Lecture Note of General Formula

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








General Formula
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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2.3 – General Formula
This type of interpolation can be used for any set of data and 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_n (x)=b_0+b_1 ?(x-x?_0)+b_2 ?(x-x?_0)?(x-x?_1)+?+b_n ?(x-x?_0)?(x-x?_1)×…×?(x-x?_(n-1))
Where:
b_0= f(x_0 )
b_1=f(x_(1,),x_0 )
b_2=f(?x_(2,),x?_(1,),x_0 )
b_n=f(?x_(n,),x?_(n-1,),?…,x?_0 )
b_0,b_1,b_2,…,b_n can be calculated as following:
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b_0 b_1 b_2 b_3
x_0 f(x_0)
(f(x_0 )-f(x_1))/(x_0-x_1 )
x_1 f(x_1) ((f(x_0 )-f(x_1))/(x_0-x_1 )-(f(x_1 )-f(x_2))/(x_1-x_2 ))/(x_0-x_2 )
(f(x_1 )-f(x_2))/(x_1-x_2 ) (((f(x_0 )-f(x_1))/(x_0-x_1 )-(f(x_1 )-f(x_2))/(x_1-x_2 ))/(x_0-x_2 )-((f(x_1 )-f(x_2))/(x_1-x_2 )-(f(x_2 )-f(x_3))/(x_2-x_3 ))/(x_1-x_3 ))/(x_0-x_3 )
x_2 f(x_2) ((f(x_1 )-f(x_2))/(x_1-x_2 )-(f(x_2 )-f(x_3))/(x_2-x_3 ))/(x_1-x_3 )
(f(x_2 )-f(x_3))/(x_2-x_3 )
x_3 f(x_3)

Ex1: The data in Table 1 was obtained by observation, estimate the value of y at x=2.
x y
0.5 -0.693
1 0
2.718 1
7.388 2

Solution:
f_n (x)=b_0+b_1 ?(x-x?_0)+b_2 ?(x-x?_0)?(x-x?_1)+?+b_n ?(x-x?_0)?(x-x?_1)×…×?(x-x?_(n-1))
b_0 b_1 b_2 b_3
0.5 -0.693
((-0.693)-0)/(0.5-1)=1.386
1 0 (1.386-0.582)/(0.5-2.718)=-0.362
(0-1)/(1-2.718)=0.582 (-0.362-(-0.0576))/(0.5-7.388)=0.0442
2.718 1 (0.582-0.214)/(1-7.388)=-0.0576
(1-2)/(2.718-7.388)=0.214
7.388 2

f_3 (2)=(-0.693)+1.386(2-0.5)+(-0.362)(2-0.5)(2-1)+(0.0442)(2-0.5)(2-1)(2-2.718)=0.795

Ex2: If the temperature (T) of a lake is given in the following table as a function of the lake depth (d), estimate the lake temperature at d = -7.5 m.
d (m) T (oC)
0 19.1
-1 19.1
-2 19.0
-3 18.8
-4 18.7
-5 18.3
-6 18.2
-7 17.6
-8 11.7
-9 9.9
-10 9.1
Solution:
For simplicity let’s do third order polynomial:
Choose only four points (from -6 to -9)
f_n (x)=b_0+b_1 ?(x-x?_0)+b_2 ?(x-x?_0)?(x-x?_1)+?+b_n ?(x-x?_0)?(x-x?_1)×…×?(x-x?_(n-1))
b_0 b_1 b_2 b_3
-6 18.2
(18.2-17.6)/((-6)-(-7) )=0.6
-7 17.6 (0.6-5.9)/((-6)-(-8) )=-2.65
(17.6-11.7)/((-7)-(-8) )=5.9 ((-2.65)-2.05)/((-6)-(-9) )=-1.567
-8 11.7 (5.9-1.8)/((-7)-(-9) )=2.05
(11.7-9.9)/((-8)-(-9) )=1.8
-9 9.9

f_3 (-7.5)=18.2+0.6(-7.5-(-6))+(-2.65)(-7.5-(-6))(-7.5-(-7))+(-1.567)(-7.5-(-6))(-7.5-(-7))(-7.5-(-8))=14.725 oC

Ex3: The velocity (v) of a rocket is given in the following table as a function of the time (t), estimate the rocket velocity when t =16 seconds.
t (s) v (m/s)
0 0
10 227.04
15 362.78
20 517.35
22.5 602.97
30 901.67

Solution:
For simplicity let’s do third order polynomial:
Choose only four points (from 10 to 22.5)
f_n (x)=b_0+b_1 ?(x-x?_0)+b_2 ?(x-x?_0)?(x-x?_1)+?+b_n ?(x-x?_0)?(x-x?_1)×…×?(x-x?_(n-1))
b_0 b_1 b_2 b_3
10 227.04
(227.04-362.78)/(10-15)=27.148
15 362.78 (27.148-30.914)/(10-20)=0.377
(362.78-517.35)/(15-20)=30.914 (0.377-0.445)/(10-22.5)=0.005
20 517.35 (30.914-34.248)/(15-22.5)=0.445
(517.35-602.97)/(20-22.5)=34.248
22.5 602.97

f_n (16)=227.04+27.148(16-10)+0.377(16-10)(16-15)+0.005(16-10)(16-15)(16-20)=392.07 m/s
Homework 8
For the data set, shown in the following table, estimate f(3) using:
1- Third degree polynomial.
2- Fourth degree polynomial.
i x f(x)
0 -1 13
1 1 15
2 2 13
3 4 33
4 5 64







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