## Convergence analysis

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

$\begin{array}{c}\hfill \left\{\begin{array}{c}{y}_{m}={x}_{m}-\frac{f\left({x}_{m}\right)}{{f}^{\prime }\left({x}_{m}\right)},\hfill \\ {z}_{m}={y}_{m}-G\left(\frac{f\left({y}_{m}\right)}{f\left({x}_{m}\right)}\right)\phantom{\rule{0.166667em}{0ex}}\frac{f\left({y}_{m}\right)}{{f}^{\prime }\left({x}_{m}\right)},\hfill \\ {x}_{m+1}={z}_{m}-\frac{\mathit{\mu }}{\mathit{\lambda }+\mathit{\nu }{q}_{m}^{2}}\phantom{\rule{0.166667em}{0ex}}\frac{f\left({z}_{m}\right)}{K-C\phantom{\rule{0.166667em}{0ex}}\left({y}_{m}-{z}_{m}\right)-D\phantom{\rule{0.166667em}{0ex}}{\left({y}_{m}-{z}_{m}\right)}^{2}},\hfill \end{array}\right\\end{array}$
(1)

where

$\begin{array}{c}\hfill \left\{\begin{array}{c}H=\frac{f\left({x}_{m}\right)-f\left({y}_{m}\right)}{{x}_{m}-{y}_{m}},\hfill \\ K=\frac{f\left({y}_{m}\right)-f\left({z}_{m}\right)}{{y}_{m}-{z}_{m}},\hfill \\ D=\frac{{f}^{\prime }\left({x}_{m}\right)-H}{\left({x}_{m}-{y}_{m}\right)\left({x}_{m}-{z}_{m}\right)}-\frac{H-K}{{\left({x}_{m}-{z}_{m}\right)}^{2}},\hfill \\ C=\frac{H-K}{\left({x}_{m}-{y}_{m}\right)\left({x}_{m}-{z}_{m}\right)}-D\phantom{\rule{0.166667em}{0ex}}\left({x}_{m}+{y}_{m}-2{z}_{m}\right),\hfill \\ {q}_{m}=\frac{f\left({z}_{m}\right)}{f\left({x}_{m}\right)},\hfill \end{array}\right\\end{array}$
(2)

and $\mathit{\lambda },\phantom{\rule{0.166667em}{0ex}}\mathit{\mu },\phantom{\rule{0.166667em}{0ex}}\mathit{\nu }\in R$ and $G\left(t\right)$ 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 $f:\phantom{\rule{0.166667em}{0ex}}D\subset R\to R$ 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 $\left\{{x}_{m}\right\}$ generated by any method of the family (1.1) converges to ${x}^{\ast }$. If $G$ is any function with ${M}_{0}=G\left(0\right)=1$, ${M}_{1}={G}^{\prime }\left(0\right)=2$, ${M}_{2}={G}^{\prime \prime }\left(0\right)<\infty$ 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 ${c}_{k}=\frac{{f}^{\left(k\right)}\left({x}^{\ast }\right)}{k!{f}^{\prime }\left({x}^{\ast }\right)},\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}k=0,1,2,\dots$. Since $f\left({x}^{\ast }\right)=0\ne {f}^{\prime }\left({x}^{\ast }\right)$, we can write

$\begin{array}{cc}& f\left(x\right)={f}^{\prime }\left({x}^{\ast }\right)\sum _{k=1}^{8}{c}_{k}{e}^{k}+O\left({e}^{9}\right),\hfill \\ \hfill & {f}^{\prime }\left(x\right)={f}^{\prime }\left({x}^{\ast }\right)\sum _{k=1}^{7}k{c}_{k}{e}^{k-1}+O\left({e}^{8}\right).\hfill \end{array}$
(3)

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

$\begin{array}{cc}\hfill {e}_{y}& =e-\frac{f\left(x\right)}{{f}^{\prime }\left(x\right)}\hfill \\ \hfill & ={c}_{2}{e}^{2}+\left(2{c}_{3}-2{c}_{2}^{2}\right){e}^{3}+\left(4{c}_{2}^{3}-7{c}_{3}{c}_{2}+3{c}_{4}\right){e}^{4}\phantom{\rule{1.em}{0ex}}+\left(-8{c}_{2}^{4}+20{c}_{3}{c}_{2}^{2}-10{c}_{4}{c}_{2}-6{c}_{3}^{2}\right){e}^{5}+O\left({e}^{6}\right).\hfill \end{array}$
(4)

Set,

$t=\frac{f\left(y\right)}{f\left(x\right)}={c}_{2}e+\left(2{c}_{3}-3{c}_{2}^{2}\right){e}^{2}+\left(8{c}_{2}^{3}-10{c}_{3}{c}_{2}+3{c}_{4}\right){e}^{3}$
(5)
$\begin{array}{cc}& \phantom{\rule{1.em}{0ex}}+\left(-20{c}_{2}^{4}+37{c}_{3}{c}_{2}^{2}-14{c}_{4}{c}_{2}-8{c}_{3}^{2}\right){e}^{4}\hfill \\ \hfill & \phantom{\rule{1.em}{0ex}}+\left(48{c}_{2}^{5}-118{c}_{3}{c}_{2}^{3}+51{c}_{4}{c}_{2}^{2}+55{c}_{3}^{2}{c}_{2}-22{c}_{3}{c}_{4}\right){e}^{5}+O\left({e}^{6}\right).\hfill \end{array}$
(6)

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

$\begin{array}{c}\hfill G\left(t\right)={M}_{0}+{M}_{1}t+{M}_{2}t2.\end{array}$
(7)

Then,

$\begin{array}{cc}\hfill {e}_{z}=z-{x}^{\ast }& ={e}_{y}-G\left(t\right)\frac{f\left(y\right)}{{f}^{\prime }\left(x\right)}\hfill \\ \hfill & \phantom{\rule{1.em}{0ex}}-\left({c}_{2}\left({M}_{0}-1\right)\right){e}^{2}\phantom{\rule{1.em}{0ex}}+\left({c}_{2}^{2}\left(4{M}_{0}-{M}_{1}-2\right)-2{c}_{3}\left({M}_{0}-1\right)\right){e}^{3}\hfill \\ \hfill & \phantom{\rule{1.em}{0ex}}+\left({c}_{2}^{3}\left(-13{M}_{0}+7{M}_{1}-\frac{{M}_{2}}{2}+4\right)+{c}_{3}{c}_{2}\left(14{M}_{0}-4{M}_{1}-7\right)-3{c}_{4}\left({M}_{0}-1\right)\right){e}^{4}\hfill \\ \hfill & \phantom{\rule{1.em}{0ex}}+\left({c}_{3}{c}_{2}^{2}\left(-64{M}_{0}+38{M}_{1}-3{M}_{2}+20\right)+2{c}_{4}{c}_{2}\left(10{M}_{0}-3{M}_{1}-5\right)+2{c}_{3}^{2}\left(6{M}_{0}-2{M}_{1}-3\right)\hfill \\ \hfill & \phantom{\rule{1.em}{0ex}}+{c}_{2}^{4}\left(38{M}_{0}-33{M}_{1}+5{M}_{2}-8\right)\right){e}^{5}+O\left({e}^{6}\right).\hfill \end{array}$
(8)

If ${M}_{0}=1$ and ${M}_{1}=2$, then

$\begin{array}{c}\hfill {e}_{z}=\left({c}_{2}^{3}\left(5-\frac{{M}_{2}}{2}\right)-{c}_{2}{c}_{3}\right){e}^{4}+\left({c}_{2}^{4}\left(5{M}_{2}-36\right)+\phantom{\rule{0.166667em}{0ex}}{c}_{3}{c}_{2}^{2}\left(32-3{M}_{2}\right)-2{c}_{4}{c}_{2}-2{c}_{3}^{2}\right){e}^{5}+O\left({e}^{6}\right).\end{array}$
(9)

Now, assume that $H,\phantom{\rule{0.166667em}{0ex}}K\phantom{\rule{0.166667em}{0ex}}D$, and $C$ are given by (1.2). Also, let $q=\frac{f\left(z\right)}{f\left(x\right)}$ and $\mathit{\lambda }=\mathit{\mu }\ne 0$. Consequently, the final error equation, i.e., $\stackrel{^}{e}$, for the method (1.1) is obtained as follows

$\begin{array}{cc}\hfill \stackrel{^}{e}& ={e}_{z}-\frac{\mathit{\mu }}{\mathit{\lambda }+\mathit{\nu }{q}^{2}}\phantom{\rule{0.166667em}{0ex}}\frac{f\left(z\right)}{K-C\left(y-z\right)-D{\left(y-z\right)}^{2}}\hfill \\ \hfill & =\frac{1}{2}\left({c}_{2}^{5}\left({M}_{2}-10\right)+2{c}_{3}{c}_{2}^{3}\right){e}^{5}+O\left({e}^{6}\right),\hfill \end{array}$
(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 $G\left(t\right)=\frac{1}{1-2t+\mathit{\omega }{t}^{2}}$, where $\mathit{\omega }\in R$. This is the Method 1 in [1] (see Equations 21 and 22 in Section 5 there.)

$\begin{array}{c}\hfill \left\{\begin{array}{c}{y}_{m}={x}_{m}-\frac{f\left({x}_{m}\right)}{{f}^{\prime }\left({x}_{m}\right)},\hfill \\ {z}_{m}={y}_{m}-\frac{{f}^{2}\left({x}_{m}\right)}{{f}^{2}\left({x}_{m}\right)-2f\left({x}_{m}\right)f\left({y}_{m}\right)+\mathit{\omega }{f}^{2}\left({y}_{m}\right)}\phantom{\rule{0.166667em}{0ex}}\frac{f\left({y}_{m}\right)}{{f}^{\prime }\left({x}_{m}\right)},\hfill \\ {x}_{m+1}={z}_{m}-\frac{1}{1+\mathit{\nu }{q}_{m}^{2}}\phantom{\rule{0.166667em}{0ex}}\frac{f\left({z}_{m}\right)}{K-C\phantom{\rule{0.166667em}{0ex}}\left({y}_{m}-{z}_{m}\right)-D\phantom{\rule{0.166667em}{0ex}}{\left({y}_{m}-{z}_{m}\right)}^{2}},\hfill \end{array}\right\\end{array}$
(11)

where

$\begin{array}{c}\hfill \left\{\begin{array}{c}H=\frac{f\left({x}_{m}\right)-f\left({y}_{m}\right)}{{x}_{m}-{y}_{m}},\hfill \\ K=\frac{f\left({y}_{m}\right)-f\left({z}_{m}\right)}{{y}_{m}-{z}_{m}},\hfill \\ D=\frac{{f}^{\prime }\left({x}_{m}\right)-H}{\left({x}_{m}-{y}_{m}\right)\left({x}_{m}-{z}_{m}\right)}-\frac{H-K}{{\left({x}_{m}-{z}_{m}\right)}^{2}},\hfill \\ C=\frac{H-K}{\left({x}_{m}-{y}_{m}\right)\left({x}_{m}-{z}_{m}\right)}-D\phantom{\rule{0.166667em}{0ex}}\left({x}_{m}+{y}_{m}-2{z}_{m}\right),\hfill \\ {q}_{m}=\frac{f\left({z}_{m}\right)}{f\left({x}_{m}\right)}.\hfill \end{array}\right\\end{array}$
(12)

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

$\begin{array}{c}\hfill rho\approx \frac{ln\left(|{x}_{m+1}-{x}_{m}|\phantom{\rule{0.166667em}{0ex}}|{x}_{m}-{x}_{m-1}{|}^{-1}\right)}{ln\left(|{x}_{m}-{x}_{m-1}|\phantom{\rule{0.166667em}{0ex}}|{x}_{m-1}-{x}_{m-2}{|}^{-1}\right)}.\end{array}$

Among many test problems, the following four examples are considered

$\begin{array}{cc}\hfill {f}_{1}\left(x\right)& ={x}^{2}-9,\phantom{\rule{1.em}{0ex}}{x}^{\ast }=3,\phantom{\rule{0.166667em}{0ex}}{x}_{0}=2.6\hfill \\ \hfill {f}_{2}\left(x\right)& =\left(x-2\right)\left({x}^{6}+{x}^{3}+1\right){e}^{-{x}^{2}},\phantom{\rule{1.em}{0ex}}{x}^{\ast }=2,\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{x}_{0}=1.8,\hfill \\ \hfill {f}_{3}\left(x\right)& =\prod _{k=1}^{12}\left(x-k\right),\phantom{\rule{1.em}{0ex}}{x}^{\ast }=5,\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{x}_{0}=5.3.\hfill \\ \hfill \end{array}$

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.