Presentation on application of numerical method in our life

Presented By
SHIVAM KUMAR 15/632
ASFARUL HAQ SULTAN 15/839
MANISH KUMAR SINGH 15/1208
AKASH 15/1508

The Application of Numerical
Methods in Real Life
• 1. Estimation of ocean currents
• 2. Modeling combustion flow in a coal power plant
• 3. Airflow patterns in the respiratory tract (and diff. eqs.)
• 4. Regional uptake of inhaled materials by respiratory tract
• 5. Transport and disposition of chemicals through the body (and ODEs +
PDEs)
• 6. Molecular and cellular mechanisms of toxicity (and ODEs + PDEs)
• 7. Reentry simulations for the Space Shuttle
• 8. Trajectory prescribed path control and optimal control problems
• 9. Shuttle/tank separation
• 10. Scientific programming
• 11. Modeling of airflow over airplane bodies
• 12. Electromagnetics analysis for detection by radar
• 13. Design and analysis of control systems for aircraft
• 14. Electromagnetics
• 15. Large scale shock wave physics code development
• 16. Curve fitting of tabular data

Definition
• The Bisection method in mathematics is a root
finding method which repeatedly bisects an
interval and then selects a subinterval in
which a root must lie for further processing.

6
Example
The floating ball has a specific gravity of 0.6 and has a radius
of 5.5 cm. we are asked to find the depth to which the ball is
submerged when floating in water.
Figure 6 Diagram of the floating ball

7
The equation that gives the depth x to which the ball is
submerged under water is given by
a) Use the bisection 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.
b) Find the absolute relative approximate error at the end of
each iteration, and the number of significant digits at least
correct at the end of each iteration.
010993.3165.0 423
 
xx

8
From the physics of the problem, the ball would be submerged
between x = 0 and x = 2R,
where R = radius of the ball,
that is
 
11.00
055.020
20



x
x
Rx
Figure 6 Diagram of the floating ball

To aid in the understanding of how this
method works to find the root of an
equation, the graph of f(x) is shown to the
right,
where
9
  423
1099331650 -
.x.xxf 
Figure 7 Graph of the function f(x)
Solution-

10
Let us assume
11.0
00.0


ux
x
Check if the function changes sign between x and xu .
       
        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
So there is at least on root between x and xu, that is between 0 and 0.11

11
055.0
2
11.00
2




 u
m
xx
x 
       
           010655.610993.3055.00
10655.610993.3055.0165.0055.0055.0
54
5423




ffxfxf
fxf
ml
m
Iteration 1
The estimate of the root is
Hence the root is bracketed between xm and xu, that is, between 0.055 and
0.11. So, the lower and upper limits of the new bracket are
At this point, the absolute relative approximate error cannot be
calculated as we do not have a previous approximation.
11.0,055.0  ul xx
a

12
0825.0
2
11.0055.0
2




 u
m
xx
x 
       
         010655.610622.1)0825.0(055.0
10622.110993.30825.0165.00825.00825.0
54
4423




ffxfxf
fxf
ml
m
Iteration 2
Hence the root is bracketed between x and xm, that is, between 0.055 and
0825.0,055.0  ul xx

13
The absolute relative approximate error at the end of Iteration 2 isa
%333.33
100
0825.0
055.00825.0
100






 new
m
old
m
new
m
a
x
xx
None of the significant digits are at least correct in the estimate root of xm =
0.0825 because the absolute relative approximate error is greater than 5%.

14
06875.0
2
0825.0055.0
2




 u
m
xx
x 
       
           010563.510655.606875.0055.0
10563.510993.306875.0165.006875.006875.0
55
5423




ffxfxf
fxf
ml
m
Iteration 3
Hence the root is bracketed between x and xm, that is, between 0.055 and
06875.0,055.0  ul xx

15
The absolute relative approximate error at the end of Iteration 3 is
a
%20
100
06875.0
0825.006875.0
100






 new
m
old
m
new
m
a
x
xx
Still none of the significant digits are at least correct in the estimated root of
the equation as the absolute relative approximate error is greater than 5%.
Seven more iterations were conducted and these iterations are shown in Table
1.

16
Table 1
Root of f(x)=0 as function of number of iterations for bisection method.
Iteration x xu xm a % f(xm)
1
2
3
4
5
6
7
8
9
10
0.00000
0.055
0.055
0.055
0.06188
0.06188
0.06188
0.06188
0.0623
0.0623
0.11
0.11
0.0825
0.06875
0.06875
0.06531
0.06359
0.06273
0.06273
0.06252
0.055
0.0825
0.06875
0.06188
0.06531
0.06359
0.06273
0.0623
0.06252
0.06241
----------
33.33
20.00
11.11
5.263
2.702
1.370
0.6897
0.3436
0.1721
6.655×10−5
−1.622×10−4
−5.563×10−5
4.484×10−6
−2.593×10−5
−1.0804×10−5
−3.176×10−6
6.497×10−7
−1.265×10−6
−3.0768×10−7

Application-1
• Finding the value of resistance-
Thermistors are temperature-measuring devices based on
the principle that the thermistor material exhibits a change in
electrical resistance with a change in temperature. By
measuring the resistance of the thermistor material, one can
then determine the temperature.

• For a 10K3A Betatherm thermistor, the
relationship between the resistance ‘R’ of
the thermistor and the temperature is given
by
where note that T is in Kelvin and R is in ohms.
  3833
ln10775468.8)ln(10341077.210129241.1
1
RxRxx
T



Solution- Graph of function f(x)
   3383
10293775.2ln10775468.8)ln(10341077.2)( 
 xRxRxRf
1 1.5 2 2.5 3 3.5 4
0.003
0.002
0.001
0
0.001
f(x)
9.51881 10
4

2.29378 10
3

0
f x( )
41 x

Choose the bracket
 
  4
3
1051881.94
10293775.21
4
1






xf
xf
R
R
u

0.5 1 1.5 2 2.5 3 3.5 4 4.5
0.004
0.003
0.002
0.001
0
0.001
f(x)
xu (upper guess)
xl (lower guess)
1.22768 10
3

3.91652 10
3

0
f x( )
f x( )
f x( )
4.50.5 x x u x l
Entered function on given interval with initial upper and lower guesses

5.2
2
41
4,1




m
u
R
RR
 
 
  4
4
3
10486.15.2
1051881.94
10293775.21






xf
xf
xf
4
5.2


uR
R
0.5 1 1.5 2 2.5 3 3.5 4 4.5
0.004
0.003
0.002
0.001
0
0.001
f(x)
xu (upper guess)
xl (lower guess)
new guess
1.22768 10
3

3.91652 10
3

0
f x( )
f x( )
f x( )
f x( )
4.50.5 x x u x l x r
First iteration-

Second iteration-
%07.23
25.3
2
45.2
4,5.2





a
m
u
x
RR
 
 
 
25.3,5.2
106569.425.3
1051881.94
10486.15.2
4
4
4







uRR
xf
xf
xf

0.5 1 1.5 2 2.5 3 3.5 4 4.5
0.004
0.003
0.002
0.001
0
0.001
f(x)
xu (upper guess)
xl (lower guess)
new guess
1.22768 10
3

3.91652 10
3

0
f x( )
f x( )
f x( )
f x( )
4.50.5 x x u x l x r

Third iteration-
875.2
2
25.35.2
25.3,5.2




m
u
R
RR
 
 
  4
4
4
107863.1875.2
106569.425.3
10486.15.2
%0435.13







xf
xf
xf
a
0.5 1 1.5 2 2.5 3 3.5 4 4.5
0.004
0.003
0.002
0.001
0
0.001
f(x)
xu (upper guess)
xl (lower guess)
new guess
1.22768 10
3

3.91652 10
3

0
f x( )
f x( )
f x( )
f x( )
4.50.5 x x u x l x r

Table 1: Root of f(R)=0 as function of number of iterations
for bisection method.
a
Iteration Rl Ru Rm % f(Rm)
1
2
3
4
5
6
7
8
9
10
1
2.5
2.5
2.5
2.5
2.59375
2.64063
2.64063
2.65234
2.6582
4
4
3.25
2.875
2.6875
2.6875
2.6875
2.66406
2.66406
2.66406
2.5
3.25
2.875
2.6875
2.59375
2.64063
2.66406
2.65234
2.6582
2.6611
----------
23.077
13.0435
6.97674
3.61446
1.77515
0.87977
0.44183
0.22043
0.11009
-1.486x10-4
4.6567x10-4
1.7863x10-4
2.07252x10-5
-6.24075x105
-2.04723x10-5
2.17037x10-7
-1.01048x10-5
-4.9382x10-6
-2.3592x10-6

2525
Drawbacks
 It may take many iterations.
 If one of the initial guesses is close to the
root, the convergence is slower.
 The method can not find complex roots of
polynomials
The bisection method only finds root when
the function crosses the x axis.

Conclusion
This bisection method is a very simple and a robust
method and it is one of the first numerical methods
developed to find root of a non-linear equation .But
at the same time it is relatively very slow method.
We can use this method for various purpose related
to non linear continuous functions. Though it is a
slow one ,but it is one of the most reliable methods
for finding the root of a non-linear equation.

The ta
x 0 1 3 4 7
f 1 3 49 129 813
x 0 1 3 4 7
F 1 3 49 129 813
Suppose initially a baby plant of height 1 inch is
planted and its growth in successive month is given
in table. If we want to find its height after 8 months
then this can be done using numerical method.
Use of Newton's forward divided difference method.

Divided difference table
xi fi 1st 2nd 3rd
x0
0 1
2
x1
1 3 7
23 3
x2
3 49 19
80 3
x3
4 129 37
228
x4
7 813
This can be solved using Newton's divided
difference formula.

f(x) = 3x3 - 5x2 + 4x +1
By putting the values of x0, x1, x2, x3 in above eq. we get
For height after 8 months, we put x=8 in above equation
i.e.
f(8) = 3(8)3 - 5(8)2 + 4(8) +1 = 1249
Therefore height of tree after 8 months is 1249 inch.
Now Newton's divided difference formula is
f(x) = f [x0] + (x - x0) f [x0, x1] + (x - x0) (x - x1) f [x0, x1, x2] + (x - x0)
(x - x1) (x - x2)f [x0, x1, x2, x3]

REFERENCES
Numerical Methods for scientific and engineering
computation : Jain and Iyengar
http://numericalmethods.eng.usf.edu

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