範例1-3 通過下面4點的Lagrange內插法求P(1.5)之值?

範例1-3
通過下面4點的Lagrange內插法求P(1.5)之值?

(1 , 0.0)  (2 , 0.693)  (3 , 1.099)  (4 , 1.386)

x               f(x)
==================
1.0              0.0
2.0              0.693
3.0              1.099
4.0              1.386
===================



# Lagrange Interpolation  Code
import math

def Lagrange( n, k,val):
    result = 1.0
    for i in range  (0 , n):
        if (i!=k):
            result = result * ( (val - x[i])  /  (x[k] - x[i]))
    return result

def Calculate(n, x):
    intPolasyon = 0.0
    L=  [  0.0 for i in range(size) ]
    for i in range (0 ,  n):
        L[i] = Lagrange(n, i, x)
        intPolasyon += fx[i] * L[i]
        print("L(%d) = (%5.3f)\n" %( i, L[i]))

    print("\n")
    return intPolasyon




size=4
print("Please enter function's point count:",size)
fx= [  0.0 for i in range(size) ]
x=  [  0.0 for i in range(size) ]
x=[1.0  , 2.0  , 3.0  ,4.0]
fx=[0.0 , 0.693 , 1.099 , 1.386]
for i  in range (0,len(x)) :   
    print(  '\t',round(x[i],4),"\t\t",round( fx[i],4))
print()

question=1.5
print("Please enter point to calculate:", question)
print("\n");
print("Result =====> f(%4.2f) = (%5.4f)\n" %(question, Calculate(size, question)) )

========= RESTART: F:/2018-09勤益科大數值分析/數值分析/PYTHON/EX1-3.py ============
Please enter function's point count: 4
1.0 0.0
2.0 0.693
3.0 1.099
4.0 1.386

Please enter point to calculate: 1.5


L(0) = (0.312)

L(1) = (0.938)

L(2) = (-0.312)

L(3) = (0.062)



Result =====> f(1.50) = (0.3929)

>>>




源自於
https://mat.iitm.ac.in/home/sryedida/public_html/caimna/interpolation/lagrange.html
LAGRANGE'S INTERPOLATION FORMULA

This is again an N^th degree polynomial approximation formula to the function f(x), which is known at discrete points x_i, i = 0, 1, 2 . . . N^th. The formula can be derived from the Vandermonds determinant but a much simpler way of deriving this is from Newton's divided difference formula. If f(x) is approximated with an N^th degree polynomial then the N^thdivided difference of f(x) constant and (N+1)^th divided difference is zero. That is

f [x₀, x₁, . . . x_n, x] = 0

From the second property of divided difference we can write

f0	+	fn		fx	=  0
			+ . . . +
(x0 - x1) . . . (x0 - xn)(x0- x)		(xn - x0) . . . (xn - xn-1)(xn- x)		(x - x0) . . . (x - xn)

or

	(x - x1) . . . (x - xn)		(x - x0) .  .  . (x - xn-1)
f(x)  =		f0  + .   .   . +		fn
	(x0 - x1) . . . (x0 - xn)		(xn - x0) .  .  . (xn - xn-1)

 

n	(	n			)fi
S		|  |		x - xj
		j = 0		(xi - xj)
i = 0		j ¹ 1

Since Lagrange's interpolation is also an N^th degree polynomial approximation to f(x) and the N^th degree polynomial passing through (N+1) points is unique hence the Lagrange's and Newton's divided difference approximations are one and the same. However, Lagrange's formula is more convinent to use in computer programming and Newton's divided difference formula is more suited for hand calculations.

Example : Compute f(0.3) for the data

x	0	1	3	4	7
f	1	3	49	129	813

using Lagrange's interpolation formula (Analytic value is 1.831)

 

	(x - x1) (x - x2)(x- x3)(x - x4)		(x - x0)(x - x1) (x - x2)(x - x3)
f(x)  =		f0+ . . . +		f4
	(x0 - x1) (x0 - x2)(x0 - x3)(x0 - x4)		(x4 - x0)(x4 - x1)(x4 - x2)(x4 - x3)

 

	(0.3 - 1)(0.3 - 3)(0.3 - 4)(0.3 - 7)		(0.3 - 0)(0.3 - 3)(0.3 - 4)(0.3 - 7)
=		1+		3 +
	(-1) (-3)(-4)(-7)		1 x (-2)(-3)(-6)

  

(0.3 - 0)(0.3 - 1)(0.3 - 4)(0.3 - 7)		(0.3 - 0)(0.3 - 1)(0.3 - 3)(0.3 - 7)
	49 +		129 +
3 x 2 x (-1)(-4)		4 x 3 x 1 (-3)

  

(0.3 - 0)(0.3 - 1)(0.3 - 3)(0.3 - 4)
	813
7 x 6 x 4 x 3

         = 1.831