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)

外部参考资料

1. 原文