n=3
xa=1.5
1.0 0.0
2.0 0.693
3.0 1.099
4.0 1.386
/* ex1-4.c: 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])
*/
'''
xa=1.5
x= list()
x.extend([1.0,2.0,3.0,4.0])
f= list()
f.extend([0.0,0.693,1.099,1.386])
result=0.0
n=3
for k in range (0,n):
print(x)
print(f)
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)
Lagrange Interpolation Algorithm
[1.0, 2.0, 3.0, 4.0]
[0.0, 0.693, 1.099, 1.386]
The value of p1.5= 0.392875...
...Program finished with exit code 0
Press ENTER to exit console.
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