A note on the paper “A new general eighth-order family of iterative methods for solving nonlinear equations”

Abstract

In this work, we briefly talk about the incorrect convergence order presented in the paper Khan et al. (Appl. Math. Lett. 25:2262–2266, 2012). Accordingly, we first show that their convergence theorem includes major errors, and it is attempted to provide the correct error equation of their method having fifth-order not eighth-order convergence. Finally, to support our assertion, some numerical examples are tested.

Convergence analysis

Khan et al. [1], proposed the following three-step iterative method

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{y}_m=x_m-\frac{f(x_m)}{f^{\prime }(x_m)},\hfill \\ z_m=y_m-G(\frac{f(y_m)}{f(x_m)})\frac{f(y_m)}{f^{\prime }(x_m)},\hfill \\ x_{m+1}=z_m-\frac{\mathit{\mu }}{\mathit{\lambda }+\mathit{\nu }{q}_m^2}\frac{f(z_m)}{K-C(y_m-z_m)-D{(y_m-z_m)}^2},\hfill \end{array}\right.\end{array}$$

(1)

where

$$\begin{array}{c}\hfill \left\{\begin{array}{c}H=\frac{f(x_m)-f(y_m)}{x_m-y_m},\hfill \\ K=\frac{f(y_m)-f(z_m)}{y_m-z_m},\hfill \\ D=\frac{f^{\prime }(x_m)-H}{(x_m-y_m)(x_m-z_m)}-\frac{H-K}{{(x_m-z_m)}^2},\hfill \\ C=\frac{H-K}{(x_m-y_m)(x_m-z_m)}-D(x_m+y_m-2z_m),\hfill \\ q_m=\frac{f(z_m)}{f(x_m)},\hfill \end{array}\right.\end{array}$$

(2)

and $\mathit{\lambda },\mathit{\mu },\mathit{\nu }\in R$ and $G(t)$ represent a real-value function.

Khan et al. assert that if the conditions of the following theorem hold, then the iterative method (1.1) has convergence order eight (see Theorem in Section 4 in [1]). However, we prove that this is not true and prove that its convergence order is five.

Theorem 1.0.1

Let $f:D\subset R\to R$ have a single root $x^{\ast }\in D$, for an open interval $D$. If the initial point $x_0$ is sufficiently close to $x^{\ast }$, then the sequence $\{x_m\}$ generated by any method of the family (1.1) converges to $x^{\ast }$. If $G$ is any function with $M_0=G(0)=1$, $M_1=G^{\prime }(0)=2$, $M_2=G^{\prime \prime }(0)<\infty$ and $\mathit{\lambda }=\mathit{\mu }\ne 0$, then the methods defined by (1.1) have convergence order of at least 5.

Proof

For the sake of simplicity, we drop the iterative index $m$. Let $e=x-x^{\ast }$, and $c_k=\frac{f^{(k)}(x^{\ast })}{k!f^{\prime }(x^{\ast })},k=0,1,2,\dots$. Since $f(x^{\ast })=0\ne f^{\prime }(x^{\ast })$, we can write

$$\begin{array}{cc}& f(x)=f^{\prime }(x^{\ast })\sum _{k=1}^8{c}_k{e}^k+O(e^9),\hfill \\ \hfill & f^{\prime }(x)=f^{\prime }(x^{\ast })\sum _{k=1}^{7}k{c}_k{e}^{k-1}+O(e^8).\hfill \end{array}$$

(3)

Let $e_y=y-x^{\ast }$. Then,

$$\begin{array}{cc}\hfill e_y& =e-\frac{f(x)}{f^{\prime }(x)}\hfill \\ \hfill & =c_2{e}^2+\left(2c_3-2c_2^2\right)e^3+\left(4c_2^3-7c_3{c}_2+3c_4\right)e^4+\left(-8c_2^4+20c_3{c}_2^2-10c_4{c}_2-6c_3^2\right)e^5+O(e^6).\hfill \end{array}$$

(4)

Set,

$$t=\frac{f(y)}{f(x)}=c_{2}e+\left(2c_3-3c_2^2\right)e^2+\left(8c_2^3-10c_3{c}_2+3c_4\right)e^3$$

(5)

$$\begin{array}{cc}& +\left(-20c_2^4+37c_3{c}_2^2-14c_4{c}_2-8c_3^2\right)e^4\hfill \\ \hfill & +\left(48c_2^5-118c_3{c}_2^3+51c_4{c}_2^2+55c_3^2{c}_2-22c_3{c}_4\right)e^5+O(e^6).\hfill \end{array}$$

(6)

Now, define the weight function $G(t)$ by

$$\begin{array}{c}\hfill G(t)=M_0+M_{1}t+M_{2}t2.\end{array}$$

(7)

Then,

$$\begin{array}{cc}\hfill e_z=z-x^{\ast }& =e_y-G(t)\frac{f(y)}{f^{\prime }(x)}\hfill \\ \hfill & -\left(c_2\left(M_0-1\right)\right)e^2+\left(c_2^2\left(4M_0-M_1-2\right)-2c_3\left(M_0-1\right)\right)e^3\hfill \\ \hfill & +\left(c_2^3\left(-13M_0+7M_1-\frac{M_2}{2}+4\right)+c_3{c}_2\left(14M_0-4M_1-7\right)-3c_4\left(M_0-1\right)\right)e^4\hfill \\ \hfill & +(c_3{c}_2^2\left(-64M_0+38M_1-3M_2+20\right)+2c_4{c}_2\left(10M_0-3M_1-5\right)+2c_3^2\left(6M_0-2M_1-3\right)\hfill \\ \hfill & +c_2^4\left(38M_0-33M_1+5M_2-8\right))e^5+O(e^6).\hfill \end{array}$$

(8)

If $M_0=1$ and $M_1=2$, then

$$\begin{array}{c}\hfill e_z=(c_2^3\left(5-\frac{M_2}{2}\right)-c_2{c}_3)e^4+\left(c_2^4\left(5M_2-36\right)+c_3{c}_2^2\left(32-3M_2\right)-2c_4{c}_2-2c_3^2\right)e^5+O(e^6).\end{array}$$

(9)

Now, assume that $H,KD$, and $C$ are given by (1.2). Also, let $q=\frac{f(z)}{f(x)}$ and $\mathit{\lambda }=\mathit{\mu }\ne 0$. Consequently, the final error equation, i.e., $\widehat{e}$, for the method (1.1) is obtained as follows

$$\begin{array}{cc}\hfill \widehat{e}& =e_z-\frac{\mathit{\mu }}{\mathit{\lambda }+\mathit{\nu }{q}^2}\frac{f(z)}{K-C(y-z)-D{(y-z)}^2}\hfill \\ \hfill & =\frac{1}{2}\left(c_2^5\left(M_2-10\right)+2c_3{c}_2^3\right)e^5+O(e^6),\hfill \end{array}$$

(10)

which completes the proof. □

Numerical performances

This section concerns with numerical results of the proposed methods (1.1). We take the derived method from it by considering $\mathit{\lambda }=\mathit{\mu }=1$ and $G(t)=\frac{1}{1-2t+\mathit{\omega }{t}^2}$, where $\mathit{\omega }\in R$. This is the Method 1 in [1] (see Equations 21 and 22 in Section 5 there.)

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{y}_m=x_m-\frac{f(x_m)}{f^{\prime }(x_m)},\hfill \\ z_m=y_m-\frac{f^2(x_m)}{f^2(x_m)-2f(x_m)f(y_m)+\mathit{\omega }{f}^2(y_m)}\frac{f(y_m)}{f^{\prime }(x_m)},\hfill \\ x_{m+1}=z_m-\frac{1}{1+\mathit{\nu }{q}_m^2}\frac{f(z_m)}{K-C(y_m-z_m)-D{(y_m-z_m)}^2},\hfill \end{array}\right.\end{array}$$

(11)

where

$$\begin{array}{c}\hfill \left\{\begin{array}{c}H=\frac{f(x_m)-f(y_m)}{x_m-y_m},\hfill \\ K=\frac{f(y_m)-f(z_m)}{y_m-z_m},\hfill \\ D=\frac{f^{\prime }(x_m)-H}{(x_m-y_m)(x_m-z_m)}-\frac{H-K}{{(x_m-z_m)}^2},\hfill \\ C=\frac{H-K}{(x_m-y_m)(x_m-z_m)}-D(x_m+y_m-2z_m),\hfill \\ q_m=\frac{f(z_m)}{f(x_m)}.\hfill \end{array}\right.\end{array}$$

(12)

Numerical results have been carried out using Mathematica 9 with 200 digits of precision. $a(-b)$ means $a\times {10}^b$. In each table, COC stands for computational order of convergence (see [1]) which is given by

$$\begin{array}{c}\hfill rho\approx \frac{ln(|x_{m+1}-x_m||x_m-x_{m-1}{|}^{-1})}{ln(|x_m-x_{m-1}||x_{m-1}-x_{m-2}{|}^{-1})}.\end{array}$$

Among many test problems, the following four examples are considered

$$\begin{array}{cc}\hfill f_1(x)& =x^2-9,x^{\ast }=3,x_0=2.6\hfill \\ \hfill f_2(x)& =(x-2)(x^6+x^3+1)e^{-x^2},x^{\ast }=2,x_0=1.8,\hfill \\ \hfill f_3(x)& =\prod _{k=1}^{12}(x-k),x^{\ast }=5,x_0=5.3.\hfill \\ \hfill \end{array}$$

Table 1 Numerical results for $\mathit{\nu }=0=\mathit{\omega }$

To sum up, it can be concluded the that the method (1.1) has fifth-order convergence. Therefore, authors’ claim is not true that they have presented a family of iterative methods for solving nonlinear equations with eighth-order convergence.