Lecture Note of Lagrange Interpolation

University of Babylon
College of Engineering
Department of Environmental Engineering
Engineering Analysis I (ENAN 103)








Lagrange Interpolation
Undergraduate Leve, 3th Stage



Mr. Waleed Ali Tameemi
Engineer/ College of Engineering/ Babylon University
M.Sc. Civil Engineering/ the University of Kansas/ USA



2016-2017
?
4.0 – Lagrange Interpolation
If the data set in the following table was obtained by experiments or by observation, the value corresponding to any x can be estimated as follows:
x_i f(x_i)
x_0 f(x_0)
x_1 f(x_1)
x_2 f(x_2)
x_3 f(x_3)
. .
. .
x_(n-1) f(x_(n-1))
x_n f(x_n)

f(x)=((x-x_1 )(x-x_2 )…(x-x_n))/((x_0-x_1 )(x_0-x_2 )…(x_0-x_n ) ) f(x_0 )+((x-x_0 )(x-x_2 )…(x-x_n))/((x_1-x_0 )(x_1-x_2 )…(x_1-x_n ) ) f(x_1 )+?+((x-x_0 )(x-x_1 )…(x-x_(n-1)))/((x_n-x_0 )(x_n-x_1 )…(x_n-x_(n-1) ) ) f(x_n )

Ex1: Find the value of f(1.5) for the data set shown:
i x y
0 1 1
1 2 4
2 3 8

Solution:
f(1.5)=(1.5-2)(1.5-3)/(1-2)(1-3) ×1+(1.5-1)(1.5-3)/(2-1)(2-3) ×4+(1.5-1)(1.5-2)/(3-1)(3-2) ×8=-0.375+3+(-1)=1.625

?
Homework 10
For the data set, shown in the following table, estimate f(4.3) using Lagrange Polynomial:
i x f(x)
1 0 0
2 1 0.569
3 2 0.791
4 3.8 0.224
5 5 -0.185
?