Lecture Note of Gregory-Newton Polynomial Interpolation

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








Gregory-Newton Polynomial 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
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3.0 – Gregory-Newton Polynomial Interpolation
This type is special version of Newton’s Interpolation. This interpolation can be used only for a set of data when the values of x are equally spaced (h).
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)

h=x_(i+1)-x_i
r=(x-x_0)/h

f_n (x)=f(x_0 )+r?f_0+r(r-1) (?^2 f_0)/2!+r(r-1)(r-2) (?^3 f_0)/3!+?+r(r-1)…(r-(n-1)) (?^n f_0)/n!
?f_0,?^2 f_0,?^3 f_0,…,?^n f_0 can be calculated as following:
f(x_0) ?f_0 ?^2 f_0 ?^3 f_0
x_0 f(x_0)
f(x_1 )-f(x_0)
x_1 f(x_1) [f(x_2 )-f(x_1)]-[f(x_1 )-f(x_0)]
f(x_2 )-f(x_1) {[f(x_3 )-f(x_2)]-[f(x_2 )-f(x_1)]}-{[f(x_2 )-f(x_1)]-[f(x_1 )-f(x_0)]}
x_2 f(x_2) [f(x_3 )-f(x_2)]-[f(x_2 )-f(x_1)]
f(x_3 )-f(x_2)
x_3 f(x_3)

Ex1: Find the value of f(1.83) for the data set shown:
x y
1 0
3 1.0986
5 1.6094
7 1.9459
9 2.1972

Solution:
h=x_(i+1)-x_i=3-1=2
r=(x-x_0)/h=(1.83-1)/2=0.415


f(x_0) ?f_0 ?^2 f_0 ?^3 f_0 ?^4 f_0
1 0
1.0986-0=1.0986
3 1.0986 0.3365-1.0986=-0.5878
1.6094-1.0986=0.5108 (-0.1743)-(-0.5878)=0.4135
5 1.6094 0.3365-0.5108=-0.1743 0.0891-0.4135=-0.3244
1.9459-1.6094=0.3365 (-0.0852)-(-0.1743)=0.0891
7 1.9459 0.2513-0.3365=-0.0852
2.1972-1.9459=0.2513
9 2.1972

f_n (1.83)=0+0.415×(1.0986)+0.415×(0.415-1) (-0.5108)/2!+0.415×(0.415-1)(0.415-2) 0.4135/3!+0.415×(0.415-1)(0.415-2)(0.415-3) (-0.3244)/4!=0.5676

Ex2: Find the value of f(4.12) for the data set shown:
x y
0 1
1 2
2 4
3 8
4 16
5 32

Solution:
h=x_(i+1)-x_i=1-0=1
r=(x-x_0)/h=(4.12-0)/1=4.12

f(x_0) ?f_0 ?^2 f_0 ?^3 f_0 ?^4 f_0 ?^5 f_0
0 1
2-1=1
1 2 2-1=1
4-2=2 2-1=1
2 4 4-2=2 2-1=1
8-4=4 4-2=2 2-1=1
3 8 8-4=4 4-2=2
16-8-=8 8-4 =4
4 16 16-8=8
32-16=16
5 32

f_n (4.12)=1+4.12×(1)+4.12(4.12-1) 1/2!+4.12(4.12-1)(4.12-2) 1/3!+4.12(4.12-1)(4.12-2)(4.12-3) 1/3!+4.12(4.12-1)(4.12-2)(4.12-3)(4.12-4) 1/4!=17.3913



Homework 9
For the data set, shown in the following table, estimate f(1.65) using Gregory-Newton Polynomial:
i x f(x)
1 1.5 2.25
2 1.6 2.56
3 1.7 2.89
4 1.8 3.24





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