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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**Example text**

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 + ...