have a limit $g'(x)$, were -we to let $h$ go to $0$. - -It isn't obvious, but the left part of the right side has the limit -$f'(g(x))$. This would be clear if *only* $g(x+h) = g(x) + h$, for -then the expression would be exactly the limit expression with -$c=g(x)$. But, alas, except to some hopeful students and some special -cases, it is definitely not the case in general that $g(x+h) = g(x) + h$ - that -right parentheses actually means something. However, it is *nearly* -the case that $g(x+h) = g(x) + kh$ for some $k$ and this can be used to formulate a -proof (one of the two detailed -[here](http://en.wikipedia.org/wiki/Chain_rule#Proofs) and [here](http://kruel.co/math/chainrule.pdf)). - - -Combined, we would end up with: - - -> *The chain rule*: $[f(g(x))]' = f'(g(x)) \cdot g'(x)$. That is the -> derivative of the outer function evaluated at the inner function -> times the derivative of the inner function. - - -To see that this works in our specific case, we assume the general -power rule that $[x^n]' = n x^{n-1}$ to get: - -```math -\begin{align*} -f(x) &= x^2 & g(x) &= \sqrt{x}\\ -f'(\square) &= 2(\square) & g'(x) &= \frac{1}{2}x^{-1/2} -\end{align*} -``` - -We use ``\square`` for the argument of `f'` to emphasize that ``g(x)`` is the needed value, not just ``x``: - -```math -\begin{align*} -[(\sqrt{x})^2]' &= [f(g(x)]'\\ -&= f'(g(x)) \cdot g'(x) \\ -&= 2(\sqrt{x}) \cdot \frac{1}{2}x^{-1/2}\\ -&= \frac{2\sqrt{x}}{2\sqrt{x}}\\ -&=1 -\end{align*} -``` - - -This is the same as the derivative of $x$ found by first evaluating the composition. For this problem, the chain rule is not necessary, but typically it is a needed rule to fully differentiate a function. - -##### Examples - -Find the derivative of ``f(x) = \sqrt{1 - x^2}``. We identify the composition of ``\sqrt{x}`` and ``(1-x^2)``. We set the functions and their derivatives into a pattern to emphasize the pieces in the chain-rule formula: - -```math -\begin{align*} -f(x) &=\sqrt{x} = x^{1/2} & g(x) &= 1 - x^2 \\ -f'(\square) &=(1/2)(\square)^{-1/2} & g'(x) &= -2x -\end{align*} -``` - -Then: - -```math -[f(g(x))]' = (1/2)(1-x^2)^{-1/2} \cdot (-2x). -``` - ----- - -Find the derivative of ``\log(2 + \sin(x))``. This is a composition ``\log(x)`` -- with derivative ``1/x`` and ``2 + \sin(x)`` -- with derivative ``\cos(x)``. We get ``(1/\sin(x)) \cos(x)``. - -In general, - -```math -[\log(f(x))]' \frac{f'(x)}{f(x)}. -``` - ----- - -Find the derivative of ``e^{f(x)}``. The inner function has derivative ``f'(x)``, the outer function has derivative ``e^x`` (the same as the outer function itself). We get for a derivative - -```math -[e^{f(x)}]' = e^{f(x)} \cdot f'(x). -``` - -This is a useful rule to remember for expressions involving exponentials. - ----- - -Find the derivative of ``\sin(x)\cos(2x)`` at ``x=\pi``. - -```math -[\sin(x)\cos(2x)]'\big|_{x=\pi} = -(\cos(x)\cos(2x) + \sin(x)(-\sin(2x)\cdot 2))\big|_{x=\pi} = -((-1)(1) + (0)(-0)(2)) = -1. -``` - -##### Proof of the Chain Rule - -A function is *differentiable* at $a$ if the following limit exists $\lim_{h \rightarrow 0}(f(a+h)-f(a))/h$. Reexpressing this as: $f(a+h) - f(a) - f'(a)h = \epsilon_f(h) h$ where as $h\rightarrow 0$, $\epsilon_f(h) \rightarrow 0$. Then, we have: - -```math -g(a+h) = g(a) + g'(a)h + \epsilon_g(h) h = g(a) + h', -``` - -Where $h' = (g'(a) + \epsilon_g(h))h \rightarrow 0$ as $h \rightarrow 0$ will be used to simplify the following: - - - -```math -\begin{align} -f(g(a+h)) - f(g(a)) &= -f(g(a) + g'(a)h + \epsilon_g(h)h) - f(g(a)) \\ -&= f(g(a)) + f'(g(a)) (g'(a)h + \epsilon_g(h)h) + \epsilon_f(h')(h') - f(g(a))\\ -&= f'(g(a)) g'(a)h + f'(g(a))(\epsilon_g(h)h) + \epsilon_f(h')(h'). -\end{align} -``` - -Rearranging: - -```math -f(g(a+h)) - f(g(a)) - f'(g(a)) g'(a) h = f'(g(a))\epsilon_g(h))h + \epsilon_f(h')(h') = -(f'(g(a)) \epsilon_g(h) + \epsilon_f(h')( (g'(a) + \epsilon_g(h))))h = -\epsilon(h)h, -``` - -where $\epsilon(h)$ combines the above terms which go to zero as $h\rightarrow 0$ into one. This is -the alternative definition of the derivative, showing $(f\circ g)'(a) = f'(g(a)) g'(a)$ when $g$ is differentiable at $a$ and $f$ is differentiable at $g(a)$. - -##### The "chain" rule - -The chain rule name could also be simply the "composition rule," as that is the operation the rule works for. However, in practice, there are usually *multiple* compositions, and the "chain" rule is used to chain together the different pieces. To get a sense, consider a triple composition ``u(v(w(x())))``. This will have derivative: - -```math -\begin{align*} -[u(v(w(x)))]' &= u'(v(w(x))) \cdot [v(w(x))]' \\ -&= u'(v(w(x))) \cdot v'(w(x)) \cdot w'(x) -\end{align*} -``` - -The answer can be viewed as a repeated peeling off of the outer -function, a view with immediate application to many compositions. To -see that in action with an expression, consider this derivative -problem, shown in steps: - -```math -\begin{align*} -[\sin(e^{\cos(x^2-x)})]' -&= \cos(e^{\cos(x^2-x)}) \cdot [e^{\cos(x^2-x)}]'\\ -&= \cos(e^{\cos(x^2-x)}) \cdot e^{\cos(x^2-x)} \cdot [\cos(x^2-x)]'\\ -&= \cos(e^{\cos(x^2-x)}) \cdot e^{\cos(x^2-x)} \cdot (-\sin(x^2-x)) \cdot [x^2-x]'\\ -&= \cos(e^{\cos(x^2-x)}) \cdot e^{\cos(x^2-x)} \cdot (-\sin(x^2-x)) \cdot (2x-1)\\ -\end{align*} -``` - - -##### More examples of differentiation - -Find the derivative of $x^5 \cdot \sin(x)$. - -This is a product of functions, using $[u\cdot v]' = u'v + uv'$ we get: - -```math -5x^4 \cdot \sin(x) + x^5 \cdot \cos(x) -``` ----- - -Find the derivative of $x^5 / \sin(x)$. - -This is a quotient of functions. Using $[u/v]' = (u'v - uv')/v^2$ we get - -```math -(5x^4 \cdot \sin(x) - x^5 \cdot \cos(x)) / (\sin(x))^2. -``` - ----- - -Find the derivative of $\sin(x^5)$. This is a composition of -functions $u(v(x))$ with $v(x) = x^5$. The chain rule says find the -derivative of $u$ ($\cos(x)$) and evaluate at $v(x)$ ($\cos(x^5)$) -then multiply by the derivative of $v$: - -```math -\cos(x^5) \cdot 5x^4. -``` - ----- - -Similarly, but differently, find the derivative of $\sin(x)^5$. Now -$v(x) = \sin(x)$, so the derivative of $u(x)$ ($5x^4$) evaluated at -$v(x)$ is $5(\sin(x))^4$ so multiplying by $v'$ gives: - -```math -5(\sin(x))^4 \cdot \cos(x) -``` - ----- - -We can verify these with `SymPy`. Rather than take a limit, we will -use `SymPy`'s `diff` function to compute derivatives. - -```julia -diff(x^5 * sin(x)) -``` - -```julia -diff(x^5/sin(x)) -``` - -```julia -diff(sin(x^5)) -``` - -and finally, - -```julia -diff(sin(x)^5) -``` - -!!! note - The `diff` function can be called as `diff(ex)` when there is - just one free variable, as in the above examples; as `diff(ex, - var)` when there are parameters in the expression. - ----- - -The general product rule: For any $n$ - not just integer values - we can re-express $x^n$ using $e$: $x^n = e^{n \log(x)}$. Now the chain rule can be applied: - -```math -[x^n]' = [e^{n\log(x)}]' = e^{n\log(x)} \cdot (n \frac{1}{x}) = n x^n \cdot \frac{1}{x} = n x^{n-1}. -``` - ----- - -Find the derivative of $f(x) = x^3 (1-x)^2$ using either the power rule or the sum rule. - -The power rule expresses $f=u\cdot v$. With $u(x)=x^3$ and $v(x)=(1-x)^2$ we get: - -```math -u'(x) = 3x^2, \quad v'(x) = 2 \cdot (1-x)^1 \cdot (-1), -``` - -the last by the chain rule. Combining with $u' v + u v'$ we get: -$f'(x) = (3x^2)\cdot (1-x)^2 + x^3 \cdot (-2) \cdot (1-x)$. - -Otherwise, the polynomial can be expanded to give -$f(x)=x^5-2x^4+x^3$ which has derivative $f'(x) = 5x^4 - 8x^3 + 3x^2$. - ----- - -Find the derivative of $f(x) = x \cdot e^{-x^2}$. - -Using the product rule and then the chain rule, we have: - -```math -\begin{align} -f'(x) &= [x \cdot e^{-x^2}]'\\ -&= [x]' \cdot e^{-x^2} + x \cdot [e^{-x^2}]'\\ -&= 1 \cdot e^{-x^2} + x \cdot (e^{-x^2}) \cdot [-x^2]'\\ -&= e^{-x^2} + x \cdot e^{-x^2} \cdot (-2x)\\ -&= e^{-x^2} (1 - 2x^2). -\end{align} -``` - ----- - -Find the derivative of $f(x) = e^{-ax} \cdot \sin(x)$. - -Using the product rule and then the chain rule, we have: - -```math -\begin{align} -f'(x) &= [e^{-ax} \cdot \sin(x)]'\\ -&= [e^{-ax}]' \cdot \sin(x) + e^{-ax} \cdot [\sin(x)]'\\ -&= e^{-ax} \cdot [-ax]' \cdot \sin(x) + e^{-ax} \cdot \cos(x)\\ -&= e^{-ax} \cdot (-a) \cdot \sin(x) + e^{-ax} \cos(x)\\ -&= e^{-ax}(\cos(x) - a\sin(x)). -\end{align} -``` - ----- - -Find the derivative of ``e^{-x^2/2}`` at ``x=1``. - -```math -[e^{-x^2/2}]'\big|_{x=1} = -(e^{-x^2/2} \cdot \frac{-2x}{2}) \big|_{x=1} = -e^{-1/2} \cdot (-1) = -e^{-1/2}. -``` - -##### Example: derivative of inverse functions - - -Suppose we knew that $\log(x)$ had derivative of $1/x$, but didn't know the derivative of $e^x$. From their inverse relation, we have: $x=\log(e^x)$, so taking derivatives of both sides would yield: - -```math -1 = (\frac{1}{e^x}) \cdot [e^x]'. -``` - -Or solving, $[e^x]' = e^x$. This is a general strategy to find the -derivative of an *inverse* function. - - -The graph of an inverse function is related to the graph of the function through the symmetry ``y=x``. - -For example, the graph of ``e^x`` and ``\log(x)`` have this symmetry, emphasized below: - -```julia;hold=true;echo=false; -f(x) = exp(x) -f′(x) = exp(x) -f⁻¹(x) = log(x) # using Unicode typed with "f^\-^\1" -xs = range(0, 2, length=25) -ys = f.(xs) -plot(f, 0, 2, aspect_ratio=:equal, xlim=(0,8), ylim=(0,8), legend=false) -scatter!(xs, ys) -plot!(f⁻¹, extrema(ys)...) -scatter!(ys, xs, color=:blue) -plot!(identity, linestyle=:dot) # the line y=x -x₀, y₀ = xs[13], ys[13] -plot!([x₀, y₀],[y₀, x₀], linestyle=:dash) -ys′ = @. y₀ + f(x₀)*(xs - x₀) -plot!(xs, ys′, linestyle=:dash) -g(y) = 1/f′(f⁻¹(y)) -xs′ = @. x₀ + g(y₀) * (ys - y₀) -plot!(ys, xs′, linestyle=:dash) -``` - -The point ``(1, e)`` on the graph of ``e^x`` matches the point ``(e, 1)`` on the graph of the inverse function, ``\log(x)``. The slope of the tangent line at ``x=1`` to ``e^x`` is given by ``e`` as well. What is the slope of the tangent line to ``\log(x)`` at ``x=e``? - -As seen, the value can be computed, but how? - -Finding the derivative of the inverse function can be achieved from the chain rule using the identify ``f^{-1}(f(x)) = x`` for all ``x`` in the domain of ``f``. - -The chain rule applied to both sides, yields: - -```math -1 = [f^{-1}]'(f(x)) \cdot f'(x) -``` - -Solving, we see that ``[f^{-1}]'(f(x)) = 1/f'(x)``. To emphasize the evaluation of the derivative of the inverse function at ``f(x)`` we might write: - -```math -\frac{d}{du} (f^{-1}(u)) \big|_{u=f(x)} = \frac{1}{f'(x)} -``` - -So the reciprocal of the slope of the tangent line of ``f`` at the mirror image point. In the above, we see if the slope of the tangent line at ``(1,e)`` to ``f`` is ``e``, then the slope of the tangent line to ``f^{-1}(x)`` at ``(e,1)`` would be ``1/e``. - -#### Rules of derivatives and some sample functions - -This table summarizes the rules of derivatives that allow derivatives -of more complicated expressions to be computed with the derivatives of -their pieces. - -```julia; hold=true; echo=false -nm = ["Power rule", "constant", "sum/difference", "product", "quotient", "chain"] -rule = [L"[x^n]' = n\cdot x^{n-1}", - L"[cf(x)]' = c \cdot f'(x)", - L"[f(x) \pm g(x)]' = f'(x) \pm g'(x)", - L"[f(x) \cdot g(x)]' = f'(x)\cdot g(x) + f(x) \cdot g'(x)", - L"[f(x)/g(x)]' = (f'(x) \cdot g(x) - f(x) \cdot g'(x)) / g(x)^2", - L"[f(g(x))]' = f'(g(x)) \cdot g'(x)"] -d = DataFrame(Name=nm, Rule=rule) -table(d) -``` - -This table gives some useful derivatives: - -```julia; hold=true; echo=false -fn = [L"x^n (\text{ all } n)", -L"e^x", -L"\log(x)", -L"\sin(x)", -L"\cos(x)"] -a = [L"nx^{n-1}", -L"e^x", -L"1/x", -L"\cos(x)", -L"-\sin(x)"] -d = DataFrame(Function=fn, Derivative=a) -table(d) -``` - -## Higher-order derivatives - -The derivative of a function is an operator, it takes a function and -returns a new, derived, function. We could repeat this -operation. The result is called a higher-order derivative. The -Lagrange notation uses additional "primes" to indicate how many. So -$f''(x)$ is the second derivative and $f'''(x)$ the third. For even -higher orders, sometimes the notation is $f^{(n)}(x)$ to indicate an -$n$th derivative. - - -##### Examples - -Find the first ``3`` derivatives of ``f(x) = ax^3 + bx^2 + cx + d``. - -Differentiating a polynomial is done with the sum rule, here we repeat three times: - -```math -\begin{align} -f(x) &= ax^3 + bx^2 + cx + d\\ -f'(x) &= 3ax^2 + 2bx + c \\ -f''(x) &= 3\cdot 2 a x + 2b \\ -f'''(x) &= 6a -\end{align} -``` - -We can see, the fourth derivative -- and all higher order ones -- would be identically ``0``. This is part of a general phenomenon: an ``n``th degree polynomial has only ``n`` non-zero derivatives. - - ----- - -Find the first ``5`` derivatives of ``\sin(x)``. - -```math -\begin{align} -f(x) &= \sin(x) \\ -f'(x) &= \cos(x) \\ -f''(x) &= -\sin(x) \\ -f'''(x) &= -\cos(x) \\ -f^{(4)} &= \sin(x) \\ -f^{(5)} &= \cos(x) -\end{align} -``` - -We see the derivatives repeat themselves. (We also see alternative notation for higher order derivatives.) - - ----- - -Find the second derivative of $e^{-x^2}$. - -We need the chain rule *and* the product rule: - -```math -[e^{-x^2}]'' = [e^{-x^2} \cdot (-2x)]' = \left(e^{-x^2} \cdot (-2x)\right) \cdot(-2x) + e^{-x^2} \cdot (-2) = -e^{-x^2}(4x^2 - 2). -``` - -This can be verified: - -```julia -diff(diff(exp(-x^2))) |> simplify -``` - -Having to iterate the use of `diff` is cumbersome. An alternate notation is either specifying the variable twice: `diff(ex, x, x)` or using a number after the variable: `diff(ex, x, 2)`: - - -```julia -diff(exp(-x^2), x, x) |> simplify -``` - -Higher-order derivatives can become involved when the product or quotient rules becomes involved. - - -## Questions - -###### Question - -The derivative at $c$ is the slope of the tangent line at $x=c$. Answer the following based on this graph: - -```julia -fn = x -> -x*exp(x)*sin(pi*x) -plot(fn, 0, 2) -``` - -At which of these points $c= 1/2, 1, 3/2$ is the derivative negative? - -```julia; hold=true; echo=false -choices = ["``1/2``", "``1``", "``3/2``"] -answ = 1 -radioq(choices, answ, keep_order=true) -``` - -Which value looks bigger from reading the graph: - -```julia; hold=true; echo=false -choices = ["``f(1)``", "``f(3/2)``"] -answ = 2 -radioq(choices, answ, keep_order=true) -``` - -At $0.708 \dots$ and $1.65\dots$ the derivative has a common value. What is it? - -```julia; hold=true; echo=false -numericq(0, 1e-2) -``` - -###### Question - -Consider the graph of the `airyai` function (from `SpecialFunctions`) over $[-5, 5]$. - -```julia; hold=true;echo=false -plot(airyai, -5, 5) -``` - -At $x = -2.5$ the derivative is postive or negative? - -```julia; hold=true; echo=false -choices = ["positive", "negative"] -answ = 1 -radioq(choices, answ, keep_order=true) -``` - - -At $x=0$ the derivative is postive or negative? - -```julia; hold=true; echo=false -choices = ["positive", "negative"] -answ = 2 -radioq(choices, answ, keep_order=true) -``` - -At $x = 2.5$ the derivative is postive or negative? - -```julia; hold=true; echo=false -choices = ["positive", "negative"] -answ = 2 -radioq(choices, answ, keep_order=true) -``` - -###### Question - -Compute the derivative of $e^x$ using `limit`. What do you get? - -```julia; hold=true; echo=false -choices = ["``e^x``", "``x^e``", "``(e-1)x^e``", "``e x^{(e-1)}``", "something else"] -answ = 1 -radioq(choices, answ, keep_order=true) -``` - -###### Question - -Compute the derivative of $x^e$ using `limit`. What do you get? - -```julia; hold=true; echo=false -choices = ["``e^x``", "``x^e``", "``(e-1)x^e``", "``e x^{(e-1)}``", "something else"] -answ = 5 -radioq(choices, answ, keep_order=true) -``` - -###### Question - -Compute the derivative of $e^{e\cdot x}$ using `limit`. What do you get? - -```julia; hold=true; echo=false -choices = ["``e^x``", "``x^e``", "``(e-1)x^e``", "``e x^{(e-1)}``", "``e \\cdot e^{e\\cdot x}``", "something else"] -answ = 5 -radioq(choices, answ, keep_order=true) -``` - -###### Question - -In the derivation of the derivative of $\sin(x)$, the following limit is needed: - -```math -L = \lim_{h \rightarrow 0} \frac{\cos(h) - 1}{h}. -``` - -This is - -```julia; hold=true; echo=false -choices = [ -L" $1$, as this is clearly the analog of the limit of $\sin(h)/h$.", -L"Does not exist. The answer is $0/0$ which is undefined", -L" $0$, as this expression is the derivative of cosine at $0$. The answer follows, as cosine clearly has a tangent line with slope $0$ at $x=0$."] -answ = 3 -radioq(choices, answ) -``` - -###### Question - -Let $f(x) = (e^x + e^{-x})/2$ and $g(x) = (e^x - e^{-x})/2$. Which is true? - -```julia; hold=true; echo=false -choices = [ -"``f'(x) = g(x)``", -"``f'(x) = -g(x)``", -"``f'(x) = f(x)``", -"``f'(x) = -f(x)``" -] -answ= 1 -radioq(choices, answ) -``` - - - -###### Question - -Let $f(x) = (e^x + e^{-x})/2$ and $g(x) = (e^x - e^{-x})/2$. Which is true? - -```julia; hold=true; echo=false -choices = [ -"``f''(x) = g(x)``", -"``f''(x) = -g(x)``", -"``f''(x) = f(x)``", -"``f''(x) = -f(x)``"] -answ= 3 -radioq(choices, answ) -``` - - - - - -###### Question - -Consider the function $f$ and its transformation $g(x) = a + f(x)$ -(shift up by $a$). Do $f$ and $g$ have the same derivative? - -```julia; hold=true; echo=false -yesnoq("yes") -``` - -Consider the function $f$ and its transformation $g(x) = f(x - a)$ -(shift right by $a$). Do $f$ and $g$ have the same derivative? - -```julia; hold=true; echo=false -yesnoq("no") -``` - - - -Consider the function $f$ and its transformation $g(x) = f(x - a)$ -(shift right by $a$). Is $f'$ at $x$ equal to $g'$ at $x-a$? - -```julia; hold=true; echo=false -yesnoq("yes") -``` - -Consider the function $f$ and its transformation $g(x) = c f(x)$, $c > -1$. Do $f$ and $g$ have the same derivative? - -```julia; hold=true; echo=false -yesnoq("no") -``` - -Consider the function $f$ and its transformation $g(x) = f(x/c)$, $c > -1$. Do $f$ and $g$ have the same derivative? - -```julia; hold=true; echo=false -yesnoq("no") -``` - -Which of the following is true? - -```julia; hold=true; echo=false -choices = [ -L"If the graphs of $f$ and $g$ are translations up and down, the tangent line at corresponding points is unchanged.", -L"If the graphs of $f$ and $g$ are rescalings of each other through $g(x)=f(x/c)$, $c > 1$. Then the tangent line for corresponding points is the same.", -L"If the graphs of $f$ and $g$ are rescalings of each other through $g(x)=cf(x)$, $c > 1$. Then the tangent line for corresponding points is the same." -] -answ = 1 -radioq(choices, answ) -``` - - - -###### Question - -The rate of change of volume with respect to height is $3h$. The rate of change of height with respect to time is $2t$. At at $t=3$ the height is $h=14$ what is the rate of change of volume with respect to time when $t=3$? - -```julia; hold=true; echo=false -## dv/dt = dv/dh * dh/dt = 3h * 2t -h = 14; t=3 -val = (3*h) * (2*t) -numericq(val) -``` - - -###### Question - -Which equation below is $f(x) = \sin(k\cdot x)$ a solution of ($k > 1$)? - -```julia; hold=true; echo=false -choices = [ -"``f'(x) = k^2 \\cdot f(x)``", -"``f'(x) = -k^2 \\cdot f(x)``", -"``f''(x) = k^2 \\cdot f(x)``", -"``f''(x) = -k^2 \\cdot f(x)``"] -answ = 4 -radioq(choices, answ) -``` - -###### Question - -Let $f(x) = e^{k\cdot x}$, $k > 1$. Which equation below is $f(x)$ a solution of? - - -```julia; hold=true; echo=false -choices = [ -"``f'(x) = k^2 \\cdot f(x)``", -"``f'(x) = -k^2 \\cdot f(x)``", -"``f''(x) = k^2 \\cdot f(x)``", -"``f''(x) = -k^2 \\cdot f(x)``"] -answ = 3 -radioq(choices, answ) -``` - -##### Question - -Their are ``6`` trig functions. The derivatives of ``\sin(x)`` and ``\cos(x)`` should be memorized. The others can be derived if not memorized using the quotient rule or chain rule. - -What is ``[\tan(x)]'``? (Use ``\tan(x) = \sin(x)/\cos(x)``.) - -```julia; echo=false -trig_choices = [ -"``\\sec^2(x)``", -"``\\sec(x)\\tan(x)``", -"``-\\csc^2(x)``", -"``-\\csc(x)\\cot(x)``" -] -radioq(trig_choices, 1) -``` - -What is ``[\cot(x)]'``? (Use ``\tan(x) = \cos(x)/\sin(x)``.) - -```julia; echo=false -radioq(trig_choices, 3) -``` - -What is ``[\sec(x)]'``? (Use ``\sec(x) = 1/\cos(x)``.) - -```julia; echo=false -radioq(trig_choices, 2) -``` - -What is ``[\csc(x)]'``? (Use ``\csc(x) = 1/\sin(x)``.) - -```julia; echo=false -radioq(trig_choices, 4) -``` - -##### Question - -Consider this picture of composition: - -```julia; hold=true; echo=false -f(x) = sin(x) -g(x) = exp(x) -a,b = 0, 1.55 - -xs = range(a, b, length=100) -ys = g.(xs) -us = range(extrema(ys)..., length=100) -vs = f.(us) -pf = plot(vs, us, ylim=extrema(ys), ymirror=true, legend=false) -xs′ = range(0.5, 1.5, length=100) -us′ = g.(xs′) -vs′ = [f(g(1) + g'(1)*(x-1)) for x ∈ xs′] -plot!(pf, vs′, us′) - -plot!(pf, [1, f(g(1)), f(g(1))], [g(1), g(1), 1]) -quiver!(pf, [.75, f(g(1))], [g(1), 2], quiver=([-.01, 0],[0, -0.1])) -pg = plot(xs, ys, ylim=extrema(ys), legend=false) -plot!(pg, [1,1,0],[1, g(1),g(1)]) -quiver!(pg, [1, 0.5], [2, g(1)], quiver=([0, -0.1],[0.1, 0])) -plot!(tangent(g,1)) -l = @layout [a b] -plot(pf, pg, layout=l) -``` - -The right graph is of ``g(x) = \exp(x)`` at ``x=1``, the left graph of ``f(x) = \sin(x)`` *rotated* ``90`` degrees counter-clockwise. Chasing the arrows shows graphically how ``f(g(1))`` can be computed. The nearby values ``f(g(1+h))`` are -- using the tangent line of ``g`` at ``x-1`` -- approximated by ``f(g(1) + g'(1)\cdot h)``, as shown in the graph segment on the left. - -Assuming the approximation gets better for ``h`` close to ``0``, as it visually does, the derivative at ``1`` for ``f(g(x))`` should be given by this limit: - - -```math -\begin{align*} -\frac{d(f\circ g)}{dx}\mid_{x=1} -&= \lim_{h\rightarrow 0} \frac{f(g(1) + g'(1)h)-f(g(1))}{h}\\ -&= \lim_{h\rightarrow 0} \frac{f(g(1) + g'(1)h)-f(g(1))}{h}\\ -&= \lim_{h\rightarrow 0} \frac{f(g(1) + g'(1)h)-f(g(1))}{g'(1)h} \cdot g'(1)\\ -&= \lim_{h\rightarrow 0} (f\circ g)'(1) \cdot g'(1). -\end{align*} -``` - -What limit law, described below assuming all limits exist. allows the last equals sign? - -```julia; hold=true; echo=false -choices = [ -raw""" -The limit of a sum is the sum of the limits: -``\lim_{x\rightarrow c}(au(x)+bv(x)) = a\lim_{x\rightarrow c}u(x) + b\lim_{x\rightarrow c}v(x)`` -""", -raw""" -The limit of a product is the product of the limits: -``\lim_{x\rightarrow c}(u(x)\cdot v(x)) = \lim_{x\rightarrow c}u(x) \cdot \lim_{x\rightarrow c}v(x)`` -""", -raw""" -The limit of a composition (under assumptions on ``v``): -``\lim_{x \rightarrow c}u(v(x)) = \lim_{w \rightarrow \lim_{x \rightarrow c}v(x)} u(w)``. -""" -] -radioq(choices, 3, keep_order=true) -``` diff --git a/CwJ/derivatives/figures/extrema-ladder.png b/CwJ/derivatives/figures/extrema-ladder.png deleted file mode 100644 index 8654ffed1734a83a359072c74ae9c5b8d78ded42..0000000000000000000000000000000000000000 GIT binary patch literal 0 HcmV?d00001 literal 15077 zcmeIZcRbbq`#*jhJK05&y`oSmbjTQWD}W1vW4&U zJg3+D{r-GzzrTOq?;mfsn?H`8=ka`u>wdrP*Zq1FqM@cpdFc2d6bePDbX86Zg(8?h ze#uDSFFx4DXcUTw+xE&84W%nr*fs9mv9WcqMxn%MAL)+Y8P+)>6>>)_lR=u@KruM< zLy4}o@oVPeMIT9&BC~!dGi1#V3UE(j#6t3zNr&gF^3}Zp#7O=S$b2QFzADKqc{hl6J5_zS32%x%bo@yg!I_C+l3V^J%{~S>I{9IDT$$Q*{K6xGn4a;yp=wqRaO@I_k(ozR=3o1Kf3ETLp=KVJOYSKHuf@_aJ#C4^JG 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Redo in Julia?? -plot.new() -plot.window(xlim=c(0,1), ylim=c(-5, 1.1)) -x <- seq(.1, .9, length=9) -y <- c(-4.46262,-4.46866, -4.47268, -4.47469, -4.47468, -4.47267, -4.46864, -4.4626 , -4.45454) -lines(c(0, x[3], 1), c(0, y[3], 1)) -points(c(0,1), c(0,1), pch=16, cex=2) -text(c(0,1), c(0,1), c("(0,0)", c("(a,b)")), pos=3) - -lines(c(0, x[3], x[3]), c(0, 0, y[3]), cex=2, col="gray") -lines(c(1, x[3], x[3]), c(1, 1, y[3]), cex=2, col="gray") -text(x[3]/2, 0, "x", pos=1) -text(x[3], y[3]/2, "|y|", pos=2) -text(x[3], (1 + y[3])/2, "b-y", pos=4) -text((x[3] + 1)/2, 1, "a-x", pos=1) - -text(x[3], y[3], "0", cex=4, col="gold") diff --git a/CwJ/derivatives/figures/extrema-ring-string.png b/CwJ/derivatives/figures/extrema-ring-string.png deleted file mode 100644 index bd422b10ddd31209718df2b4b2af755a3262b347..0000000000000000000000000000000000000000 GIT binary patch literal 0 HcmV?d00001 literal 18840 zcmeIaWms10w>=Dx4-zUVC<4-5D$)|tjev9uf=GA6gQ6fPT`D0h(t;oj7TqnW(k&qk 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i;2J%@x_9UC-MfF*Bu-hJ=dWNH@#oHH_5XJ&tN#zRRZ$KA diff --git a/CwJ/ODEs/differential_equations.jmd b/CwJ/ODEs/differential_equations.jmd deleted file mode 100644 index 38987c7..0000000 --- a/CwJ/ODEs/differential_equations.jmd +++ /dev/null @@ -1,376 +0,0 @@ -# The `DifferentialEquations` suite - -This section uses these add-on packages: - -```julia -using OrdinaryDiffEq -using Plots -using ModelingToolkit -``` - -```julia; echo=false; results="hidden" -using CalculusWithJulia.WeaveSupport - -const frontmatter = ( - title = "The `DifferentialEquations` suite", - description = "Calculus with Julia: The `DifferentialEquations` suite", - tags = ["CalculusWithJulia", "odes", "the `differentialequations` suite"], -); -fig_size = (800, 600) -nothing -``` - ----- - -The -[`DifferentialEquations`](https://github.com/SciML/DifferentialEquations.jl) -suite of packages contains solvers for a wide range of various -differential equations. This section just briefly touches touch on -ordinary differential equations (ODEs), and so relies only on -`OrdinaryDiffEq` part of the suite. For more detail on this type and -many others covered by the suite of packages, there are many other -resources, including the -[documentation](https://diffeq.sciml.ai/stable/) and accompanying -[tutorials](https://github.com/SciML/SciMLTutorials.jl). - -## SIR Model - -We follow along with an introduction to the SIR model for the spread of disease by [Smith and Moore](https://www.maa.org/press/periodicals/loci/joma/the-sir-model-for-spread-of-disease-introduction). This model received a workout due to the COVID-19 pandemic. - -The basic model breaks a population into three cohorts: The **susceptible** individuals, the **infected** individuals, and the **recovered** individuals. These add to the population size, ``N``, which is fixed, but the cohort sizes vary in time. We name these cohort sizes ``S(t)``, ``I(t)``, and ``R(t)`` and define ``s(t)=S(t)/N``, ``i(t) = I(t)/N`` and ``r(t) = R(t)/N`` to be the respective proportions. - -The following *assumptions* are made about these cohorts by Smith and Moore: - -> No one is added to the susceptible group, since we are ignoring births and immigration. The only way an individual leaves the susceptible group is by becoming infected. - - -This implies the rate of change in time of ``S(t)`` depends on the current number of susceptibles, and the amount of interaction with the infected cohorts. The model *assumes* each infected person has ``b`` contacts per day that are sufficient to spread the disease. Not all contacts will be with susceptible people, but if people are assumed to mix within the cohorts, then there will be on average ``b \cdot S(t)/N`` contacts with susceptible people per infected person. As each infected person is modeled identically, the time rate of change of ``S(t)`` is: - -```math -\frac{dS}{dt} = - b \cdot \frac{S(t)}{N} \cdot I(t) = -b \cdot s(t) \cdot I(t) -``` - -It is negative, as no one is added, only taken off. After dividing by -``N``, this can also be expressed as ``s'(t) = -b s(t) i(t)``. - -> assume that a fixed fraction ``k`` of the infected group will recover during any given day. - -This means the change in time of the recovered depends on ``k`` and the number infected, giving rise to the equation - -```math -\frac{dR}{dt} = k \cdot I(t) -``` - -which can also be expressed in proportions as ``r'(t) = k \cdot i(t)``. - -Finally, from ``S(t) + I(T) + R(t) = N`` we have ``S'(T) + I'(t) + R'(t) = 0`` or ``s'(t) + i'(t) + r'(t) = 0``. - - -Combining, it is possible to express the rate of change of the infected population through: - -```math -\frac{di}{dt} = b \cdot s(t) \cdot i(t) - k \cdot i(t) -``` - -The author's apply this model to flu statistics from Hong Kong where: - -```math -\begin{align*} -S(0) &= 7,900,000\\ -I(0) &= 10\\ -R(0) &= 0\\ -\end{align*} -``` - -In `Julia` we define these, `N` to model the total population, and `u0` to be the proportions. - -```julia -S0, I0, R0 = 7_900_000, 10, 0 -N = S0 + I0 + R0 -u0 = [S0, I0, R0]/N # initial proportions -``` - -An *estimated* set of values for ``k`` and ``b`` are ``k=1/3``, coming from the average period of infectiousness being estimated at three days and ``b=1/2``, which seems low in normal times, but not for an infected person who may be feeling quite ill and staying at home. (The model for COVID would certainly have a larger ``b`` value). - -Okay, the mathematical modeling is done; now we try to solve for the unknown functions using `DifferentialEquations`. - -To warm up, if ``b=0`` then ``i'(t) = -k \cdot i(t)`` describes the infected. (There is no circulation of people in this case.) The solution would be achieved through: - -```julia; hold=true -k = 1/3 - -f(u,p,t) = -k * u # solving u′(t) = - k u(t) -time_span = (0.0, 20.0) - -prob = ODEProblem(f, I0/N, time_span) -sol = solve(prob, Tsit5(), reltol=1e-8, abstol=1e-8) - -plot(sol) -``` - -The `sol` object is a set of numbers with a convenient `plot` method. As may have been expected, this graph shows exponential decay. - - -A few comments are in order. The problem we want to solve is - -```math -\frac{di}{dt} = -k \cdot i(t) = F(i(t), k, t) -``` - -where ``F`` depends on the current value (``i``), a parameter (``k``), and the time (``t``). We did not utilize ``p`` above for the parameter, as it was easy not to, but could have, and will in the following. The time variable ``t`` does not appear by itself in our equation, so only `f(u, p, t) = -k * u` was used, `u` the generic name for a solution which in this case is ``i``. - -The problem we set up needs an initial value (the ``u0``) and a time span to solve over. Here we want time to model real time, so use floating point values. - -The plot shows steady decay, as there is no mixing of infected with others. - -Adding in the interaction requires a bit more work. We now have what is known as a *system* of equations: - -```math -\begin{align*} -\frac{ds}{dt} &= -b \cdot s(t) \cdot i(t)\\ -\frac{di}{dt} &= b \cdot s(t) \cdot i(t) - k \cdot i(t)\\ -\frac{dr}{dt} &= k \cdot i(t)\\ -\end{align*} -``` - -Systems of equations can be solved in a similar manner as a single ordinary differential equation, though adjustments are made to accommodate the multiple functions. - -We use a style that updates values in place, and note that `u` now holds ``3`` different functions at once: - -```julia -function sir!(du, u, p, t) - k, b = p - s, i, r = u[1], u[2], u[3] - - ds = -b * s * i - di = b * s * i - k * i - dr = k * i - - du[1], du[2], du[3] = ds, di, dr -end -``` - -The notation `du` is suggestive of both the derivative and a small increment. The mathematical formulation follows the derivative, the numeric solution uses a time step and increments the solution over this time step. The `Tsit5()` solver, used here, adaptively chooses a time step, `dt`; were the `Euler` method used, this time step would need to be explicit. - -!!! note "Mutation not re-binding" - The `sir!` function has the trailing `!` indicating -- by convention -- it *mutates* its first value, `du`. In this case, through an assignment, as in `du[1]=ds`. This could use some explanation. The *binding* `du` refers to the *container* holding the ``3`` values, whereas `du[1]` refers to the first value in that container. So `du[1]=ds` changes the first value, but not the *binding* of `du` to the container. That is, `du` mutates. This would be quite different were the call `du = [ds,di,dr]` which would create a new *binding* to a new container and not mutate the values in the original container. - -With the update function defined, the problem is setup and a solution found with in the same manner: - -```julia; -p = (k=1/3, b=1/2) # parameters -time_span = (0.0, 150.0) # time span to solve over, 5 months - -prob = ODEProblem(sir!, u0, time_span, p) -sol = solve(prob, Tsit5()) - -plot(sol) -plot!(x -> 0.5, linewidth=2) # mark 50% line -``` - -The lower graph shows the number of infected at each day over the five-month period displayed. The peak is around 6-7% of the population at any one time. However, over time the recovered part of the population reaches over 50%, meaning more than half the population is modeled as getting sick. - - -Now we change the parameter ``b`` and observe the difference. We passed in a value `p` holding our two parameters, so we just need to change that and run the model again: - -```julia; hold=true -p = (k=1/2, b=2) # change b from 1/2 to 2 -- more daily contact -prob = ODEProblem(sir!, u0, time_span, p) -sol = solve(prob, Tsit5()) - -plot(sol) -``` - -The graphs are somewhat similar, but the steady state is reached much more quickly and nearly everyone became infected. - - -What about if ``k`` were bigger? - -```julia; hold=true -p = (k=2/3, b=1/2) -prob = ODEProblem(sir!, u0, time_span, p) -sol = solve(prob, Tsit5()) - -plot(sol) -``` - - -The graphs show that under these conditions the infections never take off; we have ``i' = (b\cdot s-k)i = k\cdot((b/k) s - 1) i`` which is always negative, since `(b/k)s < 1`, so infections will only decay. - - -The solution object is indexed by time, then has the `s`, `i`, `r` estimates. We use this structure below to return the estimated proportion of recovered individuals at the end of the time span. - -```julia -function recovered(k,b) - prob = ODEProblem(sir!, u0, time_span, (k,b)); - sol = solve(prob, Tsit5()); - s,i,r = last(sol) - r -end -``` - -This function makes it easy to see the impact of changing the parameters. For example, fixing ``k=1/3`` we have: - -```julia -f(b) = recovered(1/3, b) -plot(f, 0, 2) -``` - -This very clearly shows the sharp dependence on the value of ``b``; below some level, the proportion of people who are ever infected (the recovered cohort) remains near ``0``; above that level it can climb quickly towards ``1``. - -The function `recovered` is of two variables returning a single value. In subsequent sections we will see a few ``3``-dimensional plots that are common for such functions, here we skip ahead and show how to visualize multiple function plots at once using "`z`" values in a graph. - -```julia; hold=true -k, ks = 0.1, 0.2:0.1:0.9 # first `k` and then the rest -bs = range(0, 2, length=100) -zs = recovered.(k, bs) # find values for fixed k, each of bs -p = plot(bs, k*one.(bs), zs, legend=false) # k*one.(ks) is [k,k,...,k] -for k in ks - plot!(p, bs, k*one.(bs), recovered.(k, bs)) -end -p -``` - -The 3-dimensional graph with `plotly` can have its viewing angle -adjusted with the mouse. When looking down on the ``x-y`` plane, which -code `b` and `k`, we can see the rapid growth along a line related to -``b/k``. - - -Smith and Moore point out that ``k`` is roughly the reciprocal of the number of days an individual is sick enough to infect others. This can be estimated during a breakout. However, they go on to note that there is no direct way to observe ``b``, but there is an indirect way. - -The ratio ``c = b/k`` is the number of close contacts per day times the number of days infected which is the number of close contacts per infected individual. - -This can be estimated from the curves once steady state has been reached (at the end of the pandemic). - -```math -\frac{di}{ds} = \frac{di/dt}{ds/dt} = \frac{b \cdot s(t) \cdot i(t) - k \cdot i(t)}{-b \cdot s(t) \cdot i(t)} = -1 + \frac{1}{c \cdot s} -``` - -This equation does not depend on ``t``; ``s`` is the dependent variable. It could be solved numerically, but in this case affords an algebraic solution: ``i = -s + (1/c) \log(s) + q``, where ``q`` is some constant. The quantity ``q = i + s - (1/c) \log(s)`` does not depend on time, so is the same at time ``t=0`` as it is as ``t \rightarrow \infty``. At ``t=0`` we have ``s(0) \approx 1`` and ``i(0) \approx 0``, whereas ``t \rightarrow \infty``, ``i(t) \rightarrow 0`` and ``s(t)`` goes to the steady state value, which can be estimated. Solving with ``t=0``, we see ``q=0 + 1 - (1/c)\log(1) = 1``. In the limit them ``1 = 0 + s_{\infty} - (1/c)\log(s_\infty)`` or ``c = \log(s_\infty)/(1-s_\infty)``. - - -## Trajectory with drag - -We now solve numerically the problem of a trajectory with a drag force from air resistance. - -The general model is: - -```math -\begin{align*} -x''(t) &= - W(t,x(t), x'(t), y(t), y'(t)) \cdot x'(t)\\ -y''(t) &= -g - W(t,x(t), x'(t), y(t), y'(t)) \cdot y'(t)\\ -\end{align*} -``` - -with initial conditions: ``x(0) = y(0) = 0`` and ``x'(0) = v_0 \cos(\theta), y'(0) = v_0 \sin(\theta)``. - -This into an ODE by a standard trick. Here we define our function for updating a step. As can be seen the vector `u` contains both ``\langle x,y \rangle`` -and ``\langle x',y' \rangle`` - -```julia -function xy!(du, u, p, t) - g, γ = p.g, p.k - x, y = u[1], u[2] - x′, y′ = u[3], u[4] # unicode \prime[tab] - - W = γ - - du[1] = x′ - du[2] = y′ - du[3] = 0 - W * x′ - du[4] = -g - W * y′ -end -``` - -This function ``W`` is just a constant above, but can be easily modified as desired. - -!!! note "A second-order ODE is a coupled first-order ODE" - The "standard" trick is to take a second order ODE like ``u''(t)=u`` and turn this into two coupled ODEs by using a new name: ``v=u'(t)`` and then ``v'(t) = u(t)``. In this application, there are ``4`` equations, as we have *both* ``x''`` and ``y''`` being so converted. The first and second components of ``du`` are new variables, the third and fourth show the original equation. - -The initial conditions are specified through: - -```julia -θ = pi/4 -v₀ = 200 -xy₀ = [0.0, 0.0] -vxy₀ = v₀ * [cos(θ), sin(θ)] -INITIAL = vcat(xy₀, vxy₀) -``` - -The time span can be computed using an *upper* bound of no drag, for which the classic physics formulas give (when ``y_0=0``) ``(0, 2v_{y0}/g)`` - -```julia -g = 9.8 -TSPAN = (0, 2*vxy₀[2] / g) -``` - -This allows us to define an `ODEProblem`: - -```julia -trajectory_problem = ODEProblem(xy!, INITIAL, TSPAN) -``` - -When ``\gamma = 0`` there should be no drag and we expect to see a parabola: - -```julia; hold=true -ps = (g=9.8, k=0) -SOL = solve(trajectory_problem, Tsit5(); p = ps) - -plot(t -> SOL(t)[1], t -> SOL(t)[2], TSPAN...; legend=false) -``` - -The plot is a parametric plot of the ``x`` and ``y`` parts of the solution over the time span. We can see the expected parabolic shape. - -On a *windy* day, the value of ``k`` would be positive. Repeating the above with ``k=1/4`` gives: - -```julia; hold=true -ps = (g=9.8, k=1/4) -SOL = solve(trajectory_problem, Tsit5(); p = ps) - -plot(t -> SOL(t)[1], t -> SOL(t)[2], TSPAN...; legend=false) -``` - -We see that the ``y`` values have gone negative. The `DifferentialEquations` package can adjust for that with a *callback* which terminates the problem once ``y`` has gone negative. This can be implemented as follows: - - -```julia; hold=true -condition(u,t,integrator) = u[2] # called when `u[2]` is negative -affect!(integrator) = terminate!(integrator) # stop the process -cb = ContinuousCallback(condition, affect!) - -ps = (g=9.8, k = 1/4) -SOL = solve(trajectory_problem, Tsit5(); p = ps, callback=cb) - -plot(t -> SOL(t)[1], t -> SOL(t)[2], TSPAN...; legend=false) -``` - - -Finally, we note that the `ModelingToolkit` package provides symbolic-numeric computing. This allows the equations to be set up symbolically, as in `SymPy` before being passed off to `DifferentialEquations` to solve numerically. The above example with no wind resistance could be translated into the following: - -```julia; hold=true -@parameters t γ g -@variables x(t) y(t) -D = Differential(t) - -eqs = [D(D(x)) ~ -γ * D(x), - D(D(y)) ~ -g - γ * D(y)] - -@named sys = ODESystem(eqs) -sys = ode_order_lowering(sys) # turn 2nd order into 1st - -u0 = [D(x) => vxy₀[1], - D(y) => vxy₀[2], - x => 0.0, - y => 0.0] - -p = [γ => 0.0, - g => 9.8] - -prob = ODEProblem(sys, u0, TSPAN, p, jac=true) -sol = solve(prob,Tsit5()) - -plot(t -> sol(t)[3], t -> sol(t)[4], TSPAN..., legend=false) -``` - -The toolkit will automatically generate fast functions and can perform transformations (such as is done by `ode_order_lowering`) before passing along to the numeric solves. diff --git a/CwJ/ODEs/euler.jmd b/CwJ/ODEs/euler.jmd deleted file mode 100644 index d126ff6..0000000 --- a/CwJ/ODEs/euler.jmd +++ /dev/null @@ -1,834 +0,0 @@ -# Euler's method - -This section uses these add-on packages: - -```julia -using CalculusWithJulia -using Plots -using SymPy -using Roots -``` - -```julia; echo=false; results="hidden" -using CalculusWithJulia.WeaveSupport - - -const frontmatter = ( - title = "Euler's method", - description = "Calculus with Julia: Euler's method", - tags = ["CalculusWithJulia", "odes", "euler's method"], -); -fig_size = (800, 600) -nothing -``` - ----- - -The following section takes up the task of numerically approximating solutions to differential equations. `Julia` has a huge set of state-of-the-art tools for this task starting with the [DifferentialEquations](https://github.com/SciML/DifferentialEquations.jl) package. We don't use that package in this section, focusing on simpler methods and implementations for pedagogical purposes, but any further exploration should utilize the tools provided therein. A brief introduction to the package follows in an upcoming [section](./differential_equations.html). - - ----- - - - -Consider the differential equation: - -```math -y'(x) = y(x) \cdot x, \quad y(1)=1, -``` - -which can be solved with `SymPy`: - -```julia; -@syms x, y, u() -D = Differential(x) -x0, y0 = 1, 1 -F(y,x) = y*x - -dsolve(D(u)(x) - F(u(x), x)) -``` - -With the given initial condition, the solution becomes: - -```julia; -out = dsolve(D(u)(x) - F(u(x),x), u(x), ics=Dict(u(x0) => y0)) -``` - - -Plotting this solution over the slope field - -```julia; -p = plot(legend=false) -vectorfieldplot!((x,y) -> [1, F(x,y)], xlims=(0, 2.5), ylims=(0, 10)) -plot!(rhs(out), linewidth=5) -``` - - -we see that the vectors that are drawn seem to be tangent to the graph -of the solution. This is no coincidence, the tangent lines to integral -curves are in the direction of the slope field. - - -What if the graph of the solution were not there, could we use this -fact to *approximately* reconstruct the solution? - -That is, if we stitched together pieces of the slope field, would we -get a curve that was close to the actual answer? - -```julia; hold=true; echo=false; cache=true -## {{{euler_graph}}} -function make_euler_graph(n) - x, y = symbols("x, y") - F(y,x) = y*x - x0, y0 = 1, 1 - - h = (2-1)/5 - xs = zeros(n+1) - ys = zeros(n+1) - xs[1] = x0 # index is off by 1 - ys[1] = y0 - for i in 1:n - xs[i + 1] = xs[i] + h - ys[i + 1] = ys[i] + h * F(ys[i], xs[i]) - end - - p = plot(legend=false) - vectorfieldplot!((x,y) -> [1, F(y,x)], xlims=(1,2), ylims=(0,6)) - - ## Add Euler soln - plot!(p, xs, ys, linewidth=5) - scatter!(p, xs, ys) - - ## add function - out = dsolve(u'(x) - F(u(x), x), u(x), ics=(u, x0, y0)) - plot!(p, rhs(out), x0, xs[end], linewidth=5) - - p -end - - - - -n = 5 -anim = @animate for i=1:n - make_euler_graph(i) -end - -imgfile = tempname() * ".gif" -gif(anim, imgfile, fps = 1) - - -caption = """ -Illustration of a function stitching together slope field lines to -approximate the answer to an initial-value problem. The other function drawn is the actual solution. -""" - -ImageFile(imgfile, caption) -``` - -The illustration suggests the answer is yes, let's see. The solution -is drawn over $x$ values $1$ to $2$. Let's try piecing together $5$ -pieces between $1$ and $2$ and see what we have. - -The slope-field vectors are *scaled* versions of the vector `[1, F(y,x)]`. The `1` -is the part in the direction of the $x$ axis, so here we would like -that to be $0.2$ (which is $(2-1)/5$. So our vectors would be `0.2 * -[1, F(y,x)]`. To allow for generality, we use `h` in place of the -specific value $0.2$. - -Then our first pieces would be the line connecting $(x_0,y_0)$ to - -```math -\langle x_0, y_0 \rangle + h \cdot \langle 1, F(y_0, x_0) \rangle. -``` - -The above uses vector notation to add the piece scaled by $h$ to the -starting point. Rather than continue with that notation, we will use -subscripts. Let $x_1$, $y_1$ be the postion of the tip of the -vector. Then we have: - -```math -x_1 = x_0 + h, \quad y_1 = y_0 + h F(y_0, x_0). -``` - -With this notation, it is easy to see what comes next: - -```math -x_2 = x_1 + h, \quad y_2 = y_1 + h F(y_1, x_1). -``` - -We just shifted the indices forward by $1$. But graphically what is -this? It takes the tip of the first part of our "stitched" together -solution, finds the slope filed there (`[1, F(y,x)]`) and then uses -this direction to stitch together one more piece. - -Clearly, we can repeat. The $n$th piece will end at: - -```math -x_{n+1} = x_n + h, \quad y_{n+1} = y_n + h F(y_n, x_n). -``` - -For our example, we can do some numerics. We want $h=0.2$ and $5$ -pieces, so values of $y$ at $x_0=1, x_1=1.2, x_2=1.4, x_3=1.6, -x_4=1.8,$ and $x_5=2$. - -Below we do this in a loop. We have to be a bit careful, as in `Julia` -the vector of zeros we create to store our answers begins indexing at -$1$, and not $0$. - -```julia; -n=5 -h = (2-1)/n -xs = zeros(n+1) -ys = zeros(n+1) -xs[1] = x0 # index is off by 1 -ys[1] = y0 -for i in 1:n - xs[i + 1] = xs[i] + h - ys[i + 1] = ys[i] + h * F(ys[i], xs[i]) -end -``` - -So how did we do? Let's look graphically: - -```julia; -plot(exp(-1/2)*exp(x^2/2), x0, 2) -plot!(xs, ys) -``` - -Not bad. We wouldn't expect this to be exact - due to the concavity -of the solution, each step is an underestimate. However, we see it is -an okay approximation and would likely be better with a smaller $h$. A -topic we pursue in just a bit. - -Rather than type in the above command each time, we wrap it all up in -a function. The inputs are $n$, $a=x_0$, $b=x_n$, $y_0$, and, most -importantly, $F$. The output is massaged into a function through a -call to `linterp`, rather than two vectors. The `linterp` function we define below just -finds a function that linearly interpolates between the points and is -`NaN` outside of the range of the $x$ values: - -```julia; -function linterp(xs, ys) - function(x) - ((x < xs[1]) || (x > xs[end])) && return NaN - for i in 1:(length(xs) - 1) - if xs[i] <= x < xs[i+1] - l = (x-xs[i]) / (xs[i+1] - xs[i]) - return (1-l) * ys[i] + l * ys[i+1] - end - end - ys[end] - end -end -``` - -With that, here is our function to find an approximate solution to $y'=F(y,x)$ with initial condition: - -```julia; -function euler(F, x0, xn, y0, n) - h = (xn - x0)/n - xs = zeros(n+1) - ys = zeros(n+1) - xs[1] = x0 - ys[1] = y0 - for i in 1:n - xs[i + 1] = xs[i] + h - ys[i + 1] = ys[i] + h * F(ys[i], xs[i]) - end - linterp(xs, ys) -end -``` - -With `euler`, it becomes easy to explore different values. - -For example, we thought the solution would look better with a smaller $h$ (or larger $n$). Instead of $n=5$, let's try $n=50$: - -```julia; -u₁₂ = euler(F, 1, 2, 1, 50) -plot(exp(-1/2)*exp(x^2/2), x0, 2) -plot!(u₁₂, x0, 2) -``` - -It is more work for the computer, but not for us, and clearly a much better approximation to the actual answer is found. - - -## The Euler method - - -```julia; hold=true; echo=false -imgfile ="figures/euler.png" -caption = """ -Figure from first publication of Euler's method. From [Gander and Wanner](http://www.unige.ch/~gander/Preprints/Ritz.pdf). -""" - -ImageFile(:ODEs, imgfile, caption) -``` - - -The name of our function reflects the [mathematician](https://en.wikipedia.org/wiki/Leonhard_Euler) associated with the iteration: - -```math -x_{n+1} = x_n + h, \quad y_{n+1} = y_n + h \cdot F(y_n, x_n), -``` - -to approximate a solution to the first-order, ordinary differential -equation with initial values: $y'(x) = F(y,x)$. - - -[The Euler method](https://en.wikipedia.org/wiki/Euler_method) uses -linearization. Each "step" is just an approximation of the function -value $y(x_{n+1})$ with the value from the tangent line tangent to the -point $(x_n, y_n)$. - - -Each step introduces an error. The error in one step is known as the -*local truncation error* and can be shown to be about equal to $1/2 -\cdot h^2 \cdot f''(x_{n})$ assuming $y$ has ``3`` or more derivatives. - -The total error, or more commonly, *global truncation error*, is the -error between the actual answer and the approximate answer at the end -of the process. It reflects an accumulation of these local errors. This -error is *bounded* by a constant times $h$. Since it gets smaller as -$h$ gets smaller in direct proportion, the Euler method is called -*first order*. - -Other, somewhat more complicated, methods have global truncation errors that -involve higher powers of $h$ - that is for the same size $h$, the -error is smaller. In analogy is the fact that Riemann sums have -error that depends on $h$, whereas other methods of approximating the -integral have smaller errors. For example, Simpson's rule had error -related to $h^4$. So, the Euler method may not be employed if there -is concern about total resources (time, computer, ...), it is -important for theoretical purposes in a manner similar to the role of the Riemann -integral. - -In the examples, we will see that for many problems the simple Euler -method is satisfactory, but not always so. The task of numerically -solving differential equations is not a one-size-fits-all one. In the -following, a few different modifications are presented to the basic -Euler method, but this just scratches the surface of the topic. - -#### Examples - -##### Example - - -Consider the initial value problem $y'(x) = x + y(x)$ with initial -condition $y(0)=1$. This problem can be solved exactly. Here we -approximate over $[0,2]$ using Euler's method. - -```julia; -𝑭(y,x) = x + y -𝒙0, 𝒙n, 𝒚0 = 0, 2, 1 -𝒇 = euler(𝑭, 𝒙0, 𝒙n, 𝒚0, 25) -𝒇(𝒙n) -``` - -We graphically compare our approximate answer with the exact one: - -```julia; -plot(𝒇, 𝒙0, 𝒙n) -𝒐ut = dsolve(D(u)(x) - 𝑭(u(x),x), u(x), ics = Dict(u(𝒙0) => 𝒚0)) -plot(rhs(𝒐ut), 𝒙0, 𝒙n) -plot!(𝒇, 𝒙0, 𝒙n) -``` - -From the graph it appears our value for `f(xn)` will underestimate the -actual value of the solution slightly. - -##### Example - -The equation $y'(x) = \sin(x \cdot y)$ is not separable, so need not have an -easy solution. The default method will fail. Looking at the available methods with `sympy.classify_ode(𝐞qn, u(x))` shows a power series method which -can return a power series *approximation* (a Taylor polynomial). Let's -look at comparing an approximate answer given by the Euler method to -that one returned by `SymPy`. - -First, the `SymPy` solution: - -```julia; -𝐅(y,x) = sin(x*y) -𝐞qn = D(u)(x) - 𝐅(u(x), x) -𝐨ut = dsolve(𝐞qn, hint="1st_power_series") -``` - - - -If we assume $y(0) = 1$, we can continue: - -```julia; -𝐨ut1 = dsolve(𝐞qn, u(x), ics=Dict(u(0) => 1), hint="1st_power_series") -``` - -The approximate value given by the Euler method is - -```julia; -𝐱0, 𝐱n, 𝐲0 = 0, 2, 1 - -plot(legend=false) -vectorfieldplot!((x,y) -> [1, 𝐅(y,x)], xlims=(𝐱0, 𝐱n), ylims=(0,5)) -plot!(rhs(𝐨ut1).removeO(), linewidth=5) - -𝐮 = euler(𝐅, 𝐱0, 𝐱n, 𝐲0, 10) -plot!(𝐮, linewidth=5) -``` - -We see that the answer found from using a polynomial series matches that of Euler's method for a bit, but as time evolves, the approximate solution given by Euler's method more closely tracks the slope field. - -##### Example - - -The -[Brachistochrone problem](http://www.unige.ch/~gander/Preprints/Ritz.pdf) -was posed by Johann Bernoulli in 1696. It asked for the curve between -two points for which an object will fall faster along that curve than -any other. For an example, a bead sliding on a wire will take a certain amount of time to get from point $A$ to point $B$, the time depending on the shape of the wire. Which shape will take the least amount of time? - - -```julia; hold=true; echo=false -imgfile = "figures/bead-game.jpg" -caption = """ - -A child's bead game. What shape wire will produce the shortest time for a bed to slide from a top to the bottom? - -""" -ImageFile(:ODEs, imgfile, caption) -``` - -Restrict our attention to the $x$-$y$ plane, and consider a path, -between the point $(0,A)$ and $(B,0)$. Let $y(x)$ be the distance from -$A$, so $y(0)=0$ and at the end $y$ will be $A$. - - -[Galileo](http://www-history.mcs.st-and.ac.uk/HistTopics/Brachistochrone.html) -knew the straight line was not the curve, but incorrectly thought the -answer was a part of a circle. - -```julia; hold=true; echo=false -imgfile = "figures/galileo.gif" -caption = """ -As early as 1638, Galileo showed that an object falling along `AC` and then `CB` will fall faster than one traveling along `AB`, where `C` is on the arc of a circle. -From the [History of Math Archive](http://www-history.mcs.st-and.ac.uk/HistTopics/Brachistochrone.html). -""" -ImageFile(:ODEs, imgfile, caption) -``` - - -This simulation also suggests that a curved path is better than the shorter straight one: - -```julia; hold=true; echo=false; cache=true -##{{{brach_graph}}} - -function brach(f, x0, vx0, y0, vy0, dt, n) - m = 1 - g = 9.8 - - axs = Float64[0] - ays = Float64[-g] - vxs = Float64[vx0] - vys = Float64[vy0] - xs = Float64[x0] - ys = Float64[y0] - - for i in 1:n - x = xs[end] - vx = vxs[end] - - ax = -f'(x) * (f''(x) * vx^2 + g) / (1 + f'(x)^2) - ay = f''(x) * vx^2 + f'(x) * ax - - push!(axs, ax) - push!(ays, ay) - - push!(vxs, vx + ax * dt) - push!(vys, vys[end] + ay * dt) - push!(xs, x + vxs[end] * dt)# + (1/2) * ax * dt^2) - push!(ys, ys[end] + vys[end] * dt)# + (1/2) * ay * dt^2) - end - - [xs ys vxs vys axs ays] - -end - - -fs = [x -> 1 - x, - x -> (x-1)^2, - x -> 1 - sqrt(1 - (x-1)^2), - x -> - (x-1)*(x+1), - x -> 3*(x-1)*(x-1/3) - ] - - -MS = [brach(f, 1/100, 0, 1, 0, 1/100, 100) for f in fs] - - -function make_brach_graph(n) - - p = plot(xlim=(0,1), ylim=(-1/3, 1), legend=false) - for (i,f) in enumerate(fs) - plot!(f, 0, 1) - U = MS[i] - x = min(1.0, U[n,1]) - scatter!(p, [x], [f(x)]) - end - p - -end - - - -n = 4 -anim = @animate for i=[1,5,10,15,20,25,30,35,40,45,50,55,60] - make_brach_graph(i) -end - -imgfile = tempname() * ".gif" -gif(anim, imgfile, fps = 1) - - -caption = """ -The race is on. An illustration of beads falling along a path, as can be seen, some paths are faster than others. The fastest path would follow a cycloid. See [Bensky and Moelter](https://pdfs.semanticscholar.org/66c1/4d8da6f2f5f2b93faf4deb77aafc7febb43a.pdf) for details on simulating a bead on a wire. -""" - -ImageFile(imgfile, caption) -``` - - - - -Now, the natural question is which path is best? The solution can be -[reduced](http://mathworld.wolfram.com/BrachistochroneProblem.html) to -solving this equation for a positive $c$: - -```math -1 + (y'(x))^2 = \frac{c}{y}, \quad c > 0. -``` - -Reexpressing, this becomes: - -```math -\frac{dy}{dx} = \sqrt{\frac{C-y}{y}}. -``` - -This is a separable equation and can be solved, but even `SymPy` has -trouble with this integral. However, the result has been known to be a piece of a cycloid since the insightful -Jacob Bernoulli used an analogy from light bending to approach the problem. The answer is best described parametrically -through: - -```math -x(u) = c\cdot u - \frac{c}{2}\sin(2u), \quad y(u) = \frac{c}{2}( 1- \cos(2u)), \quad 0 \leq u \leq U. -``` - -The values of $U$ and $c$ must satisfy $(x(U), y(U)) = (B, A)$. - - -Rather than pursue this, we will solve it numerically for a fixed -value of $C$ over a fixed interval to see the shape. - - -The equation can be written in terms of $y'=F(y,x)$, where - -```math -F(y,x) = \sqrt{\frac{c-y}{y}}. -``` - -But as $y_0 = 0$, we immediately would have a problem with the first step, as there would be division by $0$. - -This says that for the optimal solution, the bead picks up speed by first sliding straight down before heading off towards $B$. That's great for the physics, but runs roughshod over our Euler method, as the first step has an infinity. - -For this, we can try the *backwards Euler* method which uses the slope at $(x_{n+1}, y_{n+1})$, rather than $(x_n, y_n)$. The update step becomes: - -```math -y_{n+1} = y_n + h \cdot F(y_{n+1}, x_{n+1}). -``` - -Seems innocuous, but the value we are trying to find, $y_{n+1}$, is -now on both sides of the equation, so is only *implicitly* defined. In -this code, we use the `find_zero` function from the `Roots` package. The -caveat is, this function needs a good initial guess, and the one we -use below need not be widely applicable. - - -```julia; -function back_euler(F, x0, xn, y0, n) - h = (xn - x0)/n - xs = zeros(n+1) - ys = zeros(n+1) - xs[1] = x0 - ys[1] = y0 - for i in 1:n - xs[i + 1] = xs[i] + h - ## solve y[i+1] = y[i] + h * F(y[i+1], x[i+1]) - ys[i + 1] = find_zero(y -> ys[i] + h * F(y, xs[i + 1]) - y, ys[i]+h) - end - linterp(xs, ys) -end -``` - -We then have with $C=1$ over the interval $[0,1.2]$ the following: - -```julia; -𝐹(y, x; C=1) = sqrt(C/y - 1) -𝑥0, 𝑥n, 𝑦0 = 0, 1.2, 0 -cyc = back_euler(𝐹, 𝑥0, 𝑥n, 𝑦0, 50) -plot(x -> 1 - cyc(x), 𝑥0, 𝑥n) -``` - -Remember, $y$ is the displacement from the top, so it is -non-negative. Above we flipped the graph to make it look more like -expectation. In general, the trajectory may actually dip below the -ending point and come back up. The above won't see this, for as -written $dy/dx \geq 0$, which need not be the case, as the defining -equation is in terms of $(dy/dx)^2$, so the derivative could have any -sign. - - - -##### Example: stiff equations - -The Euler method is *convergent*, in that as $h$ goes to $0$, the -approximate solution will converge to the actual answer. However, this -does not say that for a fixed size $h$, the approximate value will be -good. For example, consider the differential equation $y'(x) = --5y$. This has solution $y(x)=y_0 e^{-5x}$. However, if we try the -Euler method to get an answer over $[0,2]$ with $h=0.5$ we don't see -this: - -```julia; -ℱ(y,x) = -5y -𝓍0, 𝓍n, 𝓎0 = 0, 2, 1 -𝓊 = euler(ℱ, 𝓍0, 𝓍n, 𝓎0, 4) # n =4 => h = 2/4 -vectorfieldplot((x,y) -> [1, ℱ(y,x)], xlims=(0, 2), ylims=(-5, 5)) -plot!(x -> y0 * exp(-5x), 0, 2, linewidth=5) -plot!(𝓊, 0, 2, linewidth=5) -``` - -What we see is that the value of $h$ is too big to capture the decay -scale of the solution. A smaller $h$, can do much better: - -```julia; -𝓊₁ = euler(ℱ, 𝓍0, 𝓍n, 𝓎0, 50) # n=50 => h = 2/50 -plot(x -> y0 * exp(-5x), 0, 2) -plot!(𝓊₁, 0, 2) -``` - -This is an example of a -[stiff equation](https://en.wikipedia.org/wiki/Stiff_equation). Such -equations cause explicit methods like the Euler one problems, as small -$h$s are needed to good results. - -The implicit, backward Euler method does not have this issue, as we can see here: - -```julia; -𝓊₂ = back_euler(ℱ, 𝓍0, 𝓍n, 𝓎0, 4) # n =4 => h = 2/4 -vectorfieldplot((x,y) -> [1, ℱ(y,x)], xlims=(0, 2), ylims=(-1, 1)) -plot!(x -> y0 * exp(-5x), 0, 2, linewidth=5) -plot!(𝓊₂, 0, 2, linewidth=5) -``` - - -##### Example: The pendulum - - -The differential equation describing the simple pendulum is - -```math -\theta''(t) = - \frac{g}{l}\sin(\theta(t)). -``` - -The typical approach to solving for $\theta(t)$ is to use the small-angle approximation that $\sin(x) \approx x$, and then the differential equation simplifies to: -$\theta''(t) = -g/l \cdot \theta(t)$, which is easily solved. - -Here we try to get an answer numerically. However, the problem, as stated, is not a first order equation due to the $\theta''(t)$ term. If we let $u(t) = \theta(t)$ and $v(t) = \theta'(t)$, then we get *two* coupled first order equations: - -```math -v'(t) = -g/l \cdot \sin(u(t)), \quad u'(t) = v(t). -``` - -We can try the Euler method here. A simple approach might be this iteration scheme: - -```math -\begin{align*} -x_{n+1} &= x_n + h,\\ -u_{n+1} &= u_n + h v_n,\\ -v_{n+1} &= v_n - h \cdot g/l \cdot \sin(u_n). -\end{align*} -``` - -Here we need *two* initial conditions: one for the initial value -$u(t_0)$ and the initial value of $u'(t_0)$. We have seen if we start at an angle $a$ and release the bob from rest, so $u'(0)=0$ we get a sinusoidal answer to the linearized model. What happens here? We let $a=1$, $L=5$ and $g=9.8$: - -We write a function to solve this starting from $(x_0, y_0)$ and ending at $x_n$: - -```julia; -function euler2(x0, xn, y0, yp0, n; g=9.8, l = 5) - xs, us, vs = zeros(n+1), zeros(n+1), zeros(n+1) - xs[1], us[1], vs[1] = x0, y0, yp0 - h = (xn - x0)/n - for i = 1:n - xs[i+1] = xs[i] + h - us[i+1] = us[i] + h * vs[i] - vs[i+1] = vs[i] + h * (-g / l) * sin(us[i]) - end - linterp(xs, us) -end -``` - -Let's take $a = \pi/4$ as the initial angle, then the approximate -solution should be $\pi/4\cos(\sqrt{g/l}x)$ with period $T = -2\pi\sqrt{l/g}$. We try first to plot then over 4 periods: - -```julia; -𝗅, 𝗀 = 5, 9.8 -𝖳 = 2pi * sqrt(𝗅/𝗀) -𝗑0, 𝗑n, 𝗒0, 𝗒p0 = 0, 4𝖳, pi/4, 0 -plot(euler2(𝗑0, 𝗑n, 𝗒0, 𝗒p0, 20), 0, 4𝖳) -``` - -Something looks terribly amiss. The issue is the step size, $h$, is -too large to capture the oscillations. There are basically only $5$ -steps to capture a full up and down motion. Instead, we try to get $20$ steps per period -so $n$ must be not $20$, but $4 \cdot 20 \cdot T \approx 360$. To this -graph, we add the approximate one: - -```julia; -plot(euler2(𝗑0, 𝗑n, 𝗒0, 𝗒p0, 360), 0, 4𝖳) -plot!(x -> pi/4*cos(sqrt(𝗀/𝗅)*x), 0, 4𝖳) -``` - -Even now, we still see that something seems amiss, though the issue is -not as dramatic as before. The oscillatory nature of the pendulum is -seen, but in the Euler solution, the amplitude grows, which would -necessarily mean energy is being put into the system. A familiar -instance of a pendulum would be a child on a swing. Without pumping -the legs - putting energy in the system - the height of the swing's -arc will not grow. Though we now have oscillatory motion, this growth -indicates the solution is still not quite right. The issue is likely -due to each step mildly overcorrecting and resulting in an overall -growth. One of the questions pursues this a bit further. - -## Questions - -##### Question - -Use Euler's method with $n=5$ to approximate $u(1)$ where - -```math -u'(x) = x - u(x), \quad u(0) = 1 -``` - -```julia; hold=true; echo=false -F(y,x) = x - y -x0, xn, y0 = 0, 1, 1 -val = euler(F, x0, xn, y0, 5)(1) -numericq(val) -``` - -##### Question - -Consider the equation - -```math -y' = x \cdot \sin(y), \quad y(0) = 1. -``` - -Use Euler's method with $n=50$ to find the value of $y(5)$. - -```julia; hold=true; echo=false -F(y, x) = x * sin(y) -x0, xn, y0 = 0, 5, 1 -n = 50 -u = euler(F, x0, xn, y0, n) -numericq(u(xn)) -``` - - -##### Question - -Consider the ordinary differential equation - -```math -\frac{dy}{dx} = 1 - 2\frac{y}{x}, \quad y(1) = 0. -``` - -Use Euler's method to solve for $y(2)$ when $n=50$. - -```julia; hold=true; echo=false -F(y, x) = 1 - 2y/x -x0, xn, y0 = 1, 2, 0 -n = 50 -u = euler(F, x0, xn, y0, n) -numericq(u(xn)) -``` - -##### Question - - -Consider the ordinary differential equation - -```math -\frac{dy}{dx} = \frac{y \cdot \log(y)}{x}, \quad y(2) = e. -``` - -Use Euler's method to solve for $y(3)$ when $n=25$. - -```julia; hold=true; echo=false -F(y, x) = y*log(y)/x -x0, xn, y0 = 2, 3, exp(1) -n = 25 -u = euler(F, x0, xn, y0, n) -numericq(u(xn)) -``` - - -##### Question - -Consider the first-order non-linear ODE - -```math -y' = y \cdot (1-2x), \quad y(0) = 1. -``` - -Use Euler's method with $n=50$ to approximate the solution $y$ over $[0,2]$. - -What is the value at $x=1/2$? - -```julia; hold=true; echo=false -F(y, x) = y * (1-2x) -x0, xn, y0 = 0, 2, 1 -n = 50 -u = euler(F, x0, xn, y0, n) -numericq(u(1/2)) -``` - -What is the value at $x=3/2$? - -```julia; hold=true; echo=false -F(y, x) = y * (1-2x) -x0, xn, y0 = 0, 2, 1 -n = 50 -u = euler(F, x0, xn, y0, n) -numericq(u(3/2)) -``` - -##### Question: The pendulum revisited. - -The issue with the pendulum's solution growing in amplitude can be -addressed using a modification to the Euler method attributed to -[Cromer](http://astro.physics.ncsu.edu/urca/course_files/Lesson14/index.html). The -fix is to replace the term `sin(us[i])` in the line `vs[i+1] = vs[i] + h * (-g / l) * -sin(us[i])` of the `euler2` function with `sin(us[i+1])`, which uses the updated angular -velocity in the ``2``nd step in place of the value before the step. - -Modify the `euler2` function to implement the Euler-Cromer method. What do you see? - -```julia; hold=true; echo=false -choices = [ -"The same as before - the amplitude grows", -"The solution is identical to that of the approximation found by linearization of the sine term", -"The solution has a constant amplitude, but its period is slightly *shorter* than that of the approximate solution found by linearization", -"The solution has a constant amplitude, but its period is slightly *longer* than that of the approximate solution found by linearization"] -answ = 4 -radioq(choices, answ, keep_order=true) -``` diff --git a/CwJ/ODEs/figures/bead-game.jpg b/CwJ/ODEs/figures/bead-game.jpg deleted file mode 100644 index 72e23898dd2c7804698c46e7abd6fd7d8617eb77..0000000000000000000000000000000000000000 GIT binary patch literal 0 HcmV?d00001 literal 7628 zcmb7pcTf{f({ShzdR1Cz!4L!sMS2y11Ze>R(ou@^4uXm_={5A;5~_4a5CoAbp#119 zNHriqLQy~zS~~(qN1i@pryG2W_o%$23BS^Hdba&n{TFDcsL3g5DgLvn%>p2!pdhE9q$Q`Jrl7kbA-kFYs933mXxQ$uYZyDw 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oM=__jR1P&=86QTw$|b1K+Q$VaOuqGi*CDQ_=d6}*{@>&O1K={{H~;_u diff --git a/CwJ/ODEs/odes.jmd b/CwJ/ODEs/odes.jmd deleted file mode 100644 index 5ec99e0..0000000 --- a/CwJ/ODEs/odes.jmd +++ /dev/null @@ -1,917 +0,0 @@ -# ODEs - -This section uses these add-on packages: - -```julia -using CalculusWithJulia -using Plots -using SymPy -``` - -```julia; echo=false; results="hidden" -using CalculusWithJulia.WeaveSupport - -const frontmatter = ( - title = "ODEs", - description = "Calculus with Julia: ODEs", - tags = ["CalculusWithJulia", "odes", "odes"], -); -nothing -``` - ----- - -Some relationships are easiest to describe in terms of rates or derivatives. For example: - -* Knowing the speed of a car and how long it has been driving can - summarize the car's location. - -* One of Newton's famous laws, $F=ma$, describes the force on an - object of mass $m$ in terms of the acceleration. The acceleration - is the derivative of velocity, which in turn is the derivative of - position. So if we know the rates of change of $v(t)$ or $x(t)$, we - can differentiate to find $F$. - -* Newton's law of [cooling](http://tinyurl.com/z4lmetp). This - describes the temperature change in an object due to a difference in - temperature with the object's surroundings. The formula being, - $T'(t) = -r \left(T(t) - T_a \right)$, where $T(t)$ is temperature at time $t$ - and $T_a$ the ambient temperature. - -* [Hooke's law](http://tinyurl.com/kbz7r8l) relates force on an object - to the position on the object, through $F = k x$. This is - appropriate for many systems involving springs. Combined with - Newton's law $F=ma$, this leads to an equation that $x$ must - satisfy: $m x''(t) = k x(t)$. - -## Motion with constant acceleration - -Let's consider the case of constant acceleration. This describes how nearby objects fall to earth, as the force due to gravity is assumed to be a constant, so the acceleration is the constant force divided by the constant mass. - -With constant acceleration, what is the velocity? - -As mentioned, we have $dv/dt = a$ for any velocity function $v(t)$, but in this case, the right hand side is assumed to be constant. How does this restrict the possible functions, $v(t)$, that the velocity can be? - -Here we can integrate to find that any answer must look like the following for some constant of integration: - -```math -v(t) = \int \frac{dv}{dt} dt = \int a dt = at + C. -``` - -If we are given the velocity at a fixed time, say $v(t_0) = v_0$, then we can use the definite integral to get: - -```math -v(t) - v(t_0) = \int_{t_0}^t a dt = at - a t_0. -``` - -Solving, gives: - -```math -v(t) = v_0 + a (t - t_0). -``` - -This expresses the velocity at time $t$ in terms of the initial velocity, the constant acceleration and the time duration. - -A natural question might be, is this the *only* possible answer? There are a few useful ways to think about this. - -First, suppose there were another, say $u(t)$. Then define $w(t)$ to be the difference: $w(t) = v(t) - u(t)$. We would have that $w'(t) = v'(t) - u'(t) = a - a = 0$. But from the mean value theorem, a function whose derivative is *continuously* $0$, will necessarily be a constant. So at most, $v$ and $u$ will differ by a constant, but if both are equal at $t_0$, they will be equal for all $t$. - -Second, since the derivative of any solution is a continuous function, it is true by the fundamental theorem of calculus that it *must* satisfy the form for the antiderivative. The initial condition makes the answer unique, as the indeterminate $C$ can take only one value. - -Summarizing, we have - -> If ``v(t)`` satisfies the equation: ``v'(t) = a``, ``v(t_0) = v_0,`` -> then the unique solution will be ``v(t) = v_0 + a (t - t_0)``. - - -Next, what about position? Here we know that the time derivative of position yields the velocity, so we should have that the unknown position function satisfies this equation and initial condition: - -```math -x'(t) = v(t) = v_0 + a (t - t_0), \quad x(t_0) = x_0. -``` - -Again, we can integrate to get an answer for any value $t$: - -```math -\begin{align*} -x(t) - x(t_0) &= \int_{t_0}^t \frac{dv}{dt} dt \\ -&= (v_0t + \frac{1}{2}a t^2 - at_0 t) |_{t_0}^t \\ -&= (v_0 - at_0)(t - t_0) + \frac{1}{2} a (t^2 - t_0^2). -\end{align*} -``` - -There are three constants: the initial value for the independent variable, $t_0$, and the two initial values for the velocity and position, $v_0, x_0$. Assuming $t_0 = 0$, we can simplify the above to get a formula familiar from introductory physics: - -```math -x(t) = x_0 + v_0 t + \frac{1}{2} at^2. -``` - -Again, the mean value theorem can show that with the initial value specified this is the only possible solution. - -## First-order initial-value problems - -The two problems just looked at can be summarized by the following. We are looking for solutions to an equation of the form (taking $y$ and $x$ as the variables, in place of $x$ and $t$): - -```math -y'(x) = f(x), \quad y(x_0) = y_0. -``` - -This is called an *ordinary differential equation* (ODE), as it is an equation involving the ordinary derivative of an unknown function, $y$. - -This is called a first-order, ordinary differential equation, as there is only the first derivative involved. - -This is called an initial-value problem, as the value at the initial point $x_0$ is specified as part of the problem. - -#### Examples - -Let's look at a few more examples, and then generalize. - -##### Example: Newton's law of cooling - -Consider the ordinary differential equation given by Newton's law of cooling: - -```math -T'(t) = -r (T(t) - T_a), \quad T(0) = T_0 -``` - -This equation is also first order, as it involves just the first derivative, but notice that on the right hand side is the function $T$, not the variable being differentiated against, $t$. - -As we have a difference on the right hand side, we rename the variable through $U(t) = T(t) - T_a$. Then, as $U'(t) = T'(t)$, we have the equation: - -```math -U'(t) = -r U(t), \quad U(0) = U_0. -``` - - -This shows that the rate of change of $U$ depends on $U$. Large postive values indicate a negative rate of change - a push back towards the origin, and large negative values of $U$ indicate a positive rate of change - again, a push back towards the origin. We shouldn't be surprised to either see a steady decay towards the origin, or oscillations about the origin. - -What will we find? This equation is different from the previous two -equations, as the function $U$ appears on both sides. However, we can -rearrange to get: - -```math -\frac{dU}{dt}\frac{1}{U(t)} = -r. -``` - - -This suggests integrating both sides, as before. Here we do the "$u$"-substitution $u = U(t)$, so $du = U'(t) dt$: - -```math --rt + C = \int \frac{dU}{dt}\frac{1}{U(t)} dt = -\int \frac{1}{u}du = \log(u). -``` - -Solving gives: $u = U(t) = e^C e^{-rt}$. Using the initial condition forces $e^C = U(t_0) = T(0) - T_a$ and so our solution in terms of $T(t)$ is: - - -```math -T(t) - T_a = (T_0 - T_a) e^{-rt}. -``` - -In words, the initial difference in temperature of the object and the environment exponentially decays to $0$. - -That is, as $t > 0$ goes to $\infty$, the right hand will go to $0$ for $r > 0$, so $T(t) \rightarrow T_a$ - the temperature of the object will reach the ambient temperature. The rate of this is largest when the difference between $T(t)$ and $T_a$ is largest, so when objects are cooling the statement "hotter things cool faster" is appropriate. - - -A graph of the solution for $T_0=200$ and $T_a=72$ and $r=1/2$ is made -as follows. We've added a few line segments from the defining formula, -and see that they are indeed tangent to the solution found for the differential equation. - -```julia; echo=false -let - T0, Ta, r = 200, 72, 1/2 - f(u, t) = -r*(u - Ta) - v(t) = Ta + (T0 - Ta) * exp(-r*t) - p = plot(v, 0, 6, linewidth=4, legend=false) - [plot!(p, x -> v(a) + f(v(a), a) * (x-a), 0, 6) for a in 1:2:5] - p -end -``` - - - - -The above is implicitly assuming that there could be no other -solution, than the one we found. Is that really the case? We will see -that there is a theorem that can answer this, but in this case, the -trick of taking the difference of two equations satisfying the -equation leads to the equation $W'(t) = r W(t), \text{ and } W(0) = -0$. This equation has a general solution of $W(t) = Ce^{rt}$ and the -initial condition forces $C=0$, so $W(t) = 0$, as before. Hence, the -initial-value problem for Newton's law of cooling has a unique -solution. - - - -In general, the equation could be written as (again using $y$ and $x$ as the variables): - -```math -y'(x) = g(y), \quad y(x_0) = y_0 -``` - - -This is called an *autonomous*, first-order ODE, as the right-hand side does not depend on $x$ (except through ``y(x)``). - -Let $F(y) = \int_{y_0}^y du/g(u)$, then a solution to the above is $F(y) = x - x_0$, assuming $1/g(u)$ is integrable. - - -##### Example: Toricelli's law - -[Toricelli's Law](http://tinyurl.com/hxvf3qp) describes the speed a jet of water will leave a vessel through an opening below the surface of the water. The formula is $v=\sqrt{2gh}$, where $h$ is the height of the water above the hole and $g$ the gravitational constant. This arises from equating the kinetic energy gained, $1/2 mv^2$ and potential energy lost, $mgh$, for the exiting water. - -An application of Torricelli's law is to describe the volume of water in a tank over time, $V(t)$. Imagine a cylinder of cross sectional area $A$ with a hole of cross sectional diameter $a$ at the bottom, Then $V(t) = A h(t)$, with $h$ giving the height. The change in volume over $\Delta t$ units of time must be given by the value $a v(t) \Delta t$, or - -```math -V(t+\Delta t) - V(t) = -a v(t) \Delta t = -a\sqrt{2gh(t)}\Delta t -``` - -This suggests the following formula, written in terms of $h(t)$ should apply: - -```math -A\frac{dh}{dt} = -a \sqrt{2gh(t)}. -``` - -Rearranging, this gives an equation - -```math -\frac{dh}{dt} \frac{1}{\sqrt{h(t)}} = -\frac{a}{A}\sqrt{2g}. -``` - -Integrating both sides yields: - -```math -2\sqrt{h(t)} = -\frac{a}{A}\sqrt{2g} t + C. -``` - -If $h(0) = h_0 = V(0)/A$, we can solve for $C = 2\sqrt{h_0}$, or - -```math -\sqrt{h(t)} = \sqrt{h_0} -\frac{1}{2}\frac{a}{A}\sqrt{2g} t. -``` - - -Setting $h(t)=0$ and solving for $t$ shows that the time to drain the tank would be $(2A)/(a\sqrt{2g})\sqrt{h_0}$. - - -##### Example - -Consider now the equation - -```math -y'(x) = y(x)^2, \quad y(x_0) = y_0. -``` - -This is called a *non-linear* ordinary differential equation, as the $y$ variable on the right hand side presents itself in a non-linear form (it is squared). These equations may have solutions that are not defined for all times. - -This particular problem can be solved as before by moving the $y^2$ to the left hand side and integrating to yield: - -```math -y(x) = - \frac{1}{C + x}, -``` - -and with the initial condition: - -```math -y(x) = \frac{y_0}{1 - y_0(x - x_0)}. -``` - -This answer can demonstrate *blow-up*. That is, in a finite range for $x$ values, the $y$ value can go to infinity. For example, if the initial conditions are $x_0=0$ and $y_0 = 1$, then $y(x) = 1/(1-x)$ is only defined for $x \geq x_0$ on $[0,1)$, as at $x=1$ there is a vertical asymptote. - - -## Separable equations - -We've seen equations of the form $y'(x) = f(x)$ and $y'(x) = g(y)$ both solved by integrating. The same tricks will work for equations of the form $y'(x) = f(x) \cdot g(y)$. Such equations are called *separable*. - -Basically, we equate up to constants - -```math -\int \frac{dy}{g(y)} = \int f(x) dx. -``` - -For example, suppose we have the equation - -```math -\frac{dy}{dx} = x \cdot y(x), \quad y(x_0) = y_0. -``` - -Then we can find a solution, $y(x)$ through: - -```math -\int \frac{dy}{y} = \int x dx, -``` - -or - -```math -\log(y) = \frac{x^2}{2} + C -``` - -Which yields: - -```math -y(x) = e^C e^{\frac{1}{2}x^2}. -``` - -Substituting in $x_0$ yields a value for $C$ in terms of the initial information $y_0$ and $x_0$. - - -## Symbolic solutions - -Differential equations are classified according to their type. Different types have different methods for solution, when a solution exists. - -The first-order initial value equations we have seen can be described generally by - -```math -\begin{align*} -y'(x) &= F(y,x),\\ -y(x_0) &= x_0. -\end{align*} -``` - -Special cases include: - -* *linear* if the function $F$ is linear in $y$; -* *autonomous* if $F(y,x) = G(y)$ (a function of $y$ alone); -* *separable* if $F(y,x) = G(y)H(x)$. - -As seen, separable equations are approached by moving the "$y$" terms to one side, the "$x$" terms to the other and integrating. This also applies to autonomous equations then. There are other families of equation types that have exact solutions, and techniques for solution, summarized at this [Wikipedia page](http://tinyurl.com/zywzz4q). - -Rather than go over these various families, we demonstrate that `SymPy` can solve many of these equations symbolically. - - -The `solve` function in `SymPy` solves equations for unknown -*variables*. As a differential equation involves an unknown *function* -there is a different function, `dsolve`. The basic idea is to describe -the differential equation using a symbolic function and then call -`dsolve` to solve the expression. - -Symbolic functions are defined by the `@syms` macro (also see `?symbols`) using parentheses to distinguish a function from a variable: - -```julia; -@syms x u() # a symbolic variable and a symbolic function -``` - - -We will solve the following, known as the *logistic equation*: - -```math -u'(x) = a u(1-u), \quad a > 0 -``` - -Before beginning, we look at the form of the equation. When $u=0$ or -$u=1$ the rate of change is $0$, so we expect the function might be -bounded within that range. If not, when $u$ gets bigger than $1$, then -the slope is negative and when $u$ gets less than $0$, the slope is -positive, so there will at least be a drift back to the range -$[0,1]$. Let's see exactly what happens. We define a parameter, -restricting `a` to be positive: - - - -```julia; -@syms a::positive -``` - - -To specify a derivative of `u` in our equation we can use `diff(u(x),x)` but here, for visual simplicity, use the `Differential` operator, as follows: - -```julia; -D = Differential(x) -eqn = D(u)(x) ~ a * u(x) * (1 - u(x)) # use l \Equal[tab] r, Eq(l,r), or just l - r -``` - -In the above, we evaluate the symbolic function at the variable `x` -through the use of `u(x)` in the expression. The equation above uses `~` to combine the left- and right-hand sides as an equation in `SymPy`. (A unicode equals is also available for this task). This is a shortcut for `Eq(l,r)`, but even just using `l - r` would suffice, as the default assumption for an equation is that it is set to `0`. - -The `Differential` operation is borrowed from the `ModelingToolkit` package, which will be introduced later. - - -To finish, we call `dsolve` to find a solution (if possible): - -```julia; -out = dsolve(eqn) -``` - -This answer - to a first-order equation - has one free constant, -`C_1`, which can be solved for from an initial condition. We can see -that when $a > 0$, as $x$ goes to positive infinity the solution goes -to $1$, and when $x$ goes to negative infinity, the solution goes to $0$ -and otherwise is trapped in between, as expected. - -The limits are confirmed by investigating the limits of the right-hand: - -```julia; -limit(rhs(out), x => oo), limit(rhs(out), x => -oo) -``` - -We can confirm that the solution is always increasing, hence trapped within ``[0,1]`` by observing that the derivative is positive when `C₁` is positive: - -```juila; -diff(rhs(out),x) -``` - - - -Suppose that $u(0) = 1/2$. Can we solve for $C_1$ symbolically? We can use `solve`, but first we will need to get the symbol for `C_1`: - -```julia; -eq = rhs(out) # just the right hand side -C1 = first(setdiff(free_symbols(eq), (x,a))) # fish out constant, it is not x or a -c1 = solve(eq(x=>0) - 1//2, C1) -``` - -And we plug in with: - -```julia; -eq(C1 => c1[1]) -``` - -That's a lot of work. The `dsolve` function in `SymPy` allows initial conditions to be specified for some equations. In this case, ours is $x_0=0$ and $y_0=1/2$. The extra arguments passed in through a dictionary to the `ics` argument: - -```julia; -x0, y0 = 0, Sym(1//2) -dsolve(eqn, u(x), ics=Dict(u(x0) => y0)) -``` - -(The one subtlety is the need to write the rational value as a symbolic expression, as otherwise it will get converted to a floating point value prior to being passed along.) - -##### Example: Hooke's law - - -In the first example, we solved for position, $x(t)$, from an assumption of constant acceleration in two steps. The equation relating the two is a second-order equation: $x''(t) = a$, so two constants are generated. That a second-order equation could be reduced to two first-order equations is not happy circumstance, as it can always be done. Rather than show the technique though, we demonstrate that `SymPy` can also handle some second-order ODEs. - -Hooke's law relates the force on an object to its position via $F=ma = -kx$, or $x''(t) = -(k/m)x(t)$. - -Suppose $k > 0$. Then we can solve, similar to the above, with: - -```julia; -@syms k::positive m::positive -D2 = D ∘ D # takes second derivative through composition -eqnh = D2(u)(x) ~ -(k/m)*u(x) -dsolve(eqnh) -``` - -Here we find two constants, as anticipated, for we would guess that -two integrations are needed in the solution. - -Suppose the spring were started by pulling it down to a bottom and -releasing. The initial position at time $0$ would be $a$, say, and -initial velocity $0$. Here we get the solution specifying initial -conditions on the function and its derivative (expressed through -`u'`): - -```julia; -dsolve(eqnh, u(x), ics = Dict(u(0) => -a, D(u)(0) => 0)) -``` - - -We get that the motion will follow -$u(x) = -a \cos(\sqrt{k/m}x)$. This is simple oscillatory behavior. As the spring stretches, the force gets large enough to pull it back, and as it compresses the force gets large enough to push it back. The amplitude of this oscillation is $a$ and the period $2\pi/\sqrt{k/m}$. Larger $k$ values mean shorter periods; larger $m$ values mean longer periods. - - -##### Example: the pendulum - -The simple gravity [pendulum](http://tinyurl.com/h8ys6ts) is an idealization of a physical pendulum that models a "bob" with mass $m$ swinging on a massless rod of length $l$ in a frictionless world governed only by the gravitational constant $g$. The motion can be described by this differential equation for the angle, $\theta$, made from the vertical: - -```math -\theta''(t) + \frac{g}{l}\sin(\theta(t)) = 0 -``` - -Can this second-order equation be solved by `SymPy`? - -```julia; -@syms g::positive l::positive theta()=>"θ" -eqnp = D2(theta)(x) + g/l*sin(theta(x)) -``` - -Trying to do so, can cause `SymPy` to hang or simply give up and repeat its input; no easy answer is forthcoming for this equation. - -In general, for the first-order initial value problem characterized by -$y'(x) = F(y,x)$, there are conditions -([Peano](http://tinyurl.com/h663wba) and -[Picard-Lindelof](http://tinyurl.com/3rbde5e)) that can guarantee the -existence (and uniqueness) of equation locally, but there may not be -an accompanying method to actually find it. This particular problem -has a solution, but it can not be written in terms of elementary -functions. - -However, as [Huygens](https://en.wikipedia.org/wiki/Christiaan_Huygens) first noted, if the angles involved are small, then we approximate the solution through the linearization $\sin(\theta(t)) \approx \theta(t)$. The resulting equation for an approximate answer is just that of Hooke: - - -```math -\theta''(t) + \frac{g}{l}\theta(t) = 0 -``` - -Here, the solution is in terms of sines and cosines, with period given by $T = 2\pi/\sqrt{k} = 2\pi\cdot\sqrt{l/g}$. The answer does not depend on the mass, $m$, of the bob nor the amplitude of the motion, provided the small-angle approximation is valid. - -If we pull the bob back an angle $a$ and release it then the initial conditions are $\theta(0) = a$ and $\theta'(a) = 0$. This gives the solution: - -```julia; -eqnp₁ = D2(u)(x) + g/l * u(x) -dsolve(eqnp₁, u(x), ics=Dict(u(0) => a, D(u)(0) => 0)) -``` - - -##### Example: hanging cables - -A chain hangs between two supports a distance $L$ apart. What shape -will it take if there are no forces outside of gravity acting on it? -What about if the force is uniform along length of the chain, like a -suspension bridge? How will the shape differ then? - -Let $y(x)$ describe the chain at position $x$, with $0 \leq x \leq L$, -say. We consider first the case of the chain with no force save -gravity. Let $w(x)$ be the density of the chain at $x$, taken below to be a constant. - -The chain is in equilibrium, so tension, $T(x)$, in the chain will be -in the direction of the derivative. Let $V$ be the vertical component -and $H$ the horizontal component. With only gravity acting on the -chain, the value of $H$ will be a constant. The value of $V$ will vary -with position. - -At a point $x$, there is $s(x)$ amount of chain with weight $w \cdot s(x)$. The tension is in the direction of the tangent line, so: - -```math -\tan(\theta) = y'(x) = \frac{w s(x)}{H}. -``` - -In terms of an increment of chain, we have: - -```math -\frac{w ds}{H} = d(y'(x)). -``` - -That is, the ratio of the vertical and horizontal tensions in the increment are in balance with the differential of the derivative. - - -But $ds = \sqrt{dx^2 + dy^2} = \sqrt{dx^2 + y'(x)^2 dx^2} = \sqrt{1 + y'(x)^2}dx$, so we can simplify to: - - -```math -\frac{w}{H}\sqrt{1 + y'(x)^2}dx =y''(x)dx. -``` - -This yields the second-order equation: - -```math -y''(x) = \frac{w}{H} \sqrt{1 + y'(x)^2}. -``` - -We enter this into `Julia`: - -```julia; -@syms w::positive H::positive y() -eqnc = D2(y)(x) ~ (w/H) * sqrt(1 + y'(x)^2) -``` - -Unfortunately, `SymPy` needs a bit of help with this problem, by breaking the problem into -steps. - -For the first step we solve for the derivative. Let $u = y'$, -then we have $u'(x) = (w/H)\sqrt{1 + u(x)^2}$: - -```julia; -eqnc₁ = subs(eqnc, D(y)(x) => u(x)) -``` - -and can solve via: - -```julia; -outc = dsolve(eqnc₁) -``` - -So $y'(x) = u(x) = \sinh(C_1 + w \cdot x/H)$. This can be solved by direct -integration as there is no $y(x)$ term on the right hand -side. - -```julia; -D(y)(x) ~ rhs(outc) -``` - -We see a simple linear transformation involving the hyperbolic sine. To avoid, `SymPy` struggling with the above equation, and knowing the hyperbolic sine is the derivative of the hyperbolic cosine, we anticipate an answer and verify it: - -```julia; -yc = (H/w)*cosh(C1 + w*x/H) -diff(yc, x) == rhs(outc) # == not \Equal[tab] -``` - -The shape is a hyperbolic cosine, known as the catenary. - - -```julia; echo=false -imgfile = "figures/verrazano-narrows-bridge-anniversary-historic-photos-2.jpeg" -caption = """ -The cables of an unloaded suspension bridge have a different shape than a loaded suspension bridge. As seen, the cables in this [figure](https://www.brownstoner.com/brooklyn-life/verrazano-narrows-bridge-anniversary-historic-photos/) would be modeled by a catenary. -""" -ImageFile(:ODEs, imgfile, caption) -``` - - ----- - -If the chain has a uniform load -- like a suspension bridge with a deck -- sufficient to make the weight of the chain negligible, then how does the above change? Then the vertical tension comes from $Udx$ and not $w ds$, so the equation becomes instead: - -```math -\frac{Udx}{H} = d(y'(x)). -``` - -This $y''(x) = U/H$, a constant. So it's answer will be a parabola. - - - -##### Example: projectile motion in a medium - - -The first example describes projectile motion without air resistance. If we use $(x(t), y(t))$ to describe position at time $t$, the functions satisfy: - -```math -x''(t) = 0, \quad y''(t) = -g. -``` - -That is, the $x$ position - where no forces act - has $0$ acceleration, and the $y$ position - where the force of gravity acts - has constant acceleration, $-g$, where $g=9.8m/s^2$ is the gravitational constant. These equations can be solved to give: - -```math -x(t) = x_0 + v_0 \cos(\alpha) t, \quad y(t) = y_0 + v_0\sin(\alpha)t - \frac{1}{2}g \cdot t^2. -``` - - -Furthermore, we can solve for $t$ from $x(t)$, to get an equation describing $y(x)$. Here are all the steps: - -```julia; hold=true -@syms x0::real y0::real v0::real alpha::real g::real -@syms t x u() -a1 = dsolve(D2(u)(x) ~ 0, u(x), ics=Dict(u(0) => x0, D(u)(0) => v0 * cos(alpha))) -a2 = dsolve(D2(u)(x) ~ -g, u(x), ics=Dict(u(0) => y0, D(u)(0) => v0 * sin(alpha))) -ts = solve(t - rhs(a1), x)[1] -y = simplify(rhs(a2)(t => ts)) -sympy.Poly(y, x).coeffs() -``` - -Though `y` is messy, it can be seen that the answer is a quadratic polynomial in $x$ yielding the familiar -parabolic motion for a trajectory. The output shows the coefficients. - - -In a resistive medium, there are drag forces at play. If this force is -proportional to the velocity, say, with proportion $\gamma$, then the -equations become: - -```math -\begin{align*} -x''(t) &= -\gamma x'(t), & \quad y''(t) &= -\gamma y'(t) -g, \\ -x(0) &= x_0, &\quad y(0) &= y_0,\\ -x'(0) &= v_0\cos(\alpha),&\quad y'(0) &= v_0 \sin(\alpha). -\end{align*} -``` - -We now attempt to solve these. - -```julia -@syms alpha::real, γ::postive, t::positive, v() -@syms x_0::real y_0::real v_0::real -Dₜ = Differential(t) -eq₁ = Dₜ(Dₜ(u))(t) ~ - γ * Dₜ(u)(t) -eq₂ = Dₜ(Dₜ(v))(t) ~ -g - γ * Dₜ(v)(t) - -a₁ = dsolve(eq₁, ics=Dict(u(0) => x_0, Dₜ(u)(0) => v_0 * cos(alpha))) -a₂ = dsolve(eq₂, ics=Dict(v(0) => y_0, Dₜ(v)(0) => v_0 * sin(alpha))) - -ts = solve(x - rhs(a₁), t)[1] -yᵣ = rhs(a₂)(t => ts) -``` - - -This gives $y$ as a function of $x$. - -There are a lot of symbols. Lets simplify by using constants $x_0=y_0=0$: - -```julia; -yᵣ₁ = yᵣ(x_0 => 0, y_0 => 0) -``` - - -What is the trajectory? We see -that the `log` function part will have issues when -$-\gamma x + v_0 \cos(\alpha) = 0$. - -If we fix some parameters, we can plot. - -```julia; -v₀, γ₀, α = 200, 1/2, pi/4 -soln = yᵣ₁(v_0=>v₀, γ=>γ₀, alpha=>α, g=>9.8) -plot(soln, 0, v₀ * cos(α) / γ₀ - 1/10, legend=false) -``` - -We can see that the resistance makes the path quite non-symmetric. - -## Visualizing a first-order initial value problem - -The solution, $y(x)$, is known through its derivative. A useful tool to visualize the solution to a first-order differential equation is the [slope field](http://tinyurl.com/jspzfok) (or direction field) plot, which at different values of $(x,y)$, plots a vector with slope given through $y'(x)$.The `vectorfieldplot` of the `CalculusWithJulia` package can be used to produce these. - - - -For example, in a previous example we found a solution to $y'(x) = x\cdot y(x)$, coded as - -```julia -F(y, x) = y*x -``` - -Suppose $x_0=1$ and $y_0=1$. Then a direction field plot is drawn through: - -```julia; hold=true -@syms x y -x0, y0 = 1, 1 - -plot(legend=false) -vectorfieldplot!((x,y) -> [1, F(y,x)], xlims=(x0, 2), ylims=(y0-5, y0+5)) - -f(x) = y0*exp(-x0^2/2) * exp(x^2/2) -plot!(f, linewidth=5) -``` - -In general, if the first-order equation is written as $y'(x) = F(y,x)$, then we plot a "function" that takes $(x,y)$ and returns an $x$ value of $1$ and a $y$ value of $F(y,x)$, so the slope is $F(y,x)$. - -!!! note - The order of variables in $F(y,x)$ is conventional with the equation $y'(x) = F(y(x),x)$. - - -The plots are also useful for illustrating solutions for different initial conditions: - - -```julia; hold=true -p = plot(legend=false) -x0, y0 = 1, 1 - -vectorfieldplot!((x,y) -> [1,F(y,x)], xlims=(x0, 2), ylims=(y0-5, y0+5)) -for y0 in -4:4 - f(x) = y0*exp(-x0^2/2) * exp(x^2/2) - plot!(f, x0, 2, linewidth=5) -end -p -``` - -Such solutions are called [integral -curves](https://en.wikipedia.org/wiki/Integral_curve). -These graphs illustrate the fact that the slope field is tangent to the graph of any -integral curve. - - - -## Questions - -##### Question - -Using `SymPy` to solve the differential equation - -```math -u' = \frac{1-x}{u} -``` - -gives - -```julia; hold=true -@syms x u() -dsolve(D(u)(x) - (1-x)/u(x)) -``` - -The two answers track positive and negative solutions. For the initial condition, $u(-1)=1$, we have the second one is appropriate: $u(x) = \sqrt{C_1 - x^2 + 2x}$. At $-1$ this gives: $1 = \sqrt{C_1-3}$, so $C_1 = 4$. - -This value is good for what values of $x$? - -```julia; hold=true; echo=false -choices = [ -"``[-1, \\infty)``", -"``[-1, 4]``", -"``[-1, 0]``", -"``[1-\\sqrt{5}, 1 + \\sqrt{5}]``"] -answ = 4 -radioq(choices, answ) -``` - - -##### Question - -Suppose $y(x)$ satisfies - -```math -y'(x) = y(x)^2, \quad y(1) = 1. -``` - -What is $y(3/2)$? - -```julia; hold=true; echo=false -@syms x u() -out = dsolve(D(u)(x) - u(x)^2, u(x), ics=Dict(u(1) => 1)) -val = N(rhs(out(3/2))) -numericq(val) -``` - -##### Question - -Solve the initial value problem - -```math -y' = 1 + x^2 + y(x)^2 + x^2 y(x)^2, \quad y(0) = 1. -``` - -Use your answer to find $y(1)$. - -```julia; hold=true; echo=false -eqn = D(u)(x) - (1 + x^2 + u(x)^2 + x^2 * u(x)^2) -out = dsolve(eqn, u(x), ics=Dict(u(0) => 1)) -val = N(rhs(out)(1).evalf()) -numericq(val) -``` - -##### Question - -A population is modeled by $y(x)$. The rate of population growth is generally proportional to the population ($k y(x)$), but as the population gets large, the rate is curtailed $(1 - y(x)/M)$. - -Solve the initial value problem - -```math -y'(x) = k\cdot y(x) \cdot (1 - \frac{y(x)}{M}), -``` - -when $k=1$, $M=100$, and $y(0) = 20$. Find the value of $y(5)$. - -```julia; hold=true;echo=false -k, M = 1, 100 -eqn = D(u)(x) - k * u(x) * (1 - u(x)/M) -out = dsolve(eqn, u(x), ics=Dict(u(0) => 20)) -val = N(rhs(out)(5)) -numericq(val) -``` - - -##### Question - -Solve the initial value problem - -```math -y'(t) = \sin(t) - \frac{y(t)}{t}, \quad y(\pi) = 1 -``` - -Find the value of the solution at $t=2\pi$. - -```julia; hold=true; echo=false -eqn = D(u)(x) - (sin(x) - u(x)/x) -out = dsolve(eqn, u(x), ics=Dict(u(PI) => 1)) -val = N(rhs(out(2PI))) -numericq(val) -``` - - -##### Question - -Suppose $u(x)$ satisfies: - -```math -\frac{du}{dx} = e^{-x} \cdot u(x), \quad u(0) = 1. -``` - -Find $u(5)$ using `SymPy`. - -```julia; hold=true; echo=false -eqn = D(u)(x) - exp(-x)*u(x) -out = dsolve(eqn, u(x), ics=Dict(u(0) => 1)) -val = N(rhs(out)(5)) -numericq(val) -``` - -##### Question - -The differential equation with boundary values - -```math -\frac{r^2 \frac{dc}{dr}}{dr} = 0, \quad c(1)=2, c(10)=1, -``` - -can be solved with `SymPy`. What is the value of $c(5)$? - -```julia; hold=true; echo=false -@syms x u() -eqn = diff(x^2*D(u)(x), x) -out = dsolve(eqn, u(x), ics=Dict(u(1)=>2, u(10) => 1)) |> rhs -out(5) # 10/9 -choices = ["``10/9``", "``3/2``", "``9/10``", "``8/9``"] -answ = 1 -radioq(choices, answ) -``` - - -##### Question - -The example with projectile motion in a medium has a parameter -$\gamma$ modeling the effect of air resistance. If `y` is the -answer - as would be the case if the example were copy-and-pasted -in - what can be said about `limit(y, gamma=>0)`? - -```julia; hold=true; echo=false -choices = [ -"The limit is a quadratic polynomial in `x`, mirroring the first part of that example.", -"The limit does not exist, but the limit to `oo` gives a quadratic polynomial in `x`, mirroring the first part of that example.", -"The limit does not exist -- there is a singularity -- as seen by setting `gamma=0`." -] -answ = 1 -radioq(choices, answ) -``` diff --git a/CwJ/ODEs/process.jl b/CwJ/ODEs/process.jl deleted file mode 100644 index 067dff1..0000000 --- a/CwJ/ODEs/process.jl +++ /dev/null @@ -1,27 +0,0 @@ -using WeavePynb -using Mustache - -mmd(fname) = mmd_to_html(fname, BRAND_HREF="../toc.html", BRAND_NAME="Calculus with Julia") -## uncomment to generate just .md files -mmd(fname) = mmd_to_md(fname, BRAND_HREF="../toc.html", BRAND_NAME="Calculus with Julia") - -fnames = [ - "odes", - "euler" - ] - - - -function process_file(nm, twice=false) - include("$nm.jl") - mmd_to_md("$nm.mmd") - markdownToHTML("$nm.md") - twice && markdownToHTML("$nm.md") -end - -process_files(twice=false) = [process_file(nm, twice) for nm in fnames] - -""" -## TODO ODEs - -""" diff --git a/CwJ/ODEs/solve.jmd b/CwJ/ODEs/solve.jmd deleted file mode 100644 index 8b63729..0000000 --- a/CwJ/ODEs/solve.jmd +++ /dev/null @@ -1,248 +0,0 @@ -# The problem-algorithm-solve interface - -This section uses these add-on packages: - -```julia -using Plots -using MonteCarloMeasurements -``` - -```julia; echo=false; results="hidden" -using CalculusWithJulia.WeaveSupport - -const frontmatter = ( - title = "The problem-algorithm-solve interface", - description = "Calculus with Julia: The problem-algorithm-solve interface", - tags = ["CalculusWithJulia", "odes", "the problem-algorithm-solve interface"], -); -fig_size = (800, 600) -nothing -``` - ----- - - -The [DifferentialEquations.jl](https://github.com/SciML) package is an entry point to a suite of `Julia` packages for numerically solving differential equations in `Julia` and other languages. A common interface is implemented that flexibly adjusts to the many different problems and algorithms covered by this suite of packages. In this section, we review a very informative [post](https://discourse.julialang.org/t/function-depending-on-the-global-variable-inside-module/64322/10) by discourse user `@genkuroki` which very nicely demonstrates the usefulness of the problem-algorithm-solve approach used with `DifferentialEquations.jl`. We slightly modify the presentation below for our needs, but suggest a perusal of the original post. - -##### Example: FreeFall - -The motion of an object under a uniform gravitational field is of interest. - -The parameters that govern the equation of motions are the gravitational constant, `g`; the initial height, `y0`; and the initial velocity, `v0`. The time span for which a solution is sought is `tspan`. - -A problem consists of these parameters. Typical `Julia` usage would be to create a structure to hold the parameters, which may be done as follows: - -```julia -struct Problem{G, Y0, V0, TS} - g::G - y0::Y0 - v0::V0 - tspan::TS -end - -Problem(;g=9.80665, y0=0.0, v0=30.0, tspan=(0.0,8.0)) = Problem(g, y0, v0, tspan) -``` - -The above creates a type, `Problem`, *and* a default constructor with default values. (The original uses a more sophisticated setup that allows the two things above to be combined.) - -Just calling `Problem()` will create a problem suitable for the earth, passing different values for `g` would be possible for other planets. - - -To solve differential equations there are many different possible algorithms. Here is the construction of two types to indicate two algorithms: - -```julia -struct EulerMethod{T} - dt::T -end -EulerMethod(; dt=0.1) = EulerMethod(dt) - -struct ExactFormula{T} - dt::T -end -ExactFormula(; dt=0.1) = ExactFormula(dt) -``` - -The above just specifies a type for dispatch --- the directions indicating what code to use to solve the problem. As seen, each specifies a size for a time step with default of `0.1`. - -A type for solutions is useful for different `show` methods or other methods. One can be created through: - - -```julia -struct Solution{Y, V, T, P<:Problem, A} - y::Y - v::V - t::T - prob::P - alg::A -end -``` - -The different algorithms then can be implemented as part of a generic `solve` function. Following the post we have: - -```julia -solve(prob::Problem) = solve(prob, default_algorithm(prob)) -default_algorithm(prob::Problem) = EulerMethod() - -function solve(prob::Problem, alg::ExactFormula) - g, y0, v0, tspan = prob.g, prob.y0, prob.v0, prob.tspan - dt = alg.dt - t0, t1 = tspan - t = range(t0, t1 + dt/2; step = dt) - - y(t) = y0 + v0*(t - t0) - g*(t - t0)^2/2 - v(t) = v0 - g*(t - t0) - - Solution(y.(t), v.(t), t, prob, alg) -end - -function solve(prob::Problem, alg::EulerMethod) - g, y0, v0, tspan = prob.g, prob.y0, prob.v0, prob.tspan - dt = alg.dt - t0, t1 = tspan - t = range(t0, t1 + dt/2; step = dt) - - n = length(t) - y = Vector{typeof(y0)}(undef, n) - v = Vector{typeof(v0)}(undef, n) - y[1] = y0 - v[1] = v0 - - for i in 1:n-1 - v[i+1] = v[i] - g*dt # F*h step of Euler - y[i+1] = y[i] + v[i]*dt # F*h step of Euler - end - - Solution(y, v, t, prob, alg) -end -``` - -The post has a more elegant means to unpack the parameters from the structures, but for each of the above, the parameters are unpacked, and then the corresponding algorithm employed. As of version `v1.7` of `Julia`, the syntax `(;g,y0,v0,tspan) = prob` could also be employed. - - -The exact formulas, ` y(t) = y0 + v0*(t - t0) - g*(t - t0)^2/2` and `v(t) = v0 - g*(t - t0)`, follow from well-known physics formulas. Each answer is wrapped in a `Solution` type so that the answers found can be easily extracted in a uniform manner. - -For example, plots of each can be obtained through: - -```julia -earth = Problem() -sol_euler = solve(earth) -sol_exact = solve(earth, ExactFormula()) - -plot(sol_euler.t, sol_euler.y; - label="Euler's method (dt = $(sol_euler.alg.dt))", ls=:auto) -plot!(sol_exact.t, sol_exact.y; label="exact solution", ls=:auto) -title!("On the Earth"; xlabel="t", legend=:bottomleft) -``` - -Following the post, since the time step `dt = 0.1` is not small enough, the error of the Euler method is rather large. Next we change the algorithm parameter, `dt`, to be smaller: - -```julia -earth₂ = Problem() -sol_euler₂ = solve(earth₂, EulerMethod(dt = 0.01)) -sol_exact₂ = solve(earth₂, ExactFormula()) - -plot(sol_euler₂.t, sol_euler₂.y; - label="Euler's method (dt = $(sol_euler₂.alg.dt))", ls=:auto) -plot!(sol_exact₂.t, sol_exact₂.y; label="exact solution", ls=:auto) -title!("On the Earth"; xlabel="t", legend=:bottomleft) -``` - -It is worth noting that only the first line is modified, and only the method requires modification. - -Were the moon to be considered, the gravitational constant would need adjustment. This parameter is part of the problem, not the solution algorithm. - -Such adjustments are made by passing different values to the `Problem` -constructor: - -```julia -moon = Problem(g = 1.62, tspan = (0.0, 40.0)) -sol_eulerₘ = solve(moon) -sol_exactₘ = solve(moon, ExactFormula(dt = sol_euler.alg.dt)) - -plot(sol_eulerₘ.t, sol_eulerₘ.y; - label="Euler's method (dt = $(sol_eulerₘ.alg.dt))", ls=:auto) -plot!(sol_exactₘ.t, sol_exactₘ.y; label="exact solution", ls=:auto) -title!("On the Moon"; xlabel="t", legend=:bottomleft) -``` - -The code above also adjusts the time span in addition to the -graviational constant. The algorithm for exact formula is set to use -the `dt` value used in the `euler` formula, for easier -comparison. Otherwise, outside of the labels, the patterns are the -same. Only those things that need changing are changed, the rest comes -from defaults. - -The above shows the benefits of using a common interface. Next, the post illustrates how *other* authors could extend this code, simply by adding a *new* `solve` method. For example, - -```julia -struct Symplectic2ndOrder{T} - dt::T -end -Symplectic2ndOrder(;dt=0.1) = Symplectic2ndOrder(dt) - -function solve(prob::Problem, alg::Symplectic2ndOrder) - g, y0, v0, tspan = prob.g, prob.y0, prob.v0, prob.tspan - dt = alg.dt - t0, t1 = tspan - t = range(t0, t1 + dt/2; step = dt) - - n = length(t) - y = Vector{typeof(y0)}(undef, n) - v = Vector{typeof(v0)}(undef, n) - y[1] = y0 - v[1] = v0 - - for i in 1:n-1 - ytmp = y[i] + v[i]*dt/2 - v[i+1] = v[i] - g*dt - y[i+1] = ytmp + v[i+1]*dt/2 - end - - Solution(y, v, t, prob, alg) -end -``` - -Had the two prior methods been in a package, the other user could still extend the interface, as above, with just a slight standard modification. - - -The same approach works for this new type: - -```julia - -earth₃ = Problem() -sol_sympl₃ = solve(earth₃, Symplectic2ndOrder(dt = 2.0)) -sol_exact₃ = solve(earth₃, ExactFormula()) - -plot(sol_sympl₃.t, sol_sympl₃.y; label="2nd order symplectic (dt = $(sol_sympl₃.alg.dt))", ls=:auto) -plot!(sol_exact₃.t, sol_exact₃.y; label="exact solution", ls=:auto) -title!("On the Earth"; xlabel="t", legend=:bottomleft) -``` - -Finally, the author of the post shows how the interface can compose with other packages in the `Julia` package ecosystem. This example uses the external package `MonteCarloMeasurements` which plots the behavior of the system for perturbations of the initial value: - - -```julia - -earth₄ = Problem(y0 = 0.0 ± 0.0, v0 = 30.0 ± 1.0) -sol_euler₄ = solve(earth₄) -sol_sympl₄ = solve(earth₄, Symplectic2ndOrder(dt = 2.0)) -sol_exact₄ = solve(earth₄, ExactFormula()) - -ylim = (-100, 60) -P = plot(sol_euler₄.t, sol_euler₄.y; - label="Euler's method (dt = $(sol_euler₄.alg.dt))", ls=:auto) -title!("On the Earth"; xlabel="t", legend=:bottomleft, ylim) - -Q = plot(sol_sympl₄.t, sol_sympl₄.y; - label="2nd order symplectic (dt = $(sol_sympl₄.alg.dt))", ls=:auto) -title!("On the Earth"; xlabel="t", legend=:bottomleft, ylim) - -R = plot(sol_exact₄.t, sol_exact₄.y; label="exact solution", ls=:auto) -title!("On the Earth"; xlabel="t", legend=:bottomleft, ylim) - -plot(P, Q, R; size=(720, 600)) -``` - -The only change was in the problem, `Problem(y0 = 0.0 ± 0.0, v0 = 30.0 ± 1.0)`, where a different number type is used which accounts for uncertainty. The rest follows the same pattern. - -This example, shows the flexibility of the problem-algorithm-solver pattern while maintaining a consistent pattern for execution. diff --git a/CwJ/Project.toml b/CwJ/Project.toml deleted file mode 100644 index 81648c0..0000000 --- a/CwJ/Project.toml +++ /dev/null @@ -1 +0,0 @@ -[deps] diff --git a/CwJ/TODO/AD.md b/CwJ/TODO/AD.md deleted file mode 100644 index 8d6c12e..0000000 --- a/CwJ/TODO/AD.md +++ /dev/null @@ -1,3 +0,0 @@ -Good paper recommended here (https://discourse.julialang.org/t/learning-automatic-differentiation/56158/3) - -https://www.jmlr.org/papers/volume18/17-468/17-468.pdf diff --git a/CwJ/TODO/arrows.md b/CwJ/TODO/arrows.md deleted file mode 100644 index 1f9214a..0000000 --- a/CwJ/TODO/arrows.md +++ /dev/null @@ -1,61 +0,0 @@ -This is really just - -plot!([0,cos(θ)],[0,sin(θ)], arrow=true) - - -https://stackoverflow.com/questions/58219191/drawing-an-arrow-with-specified-direction-on-a-point-in-scatter-plot-in-julia - -https://github.com/m3g/CKP/blob/master/disciplina/codes/velocities.jl - - - - -using Plots -using LaTeXStrings - -function arch(θ₁,θ₂;radius=1.,Δθ=1.) - θ₁ = π*θ₁/180 - θ₂ = π*θ₂/180 - Δθ = π*Δθ/180 - l = round(Int,(θ₂-θ₁)/Δθ) - x = zeros(l) - y = zeros(l) - for i in 1:l - θ = θ₁ + i*Δθ - x[i] = radius*cos(θ) - y[i] = radius*sin(θ) - end - return x, y -end - -plot() - -x, y = arch(0,360) -plot(x,y,seriestype=:shape,label="",alpha=0.5) - -x, y = arch(0,360,radius=0.95) -plot!(x,y,seriestype=:shape,label="",fillcolor=:white) - -x, y = arch(0,360,radius=0.7) -plot!(x,y,seriestype=:shape,label="",alpha=0.5,fillcolor=:red) - -x, y = arch(0,360,radius=0.65) -plot!(x,y,seriestype=:shape,label="",fillcolor=:white) - -plot!([0,0],[0,1.1],arrow=true,color=:black,linewidth=2,label="") -plot!([0,1.1],[0,0],arrow=true,color=:black,linewidth=2,label="") - -x, y = arch(15,16,radius=0.65) -plot!([0,x[1]],[0,y[1]],arrow=true,color=:black,linewidth=1,label="") - -x, y = arch(35,36,radius=0.95) -plot!([0,x[1]],[0,y[1]],arrow=true,color=:black,linewidth=1,label="") - -plot!(draw_arrow=true) -plot!(showaxis=:no,ticks=nothing,xlim=[-0.1,1.1],ylim=[-0.1,1.1],) -plot!(xlabel="x",ylabel="y",size=(400,400)) - -annotate!(0.58,-0.07,text(L"\Delta v_1",10)) -annotate!(0.88,-0.07,text(L"\Delta v_2",10)) - -savefig("./velocities.pdf") diff --git a/CwJ/TODO/earth.jl b/CwJ/TODO/earth.jl deleted file mode 100644 index 252f273..0000000 --- a/CwJ/TODO/earth.jl +++ /dev/null @@ -1,30 +0,0 @@ -# Calculate the temperature of the earth using the simplest model -# @jake -# https://discourse.julialang.org/t/seven-lines-of-julia-examples-sought/50416/121 - -using Unitful, Plots -p_sun = 386e24u"W" # power output of the sun -radius_a = 6378u"km" # semi-major axis of the earth -radius_b = 6357u"km" # semi-minor axis of the earth -orbit_a = 149.6e6u"km" # distance from the sun to earth -orbit_e = 0.017 # eccentricity of r = a(1-e^2)/(1+ecos(θ)) & time ≈ 365.25 * θ / 360 where θ is in degrees -a = 0.75 # absorptivity of the sun's radiation -e = 0.6 # emmissivity of the earth (very dependent on cloud cover) -σ = 5.6703e-8u"W*m^-2*K^-4" # Stefan-Boltzman constant -temp_sky = 3u"K" # sky temperature - - - -t = (0:0.25:365.25)u"d" # day of year in 1/4 day increments -θ = 2*π/365.25u"d" .* t # approximate angle around the sun -r = orbit_a * (1-orbit_e^2) ./ (1 .+ orbit_e .* cos.(θ)) # distance from sun to earth -area_projected = π * radius_a * radius_b # area of earth facing the sun -ec = sqrt(1-radius_b^2/radius_a^2) # eccentricity of earth - -area_surface = 2*π*radius_a^2*(1 + radius_b^2/(ec*radius_b^2)*atanh(ec)) # surface area of the earth - -q_in = p_sun * a * area_projected ./ (4 * π .* r.^2) # total heat impacting the earth - -temp_earth = (q_in ./ (e*σ*area_surface) .+ temp_sky^4).^0.25 # temperature of the earth - -plot(t*u"d^-1", temp_earth*u"K^-1" .- 273.15, label = false, title = "Temperature of Earth", xlabel = "Day", ylabel = "Temperature [C]") diff --git a/CwJ/TODO/ladder-questions.md b/CwJ/TODO/ladder-questions.md deleted file mode 100644 index da1b3af..0000000 --- a/CwJ/TODO/ladder-questions.md +++ /dev/null @@ -1,115 +0,0 @@ -###### Question (Ladder [questions](http://www.mathematische-basteleien.de/ladder.htm)) - - -A ``7``meter ladder leans against wall with the base ``1.5``meters from wall at its base. At which height does the ladder touch the wall? - -```julia; hold=true; echo=false -l = 7 -adj = 1.5 -opp = sqrt(l^2 - adj^2) -numericq(opp, 1e-3) -``` - - ----- - -A ``7``meter ladder leans against the wall. Between the ladder and the wall is a ``1``m cube box. The ladder touches the wall, the box and the ground. There are two such positions, what is the height of the ladder of the more upright position? - -You might find this code of help: - -```julia; eval=false -@syms x y -l, b = 7, 1 -eq = (b+x)^2 + (b+y)^2 -eq = subs(eq, x=> b*(b/y)) # x/b = b/y -solve(eq ~ l^2, y) -``` - -What is the value `b+y` in the above? - -```julia; echo=false -radioq(("The height of the ladder", - "The height of the box plus ladder", - "The distance from the base of the ladder to the box," - "The distance from the base of the ladder to the base of the wall" - ),1) -``` - - -What is the height of the ladder - -```julia; hold=true; echo=false -numericq(6.90162289514212, 1e-3) -``` - - ----- - -A ladder of length ``c`` is to moved through a 2-dimensional hallway of width ``b`` which has a right angled bend. If ``4b=c``, when will the ladder get stuck? - -Consider this picture - -```julia; hold=true; echo=false -p = plot(; axis=nothing, legend=false, aspect_ratio=:equal) -x,y=1,2 -b = sqrt(x*y) -plot!(p, [0,0,b+x], [b+y,0,0], linestyle=:dot) -plot!(p, [0,b+x],[b,b], color=:black, linestyle=:dash) -plot!(p, [b,b],[0,b+y], color=:black, linestyle=:dash) -plot!(p, [b+x,0], [0, b+y], color=:black) -``` - - -Suppose ``b=5``, then with ``b+x`` and ``b+y`` being the lengths on the walls where it is stuck *and* by similar triangles ``b/x = y/b`` we can solve for ``x``. (In the case take the largest positive value. The answer would be the angle ``\theta`` with ``\tan(\theta) = (b+y)/(b+x)``. - -```julia; hold=true; echo=false -b = 5 -l = 4*b -@syms x y -eq = (b+x)^2 + (b+y)^2 -eq =subs(eq, y=> b^2/x) -x₀ = N(maximum(filter(>(0), solve(eq ~ l^2, x)))) -y₀ = b^2/x₀ -θ₀ = Float64(atan((b+y₀)/(b+x₀))) -numericq(θ₀, 1e-2) -``` - - ------ - -Two ladders of length ``a`` and ``b`` criss-cross between two walls of width ``x``. They meet at a height of ``c``. - -```julia; hold=true; echo=false -p = plot(; legend=false, axis=nothing, aspect_ratio=:equal) -ya,yb,x = 2,3,1 -plot!(p, [0,x],[ya,0], color=:black) -plot!(p, [0,x],[0, yb], color=:black) -plot!(p, [0,0], [0,yb], color=:blue, linewidth=5) -plot!(p, [x,x], [0,yb], color=:blue, linewidth=5) -plot!(p, [0,x], [0,0], color=:blue, linewidth=5) -xc = ya/(ya+yb) -c = yb*xc -plot!(p, [xc,xc],[0,c]) -p -``` - -Suppose ``c=1``, ``b=3``, and ``a=5``. Find ``x``. - -Introduce ``x = z + y``, and ``h`` and ``k`` the heights of the ladders along the left wall and the right wall. - -The ``z/c = x/k`` and ``y/c = x/h`` by similar triangles. As ``z + y`` is ``x`` we can solve to get - -```math -x = z + y = \frac{xc}{k} + \frac{xc}{h} - = \frac{xc}{\sqrt{b^2 - x^2}} + \frac{xc}{\sqrt{a^2 - x^2}} -``` - -With ``a,b,c`` as given, this can be solved with - -```julia; hold=true; echo=false -a,b,c = 5, 3, 1 -f(x) = x*c/sqrt(b^2 - x^2) + x*c/sqrt(a^2 - x^2) - x -find_zero(f, (0, b)) -``` - -The answer is ``2.69\dots``. diff --git a/CwJ/TODO/partialcircle.jl b/CwJ/TODO/partialcircle.jl deleted file mode 100644 index 30a1186..0000000 --- a/CwJ/TODO/partialcircle.jl +++ /dev/null @@ -1,4 +0,0 @@ -plot([0,1,1,0], [0,0,1,0], aspect_ratio=:equal, legend=false) - plot!(Plots.partialcircle(0, pi/4,100, 0.25), arrow=true) - Δ = 0.05 - plot!([1-Δ, 1-Δ, 1], [0,Δ,Δ]) diff --git a/CwJ/TODO/ti-30-image.png b/CwJ/TODO/ti-30-image.png deleted file mode 100644 index 050f0a37bf73dcc6ffe13568323f5ccbc48068ba..0000000000000000000000000000000000000000 GIT binary patch literal 0 HcmV?d00001 literal 1103158 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