Mth603 Assignment Solution

Q. No. 1. (Marks 10)

Use the Method of False Position to find the solution accurate to within10-4 for the following problem.

x - 0.8 - 0.2 sin x = 0;

é0, p ù

ê 2 ú

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 Methodup 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