def diff(xi,yi,n): ''' param xi:插值節(jié)點xi param yi:插值節(jié)點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

測試一下：

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

測試一下：

x = 1.682print(Newton(x))

結(jié)果為：

2.5944760289639732

def Newton(xi,yi,n,x): ''' param xi:插值節(jié)點xi param yi:插值節(jié)點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

到此這篇關(guān)于用Python實現(xiàn)牛頓插值法的文章就介紹到這了,更多相關(guān)python牛頓插值法內(nèi)容請搜索好吧啦網(wǎng)以前的文章或繼續(xù)瀏覽下面的相關(guān)文章希望大家以后多多支持好吧啦網(wǎng)！