牛顿插值多项式（python实现）

╰+攻爆jí腚メ 2022-05-19 22:22 235阅读 0赞

## 理论知识 ##

[牛顿插值多项式（理论知识）][Link 1]

目标函数

f(x)=11\+x2  f ( x ) = 1 1 + x 2

插值点为\[-10, 10\]上的整数点。

## 图片 ##

![这里写图片描述][70]

## 代码实现 ##

import sympy
    import numpy as np
    from matplotlib import pyplot as plt
    
    
    def f(X):
        return 1 / (X ** 2 + 1)
    
    
    def ff(X=list()):
        if len(X) < 2:
            raise ValueError('X\'s length must be bigger than 2')
        ans = 0
        for i in range(len(X)):
            temp = 1.0
            for j in range(len(X)):
                if j == i:
                    continue
                temp *= (X[i] - X[j])
            ans += (f(X[i]) / temp)
        return ans
    
    
    def draw():
        plt.rcParams['font.sans-serif'] = ['SimHei']
        plt.rcParams['axes.unicode_minus'] = False
        X = np.linspace(-10, 10, 100)
    
        TargetY = f(X)
        GetY = [Px.subs(x, i) for i in X]
    
        plt.plot(X, TargetY, label=r'$\frac{1}{x^2+1}$')
        plt.plot(X, GetY, label='$L(x)$')
        plt.legend()
        plt.show()
    
    
    def generatePx(DataX):
        ans = f(DataX[0])
        if len(DataX) == 1:
            return ans
        else:
            temp = 1
            for i in range(len(DataX) - 1):
                temp *= (x - DataX[i])
                ans += ff(DataX[:i + 2]) * temp
            return ans
    
    
    if __name__ == '__main__':
        x = sympy.symbols('x')
    
        DataX = np.linspace(-10, 10, 11)  # 插值点
    
        Px = sympy.expand(generatePx(DataX))
        draw()

[Link 1]: https://blog.csdn.net/a19990412/article/details/80961574
[70]: /images/20220520/491788e3ff874a2792b6ee4eb748c077.png