5
∫
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
R2R2 - R1
R23 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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5
∫
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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5
∫
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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