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MTH603 Numerical Analysis Assignment No. 02 Solution and Discussion Fall 2012 Due Date : 20-01-2013

Assignment #  2

 

MTH603 (Fall 2012)

          

             Total marks: 10

             Lecture # 23 - 28 

          Due date: 20-01-2013                                                               

  

 

 

 

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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.

 

 

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).

 

 

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

 

 

 

                                                                                   

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

 

 

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plz discuss here about this assignment

Our main purpose here discussion not just Solution

Please Discuss here about this assignment.Thanks

Numerical Analaysis: "Divided Difference Forumula"

Find equation of a "Cubic Curve" that passes through the points (4,-43), (7, 83), (9,327) and (12, 1053) using the "Divided Difference Formula".

Points (4,-43), (7, 83), (9,327) and (12, 1053)

Using divided difference formula

4   -43

           42

7    83           16

           122          1

9    327          24

           242

12   1053

Column 1 and 2 given data

Column 3 obtained by 83-(-43) / 7-4 = 83+43 /3 = 126/3 = 42

                      327-83/9-7 = 244/2 = 122

                      1053-327/12-9 = 726/3 = 242

Column 4 obtained by 122-42/9-4 = 80/5 = 16

                      242-122/12-7 = 120/5 = 24

Column 5 obtained by 24-16/12-4 = 8/8 = 1

Third degree approximation given by

 

f(x) = f(4) + xf(4,7) + x(x-4)f(4,7,9)+x(x-4)(x-7)f(4,7,9,12)

     = -43 + x 42 + x(x-4) 16 + x(x-4)(x-7)

     = -43 + 42x-64x+16x^2 + x(x^2-11x+28)

     = -43 -22x +16x^2 + x^3-11x^2+28x

     = -43 +6x + 5x^2 +x^3

Third degree approximation is f(x) = x^3+5x^2+6x-43

Cubical curve given above thus ans.

Question No. 02 Complete Solution

Numerical Analaysis: "Lagrange's Formula"

Find interpolation polynomial by "Lagrange’s Formula", with the help of this table:
x 1 2 3 5
f(x) 0 7 26 124

And find the value of f (6).

x 1 2 3 5
f(x) 0 7 26 124
calculating lagranges coefficeitns
L0 = (x-2)(x-3)(x-5)/(1-2)(1-3)(1-5) = (x-2)(x^2-8x+15)/8 = (x^3-10x^2+31x-30)/8
L1 = (x-1)(x-3)(x-5)/(2-1)(2-3)(2-5) = (x-1)(x^2-8x+15)/3 = (x^3-9x^2+23x-15)/3
L2 = (x-1)(x-2)(x-5)/(3-1)(3-2)(3-5) = (x-5)(x^2-3x+2)/-4 = (x^3-8x^2+17x-10)/-4
L3 = (x-2)(x-3)(x-1)/(5-1)(5-2)(5-3) = (x-2)(x^2-4x+3)/24 = (x^3-6x^2+11x-6)/24
lagranges polynomial is = 0*(x^3-10x^2+31x-30)/8 + 7/3 (x^3-9x^2+23x-15)
-26/4 (x^3-8x^2+17x-10) + 124/24 (x^3-6x^2+11x-6)
simplyfying above
poly = x^3 (7/3-26/4+124/24) + x^2(-21+52-31 ) +x (23*7/3-17*26/4+11*124/24)
+ -15*7/3 +10*26/4 -31
= x^3 -1
f(x) is = x^3-1
f(6) = 6^3-1 = 216-1 = 215 thus ans.

I have same answer.. Its right

ThaNX

Q.01 and Q.02 are complted. 

Q.03 and Q. 04 are also easy easy.... discuss here I will try to solve and post soon

All person having MTH603 must particpate here

yr ya jo solution ap na uper upload keya ha keya ya thek ha? ap keh reha ho k Ist and 2nd are complited and 3rd aur 4th easy

MTH603_Assignment#02_Solution

Attachments:

Q.03 and Q. 04 in lecture no 28.

Q.3 me forward difference use krna hei derivative find krny k ly. or Q.4 me backward

both ex. 28 lec me hen.

Q4 main backward difference  k table main aik value negtive main ati ha can any body tel is ka kya krna ha?????????????????

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:

X

Y

1st divided difference

2nd divided difference

3rd divided difference

4

-43

 

 

 

7

83

42

 

 

9

327

122

16

 

12

1053

242

24

1

                                                                        

y = f(x) = -43+(x-4)42+(x-4)(x-7)16+(x-4)(x-7)(x-9)(1)

         = -43+(x-4)42+(x2-11x+28)16+(x2-11x +28)(x-9)

         = -43+(x-4)42+(x2-11x+28)16+(x3–20x2+127x-252)

         = -43+42x-168+16x2-176x+448+x3–20x2+127x-252

         = x3–4x2-7x-15

Hence y=f(x)= x3–4x2-7x–15 is the required cubic curve.

 

or

 

 

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:

 

 

 

 

 

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:

 

x

f(x)

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

 

 

= = 2.9

 

 

Attachments:

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