1. n阶差商实现
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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:
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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
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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:
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x = 1.682 print (Newton(x)) |
结果为:
2.5944760289639732
3.完整Python代码
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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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原文链接:https://blog.csdn.net/weixin_45812669/article/details/115694394