typos
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@@ -886,7 +886,7 @@ end
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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 like surfaces stitched together.
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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 like surfaces stitched together.
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With `Makie`, many implicitly defined surfaces can be adequately represented using `countour` with the attribute `levels=[0]`. We will illustrate this technique.
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With `Makie`, many implicitly defined surfaces can be adequately represented using `contour` with the attribute `levels=[0]`. We will illustrate this technique.
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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. This package has a `plot_implicit_surface` function that does something similar to below. We don't illustrate it, as it *currently* doesn't work with the latest version of `Makie`.
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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. This package has a `plot_implicit_surface` function that does something similar to below. We don't illustrate it, as it *currently* doesn't work with the latest version of `Makie`.
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@@ -757,7 +757,7 @@ R = solve((n1, n2), (x, y))
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```
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```
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Taking limits of each term as $h$ goes to zero we have after some notation-simplfying substitution:
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Taking limits of each term as $h$ goes to zero we have after some notation-simplifying substitution:
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```{julia}
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```{julia}
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R = Dict(k => limit(R[k], ℎ=>0) for k in (x,y))
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R = Dict(k => limit(R[k], ℎ=>0) for k in (x,y))
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@@ -275,7 +275,7 @@ end
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p
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p
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```
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```
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There is no hidden line calculuation, in place we give the contour lines a transparency through the argument `alpha=0.5`.
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There is no hidden line calculation, in place we give the contour lines a transparency through the argument `alpha=0.5`.
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### Gradient and surface plots
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### Gradient and surface plots
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@@ -302,7 +302,7 @@ The main point above is that *if* the vector field is the gradient of a scalar f
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::: {.callout-note icon=false}
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::: {.callout-note icon=false}
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## Conservative vector field
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## Conservative vector field
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If $F$ is a vector field defined in an *open* region $R$; $A$ and $B$ are points in $R$ and *if* for *any* curve $C$ in $R$ connecting $A$ to $B$, the line integral of $F \cdot \vec{T}$ over $C$ depends *only* on the endpoint $A$ and $B$ and not the path, then the line integral is called *path indenpendent* and the field is called a *conservative field*.
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If $F$ is a vector field defined in an *open* region $R$; $A$ and $B$ are points in $R$ and *if* for *any* curve $C$ in $R$ connecting $A$ to $B$, the line integral of $F \cdot \vec{T}$ over $C$ depends *only* on the endpoint $A$ and $B$ and not the path, then the line integral is called *path independent* and the field is called a *conservative field*.
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:::
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:::
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@@ -165,7 +165,7 @@ Some names used for the characterizing constants are:
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### Three-term reccurence
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### Three-term recurrence
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Orthogonal polynomials, as defined above through a weight function, satisfy a *three-term recurrence*:
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Orthogonal polynomials, as defined above through a weight function, satisfy a *three-term recurrence*:
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@@ -868,7 +868,7 @@ plotly()
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nothing
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nothing
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```
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```
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Illustration of Cavalieri's first principle. The discs from the left are moved around to form the left volume, but as the volumes of each cross-sectional disc remains the same, the two valumes are equally approximated. (This figure ported from @Angenent.)
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Illustration of Cavalieri's first principle. The discs from the left are moved around to form the left volume, but as the volumes of each cross-sectional disc remains the same, the two volumes are equally approximated. (This figure ported from @Angenent.)
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:::
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:::
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@@ -1913,7 +1913,7 @@ Archimedes, in finding bounds on the value of $\pi$ used $n$-gons with sides $12
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```{julia}
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```{julia}
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#| hold: true
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#| hold: true
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x = r * tan(theta/2)
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x = r * tan(theta/2)
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n = 2PI/theta # using PI to avoid floaing point roundoff in 2pi
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n = 2PI/theta # using PI to avoid floating point roundoff in 2pi
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# C < n * 2x
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# C < n * 2x
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upper = n*2x
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upper = n*2x
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```
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```
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@@ -1004,7 +1004,7 @@ Now consider the case $p_0 > 0$. There are two possibilities either `pos(p)` is
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So there is parity between `var(p)` and `pos(p)`: if $p$ is monic and $p_0 < 0$ then both `var(p)` and `pos(p)` are both odd; and if $p_0 > 0$ both `var(p)` and `pos(p)` are both even.
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So there is parity between `var(p)` and `pos(p)`: if $p$ is monic and $p_0 < 0$ then both `var(p)` and `pos(p)` are both odd; and if $p_0 > 0$ both `var(p)` and `pos(p)` are both even.
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Descartes' rule of signs will be established if it can be shown that `var(p)` is at least as big as `pos(p)`. Supppose $r$ is a positive real root of $p$ with $p = (x-r)q$. We show that `var(p) > var(q)` which can be repeatedly applied to show that if $p=(x-r_1)\cdot(x-r_2)\cdot \cdots \cdot (x-r_l) q$, where the $r_i$s are the positive real roots, then `var(p) >= l + var(q) >= l = pos(p)`.
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Descartes' rule of signs will be established if it can be shown that `var(p)` is at least as big as `pos(p)`. Suppose $r$ is a positive real root of $p$ with $p = (x-r)q$. We show that `var(p) > var(q)` which can be repeatedly applied to show that if $p=(x-r_1)\cdot(x-r_2)\cdot \cdots \cdot (x-r_l) q$, where the $r_i$s are the positive real roots, then `var(p) >= l + var(q) >= l = pos(p)`.
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As $p = (x-c)q$ we must have the leading term is $p_nx^n = x \cdot q_{n-1} x^{n-1}$ so $q_{n-1}$ will also be `+` under our monic assumption. Looking at a possible pattern for the signs of $q$, we might see the following unfinished synthetic division table for a specific $q$:
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As $p = (x-c)q$ we must have the leading term is $p_nx^n = x \cdot q_{n-1} x^{n-1}$ so $q_{n-1}$ will also be `+` under our monic assumption. Looking at a possible pattern for the signs of $q$, we might see the following unfinished synthetic division table for a specific $q$:
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