利用Lagrange內插法滿足 f(x)=log e (x)=ln (x) -----(b)

利用Lagrange內插法滿足 f(x)=log e (x)=ln (x) -----(b)

'''
n=3
xa=1.5
1.0  0.000
1.2  0.182
1.4  0.336
2.0  0.693


/* Lagrange Interpolation Algorithm
 * Read in data file of ex1-4.dat which has n point values
 * and the value of interpolating point xa. Based on Lagrange
 * Interpolation algorithm to compute p(xa) and output its value.
 * (x[i],f[i]):given points and n+1 are number of points
 * Ln,k(x)=l=summation of (x-x[i])/(x[k]-x[i]).
 * p(x)=ff=L(x)*f(x[k])
 */
'''
print('\nLagrange Interpolation Algorithm\n')

xa=1.5
x= list()
x.extend([1.0,1.2,1.3,1.4])
f= list()
f.extend([0.0,0.182,0.336,0.693])
result=0.0
n=3

print(x)
print(f)
print('\n')

for k in range (0,n+1):    #n -->n+1
    temp=1.0;
    for i in range (0,n+1):  #n -->n+1
        if(i !=k):
            temp=temp * ( xa - x[i]) / ( x[k] - x[i])
    result=result+temp*f[k]

s = 'The value of p' + repr(xa) + '= ' + repr(result) + '...'
print(s)

輸出畫面

Lagrange Interpolation Algorithm

[1.0, 1.2, 1.3, 1.4]
[0.0, 0.182, 0.336, 0.693]


The value of p1.5= 1.3737500000000018...
>>>