f(x) = x^5 - 4*x^3 - 2*x + 1
f ' (x) = 5 x^4 - 13 x^2 - 2
f (2) = - ve whereas f (3) = +ve
Hence root must lie between 2 and 3. So let xn = 2.5 be the initial solution.
No of iterations | xn | f(xn) | f'(xn) | xn+1 =xn - f(xn) / f'(xn) | error = xn+1 - xn |
1 | 2.5 | 31.15625 | 118.3125 | 2.236661384 | -0.263338616 |
2 | 2.236661 | 7.745618 | 63.10089 | 2.113911639 | -0.122749745 |
3 | 2.113912 | 1.198912 | 44.21946 | 2.086798872 | -0.027112767 |
4 | 2.086799 | 0.049987 | 40.56159 | 2.085566493 | -0.001232379 |
5 | 2.085566 | 9.99E-05 | 40.39951 | 2.08556402 | -2.47305E-06 |
now the error < 0.001 hence the final answer = 2.08556402
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