用Python实现Newton插值法

1. n阶差商实现

def diff(xi,yi,n):    """    param xi:插值节点xi    param yi:插值节点yi    param 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        return    else:        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 = 1    for 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xi    param yi:插值节点yi    param 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        return    else:        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 = 1    for i in range(n):        r *= (x - xi[i])        v.append(r)        f += diff_quot[i][0] * v[i]    return f

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