Bt11-02

  • October 2019
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dx 1-x2

B.Sc.IT 11-02

1COMPUTER ORIENTED NUMERICAL METHODS Q1. Round off the following numbers correct to the 3 significant figures: a) 23.2927 b) 1.2399 c) 1292323 d) 982121 And also calculate Relative Error, Absolute Error and Percentage Error in each case. Ans: - EA = The absolute error ER = The relative error EP = The percentage error x = The true value of the quantity a) Number rounded off to 3 significant figures = 23.3 ∴ EA =| 23.2927 – 23.3| = 7.30*10-3 ER = EA = 7.30*10-3 = 3.13*10-4 x 23.2927 EP = ER * 100 = - 3.13 * 10-4 * 100 = 3.13*10-2 Ans b) Number rounded off to 3 significant figures = 1.24 ∴ EA =| 1.2399 – 1.24| = 1.00*10-4 ER = EA = -1.00*10-4 = 8.07*10-5 x 1.2399 EP = ER * 100 = - 8.07*10-5 * 100 = 8.07*10-3 Ans c) Number rounded off to 3 significant figures = 129 ∴EA =| 1292323 – 1290000| = 2323 ER = EA = 2323 = 1.80*10-3 x 1292323 EP = ER * 100 = 1.80*10-3 * 100 = 1.80*10-1 Ans d) Number rounded off to 3 significant figures = 982 ∴ EA =| 982121 – 982000| = 121 ER = EA = 121 = 1.23*10-4 x 982121 EP = ER * 100 = 1.23*10-4 * 100 = 1023*10-2 Ans Q2. Find the inverse of the matrix by Gauss – Jordan Method:

3 5 1 4 9 8

7 6 2

Ans. The augmented matrix is

R1 R1

1 5/3 7/3 1 4 6 9 8 2

3 5 1 4 9 8 1/3 0 0 0 1 0 0 0 1 -1-

7 6 2

1 0 0

0 1 0

0 0 1

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5

3

R2R2 - R1

R23 R2 7 R3  9R1 – R3

R3  R3 7

R3  R3 + R2

R2  5/3 R2

R1  R2 – R1



dx 1-x2

1 1 5/3 7/3 1/3 0 0 0 7/3 11/3 -1/3 1 0 9 8 2 0 0 1 1 5/3 7/3 1/3 0 0 0 1 11/3 -1/7 7/3 0 9 8 2 0 0 1 1 5/3 7/3 1/3 0 0 0 1 11/7 -1/7 7/3 0 0 7 19 3 0 1 1 5/3 7/3 1/3 0 0 0 1 11/7 -1/7 7/3 0 0 1 19/7 -3/7 0 1/7 1 5/3 7/3 1/3 0 0 0 1 11/7 -1/7 7/3 0 0 0 8/7 -4/7-7/3 1/7 1 5/3 7/3 0 5/3 55/21 0 0 8/7 1 0 6/21 0 5/3 55/21 0 0 8/7

1/3 0 0 -5/21 5/7 0 -4/7 -7/3 1/7 12/21 5/7 0 -5/21 5/7 0 -4/7 -7/3 1/7

Q3. Find the multiplicity root of the equation f(x) = x 4 + x 2 – 6x = 9. Ans. f (x) = x 4 + x 2 – 6x – 9 = 0 f (2) = -1 = -ve f (3) = 63=+ve Therefore, root lies between 2&3. Now, x 4 = 9 + 6x - x 2 Hence the iteration method can be applied and we start with x0=2.then the successive approximations are, x1=(9 + 6 * 2 - 4) 1 / 4=2.030543 Now, taking x1=x0 Repeating the process for x2, x3 x2 = 2.0323370 x3 = 2.03244 Hence x2&x3 being almost same, the root is 2.032 correct to 3 decimal places.

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dx 1-x2

Q4. From the following1table of values of values of x and y obtain dy/dx and d 2y/dx2 X= 1 2 3 4 5 Y=

1.0

3.921

22.78281

45

for x=4

675.239

Ans.

∇y

x 1 2

y 1 3.921

2.921

3 4 5

22.78281 45 675.239

18.8618 22.2171 630.239

∇ 2y

∇ 3y

∇ 4y

15.9408 3.35538 608.02181

-12. 58542 604.66642

617.25184

Using backward difference formula: At x = 4 and p = 0 ; h = 1 where h is the interval between the values of x. dy = 1 [∇ y n + 1 ∇ 2 y n + 1∇ 3 y n] dx h 2 3 dy = 22.21719 + 1 x 1.67769 + 1 x 4.19514 dx 2 3 dy = 23.89488 – 4019514 dx dy = 19.69974 dx d 2y = 1 [∇ 2 y x + ∇ 3 y x + 11 ∇ 4 y x] dx 2 h2 12 = 1 [15.9408 + (-12.58542) + 11 (617.25184)] 1 12 d 2y = [569.16957] Ans. dx 2 Q5. Compute the following by using Romberg’s method.

4



dx 1-x2

0

Given : X

0

1

2

3

4

Y

0.231

1.345

4.4896

23.47

0.0234

Ans. Taking h = 2, 1, 0.5 successively we get the result using trapezoidal rule as

4

0



dx 1-x2

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dx 2 1-x + 2 (y1+y2+y3) ]

= 1 h [(y0+ y4) 2 1 = 1 x 2 [(.231 + 0.0234) + 2 (1.345 + 4.4896 +23.47)] 2 = [0.2544 + 2 x (58.6092) ] = (58.8636) Again using h = 1

4



0 0.5 and h =

dx = 1 [58.8636] = 29.4318 2 1-x2 4

0



=1[58.8636]=14.7159

dx 4 1-x2

so (h,h/2) = 1 [4f (h/2) –f (h)] 3 =1[4*29.4318 – 58.8636] 3 =58.8636 ans.

The end 

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