Newton插值法——Python实现

最近在做数值分析的作业，作业里面的小数点让计算能力本就薄弱的我雪上加霜，为了偷个小懒快速把作业完成，所以有了这篇博客。哈哈哈哈哈，让我们一起复制copy，完成作业，哈哈哈哈

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 = 3
print(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 = 1.682
print(Newton(x))

结果为：

2.5944760289639732

【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