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Solution File Assignment #  1

 

MTH603 (Fall 2012)

          

             Total marks: 10

             Lecture # 01-08  

          Due date: 15-11-2012  

                                                            

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 Newton-Raphson 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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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

1st D.D

2nd  D.D

3rd 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

 

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