Transcript PPT

Secant Method
Mechanical Engineering Majors
Authors: Autar Kaw, Jai Paul
http://numericalmethods.eng.usf.edu
Transforming Numerical Methods Education for STEM
Undergraduates
7/12/2016
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Secant Method
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Secant Method – Derivation
Newton’s Method
f(x)
x f x 
f(xi)
i,
f(xi )
xi 1 = xi f (xi )
(1)
i
Approximate the derivative
f ( xi ) 
f(xi-1)

xi+2
xi+1
xi
X
Figure 1 Geometrical illustration of
the Newton-Raphson method.
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f ( xi )  f ( xi 1 )
xi  xi 1
(2)
Substituting Equation (2)
into Equation (1) gives the
Secant method
xi 1
f ( xi )( xi  xi 1 )
 xi 
f ( xi )  f ( xi 1 )
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Secant Method – Derivation
The secant method can also be derived from geometry:
f(x)
f(xi)
The Geometric Similar Triangles
AB DC

AE DE
B
can be written as
f ( xi )
f ( xi 1 )

xi  xi 1 xi 1  xi 1
C
f(xi-1)
xi+1
E D
xi-1
A
xi
X
Figure 2 Geometrical representation of
the Secant method.
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On rearranging, the secant
method is given as
xi 1
f ( xi )( xi  xi 1 )
 xi 
f ( xi )  f ( xi 1 )
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Algorithm for Secant Method
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Step 1
Calculate the next estimate of the root from two initial guesses
xi 1
f ( xi )( xi  xi 1 )
 xi 
f ( xi )  f ( xi 1 )
Find the absolute relative approximate error
xi 1- xi
a =
 100
xi 1
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Step 2
Find if the absolute relative approximate error is greater
than the prespecified relative error tolerance.
If so, go back to step 1, else stop the algorithm.
Also check if the number of iterations has exceeded the
maximum number of iterations.
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Example 1
A trunnion has to be cooled
before it is shrink fitted into a
steel hub
The equation that gives the
temperature x to which the
trunnion has to be cooled to
obtain the desired contraction
is given by the following
Figure 3 Trunnion to be slid through
equation.
the hub after contracting.
f x   0.50598 1010 x3  0.38292 107 x 2  0.74363 104 x  0.88318 102  0
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Example 1 Cont.
Use the secant method of finding
roots of equations
a) To find the temperature x to
which the trunnion has to be
cooled. Conduct three
iterations to estimate the root
of the above equation.
b) Find the absolute relative
approximate error at the end
of each iteration, and
c)
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the number of significant digits
at least correct at the end of
each iteration.
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Example 1 Cont.
Cool down temperature
0.02
0.015
f(x)
0.01
0.005
0
-400
-300
-200
-100
-0.005
0
100
-0.01
Temperature (F)
Figure 4 Graph of the function f(x).
f x   0.50598 1010 x3  0.38292 107 x 2  0.74363 104 x  0.88318 102  0
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Example 1 Cont.
8.83210
3
Initial guesses: x1  110, x0  130
Iteration 1
The estimate of the root is
0.01
0.008
0.006
x1  x0 
f ( x)
0.004
f ( x)
f ( x)
0.002
secant ( x)
0
x1
0.002
0.004
 4.48610
3
0.006
7.709110  130   110
 130 
7.709110  1.1825 10 
5
0
f ( x)
f x0 x0  x1 
f x0   f x1 
200
150
 200
100
50
x x 0  x 1'  x x 1
f(x)
x'1, (first guess)
x0, (previous guess)
Secant line
x1, (new guess)
Figure 5 Graph of the estimated
root after Iteration 1.
0
0
5
3
 128.78
The absolute relative approximate error
is
x1  x0
a 
x1
100  0.95051%
The number of significant digits at least correct is 1.
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Example 1 Cont.
8.83210
3
Iteration 2
The estimate of the root is
0.01
0.008
0.006
x2  x1 
f ( x)
0.004
f ( x)
f ( x)
0.002
secant ( x)
0
x2
0.002
0.004
 4.41110
3
0.006
200
150
100
 200
x x 1  x 0  x x 2
50

 1.3089 10  128.78   130
 128.78 
 1.3089 10   7.709110 
6
0
f ( x)
f x1 x1  x0 
f x1   f x0 
0
0
f(x)
x1 (guess)
x0 (previous guess)
Secant line
x2 (new guess)
Figure 6 Graph of the estimated
root after Iteration 2.
6
 128.75
The absolute relative approximate error
is
x2  x1
a 
100  0.016419%
x2
The number of significant digits at least correct is 3.
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5
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Example 1 Cont.
Entered function along given interval with current and next root and the
tangent line of the curve at the current root
8.83210
3
0.01
f x2 x2  x1 
x3  x2 
f x2   f x1 
f ( x)
0.005
f ( x)
f ( x)
secant ( x)
0
0
f ( x)
 4.41610
Iteration 3
The estimate of the root is
1.524110 128.75   128.78
 128.7548 
1.524110  1.3089 10 
9
x3
3
0.005
200
150
100
 200
x x 2  x 1  x x 3
50
0
9
6
0
f(x)
x2 (guess)
x1 (previous guess)
Secant line
x3 (new guess)
Figure 7 Graph of the estimated
root after Iteration 3.
 128.75
The absolute relative approximate error
is
x3  x2
a 
100  1.9097 105 %
x3
The number of significant digits at least correct is 6.
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Advantages


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Converges fast, if it converges
Requires two guesses that do not need to
bracket the root
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Drawbacks
2
2
1
f ( x)
f ( x)
0
0
f ( x)
1
2
2
10
5
 10
0
5
x x guess1  x guess2
f(x)
prev. guess
new guess
10
10
f x  Sinx  0
Division by zero
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Drawbacks (continued)
2
2
1
f ( x)
f ( x)
0
f ( x)
0
secant ( x)
f ( x)
1
2
2
10
5
 10
0
5
10
x x 0  x 1'  x x 1
f(x)
x'1, (first guess)
x0, (previous guess)
Secant line
x1, (new guess)
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f x   Sinx  0
Root Jumping
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Additional Resources
For all resources on this topic such as digital audiovisual
lectures, primers, textbook chapters, multiple-choice
tests, worksheets in MATLAB, MATHEMATICA, MathCad
and MAPLE, blogs, related physical problems, please
visit
http://numericalmethods.eng.usf.edu/topics/secant_me
thod.html
THE END
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