# ECE 204: Numerical Methods

## Linear regression

Given a regression $y=mx+b$ and a data set $(x_{i..n}, y_{i..n})$, the **residual** is the difference between the actual and regressed data:

$$E_i=y_i-b-mx_i$$

### Method of least squares

This method minimises the sum of the square of residuals.

$$\boxed{S_r=\sum^n_{i=1}E_i^2}$$

$m$ and $b$ can be found by taking the partial derivative and solving for them:

$$\frac{\partial S_r}{\partial m}=0, \frac{\partial S_r}{\partial b}=0$$

This returns, where $\overline y$ is the mean of the actual $y$-values:

$$
\boxed{m=\frac{n\sum^n_{i=1}x_iy_i-\sum^n_{i=1}x_i\sum^n_{i=1}y_i}{n\sum^n_{i=1}x_i^2-\left(\sum^n_{i=1}x_i\right)^2}} \\
b=\overline y-m\overline x
$$

The total sum of square around the mean is based off of the actual data:

$$\boxed{S_t=\sum(y_i-\overline y)^2}$$

Error is measured with the **coefficient of determination** $r^2$ — the closer the value is to 1, the lower the error.

$$
r^2=\frac{S_t-S_r}{S_t}
$$

If the intercept is the **origin**, $m$ reduces down to a simpler form:

$$m=\frac{\sum^n_{i=1}x_iy_i}{\sum^n_{i=1}x_i^2}$$

## Non-linear regression

### Exponential regression

Solving for the same partial derivatives returns the same values, although bisection may be required for the exponent coefficient ($e^{bx}$) Instead, linearising may make things easier (by taking the natural logarithm of both sides. Afterward, solving as if it were in the form $y=mx+b$ returns correct 

!!! example
    $y=ax^b\implies\ln y = \ln a + b\ln x$

### Polynomial regression

The residiual is the offset at the end of a polynomial.

$$y=a+bx+cx^2+E$$

Taking the relevant partial derivatives returns a system of equations which can be solved in a matrix.

## Interpolation

Interpolation ensures that every point is crossed.

### Direct method

To interpolate $n+1$ data points, you need a polynomial of a degree **up to $n$**, and points that enclose the desired value. Substituting the $x$ and $y$ values forms a system of equations for a polynomial of a degree equal to the number of points chosen - 1.

### Newton's divided difference method

This method guesses the slope to interpolate. Where $x_0$ is an existing point:

$$\boxed{f(x)=b_0+b_1(x-x_0)}$$

The constant is an existing y-value and the slope is an average.

$$
\begin{align*}
b_0&=f(x_0) \\
b_1&=\frac{f(x_1)-f(x_0)}{x_1-x_0}
\end{align*}
$$

This extends to a quadratic, where the second slope is the average of the first two slopes:

$$\boxed{f(x)=b_0+b_1(x-x_0)+b_2(x-x_0)(x-x_1)}$$

$$
b_2=\frac{\frac{f(x_2)-f(x_1)}{x_2-x_1}-\frac{f(x_1)-f(x_0)}{x_1-x_0}}{x_2-x_0}
$$

## 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.

## 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:
    
    $$\frac{y_{i+1}-2y_i+y_{i-1}{h^2}+2\frac{y_{i+1}-y_i}{h}+y_i=x_i^2$$