A new hybrid iteration method for solving algebraic equations

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Applied Mathematics and Computation 195 (2008) 772–774 www.elsevier.com/locate/amc

A new hybrid iteration method for solving algebraic equations Nasr-Al-Din Ide University of Aleppo, Faculty of Science, Department of Mathematics, Aleppo, Syria

Abstract A new hybrid iteration method (the hybrid’s name was used by Luo [Xing-Guo Luo, Applied Mathematics and Computation 171 (2) (2005) 1171–1183]) has been proposed for solving a non linear algebraic equation f(x) = 0, by using Taylor’s theorem. In this paper, we proposed a new hybrid iteration method and we show by one equation that this new hybrid iteration method is more quickly convergent than Newton’s method and hence than hybrid iteration method. Ó 2007 Elsevier Inc. All rights reserved. Keywords: Iteration method; Convergence; New hybrid iteration method

1. Introduction In the papers [1] of Luo and [2] He, a new iteration method for solving a non linear algebraic equation f(x) = 0 was proposed. It converts the non linear problem into a coupled iteration system by applying Taylor’s theorem. In the paper [1] Xing-Guo illustrated comparison of convergence for the hybrid iteration method and Newton’s method by two examples, he illustrated that the hybrid iteration is monotonically convergent and converge much more quickly than Newton’s method. 2. The principle of the new iteration method We consider the following non linear algebraic equation: f ðxÞ ¼ 0:

ð2:1Þ

We consider the Taylor’s expansion of f(xk+1) about the point xk, [3] we have 1 1 f ðxkþ1 Þ ¼ f ðxk Þ þ f 0 ðxk Þðxkþ1  xk Þ þ f 00 ðxk Þðxkþ1  xk Þ2 þ f ð3Þ ðxk Þðxkþ1  xk Þ3 þ 0½ðxkþ1  xk Þ4 : 2 6

E-mail address: [email protected] 0096-3003/$ - see front matter Ó 2007 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2007.05.022

ð2:2Þ

Nasr-Al-Din Ide / Applied Mathematics and Computation 195 (2008) 772–774

773

Also, 1 f 0 ðxkþ1 Þ ¼ f 0 ðxk Þ þ f 00 ðxk Þðxkþ1  xk Þ þ f ð3Þ ðxk Þðxkþ1  xk Þ2 þ 0½ðxkþ1  xk Þ3 : ð2:3Þ 2 Replacing Eq. (2.3) into (2.2) and neglecting the terms in ðxkþ1  xk Þ4 and higher order terms, we obtain f ðxkþ1 Þ ¼ f 00 ðxk Þx2kþ1 þ ½4f 0 ðxk Þ  2f 00 ðxk Þxk þ 2f 0 ðxkþ1 Þxk xkþ1 þ ½6f ðxk Þ  4f 0 ðxk Þxk þ f 00 ðxk Þx2k  2f 0 ðxkþ1 Þxk  ¼ 0: 0

ð2:4Þ

0

We approximate f ðxkþ1 Þ by f ðxk Þ, then we show that f ðxkþ1 Þ ¼ f 00 ðxk Þx2kþ1 þ ½6f 0 ðxk Þ  2f 00 ðxk Þxk ðxkþ1 Þ þ ½6f ðxk Þ  6f 0 ðxk Þxk þ f 00 ðxk Þx2k  ¼ 0:

ð2:5Þ

0

We next set f ðxkþ1 Þ ¼ 0 (i.e. suppose that xk=1 is a root of f(x) = 0). Therefore, we have the equation Ax2kþ1 þ Bxkþ1 þ C ¼ 0;

ð2:6Þ

where A ¼ f 00 ðxk Þ;

B ¼ 6f 0 ðxk Þ  2f 00 ðxk Þxk ;

and C ¼ 6f ðxk Þ  6f 0 ðxk Þxk þ f 00 ðxk Þx2k

solving (2.6) for xk+1 yields the following new iteration formula: pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi B  B2  4  A  C : vkþ1 ¼ 2A For obtaining a real solution, it requires that

ð2:7Þ

ð2:8Þ

B2  4  A  C P 0: 3. Example Consider the following equation: f ðxÞ ¼ x3  ex ¼ 0:

ð3:1Þ

To find a positive root near x = 1, we start with x0 = 1 and obtain x = 0.7729 by few iteration. The results obtained by Newton iteration, hybrid iteration and our new hybrid iteration are shown in Tables 1–3. The results obtained by Newton iteration and our new hybrid iteration are shown in Tables 4 and 5. 4. Discussion We have shown a comparison of convergence for the new hybrid iteration method, hybrid iteration method and Newton iteration method with numerical results. It confirms that the new hybrid iteration method seems to be of high convergence by including the second order term as well in the Taylor expansion, as a matter of fact, it converges better than Newton’s method and hybrid method.

Table 1 Newton iteration for solving f(x) = x3  ex = 0 n

xn

jf(xn)j

1 2 3 4 5 6

1 0.81230903009738120 0.77427654898550025 0.77288475620962160 0.77288295915220184 0.77288295914921012

0.63212055882855767000 0.09216677153431299100(9.2E2) 0.00314482497861333540(3.1E3) 0.00000405008554745892(4.1E6) 0.00000000000674254836(6.7E12) 0.00000000000000006456(6.5E17)

774

Nasr-Al-Din Ide / Applied Mathematics and Computation 195 (2008) 772–774

Table 2 Hybrid iteration for solving f(x) = x3  ex = 0 n

xn

jf(xn)j

1 2 3 4 5 6 7

1 0.81230903009738120 0.77597366111597321 0.77290581952315329 0.77288297294415598 0.77288295914921590 0.77288295914921012

0.63212055882855767000 0.09216677153431299100(9.2E2) 0.00698556573567983560(7.0E3) 0.00005152207246609381(5.2E5) 0.00000003109000577256(3.1E8) 0.00000000000001299950(1.3E14) 0.00000000000000006456(6.5E17)

Table 3 New hybrid iteration for solving f(x) = x3  ex = 0 n

xn

jf(xn)j

1 2 3 4 5 6 7

1 0.801305391412732989 0.773375282149371597 0.772883108807313135 0.772882959149223945 0.772882959149210113 0.772882959149210113

0.63212055882855767000 0.0657676468308537 (6.5E2) 0.0011100665084369(1.1E3) 3.37288157491135E7 3.11745529564533E14 1.62630325872826E19 1.62630325872826E19

Table 4 Newton iteration [1], for solving n

xn

1 2 3

12.601419947171719 12.566356255118672 12.566370614359174

0.03504215716101725900(3.5E2) 0.00001435924050063514(1.45) 0.00000000000000128651(1.3E15)

Table 5 New hybrid iteration for solving f(x) = sin x = 0 n

xn

1 2

12.601419947171719 12.566363437091402800

0.0000071772677701(7.1E6) 0.0000000000000008(8.0E16)

References [1] Xing-Guo Luo, A note on the new iteration method for solving algebraic equation, Applied Mathematics and Computation 171 (2) (2005) 1177–1183. [2] Ji-Huan He, A new iteration method for solving algebraic equation, Applied Mathematics and Computation 135 (1) (2003) 81–84. [3] D.K.R. Babajee, M.Z. Dauhoo, An analysis of the properties of the variants of Newton’s method with third order convergence, Applied Mathematics and Computation 183 (1) (2006) 659–684.