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Mth603 Assignment Solution

Friday, October 22, 2010 Posted In Edit This



Q. 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 Ã¹

ê    2 Ãº
ë       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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