Lecture Note of Linear Interpolation

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








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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Lecture Outline
Introduction
Newton’s Divided Difference Formula
Linear Interpolation
Quadratic Interpolation
General Formula
Gregory-Newton Polynomial Interpolation
Lagrange Interpolation
Summary













1.0 – Introduction
Interpolation is the process of producing a function that matches a given data set, which can be obtained by observations and/or experiments. This function may be utilized to approximate the data values at intermediate points (estimate y value at any given x value). In addition, this process is called extrapolation when the required y value is for a x value that located outside the given data set. The following sections will discuss different methods of interpolation.

2.0 – Newton’s Divided Difference Formula
In order to illustrate this method, linear and quadratic interpolation have to presented first.

2.1 – Linear Interpolation
If two points, (x_0,f(x_0 )) and (x_1,f(x_1 )) , are given and we asked to evaluate the value of f_1 (x) corresponding to any x that located between (x_0,f(x_0 )) and (x_1,f(x_1 )), as shown in Figure 1.
By similar triangles:
(f_1 (x)- f(x_0 ))/?x-x?_0 =(f(x_1 )- f(x_0 ))/?x_1-x?_0
By rearranging:
f_1 (x)= f(x_0 )+((f(x_1 )- f(x_0 ))/?x_1-x?_0 )×?(x-x?_0)
f_1 (x)=b_0+b_1 ?(x-x?_0)
Where:
b_0= f(x_0 )
b_1=((f(x_1 )- f(x_0 ))/?x_1-x?_0 )









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

Solution:
By linear interpolation, x=2
b_0= f(x_0 )=0
b_1=((f(x_1 )- f(x_0 ))/?x_1-x?_0 )=(1-0)/(2.718-1)=0.58
f_1 (x)=b_0+b_1 ?(x-x?_0)=0+0.58×(x-1)=0.58x-0.58
f_1 (2)=0.58×2-0.58=0.58

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:
By linear interpolation, d = -7.5m
Choose (-7,17.6) and (-8,11.7) because they are located around d = -7.5m
b_0= f(x_0 )= f(-7)=17.6
b_1=((f(x_1 )- f(x_0 ))/?x_1-x?_0 )=((11.7- 17.6)/((-8)-(-7)))=5.9
f_1 (x)=b_0+b_1 ?(x-x?_0)=17.6+5.9×(x-(-7))=5.9x+58.9
f_1 (-7.5)=5.9×(-7.5)+58.9=14.65 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:
By linear interpolation, t = 16 m/s
Choose (15 , 362.78) and (20 , 517.35) because they are located around t = 16 m/s
b_0= f(x_0 )= f(15)=362.78
b_1=((f(x_1 )- f(x_0 ))/?x_1-x?_0 )=((517.35- 362.78)/(20-15))=30.92
f_1 (x)=b_0+b_1 ?(x-x?_0)=362.78+30.92×(x-15)=30.92x-101.02
f_1 (16)=30.92×(16)-101.02=393.7 m/s
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Homework 6
For the data set, shown in the following table, estimate f(3) using:
1-Linear polynomial (First degree polynomial).
0 -1 13
1 1 15
2 2 13
3 4 33
4 5 64


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