習題1-3 利用Lagrange 寫出方程式

習題1-3 利用Lagrange 寫出方程式

'''
n=3
xa=[-0.5 ,0.5 ,1.2 ,  1.5 ,1.8, 2.5 ]

-1.0   0.3679
0.0    1.00
1.0    2.7183
2.0    7.3891

/* pr1-1-b.py : 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])
 */
'''
import math

def fx(x):
    return (math.exp(x))


print('\nLagrange Interpolation Algorithm\n')

xa=[-0.5 , 0.5 ,1.2,  1.5 ,1.8, 2.5 ]
x=[-1.0 , 0.0 , 1.0 , 2.0 ]
f=[0.3679 , 1.0 , 2.7183 , 7.3891]

n=3  # 4 point n=3


print('x=',x)
print('f=',f)
print('xa=',xa)
print('\n')


for m in range (len(xa)):
    indata=xa[m]
    result=0.0
    print("xa=",repr(indata))
    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 * ( indata - x[i]) / ( x[k] - x[i])
        result = result + temp *f[k]

    s1 = 'The value of p' + repr(indata) + '= ' + repr(result) + '   ...'
    print(s1)

    resultfx=fx(indata)
    s1 = 'The value of f(' + repr(indata) + ')= ' + repr(resultfx) + '   ...'
    print(s1)

    s1 = 'The value of  | f(' + repr(indata) + ') -  p' + repr(indata) + ' = ' + repr( abs(resultfx - result) ) + '  ...'
    print(s1)
    print("\n\n")
   



======== RESTART: H:/2018-09勤益科大數值分析/數值分析/PYTHON/pr1-1-c.py ==========

Lagrange Interpolation Algorithm

x= [-1.0, 0.0, 1.0, 2.0]
f= [0.3679, 1.0, 2.7183, 7.3891]
xa= [-0.5, 0.5, 1.2, 1.5, 1.8, 2.5]


xa= -0.5
The value of p-0.5= 0.66481875   ...
The value of f(-0.5)= 0.6065306597126334   ...
The value of  | f(-0.5) -  p-0.5 = 0.05828809028736659  ...



xa= 0.5
The value of p0.5= 1.6067312500000002   ...
The value of f(0.5)= 1.6487212707001282   ...
The value of  | f(0.5) -  p0.5 = 0.04199002070012803  ...



xa= 1.2
The value of p1.2= 3.3565384   ...
The value of f(1.2)= 3.3201169227365472   ...
The value of  | f(1.2) -  p1.2 = 0.03642147726345257  ...



xa= 1.5
The value of p1.5= 4.56799375   ...
The value of f(1.5)= 4.4816890703380645   ...
The value of  | f(1.5) -  p1.5 = 0.08630467966193578  ...



xa= 1.8
The value of p1.8= 6.1291576   ...
The value of f(1.8)= 6.0496474644129465   ...
The value of  | f(1.8) -  p1.8 = 0.07951013558705355  ...



xa= 2.5
The value of p2.5= 11.414906249999998   ...
The value of f(2.5)= 12.182493960703473   ...
The value of  | f(2.5) -  p2.5 = 0.7675877107034754  ...



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