(1)
(2)
1
Introducción
2 http://numericalmethods.eng.usf.edu
Uxf
Uxrx
Lxf
Lx
O
xf
x
Exact root
0)( xf
0)(*)( UL xfxf
Método de Bisección:
2
ULr
xxx
(3)
Figura 1 Método de Falsa-Posición
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Método de Falsa-Posición
Basedo en triángulos similares, mostrado en Figura 1, tenemos:
Ur
U
Lr
L
xx
xf
xx
xf
)()(
0;0)(
0;0)(
UrU
LrL
xxxf
xxxf
Los signos para ambos lados of Eq. (4):
(4)
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LUrULr xfxxxfxx
ULrULLU xfxfxxfxxfx
De la Eq. (4), obtenemos
De resolver la ecuación anterior obtenemos la próximapredicción de la raíz
UL
ULLUr
xfxf
xfxxfxx
rx
(5)
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De la ecuación de arriba,
UL
ULUUr
xfxf
xxxfxx
LU
LU
LLr
xx
xfxf
xfxx
o
(6)
(7)
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Step-By-Step False-Position Algorithms
as 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 xfxfLx
mx LL xx and mU xx
, then the root lies between (b) If
and ; then
0mL xfxfmx
Ux mL xx and UU xx
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, then the root is(c) If 0mL xfxf .mx
Stop 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
new
m
old
m
new
ma
x
xx
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where
= estimated root from present iteration
= estimated root from previous iteration
new
mxold
mx
.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 aradius of 5.5cm.You are asked to find the depth to which the ball issubmerged 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 xx
Use the false-position method of finding roots ofequations to find the depth to which the ball issubmerged under water. Conduct three iterations toestimate the root of the above equation. Find theabsolute relative approximate error at the end of eachiteration, and the number of significant digits at leastcorrect at the converged iteration.
x
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Solution
From the physics of the problem
11.00
)055.0(20
20
x
x
Rx
x
water
Figure 2 :Floating ballproblem
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Let us assume
11.0,0 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
0660.0
10662.210993.3
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
0611.0
101944.310993.3
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
0624.0
101944.310132.1
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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Iteration
1 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
LxUx mx %a mxf
010993.3165.0 423 xxxfTable 1: Root of
for False-Position Method.
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m
m
m
a
2
2
2
1004.0
105.002.0
105.0
3,
3979.3
)3979.1(2
)04.0log(2
2)04.0log(
mSo
m
m
m
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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References
1. S.C. Chapra, R.P. Canale, Numerical Methods for
Engineers, Fourth Edition, Mc-Graw Hill.
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Acknowledgement
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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.