Download Numerical methods: problems and solutions by Mahinder Kumar Jain, S.R.K. Iyengar, R. K. Jain PDF

By Mahinder Kumar Jain, S.R.K. Iyengar, R. K. Jain

Jain M.K., Iyengar S., Jain R. Numerical tools (New Age guides (AP),India, 2008)(ISBN 8122415342)(O)(430s)

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99998. 00. 9940 + 1 1 – 11 1 32 – 10 1 – 10 22 – 22 The deflated polynomial is x2 – 10x + 22. 5. 2905. 1903. 1683. 1683. 1673. 167. 8547 = 0. 5. 4228.

25) for solving the following equations : (i) 1 + z2 = 0, z0 = (1 + i) / 2. 36i. Solution (i) Separating the given equation into real and imaginary parts, we get u(x, y) = 1 + x2 – y2, v(x, y) = 2xy, x0 = 1 / 2, y0 = 1 / 2 ux = 2x, uy = – 2y, vx = 2y, vy = 2x. 25), we obtain xk+1 = xk – [(uvy – vuy)k] / D, yk+1 = yk – [(uxv – vxu)k] / D, D = (ux vy – uy vx)k, k = 0, 1, 2, ... 9973. (ii) We can proceed exactly as in part (i), or can use the method f ( zk ) f ′ ( zk ) directly. 3867). 42 It is required to solve the two simultaneous equations x = f (x, y), y = g(x, y) by means of an iteration sequence.

309. 88577. 8858. 12 We consider the multipoint iteration method f ( xk ) f ′ ( xk − β f ( xk ) / f ′ ( xk )) where α and β are arbitrary parameters, for solving the equation f (x) = 0. Determine α and β such that the multipoint method is of order as high as possible for finding ξ, a simple root of f (x) = 0. Solution We have xk + 1 = x k − α LM N OP Q f ( xk ) f (ξ + ε k ) c = = ε k + 2 ε 2k + ... ]–1 2 f ′ ( x k ) f ′ (ξ + ε k ) = εk – where ci = f (i) (ξ) / f ′(ξ). We also have F GH f ′ xk − β I JK 1 c ε 2 + O (εk3) 2 2 k FG H L = f ′ bξ g + M(1 − β) ε N IJ K f ( xk ) 1 = f ′ (ξ) + (1 − β) ε k + β c2 ε 2k + O(ε 3k ) 2 f ′ ( xk ) k + OP Q 1 β c2 ε 2k + ...

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