Download numerical computations

NUMERICAL ANALYSIS (MTH603)
Solution of Assignment # 1(F all 2005)
Please read the following instructions before attempting the solution of this
assignment.
(i)
(ii)
(iii)
(iv)
In order to attempt this assignment you should have full command on
first seven lectures of MTH603.
Don’t use colorful background and you should remember that if we
found the solution files of some students are same then we will reward
zero marks to all those students.
Try to make solution by yourself and protect your work from other
students, otherwise you and the student who send same solution file as
you will be given zero marks.
Try to get the concepts, consolidate your concepts and ideas from these
questions which you learn in the first seven lectures.
Also remember that you are supposed to submit your assignment in
Word format any other like scan images etc will not be accepted and we
will give zero marks correspond to these assignments.
Note:
There are two types of question in this assignment
(i)
Graded Questions
(ii) Questions for Practice.
Each Graded Question carry 10 marks and Questions for Practice
have zero marks, but you are advised to solve all the questions as
it will help you in understanding of the concepts.
1. Define the absolute error and percentage error and explain the difference
between absolute error and relative error.
2. Determine the absolute error, relative error and the error bounds of
48.242
Q2 is graded question.
Solution:
Consider z = xn ,z - ez = (x –e1)n where ez and e1 errors in z and x.
n
n
 e 
z  ez   x  e1   x n 1  1  Expanding by binomial theorem we get,
x

1
z  ez  x n  ne1 x n1 Since z  x n so we have
ez  ne1 x n1 Dividing by z  x n
ez
x n1
e
1
 ne1 n  z  ne1
z
x
z
x
By definition we have,
ez
1
e
 ne1  n 1
z
x
x
1
1
e
Hence we have RE  n 1 here we have n= and e1  103 x=48.242
2
2
x
RE 
e1 1 1 103
 .
 0.005  103
x 2 2 48.242
Now AE=RE x 6.946 = 0.034  10 3
Error Bounds = 6.946  0.034  10-3
3. Compute the difference 2.03  2 correct to five significant figures.
So relative error is RE  n
4. Evaluate the polynomial y  x 3  5x 2  6 x  0.55 at x  2.73. upto 3decimal places with truncation. Also, evaluate the percentage error.
Q4 is graded question.
Solution:
y   2.73  5  2.73  6  2.73  0.55
3
2
 20.346417  37.2645  16.38  0.55
 0.011917
up to 3 decimal places it becomes 0.012 , so we have Absolute Error=
0.011917  0.012  0.000083
Absolute Error 0.000083

 0.00696 (Up to five decimal places)
True value
0.011917
Percentage Error = 100  Re lative Error  100  0.00696  0.696% .
Relative Error =
5. Expand the function f x   Cosx in Taylor’s series and obtain a fourth-degree
polynomial to 3-decimal place when x  1.5 , x0  0 .
6. If the length l and the width w of a rectangle are in error by 1.0 and 0.5
respectively and the value of l is 10 cm and that of w is 6 cm. Give the
error bounds of the area of the rectangle and that of l .
w
Q6 is graded question.
Solution:
2
First of all note that if we take z  l
w
then relative error of z will be given
ez
e
e
 1  2 (You can prove that) here e1=1 and e2 =0.5 and we have l =10,
z
l
w
e
1
0.5

 0.1  0.083  0.183
w =6. Thus the relative error is given by RE  z 
z
10
6
AE
10
here , z 
 1.667 and RE  0.183 So we
Now as we know that RE 
z
6
have AE  z  RE  1.667  0.183  0.305061 .
Now required Error bounds can be obtained by the formula z  AE  1.667  0.305061
by RE 
7. Solve x  2  x upto 4-decimal places using Bisection method.
8. Use the bisection method to find a solution accurate to within 10 3
for x  tan x on [4,4.5].
9. Find an approximation to
method.
3
25 correct to with in 10 4 using the bisection
Q9 is graded question.
Solution:
Let x  3 25 then we have to find out the roots of the equation x3  25  0
40
30
20
10
2.5
3
3.5
4
-10
Note
that
if
we
take f ( x)  x3  25 then we have f ( x)  0 also
note
that
f (2.75)  20.797  25 0 and f (3)  27  25 0 so the real root of the equation lies
between 2 and 3 as it is also clear from the graph of the function plotted above. Also we
are required to find the approximation correct up to 10-4 decimal places which means that
1
 0.0001 that is correct up to 4
we have to find the approximation correct up to
10000
decimal places.
Take
a
b
f(a)
f(b)
f(m)
ab
m
2
2.75
3
2.875
-4.2031
2
-1.2363
3
2.875
3
2.9375
-1.2362
2
0.347412
2.875
2.9375
2.90625
-1.2362
0.347412
-0.452972
…….
……
……
……
…….
…….
Carry on in this way unless up to 4 decimal places your values became same and after
certain number of iterations your answer will be 2.924072.
NOTE: Number of iterations depends on the choice of the interval which you take, if we
pick the interval [2.75 3] then after approximately 15 iterations you will get the required
answer correct up to 4 decimal places and if you take the interval [2 3] then you may get
the required accuracy after more than fifteen iterations.
10. Find a real solution of the equation f ( x)  x 3  5 x 2  29  0 using regula
falsi method correct to 4 significant figures taking x1  5 and x2  6 .
Q10 is graded question.
Solution:
5
5.2
5.4
5.6
5.8
6
-5
-10
-15
-20
-25
Since we are given the values x1  5 and x2  6 so
f ( n)
x f ( x2 )  x2 f ( x1 ) f ( x1 )
x1
x2
f ( x2 )
n 1
f ( x2 )  f ( x1 )
5
6
5.8056
-29
7
-1.8473
5.8056
6
5.8462
-1.8492
7
-0.0033
……..
……..
……..
……
….
….
Carry on in this way unless up to 4 decimal places your values became same and after
certain number of iterations your answer will be 5.8479.
NOTE: Graph of the given equation is drawn just for your understanding so that you
would be able to realize about the root.
11. Use method of False-Position to find the solution accurate to within 10 3 for
Sinx  e  x  0 .
12. Use a fixed point iteration method to determine a solution accurate to
within 10 2 for a sin x  x  0 on [1, 2].
4
13. Use a fixed point iteration procedure to find an approximation to
accurate with in 10 4 .
3 that is
14. Use a fixed point iteration method to determine a solution accurate to within
10 2 for a sin x  x  0 on [1, 2].
15. Discuss the difference between fixed point method and Newton-Raphson
method.
16. Use Newton’s Method to find the solution accurate to with in 10 4 for the
following problem
a) ln( x  1)  cos( x  1)  0
1.2  x  2
b) e x  3x 2  0
0  x  1 and 3  x  5
17. The volume V of liquid in a spherical tank of radius r is related to the
h 2 3r  h 
depth h of the liquid by V 
. Determine h using Newton3
Raphson method, given r  1m and V  0.5m 3 upto 3-decimal places
Q17 is graded question.
Solution:
Since we are given r=1 and V=0.5 also we have   3.142 by putting these
3.142h 2  3  h 
0.5 
 1.5  9.426h 2  3.142h3
values we get
3
3
3.142h  9.426h 2  1.5  0
-1
-0.5
0.5
1
-2
-4
-6
-8
-10
First of all note that f (h)  3.142h3  9.426h2  1.5 then f (h)  9.426h2  18.852h we
can note that f(0) = 1.5 >0 and f(1)<0 so the root lies between 0 and 1. From the graph
it is quiet clear that the required root of the above polynomial lies some where near 0.5
and -0.5, so we can take the initial approximation as x0=1
5
xn 1  xn 
 4.784   1  0.508  0.492
f ( x0 )
f ( xn )
 1
For n=0 we have x1  x0 
f ( x0 )
f ( xn )
 9.426 
 0.407   1  0.058  0.434
f ( x1 )
 0.492 
f ( x1 )
 6.993
Carry on in this way unless up to 3 decimal places your values became same.
For n=1 we have x2  x1 
NOTE: Graph of the given equation is drawn just for your understanding so that you
would be able to realize about the root.
6