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Solution File Assignment # 1
MTH603 (Fall 2012)
Total marks: 10
Lecture # 0108
Due date: 15112012
Dear students as it was told there are 4 questions in the assignment but only one question will be graded. Question 2 will be graded.
Question#1 Marks 10
Perform only three iterations of secant method, round off values up to four decimal places to find the root of
Note: All the calculation should be done in the radian mode only.
Taking
This root obtained is correct upto 2 decimal places only after 3 iterations.
Question#2 Marks 10
Use Regula Falsi (method of false position) to solve the equation
Note: Perform only two iterations and round off values up to four decimal places. All the calculation should be done in the radian mode only.
Solution:
This root obtained is correct upto 1 decimal places only after 2 iterations.
Question#3 Marks 10
Apply NewtonRaphson method to determine a root of the equation.
correct to four decimal places. Only three iterations needed.
Note: All the calculation should be done in the radian mode only.
This root obtained is correct upto 3 decimal places only after 3 iterations.
Question#4 Marks 10
Find the root of the equation in the interval (0,1) by iteration method correct to four decimal places and do only three iterations.
Note: All the calculation should be done in the radian mode only.
Solution ;
x 
0 
1 
F(x)= 
1<0 
0.282>0 
So the method of iteration can be applied
Let
Hence the desired root is 0.8771
This root obtained is correct upto 1 decimal places only after 3 iterations.
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+ Click Here to Search (Looking For something at vustudents.ning.com?) + Click Here To Join (Our facebook study Group)Solution File Assignment # 2
MTH603 (Fall 2012)
Total marks: 10
Lecture # 23  28
There are 4 questions in the assignment but only question # 3 will be graded.
Question#1 Marks 10
Find an equation of a cubic curve which passes through the points
(4,43), (7, 83), (9,327) and (12, 1053) using Divided Difference Formula.
Solution:
Newton’s divided difference table is
X 
Y 
1^{st} D.D 
2^{nd} D.D 
3^{rd} D.D 
4 
43 



7 
83 
42 


9 
327 
122 
16 

12 
1053 
242 
24 
1 
Question#2 Marks 10
Find interpolation polynomial by Lagrange’s Formula, with the help of following table,
x 
1 
2 
3 
5 
f(x) 
0 
7 
26 
124 
And hence find value of f (6).
Solution:
Question#3 Marks 10
Find from the following table,
x 
0 
0.2 
0.4 
0.6 
0.8 
1.0 
1.2 
0.710 
1.175 
1.811 
2.666 
3.801 
5.292 
7.232 
Solution:
x 

0 
0.710 






0.2 
1.175 
0.465 





0.4 
1.811 
0.636 
0.171 




0.6 
2.666 
0.855 
0.219 
0.048 



0.8 
3.801 
1.135 
0.280 
0.061 
0.013 


1.0 
5.292 
1.491 
0.356 
0.076 
0.015 
0.002 

1.2 
7.232 
1.940 
0.449 
0.093 
0.017 
0.002 
0 
Since x=0 occur at the start of the table, it is appropriate to use the forward difference formula for the derivation.
Now putting the values in above formula, we get
Question#4 Marks 10
Find from the following table,
x 
1.3 
1.6 
1.9 
2.2 
2.5 
f(x) 
2.4 
2.9 
3.2 
4.7 
6.4 
Solution:
Since x=2.5 occur at the end of the table, it is appropriate to use the backward difference formula for the derivation.
Backward difference table is as





1.3 
2.4 




1.6 
2.9 
0.5 



1.9 
3.2 
0.3 
0.2 


2.2 
4.7 
1.5 
1.2 
1.4 

2.5 
6.4 
1.7 
0.2 
1 
2.4 
Ali gud keep it up
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