Numerical Methods
Part: False-Position Method of
Solving a Nonlinear Equation
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Chapter 02.02: False-Position
Method of Solving a Nonlinear
Equation
5
Introduction
f x 
f ( x)  0
(1)
In the Bisection method
f xU 
f ( xL ) * f ( xU )  0 (2)
x L  xU
xr 
2
Exact root
xL
O
6
xr
f xL 
xU
x
Figure 1 False-Position Method
(3)
1
False-Position Method
Based on two similar triangles, shown in Figure 1,
one gets:
f ( xU )
f ( xL )

xr  x L xr  xU
(4)
The signs for both sides of Eq. (4) is consistent, since:
f ( xL )  0; xr  xL  0
f ( xU )  0; xr  xU  0
7
From Eq. (4), one obtains
xr  xL  f xU   xr  xU  f xL 
xU f xL   xL f xU   xr  f xL   f xU 
The above equation can be solved to obtain the next
predicted root xr , as
xU f  x L   x L f  xU 
xr 
f  x L   f  xU 
8
(5)
The above equation,
f  xU x L  xU 
xr  xU 
f  x L   f xU 
or
9
f xL 
xr  x L 
 f xU   f xL 


 xU  xL

(6)
(7)
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Step-By-Step False-Position
Algorithms
1. Choose x L and xU as two guesses for the root such
that
f xL  f xU   0
xU f  x L   x L f xU 
2. Estimate the root, xm 
f x L   f xU 
3. Now check the following
(a) If f x L  f xm   0 , then the root lies between
and xm ; then xL  xL and xU  xm
10
xL
(b) If f xL  f xm   0 , then the root lies between xm
and xU ; then xL  xm and xU  xU
(c) If f xL  f xm   0 , then the root is
Stop the algorithm if this is true.
xm .
4. Find the new estimate of the root
xU f  x L   x L f xU 
xm 
f x L   f xU 
Find the absolute relative approximate error as
xmnew  xmold
a 
 100
new
xm
11
where
new
m =
old
m =
x
x
estimated root from present iteration
estimated root from previous iteration
3
say


10
 0.001. If a s , then go to step 3,
5.
s
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 xm ,shown in steps
#2 and 4!
12
Example 1
The 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 is
submerged under water is given by
4
x  0.165 x  3.993 10  0
Use the false-position method of finding roots of
equations to find the depth x 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
13 correct at the converged iteration. http://numericalmethods.eng.usf.edu
3
2
Solution
From the physics of the problem
0  x  2R
0  x  2(0.055)
0  x  0.11
Figure 2 :
Floating ball
problem
14
x
water
Let us assume
xL  0, xU  0.11
f  x L   f 0   0   0.1650   3.993  10 4  3.993  10 4
3
2
f  xU   f 0.11  0.11  0.1650.11  3.993  10 4  2.662  10 4
3
2
Hence,



f xL  f xU   f 0 f 0.11  3.993  10 4  2.662  10 4  0
15
Iteration 1
xU f  x L   x L f xU 
xm 
f  x L   f  xU 


0.11 3.993 10 4  0   2.662 10 4

3.993 10 4   2.662 10 4
 0.0660
3
2
f xm   f 0.0660   0.0660  0.1650.0660  3.993 10 4



 3.1944 10 5
f xL  f xm   f 0 f 0.0660     0
16
xL  0, xU  0.0660

Iteration 2
xU f  x L   x L f  xU 
xm 
f  x L   f  xU 

0.0660  3.993 10 4  0   3.1944 10 5

3.993 10 4   3.1944 10 5
 0.0611




f xm   f 0.0611  0.0611  0.1650.0611  3.993 10 4
3
 1.1320 10 5
f xL  f xm   f 0 f 0.0611      0
17
Hence,
xL  0.0611, xU  0.0660
2

0.0611  0.0660
a 
 100  8%
0.0611
Iteration 3
xU f  x L   x L f xU 
xm 
f  x L   f  xU 

0.0660  1.132  10 5  0.0611  3.1944  10 5

1.132 10 5   3.1944 10 5
 0.0624

18


f  xm   1.1313  10 7
f xL  f xm   f 0.0611 f 0.0624     0
Hence,
xL  0.0611, xU  0.0624
0.0624  0.0611
a 
100  2.05%
0.0624
19
3
2
4


f
x

x

0
.
165
x

3
.
993

10
0
Table 1: Root of
for False-Position Method.
Iteration
20
xL
xU
xm
a %
f  xm 
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
a  0.5 10 2m
0.02  0.5 10
0.04  10
2 m
2 m
log( 0.04)  2  m
m  2  log( 0.04)
m  2  (1.3979)
m  3.3979
So, m  3
The number of significant digits at least correct in the
estimated root of 0.062377619 at the end of 4th iteration
is 3.
21
References
1. S.C. Chapra, R.P. Canale, Numerical Methods for
Engineers, Fourth Edition, Mc-Graw Hill.
22
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