Secant Method
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Transcript of Secant Method
04/19/23http://
numericalmethods.eng.usf.edu 1
Secant Method
Chemical Engineering Majors
Authors: Autar Kaw, Jai Paul
http://numericalmethods.eng.usf.eduTransforming Numerical Methods Education for STEM
Undergraduates
Secant Method
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Secant Method – Derivation
)(xf
)f(x - = xx
i
iii 1
f ( x )
f ( x i )
f ( x i - 1 )
x i + 2 x i + 1 x i X
ii xfx ,
1
1 )()()(
ii
iii xx
xfxfxf
)()(
))((
1
11
ii
iiiii xfxf
xxxfxx
Newton’s Method
Approximate the derivative
Substituting Equation (2) into Equation (1) gives the Secant method
(1)
(2)
Figure 1 Geometrical illustration of the Newton-Raphson method.
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Secant Method – Derivation
)()(
))((
1
11
ii
iiiii xfxf
xxxfxx
The Geometric Similar Triangles
f(x)
f(xi)
f(xi-1)
xi+1 xi-1 xi X
B
C
E D A
11
1
1
)()(
ii
i
ii
i
xx
xf
xx
xf
DE
DC
AE
AB
Figure 2 Geometrical representation of the Secant method.
The secant method can also be derived from geometry:
can be written as
On rearranging, the secant method is given as
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Step 1
0101
1 x
- xx =
i
iia
Calculate the next estimate of the root from two initial guesses
Find the absolute relative approximate error
)()(
))((
1
11
ii
iiiii xfxf
xxxfxx
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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 1You have a spherical storage tank containing oil. The tank has a diameter of 6 ft. You are asked to calculate the height, h, to which a dipstick 8 ft long would be wet with oil when immersed in the tank when it contains 4 ft3 of oil.
h
r
Spherical Storage Tank
Dipstick
Figure 2 Spherical Storage tank problem.
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Example 1 Cont.
The equation that gives the height, h, of liquid in the spherical tank for the given volume and radius is given by
Use the secant method of finding roots of equations to find the height, h, to which the dipstick is wet with oil. Conduct three iterations to estimate the root of the above equation. Find the absolute relative approximate error at the end of each iteration and the number of significant digits at least correct at the end of each iteration.
0819739 23 .hhf(h)
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Example 1 Cont.
Figure 3 Graph of the function f(h) 0819739 23 .hhf(h)
0 1 2 3 4 5 6120
100
80
60
40
20
0
20
f(x)
3.82
104.18
0
f x( )
60 x x guess1 x guess2
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Example 1 Cont.Solution Initial guesses of the
root:
64423.0
8197.398197.39
8197.3921
31
20
30
1020
30
0
10
10001
hhhh
hhhhh
hfhf
hhhfhh
.1 and 5.0 01 hh
Iteration 1The estimate of the root is
Figure 4 Graph of the estimated root of the
equation after Iteration 1.
0 1 2 3 4 5 6120
100
80
60
40
20
0
20
f(x)x'1, (first guess)x0, (previous guess)Secant linex1, (new guess)
7.57
104.18
0
f x( )
f x( )
f x( )
secant x( )
f x( )
60 x x 0 x 1' x x 1 The absolute relative approximate error is
% 224.55
1001
01
h
hha
The number of significant digits at least correct is 0.
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Example 1 Cont.
67185.0
8197.398197.39
8197.3920
30
21
31
012
131
1
01
01112
hhhh
hhhhh
hfhf
hhhfhh
Iteration 2The estimate of the root is
Figure 5 Graph of the estimated root after Iteration 2.
0 1 2 3 4 5 6120
100
80
60
40
20
0
20
f(x)x1 (guess)x0 (previous guess)Secant linex2 (new guess)
8.559
104.18
0
f x( )
f x( )
f x( )
secant x( )
f x( )
60 x x 1 x 0 x x 2
The absolute relative approximate error is
% 1104.4
1002
12
h
hha
The number of significant digits at least correct is 1.
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Example 1 Cont.
67759.0
8197.398197.39
8197.392
131
22
32
1222
32
2
12
12223
hhhh
hhhhh
hfhf
hhhfhh
Iteration 3The estimate of the root is
The absolute relative approximate error is
% 84768.0
1002
12
h
hha
The number of significant digits at least correct is 1.
Figure 6 Graph of the estimated root after Iteration 3.
Entered function along given interval with current and next root and the tangent line of the curve at the current root
0 1 2 3 4 5 6120
100
80
60
40
20
0
20
f(x)x2 (guess)x1 (previous guess)Secant linex3 (new guess)
7.145
104.18
0
f x( )
f x( )
f x( )
secant x( )
f x( )
60 x x 2 x 1 x x 3
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Advantages
Converges fast, if it converges Requires two guesses that do not need
to bracket the root
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Drawbacks
Division by zero
10 5 0 5 102
1
0
1
2
f(x)prev. guessnew guess
2
2
0
f x( )
f x( )
f x( )
1010 x x guess1 x guess2
0 xSinxf
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Drawbacks (continued)
Root Jumping
10 5 0 5 102
1
0
1
2
f(x)x'1, (first guess)x0, (previous guess)Secant linex1, (new guess)
2
2
0
f x( )
f x( )
f x( )
secant x( )
f x( )
1010 x x 0 x 1' x x 1
0Sinxxf
Additional ResourcesFor 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_method.html