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This function will pass through all +specified N+1 interpolation points (also referred to as **data points** or **nodes**). + +The interpolation points or nodes are given as: + +$$\begin{gather*} + x_0 & f(x_0)\equiv f_0 \\ + x_1 & f(x_1)\equiv f_1 \\ + x_2 & f(x_2)\equiv f_2 \\ + ... \\ + x_N & f(x_N)\equiv f_N +\end{gather*}$$ + +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): + +$$g(x)=a_0+a_1x+a_2x^2+a_3x^3+...+a_Nx^N$$ + +where $a_i=\text{unknown coefficients}$, $i=0,N\text{(N+1 coefficients)}$. + +No matter how we derive the $N^{th}$ degree polynomial: + +* Fitting power series +* Lagrange interpolating functions +* Newton forward or backward interpolation + +The resulting polynomial will always be the same! + +## Power Series Fitting to Define Lagrange Interpolation + +g(x) must match f(x) at the selected data points: + +$$\begin{gather*} + g(x_0)=f_0 \to a_0+a_1x_0+a_2x_0^2+...+a_Nx_0^N=f_0 \\ + g(x_1)=f_1 \to a_0+a_1x_1+a_2x_1^2+...+a_Nx_1^N=f_1 \\ + ... \\ + g(x_N)=f_N \to a_0+a_1x_N+a_2x_N^2+...+a_Nx_N^N=f_N \\ +\end{gather*}$$ + +Solve set of simultaneous equations: + +$$\begin{bmatrix} + 1 & x_0 & x_0^2 & ... & x_0^N \\ + 1 & x_1 & x_1^2 & ... & x_1^N \\ + ... \\ + 1 & x_N & x_N^2 & ... & x_N^N +\end{bmatrix} \begin{bmatrix} + a_0 \\ a_1 \\ ... \\ a_N +\end{bmatrix} = \begin{bmatrix} + f_0 \\ f_1 \\ ... \\ f_N +\end{bmatrix}$$ + +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). + +## Lagrange Interpolation Using Basis Functions + +We note that in general: + +$$g(x_i)=f_i$$ + +Let + +$$g(x)=\sum_{i=0}^Nf_iV_i(x)$$ + +where $V_i(x)$ polynomial of degree associated with each node such that + +$$\begin{equation*} + V_i(x_j)\equiv \begin{cases} + 0 & i\neq j \\ + 1 & i= j + \end{cases} +\end{equation*}$$ + +For example if we have 5 interpolation points (or nodes) + +$$g(x_3)=f_0V_0(x_3)+f_1V_1(x_3)+f_2V_2(x_3)+f_3V_3(x_3)+f_4V_4(x_3)$$ + +Using the definition for $V_i(x_j)$: + +$$\begin{gather*} + V_0(x_3)=0 \\ + V_1(x_3)=0 \\ + V_2(x_3)=0 \\ + V_3(x_3)=1 \\ + V_4(x_4)=0 +\end{gather*}$$ + +we have: + +$$g(x_3)=f_3$$ + +How do we construct $V_i(x)$? + +* Degree N +* Roots at $x_0,x_1,x_2,...,x_{i-1},...,x_N$ (at all nodes except $x_i$) +* $V_i(x_i)=1$ + +Let $W_i(x)=(x-x_0)(x-x_1)(x-x_2)...(x-x_{i+1})...(x-x_N)$ + +* 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 $x_i$ +* Degree of $W_i(x)$ is N +* However $W_i(x)$ in the form presented will not equal to unity at $x_i$ + +We normalize $W_i(x)$ and define the Lagrange basis functions $V_i(x)$: + +$$V_i(x)=\frac{(x-x_0)(x-x_1)(x-x_2)...(x-x_{i-1})(x-x_{i+1})...(x-x_N)}{(x_i-x_0)(x_i-x_1)(x_i-x_2)...(x_i-x_{i-1})(x_i-x_{i+1})...(x_i-x_N)}$$ + +Now we have $V_i(x)$ such that $V_i(x_i)$ equals: + +$$\begin{gather*} +V_i(x)=\frac{(x_i-x_0)(x_i-x_1)(x_i-x_2)...(x_i-x_{i-1})(1)(x_i-x_{i+1})...(x_i-x_N)}{(x_i-x_0)(x_i-x_1)(x_i-x_2)...(x_i-x_{i-1})(x_i-x_{i+1})...(x_i-x_N)} \\ +\\ +\to V_i(x_i)=1 +\end{gather*}$$ + +We alos satisfy $V_i(x_j)=0$ for $i\neq j$ + +e.g. + +$$V_1(x_2)=\frac{(x_2-x_0)(1)(x_2-x_2)(x_2-x_3)...(x_2-x_N)}{(x_1-x_0)(1)(x_1-x_2)(x_1-x_3)...(x_1-x_N)}=0$$ + +The general form of the interpolating function g(x) with the specified form of $V_i(x)$ is: + +$$g(x)=\sum_{i=0}^{N}f_iV_i(x)$$ + +* The sum of polynomials of degree N is also polynomial of degree N +* g(x) is equivalent to fitting the power series and computing coefficients $a_0,...,a_N$ + +## Lagrange Linear Interpolation Using Basis Functions + +Linear Lagrange (N=1) is the simplest form of Lagrange Interpolation: + +$$\begin{gather*} + g(x)=\sum_{i=0}^{1}f_iV_i(x) \\ + \to g(x)=f_0V_0(x)+f_1V_1(x) +\end{gather*}$$ + +where + +$$V_0(x)=\frac{(x-x_1)}{(x_0-x_1)}=\frac{(x_1-x)}{(x_1-x_0)}$$ + +and + +$$V_1(x)=\frac{(x-x_0)}{(x_1-x_0)}$$ + +## Example + +Given the following data: + +$$\begin{gather*} + x_0=2 & f_0=1.5 \\ + x_1=5 & f_1=4.0 +\end{gather*}$$ + +Find the linear interpolating function g(x) + +Lagrange basis functions are: + +$$\begin{gather*} + V_0(x)=\frac{x-5}{-3} \\ + \\ + V_1(x)=\frac{x-2}{3} +\end{gather*}$$ + +Interpolating function g(x) is: + +$$g(x)=1.5V_0(x)+4.0V_1(x)$$ + +## 外部参考连接 + +1. [原文](https://coast.nd.edu/jjwteach/www/www/30125/pdfnotes/lecture3_6v13.pdf) diff --git a/Software/Applications/KiCAD/KiCAD_简要说明.md b/Software/Applications/KiCAD/KiCAD_简要说明.md index bd01a31..03f0824 100644 --- a/Software/Applications/KiCAD/KiCAD_简要说明.md +++ b/Software/Applications/KiCAD/KiCAD_简要说明.md @@ -1,5 +1,7 @@ # KiCAD 简要说明 +TODO: + 1. 2. 3. diff --git a/Software/Development/Language/LaTex/LaTex_数学公式.md b/Software/Development/Language/LaTex/LaTex_数学公式.md index 842c0be..ea008a7 100644 --- a/Software/Development/Language/LaTex/LaTex_数学公式.md +++ b/Software/Development/Language/LaTex/LaTex_数学公式.md @@ -8,6 +8,9 @@ - [5. 带标字母](#5-带标字母) - [6. 常用数集](#6-常用数集) - [7. 组合](#7-组合) + - [7.1. 基本组合](#71-基本组合) + - [7.2. 矩阵](#72-矩阵) + - [8. 排版](#8-排版) ## 1. 行内公式与行间公式 @@ -141,6 +144,8 @@ $$a+b$$ ## 7. 组合 +### 7.1. 基本组合 + 等式： $$\begin{equation} @@ -190,3 +195,36 @@ $$\begin{split} F_s(s)=\int_{0}^{\infty}[\sum_{n=0}^{\infty}f(nT)\delta(t-nT)]e^{-st}dt \\ =\sum_{n=0}^{\infty}f(nT)e^{-snT} \\ \end{split}$$ + +### 7.2. 矩阵 + +$$\begin{matrix} + 0 & 1 \\ 1 & 0 \\ +\end{matrix}$$ + +$$\begin{pmatrix} + 0 & 1 \\ 1 & 0 \\ +\end{pmatrix}$$ + +$$\begin{bmatrix} + 0 & 1 \\ 1 & 0 \\ +\end{bmatrix}$$ + +$$\begin{Bmatrix} + 0 & 1 \\ 1 & 0 \\ +\end{Bmatrix}$$ + +$$\begin{vmatrix} + 0 & 1 \\ 1 & 0 \\ +\end{vmatrix}$$ + +$$\begin{Vmatrix} + 0 & 1 \\ 1 & 0 \\ +\end{Vmatrix}$$ + +## 8. 排版 + +| LaTex | Description | +|-------|-------------| +| \\ | 换行 | +| & | 间隔，相当于 Tab |