範例1-9 已知6點座標,依照牛頓差除法值計算P(0.2)之值

範例1-9  已知6點座標

n=5
0.0  -6.0
0.1  -5.89483
0.3  -5.65014
0.6  -5.17788
1.0  -4.28172
1.1  -3.99583

依照牛頓差除法值計算P(0.2)之值

=============程式===============
print ("Divided Differences (Newton) 插除法")
n = 5+1

vector=[0.0 , 0.1 , 0.3 , 0.6 , 1.0 , 1.1]
matrix=[[-6.0         ,0.0, 0.0, 0.0, 0.0, 0.0, 0.0] ,
               [-5.89483 ,0.0, 0.0, 0.0, 0.0, 0.0, 0.0] ,
               [-5.65014 ,0.0, 0.0, 0.0, 0.0, 0.0, 0.0] ,
               [-5.17788 ,0.0, 0.0, 0.0, 0.0, 0.0, 0.0] ,
               [-4.28172 ,0.0, 0.0, 0.0, 0.0, 0.0, 0.0] ,
               [-3.99583 ,0.0, 0.0, 0.0, 0.0, 0.0, 0.0]]

point_offset = 0.2
for i in range(1,n):
    for j in range(i,n):
        print ("i=%5d ,   j=%5d"  %(i , j) )
        print ("(%0.6f - %0.6f ) / (%0.6f - %0.6f) ) "  % ((matrix[j][i-1])  , ( matrix[j-1][i-1]) , (vector[j]) ,( vector[j-i]) ) )
        matrix[j][i] = ( (matrix[j][i-1]-matrix[j-1][i-1]) / (vector[j]-vector[j-i]))
        print ("matrix[ %5d] [ %5d] =%0.6f "     %(j, i,  (matrix[j][i-1]-matrix[j-1][i-1])/(vector[j]-vector[j-i]))  )

for i in range(n):
    print ((matrix[i]) )
   
aprx = 0
mul = 1.0
for i in range(n):
    print ("matrix[%5d][%5d] = %0.6f"  %(i,j,matrix[i][i]) )
    mul = matrix[i][i];
    for j in range(1,i+1):
        mul = mul * (point_offset - vector[j-1])
    aprx = aprx + mul
   
print ("The approximate value of f(",point_offset,") is: %0.6f" %( aprx))

輸出結果

Divided Differences (Newton) 插除法
i=    1 ,   j=    1
(-5.894830 - -6.000000 ) / (0.100000 - 0.000000) ) 
matrix[     1] [     1] =1.051700 

i=    1 ,   j=    2
(-5.650140 - -5.894830 ) / (0.300000 - 0.100000) ) 
matrix[     2] [     1] =1.223450 

i=    1 ,   j=    3
(-5.177880 - -5.650140 ) / (0.600000 - 0.300000) ) 
matrix[     3] [     1] =1.574200 

i=    1 ,   j=    4
(-4.281720 - -5.177880 ) / (1.000000 - 0.600000) ) 
matrix[     4] [     1] =2.240400 

i=    1 ,   j=    5
(-3.995830 - -4.281720 ) / (1.100000 - 1.000000) ) 
matrix[     5] [     1] =2.858900 

i=    2 ,   j=    2
(1.223450 - 1.051700 ) / (0.300000 - 0.000000) ) 
matrix[     2] [     2] =0.572500 

i=    2 ,   j=    3
(1.574200 - 1.223450 ) / (0.600000 - 0.100000) ) 
matrix[     3] [     2] =0.701500 

i=    2 ,   j=    4
(2.240400 - 1.574200 ) / (1.000000 - 0.300000) ) 
matrix[     4] [     2] =0.951714 

i=    2 ,   j=    5
(2.858900 - 2.240400 ) / (1.100000 - 0.600000) ) 
matrix[     5] [     2] =1.237000 

i=    3 ,   j=    3
(0.701500 - 0.572500 ) / (0.600000 - 0.000000) ) 
matrix[     3] [     3] =0.215000 

i=    3 ,   j=    4
(0.951714 - 0.701500 ) / (1.000000 - 0.100000) ) 
matrix[     4] [     3] =0.278016 

i=    3 ,   j=    5
(1.237000 - 0.951714 ) / (1.100000 - 0.300000) ) 
matrix[     5] [     3] =0.356607 

i=    4 ,   j=    4
(0.278016 - 0.215000 ) / (1.000000 - 0.000000) ) 
matrix[     4] [     4] =0.063016 

i=    4 ,   j=    5
(0.356607 - 0.278016 ) / (1.100000 - 0.100000) ) 
matrix[     5] [     4] =0.078591 

i=    5 ,   j=    5
(0.078591 - 0.063016 ) / (1.100000 - 0.000000) ) 
matrix[     5] [     5] =0.014159 


[-6.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0]
[-5.89483, 1.051700000000002, 0.0, 0.0, 0.0, 0.0, 0.0]
[-5.65014, 1.223449999999997, 0.5724999999999832, 0.0, 0.0, 0.0, 0.0]
[-5.17788, 1.5742000000000012, 0.7015000000000082, 0.21500000000004166, 0.0, 0.0, 0.0]
[-4.28172, 2.2404, 0.9517142857142844, 0.2780158730158624, 0.06301587301582073, 0.0, 0.0]
[-3.99583, 2.858899999999995, 1.2369999999999892, 0.356607142857131, 0.07859126984126863, 0.014159451659498086, 0.0]

matrix[    0][    5] = -6.000000
matrix[    1][    5] = 1.051700
matrix[    2][    1] = 0.572500
matrix[    3][    2] = 0.215000
matrix[    4][    3] = 0.063016
matrix[    5][    4] = 0.014159

The approximate value of f( 0.2 ) is: -5.778599
>>>