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:
- in each small interval \([x_i,x_{i+1}]\), \(S(x)\) is 3-times polynomial.
- in each point \(S(x_i)=f(x_i)\)
- 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)
- Find the partial derivative for each x in the polynomial
- put the results into a system of equations
- calculate the result.