A note on the new iteration method for solving algebraic equations

doi:10.1016/j.amc.2005.01.124

Applied Mathematics and Computation

Volume 171, Issue 2, 15 December 2005, Pages 1177-1183

https://doi.org/10.1016/j.amc.2005.01.124 Get rights and content

Abstract

In [Ji-Huan He, A new iteration method for solving algebraic equations, Appl. Math. Comput. 135 (2003) 81–84], a new iteration method for solving algebraic equations has been proposed. In this paper, we show for one equation which given by the author and some other equations that the new iteration method fails to obtain expected results. Hence, in contrast to previous claims, it is no more quickly convergent than Newton’s method and may in fact be failed.

Introduction

In a recent paper Ji-Huan He [1] proposed a new iteration method for solving algebraic equations. It convert the nonlinear problem into a coupled iteration system by applying Taylor’s theorem. In the author’s opinion, it is obvious that the new iteration is monotonically convergent and converges much more quickly than Newton’s method. However, there is no illustrative comparison of convergence for the new iteration method and Newton’s method.

It is the purpose of this paper to show that, although the new iteration method seems to be of high convergence, the results are promising in that it requires more computation work and even be divergent. The rest of this paper will be organized as the following: The principle of the new iteration method is described in Section 2. Some examples are studied in Section 3. The general remarks are given in Section 4.

Section snippets

The principle of the new iteration method

We consider the following nonlinear algebraic equation: $f (x) = 0 .$

Applying Taylor’s theorem, we have $f (x) = f (x_{0}) + f^{'} (x_{0}) (x - x_{0}) + \frac{1}{2} f^{″} (x_{0}) (x - x_{0})^{2} + \dots,$ where x₀ is the initial approximation.

The nonlinear Eq. (2.1), therefore, can be converted into the following coupled system: $f (x_{0}) + f^{'} (x_{0}) (x - x_{0}) + \frac{1}{2} f^{″} (x_{0}) (x - x_{0})^{2} + g (x) = 0,$ $g (x) = f (x) - f (x_{0}) - f^{'} (x_{0}) (x - x_{0}) - \frac{1}{2} f^{″} (x_{0}) (x - x_{0})^{2} .$

We therefore, can construct the following iteration formulae: $f (x_{n}) + f^{'} (x_{n}) (x_{n + 1} - x_{n}) + \frac{1}{2} f^{″} (x_{n}) (x_{n + 1} - x_{n})^{2} + g (x_{n}) = 0,$ $g (x_{n}) = f (x_{n}) - f (x_{n - 1}) - f^{'} (x_{n - 1}) (x_{n} - x_{n - 1}) - \frac{1}{2}$

Some examples

In [1], two examples were illustrated simply. In the following Example 1, Example 2, we will check them in detail by the new iteration method. A comparison of convergence for the new iteration method and Newton’s method is illustrated with numerical results in Example 3.

Example 1

Consider a polynomial of second degree $f (x) = x^{2} + x - 2 = 0 .$

Supposing x₀ = 0 be one of its initial approximate solutions, by the iteration formula (2.8), the exact solutions can be immediately obtained by only one iteration step [1].

Discussion

We have shown a comparison of convergence for the new iteration method and Newton’s method with numerical results. It confirms that the new 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 slowly than Newton’s method and even be divergent. Further, we find out that in order to obtain a real solution the requirement B² − 4A(C + g(x_n)) ⩾ 0 is difficult to be satisfied.

References (2)

J.H. He
A new iteration method for solving algebraic equations
Appl. Math. Comput.
(2003)
Mamta et al., On a class of quadratically convergent iteration formulae, Appl. Math. Comput., in...

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