minor edits

quarto/derivatives/derivatives.qmd

@@ -688,12 +688,9 @@ $$
 ### 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 through two examples.
+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 through an example.
 
-The first involves working out on a treadmill. For this example, there is a presumed linear relationship between miles run and calories burned. With that, the rate of calories burned per hour would be proportional to the miles per hours
-
-
-Now 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 $f(y) = y^2.$
+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.
@@ -824,7 +821,11 @@ Find the derivative of $\sin(x)\cos(2x)$ at $x=\pi$.
 ##### 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:
+A function is *differentiable* at $a$ if the following limit exists $\lim_{h \rightarrow 0}(f(a+h)-f(a))/h$.
+This is reexpressed as: $f(a+h) - f(a) - f'(a)h = \epsilon_f(h) h$ where as $h\rightarrow 0$, $\epsilon_f(h) \rightarrow 0$.
+
+
+With that in mind, we have:
 
 
 $$
@@ -843,11 +844,12 @@ f(g(a) + g'(a)h + \epsilon_g(h)h) - f(g(a)) \\
 Rearranging:
 
 
-$$
-f(g(a+h)) - f(g(a)) - f'(g(a)) g'(a) h = f'(g(a))\epsilon_g(h)h + \epsilon_f(h')(h') =
-(f'(g(a)) \epsilon_g(h)  + \epsilon_f(h') (g'(a) + \epsilon_g(h)))h =
-\epsilon(h)h,
-$$
+\begin{align*}
+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,
+\end{align*}
 
 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 $g(a)$.
 

quarto/derivatives/numeric_derivatives.qmd

@@ -45,7 +45,7 @@ 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.
+  * 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.
@@ -70,7 +70,7 @@ We can compare to the actual with:
 ```{julia}
 @syms x
 df = diff(f(x), x)
-factual = N(df(c))
+factual = convert(Float64, df(c))
 abs(factual - fapprox)
 ```
 
@@ -136,16 +136,16 @@ 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
+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 = D(f)(c)
 ds_value, fwd, ds_value - fwd
 ```
 
@@ -153,11 +153,11 @@ Finally, `SymPy` gives an exact value we use to compare:
 
 
 ```{julia}
-𝒇𝒑 = diff(𝒇(x), x)
+fp = diff(f(x), x)
 ```
 
 ```{julia}
-actual = N(𝒇𝒑(PI/4))
+actual = convert(Float64, fp(PI/4))
 actual - ds_value, actual - fwd
 ```
 

quarto/derivatives/symbolic_derivatives.qmd

@@ -104,7 +104,7 @@ function D(::Val{:+}, ::Val{:nary}, args, var)
 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.
+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 using splatting to produce the sum.
 
 
 Subtraction must also be implemented in a similar manner, but not for the *nary* case:
@@ -195,7 +195,15 @@ function D(::Val{:cos}, ::Val{:unary}, args, var)
 end
 ```
 
-The pattern is similar for each. The `$a′` factor is needed due to the *chain rule*. The above illustrates the simple pattern necessary to add a derivative rule for a function. More could be, but for this example the above will suffice, as now the system is ready to be put to work.
+The pattern is similar for each. The `$a′` factor is needed due to the *chain rule*. The above illustrates the simple pattern necessary to add a derivative rule for a function.
+
+:::{.callout-note}
+Several automatic differentiation packages use a set of rules defined following an interface spelled out in the package `ChainRules.jl`. Leveraging multi-dimensional derivatives, the chain rule is the only rule needed of the sum, product, quotient and chain rules.
+
+:::
+
+
+More functions could be included, but for this example the above will suffice, as now the system is ready to be put to work.
 
 
 ```{julia}
@@ -223,5 +231,3 @@ 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.
-
-