ece204: add odes

docs/2a/ece204.md

@@ -171,4 +171,124 @@ For discrete data:
 
 - 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.
+
+## Integrals
+
+If you can represent a function as an $n$-th order polynomial, you can approximate the integral with the integral of that polynomial.
+
+### Trapezoidal rule
+
+The **trapezoidal rule** looks at the first order polynomial and 
+
+From $a$ to $b$, if there are $n$ trapezoidal segments, where $h=\frac{b-a}{n}$ is the width of each segment:
+
+$$\int^b_af(x)dx=\frac{b-a}{2n}[f(a)+2(\sum^{n-1}_{i=1}f(a+ih))+f(b)]$$
+
+The error for the $i$th trapezoidal segment is $|E_i|=\left|\frac{h^3}{12}\right|f''(x)$. This can be approximated with a maximum value of $f''$:
+
+$$\boxed{|E_T|\leq(b-a)\frac{h^2}{12}M}$$
+
+### Simpson's 1/3 rule
+
+This uses the second-order polynomial with **two segments**. Three points are usually used: $a,\frac{a+b}{2},b$. Thus for two segments:
+
+$$\int^b_af(x)dx\approx\frac h 3\left[f(a)+4f\left(\frac{a+b}{2}\right)+f(b)\right]$$
+
+For an arbitrary number of segments, as long as there are an **even number** of **equal** segments:
+$$\int^b_af(x)dx=\frac{b-a}{3n}\left[f(x_0)+4\sum^{n-1}_{\substack{i=1 \\ \text{i is odd}}}f(x_i)+2\sum^{n-2}_{\substack{i=2 \\ \text{i is even}}}f(x_i)+f(x_n)\right]$$
+
+The error is:
+$$|E_T|=(b-a)\frac{h^4}{180}M$$
+
+## Ordinary differential equations
+
+### Initial value problems
+
+These problems only have results for one value of $x$.
+
+**Euler's method** converts the function to the form $f(x,y)$, where $y'=f(x,y)$.
+
+!!! example
+    $y'+2y=1.3e^{-x},y(0)=5\implies f(x,y)=1.3e^{-x}-2y,y(0)=5$
+
+Where $h$ is the width of each estimation (lower is better):
+
+$$y_{n+1}=y_n+hf(x_n,y_n)$$
+
+!!! example
+    If $f(x,y)=2xy,h=0.1$, $y_{n+1}=y_n+h2x_ny_n$
     
+    $$
+    y(1.1)=y(1)+0.1×2×1×\underbrace{y(1)}_{1 via IVP}=1.2 \\
+    y(1.2)=y(1.1)+0.1×2×1.1×\underbrace{y(1.1)}_{1.2}=1.464
+    $$
+
+**Heun's method** uses Euler's formula as a predictor. Where $y^*$ is the Euler solution:
+
+$$y_{n+1}=y_n+h\frac{f(x_n,y_n)+f(x_{n+1},y^*_{n+1}}{2}$$
+
+!!! example
+    For $f(x,y)=2xy,h=0.1, y(1)=1$:
+    
+    Euler's formula returns $y^*_{n+1}=y_n+2hx_ny_n\implies y^*(1.1)=1.2$.
+    
+    Applying Heun's correction:
+    
+    \begin{align*}
+    y(1.1)&=y(1)=0.1\frac{2×1×y(1)+2×1.1×y^*(1.1)}{2} \\
+    &=1+0.1\frac{2×1×1+2×1.1×1.2}{2} \\
+    &=1.232
+    \end{align*}
+
+The **Runge-Kutta fourth-order method** is the most accurate of the three methods:
+
+$$y_n+1=y_n+\frac 1 6(k_1+2k_2+2k_3+k_4)$$
+
+- $k_1=hf(x_n,y_n)$
+- $k_2=hf(x_n+\tfrac 1 2h,y_n+\tfrac 1 2k_1)$
+- $k_3=hf(x_n+\tfrac 1 2 h, y_n+\tfrac 1 2 k_2)$
+- $k_4=hf(x_n+h,y_n+k_3)$
+
+### Higher order ODEs
+
+Higher order ODEs can be solved by reducing them to first order ODEs by creating a system of equations. For a second order ODE: Let $y'=u$.
+
+$$
+y'=u \\
+u'=f(x,y,u)
+$$
+
+For each ODE, the any method can be used:
+
+$$
+y_{n+1}=y_n+hu_n \\
+u_{n+1}=u_n+hf(x_n,y_n,u_n)
+$$
+
+!!! example
+    For $y''+xy'+y=0,y(0)=1,y'(0)=2,h=0.1$:
+    
+    \begin{align*}
+    y'&= u \\
+    u'&=-xu-y \\
+    y_1&=y_0+0.1u_0 \\
+    &=1+0.1×2 \\
+    &=1.2 \\
+    \\
+    u_1&=u_0+0.1×f(x_0,y_0,u_0) \\
+    &=u_0+0.1(-x_0u_0-y_0] \\
+    &=2+0.1(-0×2-1) \\
+    &=1.9
+    \end{align*}
+
+### Boundary value problems
+
+The **finite difference method** divides the interval between the boundary into $n$ sub-intervals, replacing derivatives with their first principles representations. Solving each $n-1$ equation returns a proper system of equations.
+
+!!! example
+    For $y''+2y'+y=x^2, y(0)=0.2,y(1)=0.8,n=4\implies h=0.25$:
+    $x_0=0,x_1=0.25,x_2=0.5,x_3=0.75,x_4=1$
+    
+    Replace with first principles:
+    
+    $$\frac{y_{i+1}-2y_i+y_{i-1}{h^2}+2\frac{y_{i+1}-y_i}{h}+y_i=x_i^2$$

docs/2a/ece205.md

@@ -225,6 +225,13 @@ Two boundary conditions are requred to solve the problem for all $t>0$ — that
 - $u(x,0)=f(x),0\leq x\leq L$
 - e.g., $u(0,t)=u(L,t)=0,t>0$
 
+Thus the general solution is:
+
+$$
+\boxed{u(x,t)=\sum^\infty_{n=1}a_ne^{-\left(\frac{n\pi a}{L}\right)^2t}\sin(\frac{n\pi x}{L})} \\
+f(x)=\sum^\infty_{n=1}a_n\sin(\frac{n\pi x}{L})
+$$
+
 ### Periodicity
 
 The **period** of a function is an increment that always returns the same value: $f(x+T)=f(x)$, and its **fundamental period** of a function is the smallest possible period.

docs/2a/ece222.md

@@ -124,3 +124,7 @@ $$\text{time}=n_{instructions}\times\underbrace{\frac{\text{cycles}}{\text{instr
 Pipelining changes the granularity of a clock cycle to be per step, instead of per-instruction. This allows multiple instructions to be processed concurrently.
 
 <img src="https://upload.wikimedia.org/wikipedia/commons/c/cb/Pipeline%2C_4_stage.svg" width=500>(Source: Wikimedia Commons)</img>
+
+### Data forwarding
+
+If data needs to be used from a prior operation, a pipeline stall would normally be required to remove the hazard and wait for the desired result (a **read-after-write** data hazard). However, a processor can mitigate this hazard by allowing the stalled instrution to read from the prior instruction's result instead.