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

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

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

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29 Determine the order of convergence of the iterative method xk + 1 = (x0 f (xk) – xk f (x0)) / ( f (xk) – f (x0)) for finding a simple root of the equation f (x) = 0. 40) we get εk + 1 = ε k − [ ε k − ε 0 ] f (ξ + ε k ) . 41) εk + 1 = ε k − 1 ε 2 ″ (ξ) + ... ] 2 k f 1 (ε 2 − ε 2 ) ″ (ξ) + ... P Q = εk − εk + 1 1 2 ε k c2 + ... × 1 + (ε k + ε 0 ) c2 + ... 2 2 = εk 1 2 ε k c2 2 k + k + ε 0 ) c2 −1 1 ε ε c + O (εk2ε0 + εkε02) 2 k 0 2 where c2 = f ″(ξ) / f ′(ξ). Thus, the method has linear rate of convergence, since ε0 is independent of k.

Calculate the smallest integer n for which the inequality | xn – ξ | < 10–6 in valid. , Sweden, BIT 13 (1973), 493) As n → ∞, the method converges to ξ4 – 8ξ3 + 112ξ – 192 = 0. Hence, the method finds a solution of the equation f (x) = x4 – 8x3 + 112x – 192 = 0. 43 Transcendental and Polynomial Equations Substituting xn = ξ + εn and xn+1 = ξ + εn+1 in the given iteration formula, we get the error equation εn+1 = FG 1 ξ − 1 ξ + 7ξ − 12IJ + FG 1 ξ − 3 ξ + 8IJ ε H 16 2 K H4 2 K F 3 3 I F 1 1I 1 ε .

259. 37 The equation x3 – 5x2 + 4x – 3 = 0 has one root near x = 4, which is to be computed by the iteration x0 = 4 3 + (k − 4) xn + 5 xn2 − xn3 , k integer k (a) Determine which value of k will give the fastest convergence. (b) Using this value of k, iterate three times and estimate the error in x3. (Royal Inst. Tech. Stockholm, Sweden, BIT 11 (1971), 125) Solution (a) Let ξ be the exact root of the given equation. 55) Since the root is near x = 4, we can choose ξ = 4 + δ. 55), we obtain kεn+1 = (k – 12)εn + O(δεn).