NUMERICAL METHODS FALSE POSITION METHOD

1. Numerical Methods
Part: False-Position Method of
Solving a Nonlinear Equation
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5. Chapter 03.06: False-Position
Method of Solving a Nonlinear
Equation
Major: All Engineering Majors
Authors: Duc Nguyen
http://numericalmethods.eng.usf.edu
Numerical Methods for STEM undergraduates
http://numericalmethods.eng.usf.edu 5
1/26/2023
Lecture # 1

6. (1)
(2)
1
Introduction
6 http://numericalmethods.eng.usf.edu
 
U
x
f
U
x
r
x
 
L
x
f
L
x
O
 
x
f
x
Exact root
0
)
( 
x
f
0
)
(
*
)
( 
U
L x
f
x
f
In the Bisection method
2
U
L
r
x
x
x

 (3)
Figure 1 False-Position Method

7. http://numericalmethods.eng.usf.edu
7
False-Position Method
Based on two similar triangles, shown in Figure 1,
one gets:
U
r
U
L
r
L
x
x
x
f
x
x
x
f



)
(
)
(
0
;
0
)
(
0
;
0
)
(






U
r
U
L
r
L
x
x
x
f
x
x
x
f
The signs for both sides of Eq. (4) is consistent, since:
(4)

8. http://numericalmethods.eng.usf.edu
8
       
L
U
r
U
L
r x
f
x
x
x
f
x
x 


       
 
U
L
r
U
L
L
U x
f
x
f
x
x
f
x
x
f
x 


From Eq. (4), one obtains
The above equation can be solved to obtain the next
predicted root
   
   
U
L
U
L
L
U
r
x
f
x
f
x
f
x
x
f
x
x



r
x , as
(5)

9. http://numericalmethods.eng.usf.edu
9
The above equation,
  
   
U
L
U
L
U
U
r
x
f
x
f
x
x
x
f
x
x




 
   










L
U
L
U
L
L
r
x
x
x
f
x
f
x
f
x
x
or
(6)
(7)

10. http://numericalmethods.eng.usf.edu
10
Step-By-Step False-Position
Algorithms
as two guesses for the root such
1. Choose L
x U
x
and
that
    0

U
L x
f
x
f
2. Estimate the root,
   
   
U
L
U
L
L
U
m
x
f
x
f
x
f
x
x
f
x
x



3. Now check the following
, then the root lies between
(a) If
and ; then
    0

m
L x
f
x
f L
x
m
x L
L x
x  and m
U x
x 
, then the root lies between
(b) If
and ; then
    0

m
L x
f
x
f m
x
U
x m
L x
x  and U
U x
x 

11. http://numericalmethods.eng.usf.edu
11
, then the root is
(c) If     0

m
L x
f
x
f .
m
x
Stop the algorithm if this is true.
4. Find the new estimate of the root
   
   
U
L
U
L
L
U
m
x
f
x
f
x
f
x
x
f
x
x



Find the absolute relative approximate error as
100



 new
m
old
m
new
m
a
x
x
x

12. http://numericalmethods.eng.usf.edu
12
where
= estimated root from present iteration
= estimated root from previous iteration
new
m
x
old
m
x
.
001
.
0
10 3


 
s
say
5. If s
a 
 , 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 ,
m
x shown in steps
#2 and 4!

13. http://numericalmethods.eng.usf.edu
13
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
0
10
993
.
3
165
.
0 4
2
3



 
x
x
Use 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

14. http://numericalmethods.eng.usf.edu
14
Solution
From the physics of the problem
11
.
0
0
)
055
.
0
(
2
0
2
0






x
x
R
x
x
water
Figure 2 :
Floating ball
problem

15. http://numericalmethods.eng.usf.edu
15
Let us assume
11
.
0
,
0 
 U
L x
x
       
        4
4
2
3
4
4
2
3
10
662
.
2
10
993
.
3
11
.
0
165
.
0
11
.
0
11
.
0
10
993
.
3
10
993
.
3
0
165
.
0
0
0



















f
x
f
f
x
f
U
L
Hence,
           0
10
662
.
2
10
993
.
3
11
.
0
0 4
4





 

f
f
x
f
x
f U
L

16. http://numericalmethods.eng.usf.edu
16
Iteration 1
   
   
 
 
0660
.
0
10
662
.
2
10
993
.
3
10
662
.
2
0
10
993
.
3
11
.
0
4
4
4
4



















U
L
U
L
L
U
m
x
f
x
f
x
f
x
x
f
x
x
         
5
4
2
3
10
1944
.
3
10
993
.
3
0660
.
0
165
.
0
0660
.
0
0660
.
0









 f
x
f m
           0
0660
.
0
0 



 f
f
x
f
x
f m
L
0660
.
0
,
0 
 U
L x
x

17. http://numericalmethods.eng.usf.edu
17
Iteration 2
   
   
 
 
0611
.
0
10
1944
.
3
10
993
.
3
10
1944
.
3
0
10
993
.
3
0660
.
0
5
4
5
4



















U
L
U
L
L
U
m
x
f
x
f
x
f
x
x
f
x
x
         
5
4
2
3
10
1320
.
1
10
993
.
3
0611
.
0
165
.
0
0611
.
0
0611
.
0








 f
x
f m
           0
0611
.
0
0 



 f
f
x
f
x
f m
L
0660
.
0
,
0611
.
0 
 U
L x
x
Hence,

18. http://numericalmethods.eng.usf.edu
18
%
8
100
0611
.
0
0660
.
0
0611
.
0




a
Iteration 3
   
   
 
 
0624
.
0
10
1944
.
3
10
132
.
1
10
1944
.
3
0611
.
0
10
132
.
1
0660
.
0
5
5
5
5



















U
L
U
L
L
U
m
x
f
x
f
x
f
x
x
f
x
x

19. http://numericalmethods.eng.usf.edu
19
  7
10
1313
.
1 



m
x
f
           0
0624
.
0
0611
.
0 



 f
f
x
f
x
f m
L
Hence,
0624
.
0
,
0611
.
0 
 U
L x
x
%
05
.
2
100
0624
.
0
0611
.
0
0624
.
0




a

20. http://numericalmethods.eng.usf.edu
20
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
L
x U
x m
x %
a
  
m
x
f
  0
10
993
.
3
165
.
0 4
2
3




 
x
x
x
f
Table 1: Root of
for False-Position Method.

21. http://numericalmethods.eng.usf.edu
21
m
m
m
a









2
2
2
10
04
.
0
10
5
.
0
02
.
0
10
5
.
0
3
,
3979
.
3
)
3979
.
1
(
2
)
04
.
0
log(
2
2
)
04
.
0
log(









m
So
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.

22. http://numericalmethods.eng.usf.edu
22
References
1. S.C. Chapra, R.P. Canale, Numerical Methods for
Engineers, Fourth Edition, Mc-Graw Hill.

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