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小編給大家分享一下如何使用Python實(shí)現(xiàn)Newton插值法,相信大部分人都還不怎么了解,因此分享這篇文章給大家參考一下,希望大家閱讀完這篇文章后大有收獲,下面讓我們一起去了解一下吧!
def diff(xi,yi,n): """ param xi:插值節(jié)點(diǎn)xi param yi:插值節(jié)點(diǎn)yi param n: 求幾階差商 return: n階差商 """ if len(xi) != len(yi): #xi和yi必須保證長(zhǎng)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
測(cè)試一下:
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))
返回的差商結(jié)果為:
[[2.53105263157897, 2.7958823529411716, 2.997142857142854], [3.0440197857724347, 1.0374252793901158], [-9.420631485362996]]
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
測(cè)試一下:
x = 1.682 print(Newton(x))
結(jié)果為:
2.5944760289639732
def Newton(xi,yi,n,x): """ param xi:插值節(jié)點(diǎn)xi param yi:插值節(jié)點(diǎn)yi param n: 求幾階差商 param x: 代求近似值 return: n階差商 """ if len(xi) != len(yi): #xi和yi必須保證長(zhǎng)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
以上是“如何使用Python實(shí)現(xiàn)Newton插值法”這篇文章的所有內(nèi)容,感謝各位的閱讀!相信大家都有了一定的了解,希望分享的內(nèi)容對(duì)大家有所幫助,如果還想學(xué)習(xí)更多知識(shí),歡迎關(guān)注億速云行業(yè)資訊頻道!
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