eifueo/docs/2a/ece204.md

# 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:

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

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