ece204: add derivatives

docs/2a/ece204.md

@@ -84,3 +84,91 @@ $$\boxed{f(x)=b_0+b_1(x-x_0)+b_2(x-x_0)(x-x_1)}$$
 $$
 b_2=\frac{\frac{f(x_2)-f(x_1)}{x_2-x_1}-\frac{f(x_1)-f(x_0)}{x_1-x_0}}{x_2-x_0}
 $$
+
+## Derivatives
+
+Derivatives are estimated based on first principles:
+
+$$f'(x)=\frac{f(x+h)-f(x)}{h}$$
+
+### Derivatives of continuous functions
+
+At a desired $x$ for $f'(x)$:
+
+1. Choose an arbitrary $h$
+2. Calculate derivative via first principles
+3. Shrink $h$ and recalculate derivative
+4. If the answer is drastically different, repeat step 3
+
+### Derivatives of discrete functions
+
+Guesses are made based on the average slope between two points.
+
+$$f'(x_i)=\frac{f(x_{i+1})-f(x_i)}{x_{i+1}-x_i}$$
+
+### Divided differences
+
+- Using the next term, or a $\Delta x > 0$ indicates a **forward divided difference (FDD)**.
+- Using the previous term, or a $\Delta x < 0$ indicates a **backward divided difference (BDD)**.
+
+The **central divided difference** averages both if $h$ or $\Delta x$ of the forward and backward DDs are equal.
+
+$$f'(x)=CDD=\frac{f(x+h)-f(x-h)}{2h}$$
+
+### Higher order derivatives
+
+Taking the Taylor expansion of the function or discrete set and then expanding it as necessary can return any order of derivative. This also applies for $x-h$ if positive and negative are alternated.
+
+$$f(x+h)=f(x)+f'(x)h+\frac{f''(x)}{2!}h^2+\frac{f'''(x)}{3!}h^3$$
+
+!!! example
+    To find second order derivatives:
+    
+    \begin{align*}
+    f''(x)&=\frac{2f(x+h)-2f(x)-2f'(x)h}{h^2} \\
+    &=\frac{2f(x+h)-2f(x)-(f(x+h)-f(x-h))}{h^2} \\
+    &=\frac{f(x+h)-2f(x)+f(x-h)}{h^2}
+    \end{align*}
+
+!!! example
+    $f''(3)$ if $f(x)=2e^{1.5x}$ and $h=0.1$:
+    
+    \begin{align*}
+    f''(3)&=\frac{f(3.1)-2\times2f(3)+f(2.9)}{0.1^2} \\
+    &=405.08
+    \end{align*}
+    
+For discrete data:
+
+- If the desired point does not exist, differentiating the surrounding points to create a polynomial interpolation of the derivative may be close enough.
+
+!!! example
+    | t | 0 | 10 | 15 | 20 | 22.5 | 30 |
+    | --- | --- | --- | --- | --- | --- | --- |
+    | v(t) | 0 | 227.04 | 362.78 | 517.35 | 602.47 | 901.67 |
+    
+    $v'(16)$ with FDD:
+    
+    Using points $t=15,t=20$:
+    
+    \begin{align*}
+    v'(x)&=\frac{f(x+h)-f(x)}{h} \\
+    &=\frac{f(15+5)-f(15)}{5} \\
+    &=\frac{517.35-362.78}{5} \\
+    &=30.914
+    \end{align*}
+    
+    $v'(16)$ with Newton's first-order interpolation:
+    
+    \begin{align*}
+    v(t)&=v(15)+\frac{v(20)-v(15)}{20-15}(t-15) \\
+    &=362.78+30.914(t-15) \\
+    &=-100.93+30.914t \\
+    v'(t)&=\frac{v(t+h)-v(t)}{2h} \\
+    &=\frac{v(16.1)-v(15.9)}{0.2} \\
+    &=30.914
+    \end{align*}
+
+- If the spacing is not equal (to make DD impossible), again creating an interpolation may be close enough.
+- If data is noisy, regressing and then solving reduces random error.
+