用Python实现Newton插值法用Python实现Newton插值法

1. n阶差商实现
2. 牛顿插值实现
3.完整Python代码

1. n阶差商实现

def diff(xi,yi,n):"""param xi:插值节点xiparam yi:插值节点yiparam n: 求几阶差商return: n阶差商"""if len(xi) != len(yi):#xi和yi必须保证长度一致returnelse:diff_quot = [[] for i in range(n)]for j in range(1,n+1):if j == 1:for i in range(n+1-j):diff_quot[j-1].append((yi[i]-yi[i+1]) / (xi[i] - xi[i + 1]))else:for i in range(n+1-j):diff_quot[j-1].append((diff_quot[j-2][i]-diff_quot[j-2][i+1]) / (xi[i] - xi[i + j]))return diff_quot

测试一下：

xi = [1.615,1.634,1.702,1.828]yi = [2.41450,2.46259,2.65271,3.03035]n = 3print(diff(xi,yi,n))

返回的差商结果为：

[[2.53105263157897, 2.7958823529411716, 2.997142857142854], [3.0440197857724347, 1.0374252793901158], [-9.420631485362996]]

2. 牛顿插值实现

def Newton(x):f = yi[0]v = []r = 1for i in range(n):r *= (x - xi[i])v.append(r)f += diff_quot[i][0] * v[i]return f

测试一下：

x = 1.682print(Newton(x))

结果为：

2.5944760289639732

3.完整Python代码

def Newton(xi,yi,n,x):"""param xi:插值节点xiparam yi:插值节点yiparam n: 求几阶差商param x: 代求近似值return: n阶差商"""if len(xi) != len(yi):#xi和yi必须保证长度一致returnelse:diff_quot = [[] for i in range(n)]for j in range(1,n+1):if j == 1:for i in range(n+1-j):diff_quot[j-1].append((yi[i]-yi[i+1]) / (xi[i] - xi[i + 1]))else:for i in range(n+1-j):diff_quot[j-1].append((diff_quot[j-2][i]-diff_quot[j-2][i+1]) / (xi[i] - xi[i + j]))print(diff_quot)f = yi[0]v = []r = 1for i in range(n):r *= (x - xi[i])v.append(r)f += diff_quot[i][0] * v[i]return f

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