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# 拉格朗日插值
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<https://coast.nd.edu/jjwteach/www/www/30125/pdfnotes/lecture3_6v13.pdf>
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Fit N+1 points with an $N^{th}$ degree polynomial
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f(x) = exact function of which only N+1 discrete values are known and used to estab-
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lish an interpolating or approximating function g(x).
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g(x) = approximating or interpolating function. This function will pass through all
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specified N+1 interpolation points (also referred to as **data points** or **nodes**).
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The interpolation points or nodes are given as:
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$$\begin{gather*}
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x_0 & f(x_0)\equiv f_0 \\
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x_1 & f(x_1)\equiv f_1 \\
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x_2 & f(x_2)\equiv f_2 \\
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... \\
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x_N & f(x_N)\equiv f_N
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\end{gather*}$$
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There exists only one $N^{th}$ degree polynomial that passes through a given set of N+1 points. It’s form is (expressed as a power series):
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$$g(x)=a_0+a_1x+a_2x^2+a_3x^3+...+a_Nx^N$$
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where $a_i=\text{unknown coefficients}$, $i=0,N\text{(N+1 coefficients)}$.
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No matter how we derive the $N^{th}$ degree polynomial:
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* Fitting power series
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* Lagrange interpolating functions
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* Newton forward or backward interpolation
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The resulting polynomial will always be the same!
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## Power Series Fitting to Define Lagrange Interpolation
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g(x) must match f(x) at the selected data points:
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$$\begin{gather*}
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g(x_0)=f_0 \to a_0+a_1x_0+a_2x_0^2+...+a_Nx_0^N=f_0 \\
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g(x_1)=f_1 \to a_0+a_1x_1+a_2x_1^2+...+a_Nx_1^N=f_1 \\
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... \\
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g(x_N)=f_N \to a_0+a_1x_N+a_2x_N^2+...+a_Nx_N^N=f_N \\
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\end{gather*}$$
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Solve set of simultaneous equations:
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$$\begin{bmatrix}
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1 & x_0 & x_0^2 & ... & x_0^N \\
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1 & x_1 & x_1^2 & ... & x_1^N \\
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... \\
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1 & x_N & x_N^2 & ... & x_N^N
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\end{bmatrix} \begin{bmatrix}
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a_0 \\ a_1 \\ ... \\ a_N
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\end{bmatrix} = \begin{bmatrix}
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f_0 \\ f_1 \\ ... \\ f_N
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\end{bmatrix}$$
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It is relatively computationally costly to solve the coefficients of the interpolating function g(x) (i.e. you need to program a solution to these equations).
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## Lagrange Interpolation Using Basis Functions
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We note that in general:
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$$g(x_i)=f_i$$
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Let
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$$g(x)=\sum_{i=0}^Nf_iV_i(x)$$
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where $V_i(x)$ polynomial of degree associated with each node such that
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$$\begin{equation*}
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V_i(x_j)\equiv \begin{cases}
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0 & i\neq j \\
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1 & i= j
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\end{cases}
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\end{equation*}$$
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For example if we have 5 interpolation points (or nodes)
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$$g(x_3)=f_0V_0(x_3)+f_1V_1(x_3)+f_2V_2(x_3)+f_3V_3(x_3)+f_4V_4(x_3)$$
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Using the definition for $V_i(x_j)$:
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$$\begin{gather*}
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V_0(x_3)=0 \\
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V_1(x_3)=0 \\
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V_2(x_3)=0 \\
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V_3(x_3)=1 \\
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V_4(x_4)=0
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\end{gather*}$$
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we have:
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$$g(x_3)=f_3$$
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How do we construct $V_i(x)$?
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* Degree N
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* Roots at $x_0,x_1,x_2,...,x_{i-1},...,x_N$ (at all nodes except $x_i$)
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* $V_i(x_i)=1$
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Let $W_i(x)=(x-x_0)(x-x_1)(x-x_2)...(x-x_{i+1})...(x-x_N)$
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* The function $W_i$ is such that we do have the required roots, i.e. it equals zeros at nodes $x_0,x_1,x_2,...,x_N$ except at node $x_i$
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* Degree of $W_i(x)$ is N
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* However $W_i(x)$ in the form presented will not equal to unity at $x_i$
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We normalize $W_i(x)$ and define the Lagrange basis functions $V_i(x)$:
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$$V_i(x)=\frac{(x-x_0)(x-x_1)(x-x_2)...(x-x_{i-1})(x-x_{i+1})...(x-x_N)}{(x_i-x_0)(x_i-x_1)(x_i-x_2)...(x_i-x_{i-1})(x_i-x_{i+1})...(x_i-x_N)}$$
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Now we have $V_i(x)$ such that $V_i(x_i)$ equals:
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$$\begin{gather*}
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V_i(x)=\frac{(x_i-x_0)(x_i-x_1)(x_i-x_2)...(x_i-x_{i-1})(1)(x_i-x_{i+1})...(x_i-x_N)}{(x_i-x_0)(x_i-x_1)(x_i-x_2)...(x_i-x_{i-1})(x_i-x_{i+1})...(x_i-x_N)} \\
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\\
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\to V_i(x_i)=1
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\end{gather*}$$
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We alos satisfy $V_i(x_j)=0$ for $i\neq j$
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e.g.
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$$V_1(x_2)=\frac{(x_2-x_0)(1)(x_2-x_2)(x_2-x_3)...(x_2-x_N)}{(x_1-x_0)(1)(x_1-x_2)(x_1-x_3)...(x_1-x_N)}=0$$
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The general form of the interpolating function g(x) with the specified form of $V_i(x)$ is:
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$$g(x)=\sum_{i=0}^{N}f_iV_i(x)$$
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* The sum of polynomials of degree N is also polynomial of degree N
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* g(x) is equivalent to fitting the power series and computing coefficients $a_0,...,a_N$
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## Lagrange Linear Interpolation Using Basis Functions
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Linear Lagrange (N=1) is the simplest form of Lagrange Interpolation:
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$$\begin{gather*}
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g(x)=\sum_{i=0}^{1}f_iV_i(x) \\
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\to g(x)=f_0V_0(x)+f_1V_1(x)
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\end{gather*}$$
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where
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$$V_0(x)=\frac{(x-x_1)}{(x_0-x_1)}=\frac{(x_1-x)}{(x_1-x_0)}$$
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and
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$$V_1(x)=\frac{(x-x_0)}{(x_1-x_0)}$$
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## Example
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Given the following data:
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$$\begin{gather*}
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x_0=2 & f_0=1.5 \\
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x_1=5 & f_1=4.0
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\end{gather*}$$
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Find the linear interpolating function g(x)
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Lagrange basis functions are:
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$$\begin{gather*}
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V_0(x)=\frac{x-5}{-3} \\
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\\
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V_1(x)=\frac{x-2}{3}
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\end{gather*}$$
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Interpolating function g(x) is:
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$$g(x)=1.5V_0(x)+4.0V_1(x)$$
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## 外部参考连接
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1. [原文](https://coast.nd.edu/jjwteach/www/www/30125/pdfnotes/lecture3_6v13.pdf)
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@ -1,5 +1,7 @@
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# KiCAD 简要说明
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TODO:
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1. <https://blog.csdn.net/weixin_44916154/article/details/124503031>
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2. <https://gitee.com/hooke6164/footprint_-ki-cad_-gg/>
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3. <https://blog.csdn.net/yannanxiu/article/details/52643410>
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@ -8,6 +8,9 @@
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- [5. 带标字母](#5-带标字母)
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- [6. 常用数集](#6-常用数集)
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- [7. 组合](#7-组合)
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- [7.1. 基本组合](#71-基本组合)
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- [7.2. 矩阵](#72-矩阵)
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- [8. 排版](#8-排版)
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## 1. 行内公式与行间公式
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@ -141,6 +144,8 @@ $$a+b$$
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## 7. 组合
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### 7.1. 基本组合
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等式：
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$$\begin{equation}
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@ -190,3 +195,36 @@ $$\begin{split}
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F_s(s)=\int_{0}^{\infty}[\sum_{n=0}^{\infty}f(nT)\delta(t-nT)]e^{-st}dt \\
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=\sum_{n=0}^{\infty}f(nT)e^{-snT} \\
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\end{split}$$
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### 7.2. 矩阵
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$$\begin{matrix}
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0 & 1 \\ 1 & 0 \\
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\end{matrix}$$
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$$\begin{pmatrix}
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0 & 1 \\ 1 & 0 \\
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\end{pmatrix}$$
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$$\begin{bmatrix}
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0 & 1 \\ 1 & 0 \\
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\end{bmatrix}$$
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$$\begin{Bmatrix}
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0 & 1 \\ 1 & 0 \\
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\end{Bmatrix}$$
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$$\begin{vmatrix}
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0 & 1 \\ 1 & 0 \\
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\end{vmatrix}$$
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$$\begin{Vmatrix}
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0 & 1 \\ 1 & 0 \\
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\end{Vmatrix}$$
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## 8. 排版
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| LaTex | Description |
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|-------|-------------|
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| \\ | 换行 |
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| & | 间隔，相当于 Tab |
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