use quarto, not Pluto to render pages

CwJ/differentiable_vector_calculus/scalar_functions_applications.jmd

@@ -8,6 +8,11 @@ using CalculusWithJulia
 using Plots
 using SymPy
 using Roots
+```
+
+And the following from the `Contour` package:
+
+```julia
 import Contour: contours, levels, level, lines, coordinates
 ```
 
@@ -109,8 +114,8 @@ pt = [a, b, 0]
 scatter!(unzip([pt])...)
 arrow!(pt, [1,0,0], linestyle=:dash)
 arrow!(pt, [0,1,0], linestyle=:dash)
-
 ```
+
 #### Alternate forms
 
 The equation for the tangent plane is often expressed in a more explicit form. For $n=2$, if we set $dx = x-a$ and $dy=y-a$, then the equation for the plane becomes:
@@ -915,8 +920,10 @@ Another might be the vertical squared distance to the line:
 
 
 ```math
-d2(\alpha, \beta) = (y_1 - l(x_1))^2 + (y_2 - l(x_2))^2 + (y_3 - l(x_3))^2 =
-(y1 - (\alpha + \beta x_1))^2 + (y3 - (\alpha + \beta x_3))^2 + (y3 - (\alpha + \beta x_3))^2
+\begin{align*}
+d2(\alpha, \beta) &= (y_1 - l(x_1))^2 + (y_2 - l(x_2))^2 + (y_3 - l(x_3))^2 \\
+&= (y1 - (\alpha + \beta x_1))^2 + (y3 - (\alpha + \beta x_3))^2 + (y3 - (\alpha + \beta x_3))^2
+\end{align*}
 ```
 
 Another might be the *shortest* distance to the line:
@@ -1011,8 +1018,8 @@ gammas₂ = [1.0]
 
 for n in 1:5
     xn = xs₂[end]
-    gamma = gammas₂[end]
-    xn1 = xn - gamma * gradient(f₂)(xn)
+    gamma₀ = gammas₂[end]
+    xn1 = xn - gamma₀ * gradient(f₂)(xn)
     dx, dy = xn1 - xn, gradient(f₂)(xn1) - gradient(f₂)(xn)
     gamman1 = abs( (dx ⋅ dy) / (dy ⋅ dy) )
 
@@ -1133,10 +1140,8 @@ fxx, d
 Consequently we have a local maximum at this critical point.
 
 
-```julia; echo=false
-note(""" The `Optim.jl` package provides efficient implementations of these two numeric methods, and others. """)
-```
-
+!!! note
+The `Optim.jl` package provides efficient implementations of these two numeric methods, and others.
 
 ## Constrained optimization, Lagrange multipliers
 
@@ -1557,11 +1562,13 @@ This theorem can be generalized to scalar functions, but the notation can be cum
 Following [Folland](https://sites.math.washington.edu/~folland/Math425/taylor2.pdf) we use *multi-index* notation. Suppose $f:R^n \rightarrow R$, and let $\alpha=(\alpha_1, \alpha_2, \dots, \alpha_n)$. Then define the following notation:
 
 ```math
-|\alpha| = \alpha_1 + \cdots + \alpha_n, \quad
-\alpha! = \alpha_1!\alpha_2!\cdot\cdots\cdot\alpha_n!,\quad
-\vec{x}^\alpha = x_1^{\alpha_1}x_2^{\alpha_2}\cdots x_n^{\alpha^n}, \quad
-\partial^\alpha f = \partial_1^{\alpha_1}\partial_2^{\alpha_2}\cdots \partial_n^{\alpha_n} f =
-\frac{\partial^{|\alpha|}f}{\partial x_1^{\alpha_1} \partial x_2^{\alpha_2} \cdots \partial x_n^{\alpha_n}}.
+\begin{align*}
+|\alpha| &= \alpha_1 + \cdots + \alpha_n, \\
+\alpha! &= \alpha_1!\alpha_2!\cdot\cdots\cdot\alpha_n!, \\
+\vec{x}^\alpha &= x_1^{\alpha_1}x_2^{\alpha_2}\cdots x_n^{\alpha^n}, \\
+\partial^\alpha f &= \partial_1^{\alpha_1}\partial_2^{\alpha_2}\cdots \partial_n^{\alpha_n} f \\
+& = \frac{\partial^{|\alpha|}f}{\partial x_1^{\alpha_1} \partial x_2^{\alpha_2} \cdots \partial x_n^{\alpha_n}}.
+\endalign*}
 ```
 
 This notation makes many formulas from one dimension carry over to higher dimensions. For example, the binomial theorem says:
@@ -1781,8 +1788,8 @@ choices = [
     raw"`` 2x + y - 2z = 1``",
     raw"`` x + 2y + 3z = 6``"
 ]
-ans = 1
-radioq(choices, ans)
+answ = 1
+radioq(choices, answ)
 ```
 
 
@@ -1798,8 +1805,8 @@ choices = [
     raw"`` y^2 + y, x^2 + x``",
     raw"`` \langle 2y + y^2, 2x + x^2``"
 ]
-ans = 1
-radioq(choices, ans)
+answ = 1
+radioq(choices, answ)
 ```
 
 Is this the Hessian of $f$?
@@ -1830,8 +1837,8 @@ choices = [
     L"The function $f$ has a saddle point, as $d  < 0$",
     L"Nothing can be said, as $d=0$"
 ]
-ans = 2
-radioq(choices, ans, keep_order=true)
+answ = 2
+radioq(choices, answ, keep_order=true)
 ```
 
 
@@ -1885,8 +1892,8 @@ choices = [
     L"Nothing can be said, as $d=0$",
     L"The test does not apply, as $\nabla{f}$ is not $0$ at this point."
 ]
-ans = 3
-radioq(choices, ans, keep_order=true)
+answ = 3
+radioq(choices, answ, keep_order=true)
 ```
 
 Which is true of $f$ at $(0, -1/2)$:
@@ -1899,8 +1906,8 @@ choices = [
     L"Nothing can be said, as $d=0$",
     L"The test does not apply, as $\nabla{f}$ is not $0$ at this point."
 ]
-ans = 1
-radioq(choices, ans, keep_order=true)
+answ = 1
+radioq(choices, answ, keep_order=true)
 ```
 
 
@@ -1914,8 +1921,8 @@ choices = [
     L"Nothing can be said, as $d=0$",
     L"The test does not apply, as $\nabla{f}$ is not $0$ at this point."
 ]
-ans = 5
-radioq(choices, ans, keep_order=true)
+answ = 5
+radioq(choices, answ, keep_order=true)
 ```
 
 
@@ -1962,8 +1969,8 @@ choices =[
     "It is the determinant of the Hessian",
     L"It isn't, $b^2-4ac$ is from the quadratic formula"
 ]
-ans = 1
-radioq(choices, ans)
+answ = 1
+radioq(choices, answ)
 ```
 
 Which condition on $a$, $b$, and $c$ will ensure a *local maximum*:
@@ -1974,8 +1981,8 @@ choices = [
     L"That $a<0$ and $ac-b^2 > 0$",
     L"That $ac-b^2 < 0$"
 ]
-ans = 2
-radioq(choices, ans, keep_order=true)
+answ = 2
+radioq(choices, answ, keep_order=true)
 ```
 
 Which condition on $a$, $b$, and $c$ will ensure a saddle point?
@@ -1987,8 +1994,8 @@ choices = [
     L"That $a<0$ and $ac-b^2 > 0$",
     L"That $ac-b^2 < 0$"
 ]
-ans = 3
-radioq(choices, ans, keep_order=true)
+answ = 3
+radioq(choices, answ, keep_order=true)
 ```
 
 
@@ -2016,8 +2023,8 @@ choices = [
     raw"`` \langle 2x, y^2\rangle``",
     raw"`` \langle x^2, 2y \rangle``"
 ]
-ans = 1
-radioq(choices, ans)
+answ = 1
+radioq(choices, answ)
 ```
 
 Due to the form of the gradient of the constraint, finding when $\nabla{f} = \lambda \nabla{g}$ is the same as identifying when this ratio $|f_x/f_y|$ is $1$. The following solves for this by checking each point on the constraint: