用Python实现Newton插值法
作者:Amiyai
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,哈哈哈哈需要的朋友可以参考下
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
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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