Mth603 Assignment Solution
Friday, October 22, 2010 Posted In Mth Edit ThisQ. No. 1. (Marks 10)
Use the Method of False Position to find the solution accurate to within 10-4 for the following problem.
x - 0.8 - 0.2 sin x = 0;é0, p ù
ê 2 úë û
| x0 | 0 | f ( x0 ) | -0.8 |
| x1 | 1.570796 | f ( x1 ) | 0.570796 |
| x = x - x1 - x0 f ( x ) 2 1 f ( x ) - f ( x ) 1 1 0 | 0.916721 | f ( x2 ) | -0.042001 |
| x = x - x2 - x1 f ( x ) 3 2 f ( x ) - f ( x ) 2 2 1 | 0.961551 | f ( x3 ) | -0.002465 |
| x = x - x3 - x1 f ( x ) 4 3 f ( x ) - f ( x ) 3 3 1 | 0.964346 | f ( x4 ) | 0.000011 |
| x = x - x4 - x1 f ( x ) 5 4 f ( x ) - f ( x ) 4 4 1 | 0.964334 | f ( x5 ) | 0 |
| x = x - x5 - x1 f ( x ) 6 5 f ( x ) - f ( x ) 5 5 1 | 0.964334 | f ( x6 ) | 0 |
Q. No. 2. (Marks 10)
Solve ex - 3x2 = 0 for 0 £ x £ 1 and 3 £ x £ 5 by using the Secant Method up to four iterations. (Note: Accuracy up to four decimal places is required)| x0 | 0 | f ( x0 ) | 1 |
| x1 | 1 | f ( x1 ) | -0.2817 |
| x = x0 f ( x1 ) - x1 f ( x0 ) 2 f ( x ) - f ( x ) 1 0 | 0.7802 | f ( x2 ) | 0.3558 |
| x = x1 f ( x2 ) - x2 f ( x1 ) 3 f ( x ) - f ( x ) 2 1 | 0.9029 | f ( x3 ) | 0.0211 |
| x = x2 f ( x3 ) - x3 f ( x2 ) 4 f ( x ) - f ( x ) 3 2 | 0.9106 | f ( x4 ) | -0.0018 |
| x = x3 f ( x4 ) - x4 f ( x3 ) 5 f ( x ) - f ( x ) 4 3 | 0.91 | f ( x5 ) | 0 |
| x0 | 3 | f ( x0 ) | -6.9145 |
| x1 | 5 | f ( x1 ) | 73.4132 |
| x = x0 f ( x1 ) - x1 f ( x0 ) 2 f ( x ) - f ( x ) 1 0 | 3.1722 | f ( x2 ) | -6.3286 |
| x = x1 f ( x2 ) - x2 f ( x1 ) 3 f ( x ) - f ( x ) 2 1 | 3.3173 | f ( x3 ) | -5.4277 |
| x2 f ( x3 ) - x3 f ( x2 ) x4 = f ( x3 ) - f ( x2 ) | 4.1915 | f ( x4 ) | 13.4159 |
| x = x3 f ( x4 ) - x4 f ( x3 ) 5 f ( x ) - f ( x ) 4 3 | 3.5691 | f ( x5 ) | -2.7308 |
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