numerical methods part: false-position method of solving a nonlinear equation
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Numerical Methods Part: False-Position Method of Solving a Nonlinear Equation http://numericalmethods.eng.usf.edu. For more details on this topic Go to http://numericalmethods.eng.usf.edu Click on Keyword Click on False-Position Method of Solving a Nonlinear Equation. You are free. - PowerPoint PPT PresentationTRANSCRIPT
Numerical Methods
Part: False-Position Method of Solving a Nonlinear Equation
http://numericalmethods.eng.usf.edu
For more details on this topic
Go to http://numericalmethods.eng.usf.edu
Click on Keyword Click on False-Position Method of
Solving a Nonlinear Equation
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Chapter 03.06: False-Position Method of Solving a Nonlinear
EquationMajor: All Engineering Majors
Authors: Duc Nguyen
http://numericalmethods.eng.usf.edu
Numerical Methods for STEM undergraduateshttp://
numericalmethods.eng.usf.edu 504/22/23
Lecture # 1
(1)
(2)
1
Introduction
6
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Uxf
Uxrx
Lxf
LxO
xf
x
Exact root
0)( xf
0)(*)( UL xfxf
In the Bisection method
2UL
rxxx
(3)
Figure 1 False-Position Method
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False-Position MethodBased on two similar triangles, shown in Figure 1, one gets:
Ur
U
Lr
L
xxxf
xxxf
)()(
0;0)(0;0)(
UrU
LrL
xxxfxxxf
The signs for both sides of Eq. (4) is consistent, since:
(4)
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LUrULr xfxxxfxx
ULrULLU xfxfxxfxxfx
From Eq. (4), one obtains
The above equation can be solved to obtain the next predicted root
UL
ULLUr xfxf
xfxxfxx
rx , as
(5)
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The above equation, UL
ULUUr xfxf
xxxfxx
LU
LU
LLr
xxxfxf
xfxx
or
(6)
(7)
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Step-By-Step False-Position Algorithmsas two guesses for the root such1. Choose Lx Uxand
that 0UL xfxf
2. Estimate the root, UL
ULLUm xfxf
xfxxfxx
3. Now check the following, then the root lies between (a) If
and ; then 0mL xfxf Lxmx LL xx and mU xx
, then the root lies between (b) Ifand ; then
0mL xfxf mxUx mL xx and UU xx
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, then the root is(c) If 0mL xfxf .mxStop the algorithm if this is true.
4. Find the new estimate of the root UL
ULLUm xfxf
xfxxfxx
Find the absolute relative approximate error as
100
newm
oldm
newm
a xxx
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where= estimated root from present iteration= estimated root from previous iteration
newmxoldmx
.001.010 3 ssay5. If sa , then go to step 3,
else stop the algorithm.Notes: The False-Position and Bisection algorithms are quite similar. The only difference is the formula used to calculate the new estimate of the root ,mx shown in steps
#2 and 4!
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Example 1The floating ball has a specific gravity of 0.6 and has a radius of 5.5cm. You are asked to find the depth to which the ball is submerged when floating in water.
The equation that gives the depth x to which the ball issubmerged under water is given by
010993.3165.0 423 xxUse the false-position method of finding roots of equations to find the depth to which the ball is submerged under water. 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 converged iteration.
x
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SolutionFrom the physics of the problem
11.00)055.0(20
20
xx
Rx
xwater
Figure 2 :Floating ball problem
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Let us assume11.0,0 UL xx
4423
4423
10662.210993.311.0165.011.011.0
10993.310993.30165.000
fxf
fxf
U
L
Hence,
010662.210993.311.00 44 ffxfxf UL
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Iteration 1
0660.010662.210993.3
10662.2010993.311.044
44
UL
ULLUm xfxf
xfxxfxx
5
423
101944.3
10993.30660.0165.00660.00660.0
fxf m
00660.00 ffxfxf mL
0660.0,0 UL xx
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Iteration 2
0611.0101944.310993.3
101944.3010993.30660.054
54
UL
ULLUm xfxf
xfxxfxx
5
423
101320.1
10993.30611.0165.00611.00611.0
fxf m
00611.00 ffxfxf mL
0660.0,0611.0 UL xxHence,
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%81000611.0
0660.00611.0
a
Iteration 3
0624.0101944.310132.1
101944.30611.010132.10660.055
55
UL
ULLUm xfxf
xfxxfxx
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7101313.1 mxf
00624.00611.0 ffxfxf mL
Hence,0624.0,0611.0 UL xx
%05.21000624.0
0611.00624.0
a
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Iteration1 0.0000 0.1100 0.0660 N/A -3.1944x10-
5
2 0.0000 0.0660 0.0611 8.00 1.1320x10-5
3 0.0611 0.0660 0.0624 2.05 -1.1313x10-
7
4 0.0611 0.0624 0.0632377619
0.02 -3.3471x10-
10
Lx Ux mx %a mxf
010993.3165.0 423 xxxfTable 1: Root offor False-Position Method.
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m
m
ma
2
2
2
1004.0
105.002.0
105.0
3,3979.3
)3979.1(2)04.0log(2
2)04.0log(
mSommm
m
The number of significant digits at least correct in the estimated root of 0.062377619 at the end of 4th iteration is 3.
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References1. S.C. Chapra, R.P. Canale, Numerical Methods for Engineers, Fourth Edition, Mc-Graw Hill.
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This material is based upon work supported by the National Science Foundation under Grant # 0717624. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.
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