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4/13/2020
Lecture # 1
Major: All Engineering Majors
Authors: Duc Nguyen http://numericalmethods.eng.usf.edu
Numerical Methods for STEM undergraduates http://numericalmethods.eng.usf.edu
5

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

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

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

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

10
1. Choose that x
L and f x
U as two guesses for the root such
   
L f
U

0
2. Estimate the root, x m

3. Now check the following x
U f f
  x
L
   
L

 x f
L f x
U
 
U
(a) If
   
L and x m
; then
(b) If f f x m
L
   
L and x
U
; then x m
L



 x x
0
0
L m
, then the root lies between x and x

U x m
, then the root lies between and x

U x x
L m

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

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

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 submerged under water is given by x to which the ball is x
3 
0 .
165 x
2 
3 .
993

10

4 
0
13
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 correct at the converged iteration.
http://numericalmethods.eng.usf.edu

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

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

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

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

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

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

20
2
3
4
Table 1: Root of f
 x
3 
0 .
165 x for False-Position Method.
2 
3 .
993

10

4 
0
Iteration
1 x
L
0.0000
x
U
0.1100
x m
0.0660
 a
% f
  m
N/A -3.1944x10
-5
0.0000
0.0611
0.0660
0.0660
0.0611
0.0624
8.00
1.1320x10
-5
2.05
-1.1313x10
-7
0.0611
0.0624
0.0632377619
0.02
-3.3471x10
-10 http://numericalmethods.eng.usf.edu

 a

0 .
5

10
2
 m
0 .
02

0 .
5

10
2
 m
0 .
04

10
2
 m log( m

0 .
04 )

2
 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 4 th iteration is 3.
21 http://numericalmethods.eng.usf.edu

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

http://numericalmethods.eng.usf.edu

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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.