daviddoji/CalculusWithJuliaNotes.jl
1
0
Fork 0
Code Issues Pull Requests Packages Projects Releases Wiki Activity

edits

Browse Source
This commit is contained in:
jverzani
2025-07-23 08:05:43 -04:00
parent 31ce21c8ad
commit c3a94878f3
50 changed files with 3711 additions and 1385 deletions
Download Patch File Download Diff File Expand all files Collapse all files

120
quarto/limits/continuity.qmd
View File

@@ -72,7 +72,7 @@ The definition says three things
* The value of the limit is the same as $f(c)$.
The defined speaks to continuity at a point, we can extend it to continuity over an interval $(a,b)$ by saying:
The definition speaks to continuity at a point, we can extend it to continuity over an interval $(a,b)$ by saying:
::: {.callout-note icon=false}
## Definition of continuity over an open interval
@@ -130,9 +130,9 @@ There are various reasons why a function may not be continuous.
$$
f(x) = \begin{cases}
-1 & x < 0 \\
0 & x = 0 \\
1 & x > 0
-1 &~ x < 0 \\
0 &~ x = 0 \\
1 &~ x > 0
\end{cases}
$$
@@ -148,25 +148,57 @@ is implemented by `Julia`'s `sign` function. It has a value at $0$, but no limit
plot([-1,-.01], [-1,-.01], legend=false, color=:black)
plot!([.01, 1], [.01, 1], color=:black)
scatter!([0], [1/2], markersize=5, markershape=:circle)
ts = range(0, 2pi, 100)
C = Shape(0.02 * sin.(ts), 0.03 * cos.(ts))
plot!(C, fill=(:white,1), line=(:black, 1))
```
is not continuous at $x=0$. It has a limit of $0$ at $0$, a function value $f(0) =1/2$, but the limit and the function value are not equal.
* The `floor` function, which rounds down to the nearest integer, is also not continuous at the integers, but is right continuous at the integers, as, for example, $\lim_{x \rightarrow 0+} f(x) = f(0)$. This graph emphasizes the right continuity by placing a point for the value of the function when there is a jump:
* The `floor` function, which rounds down to the nearest integer, is also not continuous at the integers, but is right continuous at the integers, as, for example, $\lim_{x \rightarrow 0+} f(x) = f(0)$. This graph emphasizes the right continuity by placing a filled marker for the value of the function when there is a jump and an open marker where the function is not that value.
```{julia}
#| hold: true
#| echo: false
x = [0,1]; y=[0,0]
plt = plot(x.-2, y.-2, color=:black, legend=false)
plot!(plt, x.-1, y.-1, color=:black)
plot!(plt, x.-0, y.-0, color=:black)
plot!(plt, x.+1, y.+1, color=:black)
plot!(plt, x.+2, y.+2, color=:black)
scatter!(plt, [-2,-1,0,1,2], [-2,-1,0,1,2], markersize=5, markershape=:circle)
plt = let
empty_style = (xticks=-4:4, yticks=-4:4,
framestyle=:origin,
legend=false)
axis_style = (arrow=true, side=:head, line=(:gray, 1))
text_style = (10,)
fn_style = (;line=(:black, 3))
fn2_style = (;line=(:red, 4))
mark_style = (;line=(:gray, 1, :dot))
domain_style = (;fill=(:orange, 0.35), line=nothing)
range_style = (; fill=(:blue, 0.35), line=nothing)
ts = range(0, 2pi, 100)
xys = sincos.(ts)
xys = [.1 .* xy for xy in xys]
plot(; empty_style..., aspect_ratio=:equal)
plot!([-4.25,4.25], [0,0]; axis_style...)
plot!([0,0], [-4.25, 4.25]; axis_style...)
for k in -4:4
P,Q = (k,k),(k+1,k)
plot!([P,Q], line=(:black,1))
S = Shape([k .+ xy for xy in xys])
plot!(S; fill=(:black,))
S = Shape([(k+1,k) .+ xy for xy in xys])
plot!(S; fill=(:white,), line=(:black,1))
end
current()
end
plt
```
```{julia}
#| echo: false
plotly()
nothing
```
* The function $f(x) = 1/x^2$ is not continuous at $x=0$: $f(x)$ is not defined at $x=0$ and $f(x)$ has no limit at $x=0$ (in the usual sense).
@@ -176,8 +208,8 @@ plt
$$
f(x) =
\begin{cases}
0 & \text{if } x \text{ is irrational,}\\
1 & \text{if } x \text{ is rational.}
0 &~ \text{if } x \text{ is irrational,}\\
1 &~ \text{if } x \text{ is rational.}
\end{cases}
$$
@@ -192,8 +224,8 @@ Let a function be defined by cases:
$$
f(x) = \begin{cases}
3x^2 + c & x \geq 0,\\
2x-3 & x < 0.
3x^2 + c &~ x \geq 0,\\
2x-3 &~ x < 0.
\end{cases}
$$
@@ -383,8 +415,8 @@ Let $f(x)$ be defined by
$$
f(x) = \begin{cases}
c + \sin(2x - \pi/2) & x > 0\\
3x - 4 & x \leq 0.
c + \sin(2x - \pi/2) &~ x > 0\\
3x - 4 &~ x \leq 0.
\end{cases}
$$
@@ -423,12 +455,22 @@ Consider the function $f(x)$ given by the following graph
```{julia}
#| hold: true
#| echo: false
xs = range(0, stop=2, length=50)
plot(xs, [sqrt(1 - (x-1)^2) for x in xs], legend=false, xlims=(0,4))
plot!([2,3], [1,0])
scatter!([3],[0], markersize=5)
plot!([3,4],[1,0])
scatter!([4],[0], markersize=5)
let
xs = range(0, stop=2, length=50)
plot(xs, [sqrt(1 - (x-1)^2) for x in xs];
line=(:black,1),
legend=false, xlims=(-0.1,4.1))
plot!([2,3], [1,0]; line=(:black,1))
plot!([3,4],[1,0]; line=(:black,1))
scatter!([(0,0)], markersize=5, markercolor=:black)
scatter!([(2,0)], markersize=5, markercolor=:white)
scatter!([(2, 1)], markersize=5; markercolor=:black)
scatter!([(3,0)], markersize=5; markercolor=:black)
scatter!([(3,1)], markersize=5; markercolor=:white)
scatter!([(4,0)], markersize=5; markercolor=:black)
end
```
The function $f(x)$ is continuous at $x=1$?
@@ -513,3 +555,29 @@ choices = ["Can't tell",
answ = 1
radioq(choices, answ)
```
###### Question
A parametric equation is specified by a parameterization $(f(t), g(t)), a \leq t \leq b$. The parameterization will be continuous if and only if each function is continuous.
Suppose $k_x$ and $k_y$ are positive integers and $a, b$ are positive numbers, will the [Lissajous](https://en.wikipedia.org/wiki/Parametric_equation#Lissajous_Curve) curve given by $(a\cos(k_x t), b\sin(k_y t))$ be continuous?
```{julia}
#| hold: true
#| echo: false
yesnoq(true)
```
Here is a sample graph for $a=1, b=2, k_x=3, k_y=4$:
```{julia}
#| hold: true
a,b = 1, 2
k_x, k_y = 3, 4
plot(t -> a * cos(k_x *t), t-> b * sin(k_y * t), 0, 4pi)
```

414
quarto/limits/intermediate_value_theorem.qmd
View File

@@ -17,9 +17,9 @@ using SymPy
---
![Between points M and M lies an F](figures/ivt.jpg){width=40%}
![Between points M and M lies an F for a continuous curve. [L'Hospitals](https://ia801601.us.archive.org/26/items/infinimentpetits1716lhos00uoft/infinimentpetits1716lhos00uoft.pdf) figure 55.](figures/ivt.jpg){width=40%}
Continuity for functions is a valued property which carries implications. In this section we discuss two: the intermediate value theorem and the extreme value theorem. These two theorems speak to some fundamental applications of calculus: finding zeros of a function and finding extrema of a function. [L'Hospitals](https://ia801601.us.archive.org/26/items/infinimentpetits1716lhos00uoft/infinimentpetits1716lhos00uoft.pdf) figure 55, above, suggests why.
Continuity for functions is a valued property which carries implications. In this section we discuss two: the intermediate value theorem and the extreme value theorem. These two theorems speak to some fundamental applications of calculus: finding zeros of a function and finding extrema of a function.
## Intermediate Value Theorem
@@ -38,7 +38,7 @@ If $f$ is continuous on $[a,b]$ with, say, $f(a) < f(b)$, then for any $y$ with
#| echo: false
#| cache: true
### {{{IVT}}}
plt = let
let
gr()
# IVT
empty_style = (xaxis=([], false),
@@ -68,7 +68,7 @@ plt = let
plot!([(a,0),(a,f(a))]; line=(:black, 1, :dash))
plot!([(b,0),(b,f(b))]; line=(:black, 1, :dash))
m = f(a/2 + b/2)
m = f(a/2 + b/2) + 1.5
plot!([a, b], [m,m]; line=(:black, 1, :dashdot))
δx = 0.03
@@ -76,10 +76,10 @@ plt = let
domain_style...)
plot!(Shape((a-.1) .+ 2δx * [-1,1,1,-1], [f(a),f(a),f(b), f(b)]);
range_style...)
plot!(Shape((a-.1) .+ δx/2 * [-1,1,1,-1], [y0,y0,y1,y1]);
plot!(Shape((a-.1) .+ 2δx/3 * [-1,1,1,-1], [y0,y0,y1,y1]);
range_style...)
zs = find_zeros(f, (a,b))
zs = find_zeros(x -> f(x) - m, (a,b))
c = zs[2]
plot!([(c,0), (c,f(c))]; line=(:black, 1, :dashdot))
@@ -89,52 +89,11 @@ plt = let
(c, 0, text(L"c", 12, :top)),
(a-.1, f(a), text(L"f(a)", 12, :right)),
(a-.1, f(b), text(L"f(b)", 12, :right)),
(b, m, text(L"y", 12, :left)),
(a-0.2, m, text(L"y", 12, :right)),
])
end
plt
#=
function IVT_graph(n)
f(x) = sin(pi*x) + 9x/10
a,b = [0,3]
xs = range(a,stop=b, length=50)
## cheat -- pick an x, then find a y
Δ = .2
x = range(a + Δ, stop=b - Δ, length=6)[n]
y = f(x)
plt = plot(f, a, b, legend=false, size=fig_size)
plot!(plt, [0,x,x], [f(x),f(x),0], color=:orange, linewidth=3)
plt
end
n = 6
anim = @animate for i=1:n
IVT_graph(i)
end
imgfile = tempname() * ".gif"
gif(anim, imgfile, fps = 1)
caption = L"""
Illustration of intermediate value theorem. The theorem implies that any randomly chosen $y$
value between $f(a)$ and $f(b)$ will have at least one $x$ in $[a,b]$
with $f(x)=y$.
"""
plotly()
ImageFile(imgfile, caption)
=#
```
```{julia}
#| echo: false
plotly()
@@ -145,7 +104,7 @@ Illustration of the intermediate value theorem. The theorem implies that any ran
:::
In the early years of calculus, the intermediate value theorem was intricately connected with the definition of continuity, now it is a consequence.
In the early years of calculus, the intermediate value theorem was intricately connected with the definition of continuity, now it is an important consequence.
The basic proof starts with a set of points in $[a,b]$: $C = \{x \text{ in } [a,b] \text{ with } f(x) \leq y\}$. The set is not empty (as $a$ is in $C$) so it *must* have a largest value, call it $c$ (this might seem obvious, but it requires the completeness property of the real numbers). By continuity of $f$, it can be shown that $\lim_{x \rightarrow c-} f(x) = f(c) \leq y$ and $\lim_{y \rightarrow c+}f(x) =f(c) \geq y$, which forces $f(c) = y$.
@@ -157,18 +116,6 @@ The basic proof starts with a set of points in $[a,b]$: $C = \{x \text{ in } [a,
Suppose we have a continuous function $f(x)$ on $[a,b]$ with $f(a) < 0$ and $f(b) > 0$. Then as $f(a) < 0 < f(b)$, the intermediate value theorem guarantees the existence of a $c$ in $[a,b]$ with $f(c) = 0$. This was a special case of the intermediate value theorem proved by Bolzano first. Such $c$ are called *zeros* of the function $f$.
We use this fact when building a "sign chart" of a polynomial function. Between any two consecutive real zeros the polynomial can not change sign. (Why?) So a "test point" can be used to determine the sign of the function over an entire interval.
The `sign_chart` function from `CalculusWithJulia` uses this to indicate where an *assumed* continuous function changes sign:
```{julia}
f(x) = sin(x + x^2) + x/2
sign_chart(f, -3, 3)
```
The intermediate value theorem can find the sign of the function *between* adjacent zeros, but how are the zeros identified?
Here, we use the Bolzano theorem to give an algorithm - the *bisection method* - to locate a value $c$ in $[a,b]$ with $f(c) = 0$ under the assumptions:
* $f$ is continuous on $[a,b]$
@@ -178,7 +125,7 @@ Here, we use the Bolzano theorem to give an algorithm - the *bisection method* -
::: {.callout-note}
#### Between
The bisection method is used to find a zero, $c$, of $f(x)$ *between* two values, $a$ and $b$. The method is guaranteed to work under assumptions, the most important being the continuous function having different signs at $a$ and $b$.
The bisection method is used to find a zero, $c$, of $f(x)$ *between* two values, $a$ and $b$. The method is guaranteed to work under the assumption of a continuous function having different signs at $a$ and $b$.
:::
@@ -305,7 +252,7 @@ sin(c)
(Even `1pi` itself is not a "zero" due to floating point issues.)
### The `find_zero` function.
### The `find_zero` function to solve `f(x) = 0`
The `Roots` package has a function `find_zero` that implements the bisection method when called as `find_zero(f, (a,b))` where $[a,b]$ is a bracket. Its use is similar to `simple_bisection` above. This package is loaded when `CalculusWithJulia` is. We illlustrate the usage of `find_zero` in the following:
@@ -315,8 +262,8 @@ The `Roots` package has a function `find_zero` that implements the bisection met
xstar = find_zero(sin, (3, 4))
```
:::{.callout-warning}
## Warning
:::{.callout-note}
## Action template
Notice, the call `find_zero(sin, (3, 4))` again fits the template `action(function, args...)` that we see repeatedly. The `find_zero` function can also be called through `fzero`. The use of `(3, 4)` to specify the interval is not necessary. For example `[3,4]` would work equally as well. (Anything where `extrema` is defined works.)
:::
@@ -396,6 +343,10 @@ It appears (and a plot over $[0,1]$ verifies) that there is one zero between $-2
find_zero(x^3 - x + 1, (-2, -1))
```
#### The `find_zero` function to solve `f(x) = c`
Solving `f(x) = c` is related to solving `h(x) = 0`. The key is to make a new function using the difference of the two sides: `h(x) = f(x) - c`.
##### Example
Solve for a value of $x$ where `erfc(x)` is equal to `0.5`.
@@ -415,6 +366,40 @@ find_zero(h, (-Inf, Inf)) # as wide as possible in this case
```
##### Example: Inverse functions
If $f(x)$ is *monotonic* and *continuous* over an interval $[a,b]$ then it has an *inverse function*. That is for any $y$ between $f(a)$ and $f(b)$ we can find an $x$ satisfying $y = f(x)$ with $a \leq x \leq b$. This is due, of course, to both the intermediate value theorem (which guarantees an $x$) and monotonicity (which guarantees just one $x$).
To see how we can *numerically* find an inverse function using `find_zero`, we have this function:
```{julia}
function inverse_function(f, a, b, args...; kwargs...)
fa, fb = f(a), f(b)
m, M = fa < fb ? (fa, fb) : (fb, fa)
y -> begin
@assert m ≤ y ≤ M
find_zero(x ->f(x) - y, (a,b), args...; kwargs...)
end
end
```
The check on `fa < fb` is due to the possibility that $f$ is increasing (in which case `fa < fb`) or decreasing (in which case `fa > fb`).
To see this used, we consider the monotonic function $f(x) = x - \sin(x)$ over $[0, 5\pi]$. To graph, we have:
```{julia}
f(x) = x - sin(x)
a, b = 0, 5pi
plot(inverse_function(f, a, b), f(a), f(b); aspect_ratio=:equal)
```
(We plot over the range $[f(a), f(b)]$ here, as we can guess $f(x)$ is *increasing*.)
#### The `find_zero` function to solve `f(x) = g(x)`
Solving `f(x) = g(x)` is related to solving `h(x) = 0`. The key is to make a new function using the difference of the two sides: `h(x) = f(x) - g(x)`.
##### Example
@@ -455,36 +440,6 @@ find_zero(cos(x) ~ x, (0, 2))
[![Intersection of two curves as illustrated by Canadian artist Kapwani Kiwanga.](figures/intersection-biennale.jpg)](https://www.gallery.ca/whats-on/touring-exhibitions-and-loans/around-the-world/canada-pavilion-at-the-venice-biennale/kapwani-kiwanga-trinket){width=40%}
##### Example: Inverse functions
If $f(x)$ is *monotonic* and *continuous* over an interval $[a,b]$ then it has an *inverse function*. That is for any $y$ between $f(a)$ and $f(b)$ we can find an $x$ satisfying $y = f(x)$ with $a \leq x \leq b$. This is due, of course, to both the intermediate value theorem (which guarantees an $x$) and monotonicity (which guarantees just one $x$).
To see how we can *numerically* find an inverse function using `find_zero`, we have this function:
```{julia}
function inverse_function(f, a, b, args...; kwargs...)
fa, fb = f(a), f(b)
m, M = fa < fb ? (fa, fb) : (fb, fa)
y -> begin
@assert m ≤ y ≤ M
find_zero(x ->f(x) - y, (a,b), args...; kwargs...)
end
end
```
The check on `fa < fb` is due to the possibility that $f$ is increasing (in which case `fa < fb`) or decreasing (in which case `fa > fb`).
To see this used, we consider the monotonic function $f(x) = x - \sin(x)$ over $[0, 5\pi]$. To graph, we have:
```{julia}
f(x) = x - sin(x)
a, b = 0, 5pi
plot(inverse_function(f, a, b), f(a), f(b); aspect_ratio=:equal)
```
(We plot over the range $[f(a), f(b)]$ here, as we can guess $f(x)$ is *increasing*.)
##### Example
@@ -679,7 +634,7 @@ Geometry will tell us that $\cos(x) - x/p$ for *one* $x$ in $[0, \pi/2]$ wheneve
#| hold: true
f(x, p=1) = cos(x) - x/p
I = (0, pi/2)
find_zero(f, I), find_zero(f, I, p=2)
find_zero(f, I), find_zero(f, I; p=2)
```
The second number is the solution when `p=2`.
@@ -752,7 +707,7 @@ f.(zs)
The `find_zero` function in the `Roots` package is an interface to one of several methods. For now we focus on the *bracketing* methods, later we will see others. Bracketing methods, among others, include `Roots.Bisection()`, the basic bisection method though with a different sense of "middle" than $(a+b)/2$ and used by default above; `Roots.A42()`, which will typically converge much faster than simple bisection; `Roots.Brent()` for the classic method of Brent, and `FalsePosition()` for a family of *regula falsi* methods. These can all be used by specifying the method in a call to `find_zero`.
Alternatively, `Roots` implements the `CommonSolve` interface popularized by its use in the `DifferentialEquations.jl` ecosystem, a wildly successful area for `Julia`. The basic setup involves two steps: setup a "problem;" solve the problem.
Alternatively, `Roots` implements the `CommonSolve` interface popularized by its use in the `DifferentialEquations.jl` ecosystem, a wildly successful area for `Julia`. The basic setup involves two steps: setup a "problem"; solve the problem.
To set up a problem we call `ZeroProblem` with the function and an initial interval, as in:
@@ -822,6 +777,74 @@ nothing
[![Elevation profile of the Hardrock 100 ultramarathon. Treating the elevation profile as a function, the absolute maximum is just about 14,000 feet and the absolute minimum about 7600 feet. These are of interest to the runner for different reasons. Also of interest would be each local maxima and local minima - the peaks and valleys of the graph - and the total elevation climbed - the latter so important/unforgettable its value makes it into the chart's title.
](figures/hardrock-100.jpeg)](https://hardrock100.com){width=50%}
This figure shows the two concepts as well.
::: {#fig-absolute-relative}
```{julia}
#| echo: false
plt = let
gr()
empty_style = (xaxis=([], false),
yaxis=([], false),
framestyle=:origin,
legend=false)
axis_style = (arrow=true, side=:head, line=(:gray, 1))
p(x) = (x-1)*(x-2)*(x-3)*(x-4) + x/2 + 2
a, b = 0.25, 4.5
z₁, z₂, z₃ = zs = find_zeros(x -> ForwardDiff.derivative(p,x), (a, b))
a′ = -0.0
plot(; empty_style...)
plot!(p, a, b; line=(:black, 2))
plot!([a′,b+0.25], [0,0]; axis_style...)
plot!([a′,a′] .+ .1, [-1, p(0)]; axis_style...)
δ = .5
ts = range(0, 2pi, 100)
for z in zs
plot!([z-δ,z+δ],[p(z),p(z)]; line=(:black, 1))
C = Shape(z .+ 0.03 * sin.(ts), p(z) .+ 0.3 * cos.(ts))
plot!(C; fill=(:periwinkle, 1), line=(:black, 1))
end
for z in (a,b)
C = Shape(z .+ 0.03 * sin.(ts), p(z) .+ 0.3 * cos.(ts))
plot!(C; fill=(:black, 1), line=(:black, 1))
end
κ = 0.33
annotate!([
(a, 0, text(L"a", :top)),
(b,0, text(L"b", :top)),
(a + κ/5, p(a), text(raw"absolute max", 10, :left)),
(z₁, p(z₁)-κ, text(raw"absolute min", 10, :top)),
(z₂, p(z₂) + κ, text(raw"relative max", 10, :bottom)),
(z₃, p(z₃) - κ, text(raw"relative min", 10, :top)),
(b, p(b) + κ, text(raw"endpoint", 10, :bottom))
])
current()
end
plt
```
```{julia}
#| echo: false
plotly()
nothing
```
Figure illustrating absolute and relative minima for a function $f(x)$ over $I=[a,b]$. The leftmost point has a $y$ value, $f(a)$, which is an absolute maximum of $f(x)$ over $I$. The three points highlighted between $a$ and $b$ are all relative extrema. The first one is *also* the absolute minimum over $I$. The endpoint is not considered a relative maximum for technical reasons --- there is no interval around $b$, it being on the boundary of $I$.
:::
The extreme value theorem discusses an assumption that ensures absolute maximum and absolute minimum values exist.
::: {.callout-note icon=false}
@@ -849,7 +872,7 @@ The function $f(x) = \sqrt{1-x^2}$ is continuous on the interval $[-1,1]$ (in th
##### Example
The function $f(x) = x \cdot e^{-x}$ on the closed interval $[0, 5]$ is continuous. Hence it has an absolute maximum, which a graph shows to be $0.4$. It has an absolute minimum, clearly the value $0$ occurring at the endpoint.
The function $f(x) = x \cdot e^{-x}$ on the closed interval $[0, 5]$ is continuous. Hence it has an absolute maximum, which a graph shows to be about $0.4$ and occurring near $x=1$. It has an absolute minimum, clearly the value $0$ occurring at the endpoint.
```{julia}
@@ -886,7 +909,7 @@ A New York Times [article](https://www.nytimes.com/2016/07/30/world/europe/norwa
## Continuity and closed and open sets
We comment on two implications of continuity that can be generalized to more general settings.
We comment on two implications of continuity that can be generalized.
The two intervals $(a,b)$ and $[a,b]$ differ as the latter includes the endpoints. The extreme value theorem shows this distinction can make a big difference in what can be said regarding *images* of such interval.
@@ -1212,7 +1235,7 @@ radioq(choices, answ, keep_order=true)
###### Question
The extreme value theorem has two assumptions: a continuous function and a *closed* interval. Which of the following examples fails to satisfy the consequence of the extreme value theorem because the function is not continuous?
The extreme value theorem has two assumptions: a continuous function and a *closed* interval. Which of the following examples fails to satisfy the consequence of the extreme value theorem because the function is defined on $I$ but is not continuous on $I$?
```{julia}
@@ -1227,6 +1250,170 @@ answ = 4
radioq(choices, answ, keep_order=true)
```
###### Question
The extreme value theorem is true when $f$ is a continuous function on an interval $I$ *and* $I=[a,b]$ is a *closed* interval. Which of these illustrates why it doesn't apply as $f$ is not continuous on $I$ but is defined on $I$?
```{julia}
#| hold: true
#| echo: false
let
gr()
empty_style = (xaxis=([], false),
yaxis=([], false),
framestyle=:origin,
legend=false)
axis_style = (arrow=true, side=:head, line=(:gray, 1))
ts = range(0, 2pi, 100)
# defined on I; not continuous on I
p1 = plot(;empty_style..., aspect_ratio=:equal)
title!(p1, "(a)")
plot!(p1, x -> 1 - abs(2x), -1, 1, color=:black)
plot!(p1, zero; line=(:black, 1), arrow=true, side=:head)
C = Shape(0.03 .* sin.(ts), 1 .+ 0.03 .* cos.(ts))
plot!(p1, C, fill=(:white, 1), line=(:black,1))
C = Shape(0.03 .* sin.(ts), - 0.25 .+ 0.03 .* cos.(ts))
plot!(p1, C, fill=(:black,1))
annotate!(p1, [
(-1,0,text(L"a", :top)),
(1,0,text(L"b", :top))
])
# not defined on I
p2 = plot(;empty_style...)
title!(p2, "(b)")
plot!(p2, x -> 1/(1-x), 0, .95, color=:black)
plot!(p2, x-> -1/(1-x), 1.05, 2, color=:black)
plot!(p2, zero; axis_style...)
annotate!(p2,[
(0,0,text(L"a", :top)),
(2, 0, text(L"b", :top))
])
# not continuous on I
p3 = plot(;empty_style...)
title!(p3, "(c)")
plot!(p3, x -> 1/(1-x), 0, .95, color=:black)
ylims!((-0.25, 1/(1 - 0.96)))
plot!(p3, [0,1.05],[0,0]; axis_style...)
vline!(p3, [1]; line=(:black, 1, :dash))
annotate!(p3,[
(0,0,text(L"a", :top)),
(1, 0, text(L"b", :top))
])
# continuous
p4 = plot(;empty_style...)
title!(p4, "(d)")
f(x) = x^x
a, b = 0, 2
plot!(p4, f, a, b; line=(:black,1))
ylims!(p4, (-.25, f(b)))
plot!(p4, [a-.1, b+.1], [0,0]; axis_style...)
scatter!([0,2],[ f(0),f(2)]; marker=(:circle,:black))
annotate!([
(a, 0, text(L"a", :top)),
(b, 0, text(L"b", :top))
])
l = @layout[a b; c d]
p = plot(p1, p2, p3, p4, layout=l)
imgfile = tempname() * ".png"
savefig(p, imgfile)
hotspotq(imgfile, (0,1/2), (1/2,1))
end
```
The extreme value theorem is true when $f$ is a continuous function on an interval $I$ and $I=[a,b]$ is a *closed* interval. Which of these illustrates when the theorem's assumptions are true?
```{julia}
#| hold: true
#| echo: false
## come on; save this figure...
let
gr()
empty_style = (xaxis=([], false),
yaxis=([], false),
framestyle=:origin,
legend=false)
axis_style = (arrow=true, side=:head, line=(:gray, 1))
ts = range(0, 2pi, 100)
# defined on I; not continuous on I
p1 = plot(;empty_style..., aspect_ratio=:equal)
title!(p1, "(a)")
plot!(p1, x -> 1 - abs(2x), -1, 1, color=:black)
plot!(p1, zero; line=(:black, 1), arrow=true, side=:head)
C = Shape(0.03 .* sin.(ts), 1 .+ 0.03 .* cos.(ts))
plot!(p1, C, fill=(:white, 1), line=(:black,1))
C = Shape(0.03 .* sin.(ts), - 0.25 .+ 0.03 .* cos.(ts))
plot!(p1, C, fill=(:black,1))
annotate!(p1, [
(-1,0,text(L"a", :top)),
(1,0,text(L"b", :top))
])
# not defined on I
p2 = plot(;empty_style...)
title!(p2, "(b)")
plot!(p2, x -> 1/(1-x), 0, .95, color=:black)
plot!(p2, x-> -1/(1-x), 1.05, 2, color=:black)
plot!(p2, zero; axis_style...)
annotate!(p2,[
(0,0,text(L"a", :top)),
(2, 0, text(L"b", :top))
])
# not continuous on I
p3 = plot(;empty_style...)
title!(p3, "(c)")
plot!(p3, x -> 1/(1-x), 0, .95, color=:black)
ylims!((-0.1, 1/(1 - 0.96)))
plot!(p3, [0,1.05],[0,0]; axis_style...)
vline!(p3, [1]; line=(:black, 1, :dash))
annotate!(p3,[
(0,0,text(L"a", :top)),
(1, 0, text(L"b", :top))
])
# continuous
p4 = plot(;empty_style...)
title!(p4, "(d)")
f(x) = x^x
a, b = 0, 2
ylims!(p4, (-.25, f(b)))
plot!(p4, f, a, b; line=(:black,1))
plot!(p4, [a-.1, b+.1], [0,0]; axis_style...)
scatter!([0,2],[ f(0),f(2)]; marker=(:circle,:black))
annotate!([
(a, 0, text(L"a", :top)),
(b, 0, text(L"b", :top))
])
l = @layout[a b; c d]
p = plot(p1, p2, p3, p4, layout=l)
imgfile = tempname() * ".png"
savefig(p, imgfile)
hotspotq(imgfile, (1/2,1), (0,1/2))
end
```
```{julia}
#| echo: false
plotly();
```
###### Question
@@ -1318,28 +1505,3 @@ The zeros of the equation $\cos(x) \cdot \cosh(x) = 1$ are related to vibrations
val = maximum(find_zeros(x -> cos(x) * cosh(x) - 1, (0, 6pi)))
numericq(val)
```
###### Question
A parametric equation is specified by a parameterization $(f(t), g(t)), a \leq t \leq b$. The parameterization will be continuous if and only if each function is continuous.
Suppose $k_x$ and $k_y$ are positive integers and $a, b$ are positive numbers, will the [Lissajous](https://en.wikipedia.org/wiki/Parametric_equation#Lissajous_Curve) curve given by $(a\cos(k_x t), b\sin(k_y t))$ be continuous?
```{julia}
#| hold: true
#| echo: false
yesnoq(true)
```
Here is a sample graph for $a=1, b=2, k_x=3, k_y=4$:
```{julia}
#| hold: true
a,b = 1, 2
k_x, k_y = 3, 4
plot(t -> a * cos(k_x *t), t-> b * sin(k_y * t), 0, 4pi)
```

250
quarto/limits/limits.qmd
View File

@@ -36,27 +36,38 @@ colors = [:black, :blue, :orange, :red, :green, :orange, :purple]
## Area of parabola
function make_triangle_graph(n)
title = "Area of parabolic cup ..."
n==1 && (title = "\${Area = }1/2\$")
n==2 && (title = "\${Area = previous }+ 1/8\$")
n==3 && (title = "\${Area = previous }+ 2\\cdot(1/8)^2\$")
n==4 && (title = "\${Area = previous }+ 4\\cdot(1/8)^3\$")
n==5 && (title = "\${Area = previous }+ 8\\cdot(1/8)^4\$")
n==6 && (title = "\${Area = previous }+ 16\\cdot(1/8)^5\$")
n==7 && (title = "\${Area = previous }+ 32\\cdot(1/8)^6\$")
n==1 && (title = L"Area $= 1/2$")
n==2 && (title = L"Area $=$ previous $+\; \frac{1}{8}$")
n==3 && (title = L"Area $=$ previous $+\; 2\cdot\frac{1}{8^2}$")
n==4 && (title = L"Area $=$ previous $+\; 4\cdot\frac{1}{8^3}$")
n==5 && (title = L"Area $=$ previous $+\; 8\cdot\frac{1}{8^4}$")
n==6 && (title = L"Area $=$ previous $+\; 16\cdot\frac{1}{8^5}$")
n==7 && (title = L"Area $=$ previous $+\; 32\cdot\frac{1}{8^6}$")
plt = plot(f, 0, 1, legend=false, size = fig_size, linewidth=2)
annotate!(plt, [(0.05, 0.9, text(title,:left))]) # if in title, it grows funny with gr
n >= 1 && plot!(plt, [1,0,0,1, 0], [1,1,0,1,1], color=colors[1], linetype=:polygon, fill=colors[1], alpha=.2)
n == 1 && plot!(plt, [1,0,0,1, 0], [1,1,0,1,1], color=colors[1], linewidth=2)
plt = plot(f, 0, 1;
legend=false,
size = fig_size,
linewidth=2)
annotate!(plt, [
(0.05, 0.9, text(title,:left))
]) # if in title, it grows funny with gr
n >= 1 && plot!(plt, [1,0,0,1, 0], [1,1,0,1,1];
color=colors[1], linetype=:polygon,
fill=colors[1], alpha=.2)
n == 1 && plot!(plt, [1,0,0,1, 0], [1,1,0,1,1];
color=colors[1], linewidth=2)
for k in 2:n
xs = range(0,stop=1, length=1+2^(k-1))
ys = map(f, xs)
k < n && plot!(plt, xs, ys, linetype=:polygon, fill=:black, alpha=.2)
xs = range(0, stop=1, length=1+2^(k-1))
ys = f.(xs)
k < n && plot!(plt, xs, ys;
linetype=:polygon, fill=:black, alpha=.2)
if k == n
plot!(plt, xs, ys, color=colors[k], linetype=:polygon, fill=:black, alpha=.2)
plot!(plt, xs, ys, color=:black, linewidth=2)
plot!(plt, xs, ys;
color=colors[k], linetype=:polygon, fill=:black, alpha=.2)
plot!(plt, xs, ys;
color=:black, linewidth=2)
end
end
plt
@@ -1041,18 +1052,34 @@ $$
Why? We can express the function $e^{\csc(x)}/e^{\cot(x)}$ as the above function plus the polynomial $1 + x/2 + x^2/8$. The above is then the sum of two functions whose limits exist and are finite, hence, we can conclude that $M = 0 + 1$.
### The [squeeze](http://en.wikipedia.org/wiki/Squeeze_theorem) theorem
### The squeeze theorem
Sometimes limits can be found by bounding more complicated functions by easier functions.
::: {.callout-note icon=false}
## The [squeeze theorem](http://en.wikipedia.org/wiki/Squeeze_theorem)
Fix $c$ in $I=(a,b)$. Suppose for all $x$ in $I$, except possibly $c$, there are two functions $l$ and $u$, satisfying:
We note one more limit law. Suppose we wish to compute $\lim_{x \rightarrow c}f(x)$ and we have two other functions, $l$ and $u$, satisfying:
* $l(x) \leq f(x) \leq u(x)$.
* These limits exist and are equal:
$$
L = \lim_{x \rightarrow c} l(x) = \lim_{x \rightarrow c} u(x).
$$
* for all $x$ near $c$ (possibly not including $c$) $l(x) \leq f(x) \leq u(x)$.
* These limits exist and are equal: $L = \lim_{x \rightarrow c} l(x) = \lim_{x \rightarrow c} u(x)$.
Then
$$
\lim_{x\rightarrow c} f(x) = L.
$$
Then the limit of $f$ must also be $L$.
:::
The figure shows a usage of the squeeze theorem to show $\sin(x)/x \rightarrow 1$ as $\cos(x) \leq \sin(x)x \leq 1$ for $x$ close to $0$.
```{julia}
#| hold: true
@@ -1088,8 +1115,71 @@ ImageFile(imgfile, caption)
The formal definition of a limit involves clarifying what it means for $f(x)$ to be "close to $L$" when $x$ is "close to $c$". These are quantified by the inequalities $0 < \lvert x-c\rvert < \delta$ and the $\lvert f(x) - L\rvert < \epsilon$. The second does not have the restriction that it is greater than $0$, as indeed $f(x)$ can equal $L$. The order is important: it says for any idea of close for $f(x)$ to $L$, an idea of close must be found for $x$ to $c$.
The key is identifying a value for $\delta$ for a given value of $\epsilon$.
::: {#fig-limit-e-d}
```{julia}
#| echo: false
plt = let
gr()
f(x) = (x+1)^2 -1
a, b = -1/4, 2
c = 1
L = f(c)
δ = 0.2
ϵ = 3sqrt(δ)
plot(; empty_style...)#, aspect_ratio=:equal)
plot!(f, a, b; line=(:black, 2))
plot!([a,b],[0,0]; axis_style...)
plot!([0,0], [f(a), f(2)]; axis_style...)
plot!([c, c, 0], [0, f(c), f(c)]; line=(:black, 1, :dash))
plot!([c-δ, c-δ, 0], [0, f(c-δ), f(c-δ)]; line=(:black, 1, :dashdot))
plot!([c+δ, c+δ, 0], [0, f(c+δ), f(c+δ)]; line=(:black, 1, :dashdot))
S = Shape([0,b,b,0],[L-ϵ,L-ϵ,L+ϵ,L+ϵ])
plot!(S; fill=(:lightblue, 0.25), line=(nothing,))
domain_color=:red
range_color=:blue
S = Shape([c-δ, c+δ, c+δ, c-δ], 0.05*[-1,-1,1,1])
plot!(S, fill=(domain_color,0.4), line=nothing)
m,M = f(c-δ), f(c+δ)
T = Shape(0.015 * [-1,1,1,-1], [m,m,M,M])
plot!(T, fill=(range_color, 0.4), line=nothing)
C = Plots.scale(Shape(:circle), 0.02, 0.1)
plot!(Plots.translate(C, c, L), fill=(:white,1,0), line=(:black, 1))
plot!(Plots.translate(C, c, 0), fill=(:white,1,0), line=(domain_color, 1))
plot!(Plots.translate(C, 0, L), fill=(:white,1,0), line=(range_color, 1))
annotate!([
(c, 0, text(L"c", :top)),
(c-δ, 0, text(L"c - \delta", 10, :top)),
(c+δ, 0, text(L"c + \delta", 10, :top)),
(0, L, text(L"L", :right)),
(0, L+ϵ, text(L"L + \epsilon", 10, :right)),
(0, L-ϵ, text(L"L - \epsilon", 10, :right)),
])
end
plt
```
```{julia}
#| echo: false
plotly()
nothing
```
Figure illustrating requirements of $\epsilon-\delta$ definition of the limit. The image (shaded red on $y$ axis) of the $x$ within $\delta$ of $c$ (except for $c$ and shaded blue on the $x$ axis) must stay within the bounds of $L-\epsilon$ and $L+ \epsilon$, where $\delta$ may be chosen based on $\epsilon$ but needs to be chosen for every positive $\epsilon$, not just a fixed one as in this figure.
:::
A simple case is the linear case. Consider the function $f(x) = 3x + 2$. Verify that the limit at $c=1$ is $5$.
@@ -1117,64 +1207,78 @@ These lines produce a random $\epsilon$, the resulting $\delta$, and then verify
(The random numbers are technically in $[0,1)$, so in theory `epsilon` could be `0`. So the above approach would be more solid if some guard, such as `epsilon = max(eps(), rand())`, was used. As the formal definition is the domain of paper-and-pencil, we don't fuss.)
In this case, $\delta$ is easy to guess, as the function is linear and has slope $3$. This basically says the $y$ scale is 3 times the $x$ scale. For non-linear functions, finding $\delta$ for a given $\epsilon$ can be a challenge. For the function $f(x) = x^3$, illustrated below, a value of $\delta=\epsilon^{1/3}$ is used for $c=0$:
In this case, $\delta$ is easy to guess, as the function is linear and has slope $3$. This basically says the $y$ scale is 3 times the $x$ scale. For non-linear functions, finding $\delta$ for a given $\epsilon$ can be more of a challenge.
```{julia}
#| hold: true
#| echo: false
#| cache: true
## {{{ limit_e_d }}}
gr()
function make_limit_e_d(n)
f(x) = x^3
##### Example
xs = range(-.9, stop=.9, length=50)
ys = map(f, xs)
We show using the definition that for any fixed $a$ and $n$:
$$
\lim_{x \rightarrow a} x^n = a^n.
$$
This proof uses a bound based on properties of the absolute value.
plt = plot(f, -.9, .9, legend=false, size=fig_size)
if n == 0
nothing
else
k = div(n+1,2)
epsilon = 1/2^k
delta = cbrt(epsilon)
if isodd(n)
plot!(plt, xs, 0*xs .+ epsilon, color=:orange)
plot!(plt, xs, 0*xs .- epsilon, color=:orange)
else
plot!(delta * [-1, 1], epsilon * [ 1, 1], color=:orange)
plot!(delta * [ 1, -1], epsilon * [-1,-1], color=:orange)
plot!(delta * [-1, -1], epsilon * [-1, 1], color=:red)
plot!(delta * [ 1, 1], epsilon * [-1, 1], color=:red)
end
end
plt
end
We look at $f(x) - L = x^n - a^n = (x-a)(x^{n-1} + x^{n-2}a + \cdots + x^1a^{n-1} + a^n)$.
Taking absolute values gives an inequality by the triangle inequality:
n = 11
anim = @animate for i=1:n
make_limit_e_d(i-1)
end
$$
\lvert x^n - a^n\rvert \leq \lvert x-a\rvert\cdot
\left(
\lvert x^{n-1}\rvert +
\lvert x^{n-2}\rvert\lvert a\rvert +
\cdots +
\lvert x^1\rvert\lvert a^{n-1}\rvert +
\lvert a^n\rvert
\right).
$$
imgfile = tempname() * ".gif"
gif(anim, imgfile, fps = 1)
Now, for a given $\epsilon>0$ we seek a $\delta>0$ satisfying the properties of the limit definition for $f(x) = x^n$ and $L=a^n$. For now, assume $\delta < 1$. Then we can assume $\lvert x-a\rvert < \delta$ and
$$
\lvert x\rvert = \lvert x - a + a\rvert \leq \lvert x-a\rvert + \lvert a\rvert < 1 + \lvert a\rvert
$$
caption = L"""
This says then
Demonstration of $\epsilon$-$\delta$ proof of $\lim_{x \rightarrow 0}
x^3 = 0$. For any $\epsilon>0$ (the orange lines) there exists a
$\delta>0$ (the red lines of the box) for which the function $f(x)$
does not leave the top or bottom of the box (except possibly at the
edges). In this example $\delta^3=\epsilon$.
$$
\begin{align*}
\lvert x^n - a^n\rvert
&\leq
\lvert x-a\rvert\cdot \left(
\lvert x\rvert^{n-1} +
\lvert x\rvert^{n-2}\lvert a\rvert +
\cdots +
\lvert x\rvert^1\lvert a^{n-1}\rvert +
\lvert a^n\rvert
\right)\\
%%
&\leq \lvert x - a\rvert
\cdot \left(
(\lvert a\rvert+1)^{n-1} +
(\lvert a\rvert+1)^{n-2}\lvert a\rvert
+ \cdots +
(\lvert a\rvert+1)^1 \lvert a^{n-1}\rvert +
\lvert a^n \rvert
\right)\\
&\leq \lvert x-a\rvert \cdot C,
\end{align*}
$$
where $C$ is just some constant not depending on $x$, just $a$ and $n$.
Now if $\delta < 1$ and $\delta < \epsilon/C$ and if
$0 < \lvert x - a \rvert < \delta$ then
$$
\lvert f(x) - L \rvert =
\lvert x^n - a^n\rvert \leq \lvert x-a\rvert \cdot C \leq \delta\cdot C < \frac{\epsilon}{C} \cdot C = \epsilon.
$$
With this result, the rules of limits can immediately extend this to any polynomial, $p(x),$ it follows that $\lim_{x \rightarrow c} p(x) = p(a)$. (Because $c_n x^n \rightarrow c_n a^n$ and the sum of two functions with a limit has the limit of the sums.) Based on this, we will say later that any polynomial is *continuous* for all $x$.
"""
plotly()
ImageFile(imgfile, caption)
```
## Questions
@@ -1290,26 +1394,26 @@ let
title!(p1, "(a)")
plot!(p1, x -> x^2, 0, 2, color=:black)
plot!(p1, zero, linestyle=:dash)
annotate!(p1,[(1,0,"a")])
annotate!(p1,[(1,0,text(L"a",:top))])
p2 = plot(;axis=nothing, legend=false)
title!(p2, "(b)")
plot!(p2, x -> 1/(1-x), 0, .95, color=:black)
plot!(p2, x-> -1/(1-x), 1.05, 2, color=:black)
plot!(p2, zero, linestyle=:dash)
annotate!(p2,[(1,0,"a")])
annotate!(p2,[(1,0,text(L"a",:top))])
p3 = plot(;axis=nothing, legend=false)
title!(p3, "(c)")
plot!(p3, sinpi, 0, 2, color=:black)
plot!(p3, zero, linestyle=:dash)
annotate!(p3,[(1,0,"a")])
annotate!(p3,[(1,0,text("a",:top))])
p4 = plot(;axis=nothing, legend=false)
title!(p4, "(d)")
plot!(p4, x -> x^x, 0, 2, color=:black)
plot!(p4, zero, linestyle=:dash)
annotate!(p4,[(1,0,"a")])
annotate!(p4,[(1,0,text(L"a",:top))])
l = @layout[a b; c d]
p = plot(p1, p2, p3, p4, layout=l)
@@ -1520,8 +1624,8 @@ Take
$$
f(x) = \begin{cases}
0 & x \neq 0\\
1 & x = 0
0 &~ x \neq 0\\
1 &~ x = 0
\end{cases}
$$

46
quarto/limits/limits_extensions.qmd
View File

@@ -32,22 +32,24 @@ Let's begin with a function that is just problematic. Consider
$$
f(x) = \sin(1/x)
f(x) = \sin(\frac{1}{x})
$$
As this is a composition of nice functions it will have a limit everywhere except possibly when $x=0$, as then $1/x$ may not have a limit. So rather than talk about where it is nice, let's consider the question of whether a limit exists at $c=0$.
@fig-sin-1-over-x shows the issue:
A graph shows the issue:
:::{#fig-sin-1-over-x}
```{julia}
#| hold: true
#| echo: false
f(x) = sin(1/x)
plot(f, range(-1, stop=1, length=1000))
```
Graph of the function $f(x) = \sin(1/x)$ near $0$. It oscillates infinitely many times around $0$.
:::
The graph oscillates between $-1$ and $1$ infinitely many times on this interval - so many times, that no matter how close one zooms in, the graph on the screen will fail to capture them all. Graphically, there is no single value of $L$ that the function gets close to, as it varies between all the values in $[-1,1]$ as $x$ gets close to $0$. A simple proof that there is no limit, is to take any $\epsilon$ less than $1$, then with any $\delta > 0$, there are infinitely many $x$ values where $f(x)=1$ and infinitely many where $f(x) = -1$. That is, there is no $L$ with $|f(x) - L| < \epsilon$ when $\epsilon$ is less than $1$ for all $x$ near $0$.
@@ -65,11 +67,10 @@ The following figure illustrates:
```{julia}
#| hold: true
f(x) = x * sin(1/x)
plot(f, -1, 1)
plot!(abs)
plot!(x -> -abs(x))
plot(f, -1, 1; label="f")
plot!(abs; label="|.|")
plot!(x -> -abs(x); label="-|.|")
```
The [squeeze](http://en.wikipedia.org/wiki/Squeeze_theorem) theorem of calculus is the formal reason $f$ has a limit at $0$, as both the upper function, $|x|$, and the lower function, $-|x|$, have a limit of $0$ at $0$.
@@ -181,21 +182,38 @@ Consider this funny graph:
```{julia}
#| hold: true
#| echo: false
xs = range(0,stop=1, length=50)
plot(x->x^2, -2, -1, legend=false)
let
xs = range(0,stop=1, length=50)
plot(; legend=false, aspect_ratio=true,
xticks = -4:4)
plot!([(-4, -1.5),(-2,4)]; line=(:black,1))
plot!(x->x^2, -2, -1; line=(:black,1))
plot!(exp, -1,0)
plot!(x -> 1-2x, 0, 1)
plot!(sqrt, 1, 2)
plot!(x -> 1-x, 2,3)
S = Plots.scale(Shape(:circle), 0.05)
plot!(Plots.translate(S, -4, -1.5); fill=(:black,))
plot!(Plots.translate(S, -1, (-1)^2); fill=(:white,))
plot!(Plots.translate(S, -1, exp(-1)); fill=(:black,))
plot!(Plots.translate(S, 1, 1 - 2(1)); fill=(:black,))
plot!(Plots.translate(S, 1, sqrt(1)); fill=(:white,))
plot!(Plots.translate(S, 2, sqrt(2)); fill=(:white,))
plot!(Plots.translate(S, 2, 1 - (2)); fill=(:black,))
plot!(Plots.translate(S, 3, 1 - (3)); fill=(:black,))
end
```
Describe the limits at $-1$, $0$, and $1$.
* At $-1$ we see a jump, there is no limit but instead a left limit of 1 and a right limit appearing to be $1/2$.
* At $0$ we see a limit of $1$.
* Finally, at $1$ again there is a jump, so no limit. Instead the left limit is about $-1$ and the right limit $1$.
* At $-1$ we see a jump, there is no limit but instead a left limit of 1 and a right limit appearing to be $1/2$.
* At $0$ we see a limit of $1$.
* Finally, at $1$ again there is a jump, so no limit. Instead the left limit is about $-1$ and the right limit $1$.
## Limits at infinity