NotePublic/Algorithm/DSP/拉格朗日插值.md

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post	拉格朗日插值		给出拉格朗日多项式插值的推导过程和标准公式。	通过简单的推导过程和示例程序演示拉格朗日多项式插值。	2022-09-08 12:50:00	Rick Chan	DSP Lagrange Interpolation Polynomial	Algorithm	true	true

Fit N+1 points with an N^{th} degree polynomial

f(x) = exact function of which only N+1 discrete values are known and used to estab- lish an interpolating or approximating function g(x).

g(x) = approximating or interpolating function. 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)