APS 非重点复习课程之计算方法

Lagrange Interpolation

拉格朗日插值

$P_n(x)=\sum_{i=1}^ny_i(\Pi_{j\neq i}^{1\leq j\leq n}\frac{x-x_j}{x_i-x_j})$

Using polynomials to approximate the original function.

Residuals

$R(x)=f(x)-p(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}\omega(x)$, $\xi\in(a,b)$ relies on x.

Newton Interpolation

Difference Quotient

aka 差商

$f[x_i,x_{i+1}]=\frac{f(x_{i+1})-f(x_i)}{x_{i+1}-x_i}$ first order difference quotient

$f[x_i,x_{i+1},x_{i+2}]=\frac{f[x_{i+1},x_{i+2}]-f[x_i,x_{i+1}]}{x_{i+2}-x_i}$ second order difference quotient

m-th order difference quotient

Formula

$N_n(x)=f(x_0)+f[x_0,x_1](x-x_0)+f[x_0,x_1,x_2](x-x_0)(x-x_1)+\cdots$

Residuals

$R_n(x)=f[x_0,x_1,\cdots,x_n,x](x-x_0)(x-x_1)\cdots(x-x_n)$

Runge Phenomenon

龙格现象

The more points used for interpolation, the more the result deviates from the original function.

Solutions

divided the interval into several small intervals, and the interpolation is performed in each small intervals

Spline Interpolation Cubic Polynomial

样条插值

$f(x)$ in $[a,b]$, given n+1 points $x_i$, $f(x_i)$. $S(x)$ satisfies:

1. in each small interval $[x_i,x_{i+1}]$, $S(x)$ is 3-times polynomial.
2. in each point $S(x_i)=f(x_i)$
3. in $[a,b]$, exist $S'(x)$ and $S''(x)$.

The $S(x)$ is Spline Interpolation Cubic Polynomial 样条插值三次

Ordinary Least Squares

最小二乘法

Like find the minimum of loss function (Machine Learning)

1. Find the partial derivative for each x in the polynomial
2. put the results into a system of equations
3. calculate the result.