Download Numerical methods

Chapter 1 Introduction
• The solutions of engineering problems can be obtained
using analytical methods or numerical methods.
• Analytical differentiation and integration provide a closedform derivative and integral, respectively, only for simple
functions.
• Very often analytical solutions cannot be obtained.
• Numerical methods can be used where analytical methods
are not capable of or practical for providing solutions.
1- 1
The procedure for solving a
problem
1.
2.
3.
4.
5.
Identify the problem
State the objectives
Develop alternative solutions
Evaluate the alternatives
Implement the best alternative
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Analytical VS Numerical Analysis
• To find minimum of y  x 2  3x  2
• Analytical: after differentiate 2x-3
the minimum solution occurs at x = 1.5
• Numerical: define the interval from 1 to 2 increment of 0.2,
min falls in 1.4 < x < 1.6
x
y
1.0
0
1.2 1.4 1.6 1.8 2.0
-0.16 -0.24 -0.24 -0.16 0
To improve the accuracy, we can search in 1.4 < x < 1.6
with a small increment, such as 0.04.
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Analytical VS Numerical Analysis
• Analytical techniques provide a direct solution and
will result in exact solution if one exists.
• Analytical methods are practical only for functions
that have a simple, closed-form mathematical
structure.
• Numerical methods can be used with any function.
They often require many iterations to get the true
solution.
• The numerical solution usually is not exact, and it
is also necessary to provide initial estimates of the
unknowns.
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Taylor Series
Taylor series
h2 ( 2)
h 3 ( 3)
f ( x0  h )  f ( x0 )  hf ( x0 ) 
f ( x0 ) 
f ( x0 )
2!
3!
(h n ) ( n )
 ... 
f ( x0 )  Rn  1
n!
(1)
where Rn  1 is the remainder.

or
hk (k )
f ( x0  h)  
f ( x0 )
k 0 k!
First-order approximation
Second-order approximation
Third-order approximation
f ( x0  h)  f ( x0 )  hf (1) ( x0 )
f ( x0  h)  f ( x0 )  hf
f ( x0  h)  f ( x0 )  hf
(1)
(1)
h 2 ( 2)
( x0 ) 
f ( x0 )
2!
h 2 ( 2)
h 3 ( 3)
( x0 ) 
f ( x0 ) 
f ( x0 )
2!
3!
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Taylor Series Expansion
• The importance of the individual terms of the
Taylor series depends on the nature of the function
and the distance h. The higher-order terms become
more important as the nonlinearity of the function
increases and the difference h = x  x0 increases.
• If the function is linear, only the term with the first
derivative is necessary.
• The error increases as fewer terms of the Taylor
series are included.
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Example: Nonlinear Polynomials
4th-order
polynomial
f ( x) 
1 4 1 3 1 2 1
x  x  x  x2
8
6
2
2
f (1) ( x)  0.5x3  0.5x 2  x  0.5
f ( 2) ( x)  1.5x 2  x  1
f ( 3) ( x )  3 x  1
f ( 4 ) ( x)  3
f ( 5 ) ( x)  0
f ( x)  1.95833
f (1) ( x)  0.5
For a base point x0= 1.0
distance h = 0.5
f ( 2 ) ( x)  1.5
f ( 3) ( x )  2
f ( 4 ) ( x)  3
f ( 5) ( x)  0
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Part
x
h
a. Individual 1.5 0.5
terms
2.0 1.0
2.5 1.5
3.0 2.0
3.5 2.5
4.0 3.0
b. Cumulative 1.5
function 2.0
2.5
3.0
3.5
4.0
c. Errors
1.5
2.0
2.5
3.0
3.5
4.0
hf (1) ( x0 )
h 2 ( 2)
f ( x0 )
2!
h 3 ( 3)
f ( x0 )
3!
h 4 ( 4)
f ( x0 )
4!
0.25
0.50
0.75
1.00
1.25
1.50
0.1875
0.7500
1.6875
3.0000
4.6875
6.7500
0.04167
0.33333
1.12500
2.66667
5.20833
9.00000
0.0078125
0.1250000
0.6328125
2.0000000
4.8828125
10.1250000
2.20833
2.45833
2.70833
2.95833
3.20833
3.45833
2.39583
3.20833
4.39538
5.95833
7.89583
10.20833
2.43750
3.54167
5.52083
8.62500
13.10417
19.20833
2.44531
3.66667
6.15365
10.62500
17.98698
29.33333
-0.23698
-1.20833
-3.44531
-7.66667
-14.77865
-25.87500
-0.04948
-0.45833
-1.75781
-4.66667
-10.09115
-19.12500
-0.00781
-0.12500
-0.63281
-2.00000
-4.88281
-10.12500
0.00000
0.00000
0.00000
0.00000
0.00000
0.00000
1- 8
1- 9
Taylor Series Expansion of the Square Root
The square-root function can be expressed as f ( x)  x
f (1) ( x) 
1 0.5
x
2
f
( 2)
1 1.5
( x)   x
4
f
( 3)
3  2.5
( x)  x
8
For a base point x0 = 1 and h = 0.001
1
(0.001) 2 (1) 1.5
4(2!)
3

(0.001) 3 (1)  2.5
8(3!)
f (1.001)  1.001  1  0.5(0.001)(1) 0.5 
 1  0.5  103  0.125  106  0.625  1010  1.0004999
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Example Taylor Series
2
3
k
k

x
x
x
x
e x  1  x    ...   ...  
2! 3!
k!
k 0 k!
k

( x  1)2 ( x  1)3
k 1 ( x  1)
ln( x )  ( x  1) 

 ...   ( 1)
2
3
k
k 1
for 0  x  2

1
2
3
 1  x  x  x  ...   x k
1 x
k 0
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Example Taylor Series
3
5

2 k 1
x
x
x
k
sin( x)  x    ...   (1)
3! 5!
(2k  1)!
k 0
2k

x2 x4
x
k
cos( x)  1    ...   (1)
2! 4!
(2k )!
k 0
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