Question

我试图使用牛顿插值多项式计算下一个数组的有限分差，以确定x = 8时的y。数组是

x = 0  1  2  5.5  11  13  16  18

y=  0.5  3.134  5.9  9.9  10.2  9.35  7.2  6.2

我所拥有的伪代码位于http://imgur.com/gallery/Lm2KXxA/new。我可以使用任何可用的伪代码，算法或库来告诉我答案吗？

另外我相信这是如何在matlab http://imgur.com/gallery/L9wJaEH/new中完成该程序。我只是无法弄清楚如何在python中做到这一点。

Answer 1

这是Python代码。函数coef计算有限的分割差分系数，函数Eval评估给定节点的插值。

import numpy as np
import matplotlib.pyplot as plt

def coef(x, y):
    '''x : array of data points
       y : array of f(x)  '''
    x.astype(float)
    y.astype(float)
    n = len(x)
    a = []
    for i in range(n):
        a.append(y[i])

    for j in range(1, n):

        for i in range(n-1, j-1, -1):
            a[i] = float(a[i]-a[i-1])/float(x[i]-x[i-j])

    return np.array(a) # return an array of coefficient

def Eval(a, x, r):

     ''' a : array returned by function coef()
        x : array of data points
        r : the node to interpolate at  '''

    x.astype(float)
    n = len( a ) - 1
    temp = a[n]
    for i in range( n - 1, -1, -1 ):
        temp = temp * ( r - x[i] ) + a[i]
    return temp # return the y_value interpolation

Answer 2

@Ledruid提出的解决方案是最佳的。它不需要二维数组。分割差异树的方式类似于2D阵列。一种更原始的方法（从中可以获得@Leudruid的算法）正在考虑对应于树节点的“矩阵$ F_ {ij} $。这样算法就是。

import numpy as np
from numpy import *

def coeficiente(x,y) :
    ''' x: absisas x_i 
        y : ordenadas f(x_i)'''

    x.astype(float)
    y.astype(float)
    n = len(x)
    F = zeros((n,n), dtype=float)
    b = zeros(n) 
    for i in range(0,n):
        F[i,0]=y[i]



    for j in range(1, n):
        for i in range(j,n):
            F[i,j] = float(F[i,j-1]-F[i-1,j-1])/float(x[i]-x[i-j])


    for i in range(0,n):
        b[i] = F[i,i]

    return np.array(b) # return an array of coefficient

def Eval(a, x, r):

    '''  a : retorno de la funcion coeficiente() 
         x : abcisas x_i
         r : abcisa a interpolar
    '''

    x.astype(float)
    n = len( a ) - 1
    temp = a[n]
    for i in range( n - 1, -1, -1 ):
        temp = temp * ( r - x[i] ) + a[i]
    return temp # return the y_value interpolation

Answer 3

试试这个

./configure

Answer 4

如果您不想使用任何函数定义，那么这里是简单的代码：

## Newton Divided Difference Polynomial Interpolation Method

import numpy as np

x=np.array([0,1,2,5.5,11,13,16,18],float)
y=np.array([0.5,  3.134,  5.9,  9.9,  10.2,  9.35,  7.2,  6.2],float)
n=len(x)
p=np.zeros([n,n+1])#creating a Tree table (n x n+1 array)
value =float(input("Enter the point at which you want to calculate the value of the polynomial: "))
# first two columns of the table are filled with x and y data points
for i in range(n):

        p[i,0]=x[i]
        p[i,1]=y[i]

## algorithm for tree table from column 2 two n+1        
for i in range(2,n+1): #column
  for j in range(n+1-i):# defines row
    p[j,i]=(p[j+1,i-1]-p[j,i-1])/(x[j+i-1]-x[j])#Tree Table
np.set_printoptions(suppress=True)## this suppress the scientific symbol(e) and returns values in normal digits

# print(p) ## can check the complete Tree table here for NDDP
b=p[0][1:]#This vector contains the unknown coefficients in the polynomial which are the top elements of each column. 
print("b= ",b)
print("x= ",x)
lst=[] # list where we will append the values of prouct terms

t=1
for i in range(len(x)):
    t*=(value-x[i]) ##(x-x0), (x-x0)(x-x1), (x-x0)(x-x1)(x-x2) etc..
    lst.append(t)
print("The list of product elements ",lst,end = ' ')
## creating a general function
f=b[0]
for k in range(1,len(b)):
  f+=b[k]*lst[k-1] ## important**[k-1]** not k because in list we use one step earlier element. For example for b1 we have to use (x-x0), for b2, we use (x-x0)(x-x1) for b3 we use (x-x0)(x-x1)(x2)
print("The value of polynomial: ","%.3f"%f)

牛顿的插值多项式[python]

4 个答案:

牛顿的插值多项​​式[python]

4 个答案:

牛顿的插值多项式[python]