Some New Quantum Numerical Techniques for Solving Nonlinear Equations

Abstract

In this paper, we introduce some new quantum numerical techniques of midpoint and trapezoidal type essentially by using the decomposition technique. We also check the order of convergence of our suggested iterative methods. Numerical examples demonstrate that the new $\mathrm{q}$-iterative methods can generate solutions to the nonlinear equations with acceptable accuracy.

1. Introduction

In most scientific and engineering applications, the problem of finding the solution to nonlinear equations has become an active area of research. Many researchers have explored various order iterative methods to find solutions to the nonlinear equations using various techniques such as variational iterative methods and decomposition techniques; for details, see [1,2,3,4,5,6,7,8,9,10,11,12]. Symmetry analysis is a significant tool in various areas of mathematics and physics. There are several problems in engineering and mathematical sciences that possess symmetry, which can be transformed into the nonlinear systems ${\mathrm{\Xi }}_1\left(\widetilde{⌀}\right)=0$. For details, see [13].

In the Adomian decomposition method, the solution is considered in terms of an infinite series, which converges towards an exact solution. Chun [2] and Abbasbandy [14] constructed and investigated different higher-order iterative methods by applying the decomposition technique of Adomian [15]. Darvishi and Barati [16] also applied the Adomian decomposition technique to develop Newton-type methods that are cubically convergent for the solution of the system of a nonlinear equation. Daftardar-Gejji and Jafari [17] have used different modifications of the Adomian decomposition method [15] and suggested a simple technique that does not need the derivative evaluation of the Adomian polynomial, which is the major advantage of using this technique over Adomian decomposition method.

Jackson [18] introduced the $\mathrm{q}$-Taylor’s formula. Then, Jing and Fan [19] derived $\mathrm{q}$-Taylor’s formula with its $\mathrm{q}$-remainder by using the q-differentiation approach and established results on the $\mathrm{q}$-remainder in the q-Taylor’s formula. Ernest presented the four different $\mathrm{q}$-Taylor’s formulas along with $\mathrm{q}$ integral remainder; see [20,21,22].

In this paper, we consider the well-known fixed point iterative method in which we rewrite the nonlinear equation ${\mathrm{\Xi }}_1(\widetilde{⌀})=0$ as $\widetilde{⌀}={\mathrm{\Xi }}_2(\widetilde{⌀}).$ We determine the convergence of our proposed methods. In order to illustrate the efficiency of these new methods, we present several numerical examples. We hope that the ideas and techniques of this paper will inspire interested readers working in this field.

2. Construction of q-Iterative Methods

In this section, some new different order multi-step $\mathrm{q}$-iterative methods are constructed by considering the mid point and Trapezoidal rule in the setting of $\mathrm{q}$-calculus and using the technique of decomposition [17].

2.1. Mid Point Rule

Consider the nonlinear equation

$${\mathrm{\Xi }}_1\left(\widetilde{⌀}\right)=0,$$

(1)

which is equivalent to

$$\widetilde{⌀}={\mathrm{\Xi }}_2\left(\widetilde{⌀}\right).$$

(2)

Assume that $\eta$ is the simple root of nonlinear Equation (1) and $\overline{⋋}$ is the initial guess sufficiently close to the root. Using the fundamental theorem of calculus and mid-point quadrature formula in the $\mathrm{q}$-calculus, we have

$$\widetilde{⌀}={\mathrm{\Xi }}_2\left(\overline{⋋}\right)+\left(\widetilde{⌀}-\overline{⋋}\right){\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{\widetilde{⌀}+\overline{⋋}}{2}\right).$$

(3)

Now, using the technique of He [23], the nonlinear Equation (1) can be written as an equivalent coupled system of equations

$$\widetilde{⌀}={\mathrm{\Xi }}_2\left(\overline{⋋}\right)+\left(\widetilde{⌀}-\overline{⋋}\right){\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{\widetilde{⌀}+\overline{⋋}}{2}\right)+H\left(\widetilde{⌀}\right),$$

where

$$\begin{array}{cc}\hfill H\left(\widetilde{⌀}\right)& ={\mathrm{\Xi }}_2\left(\widetilde{⌀}\right)-{\mathrm{\Xi }}_2\left(\overline{⋋}\right)-\left(\widetilde{⌀}-\overline{⋋}\right){\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{\widetilde{⌀}+\overline{⋋}}{2}\right)\hfill \\ \hfill & =\widetilde{⌀}\left(1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{\widetilde{⌀}+\overline{⋋}}{2}\right)\right)-{\mathrm{\Xi }}_2\left(\overline{⋋}\right)+\overline{⋋}{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{\widetilde{⌀}+\overline{⋋}}{2}\right),\hfill \end{array}$$

(4)

from which it follows that

$$\begin{array}{cc}\hfill \widetilde{⌀}& =\frac{H\left(\widetilde{⌀}\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{\widetilde{⌀}+\overline{⋋}}{2}\right)}+\frac{{\mathrm{\Xi }}_2\left(\overline{⋋}\right)-\overline{⋋}{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{\widetilde{⌀}+\overline{⋋}}{2}\right)}{1-{\mathrm{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{\widetilde{⌀}+\overline{⋋}}{2}\right)}\hfill \\ \hfill & =\theta +M_{\mathrm{q}}\left(\widetilde{⌀}\right),\hfill \end{array}$$

(5)

where

$$\theta =\overline{⋋},$$

(6)

and

$$M_{\mathrm{q}}\left(\widetilde{⌀}\right)=\frac{H\left(\widetilde{⌀}\right)}{1-{\mathrm{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{\widetilde{⌀}+\overline{⋋}}{2}\right)}+\frac{{\mathrm{\Xi }}_2\left(\overline{⋋}\right)-\overline{⋋}}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{\widetilde{⌀}+\overline{⋋}}{2}\right)}.$$

(7)

It is clear that $M_{\mathrm{q}}\left(\widetilde{⌀}\right)$ is a nonlinear operator. Now, we establish a sequence of higher-order iterative methods implementing the decomposition technique presented by Daftardar-Gejji and Jafari [17]. In this technique, the solution of (1) can be represented as in terms of infinite series.

$$\widetilde{⌀}=\sum _{i=0}^{\infty }{\widetilde{⌀}}_i.$$

(8)

Here, the operator $M_{\mathrm{q}}\left(\widetilde{⌀}\right)$ can be decomposed as:

$$M_{\mathrm{q}}\left(\widetilde{⌀}\right)=M_{\mathrm{q}}\left({\widetilde{⌀}}_{\circ }\right)+\sum _{i=1}^{\infty }\left\{M_{\mathrm{q}}\left(\sum _{j=0}^i{\widetilde{⌀}}_j\right)-M_{\mathrm{q}}\left(\sum _{j=0}^{i-1}{\widetilde{⌀}}_j\right)\right\}.$$

(9)

Thus from Equations (5), (8) and (9), we have

$$\sum _{i=1}^{\infty }{\widetilde{⌀}}_i=\theta +M_{\mathrm{q}}\left({\widetilde{⌀}}_{\circ }\right)+\sum _{i=1}^{\infty }\left\{M_{\mathrm{q}}\left(\sum _{j=0}^i{\widetilde{⌀}}_j\right)-M_{\mathrm{q}}\left(\sum _{j=0}^{i-1}{\widetilde{⌀}}_j\right)\right\},$$

(10)

which generates the following iterative scheme

$$\left\{\begin{array}{cc}\hfill {\widetilde{⌀}}_0& =\theta ,\hfill \\ \hfill {\widetilde{⌀}}_1& =M_{\mathrm{q}}\left({\widetilde{⌀}}_{\circ }\right),\hfill \\ \hfill {\widetilde{⌀}}_2& =M_{\mathrm{q}}\left({\widetilde{⌀}}_{\circ }+{\widetilde{⌀}}_1\right)-M_{\mathrm{q}}\left({\widetilde{⌀}}_{\circ }\right),\hfill \\ & ⋮\hfill \\ \hfill {\widetilde{⌀}}_{n+1}& =M_{\mathrm{q}}\left(\sum _{j=0}^i{\widetilde{⌀}}_j\right)-M_{\mathrm{q}}\left(\sum _{j=0}^{i-1}{\widetilde{⌀}}_j\right),n=1,2,\dots .\hfill \end{array}\right.$$

Consequently, it follows that

$${\widetilde{⌀}}_1+{\widetilde{⌀}}_2+\cdots +{\widetilde{⌀}}_{n+1}=M_{\mathrm{q}}\left({\widetilde{⌀}}_{\circ }+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2+\cdots +{\widetilde{⌀}}_n\right)$$

and

$$\widetilde{⌀}=\theta +\sum _{i=1}^{\infty }{\widetilde{⌀}}_i.$$

(11)

It is noted that $\widetilde{⌀}$ is approximated by

$$U_n={\widetilde{⌀}}_{\circ }+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2\cdots +{\widetilde{⌀}}_n$$

and

$$\underset{n\to \infty }{lim}{U}_n=\widetilde{⌀}.$$

For $n=0,$

$$\widetilde{⌀}\approx U_0={\widetilde{⌀}}_0=\theta =\overline{⋋}.$$

(12)

From (4), it can easily be computed

$$H\left({\widetilde{⌀}}_0\right)=0.$$

Using (7), we obtain

$$\begin{array}{cc}\hfill {\widetilde{⌀}}_1=M_{\mathrm{q}}\left({\widetilde{⌀}}_0\right)& =\frac{H\left({\widetilde{⌀}}_0\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{{\widetilde{⌀}}_0+\overline{⋋}}{2}\right)}+\frac{{\mathrm{\Xi }}_2\left(\overline{⋋}\right)-\overline{⋋}}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{{\widetilde{⌀}}_0+\overline{⋋}}{2}\right)}\hfill \\ \hfill & =\frac{{\mathrm{\Xi }}_2\left(\overline{⋋}\right)-\overline{⋋}}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{{\widetilde{⌀}}_0+\overline{⋋}}{2}\right)}.\hfill \end{array}$$

For $n=1,$

$$\begin{array}{cc}\hfill \widetilde{⌀}\approx U_1& ={\widetilde{⌀}}_0+{\widetilde{⌀}}_1={\widetilde{⌀}}_0+M_{\mathrm{q}}\left({\widetilde{⌀}}_0\right)\hfill \\ \hfill & =\overline{⋋}+\frac{{\mathrm{\Xi }}_2\left(\overline{⋋}\right)-\overline{⋋}}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{{\widetilde{⌀}}_0+\overline{⋋}}{2}\right)}.\hfill \end{array}$$

Using (12), we have

$$\widetilde{⌀}=\frac{{\mathrm{\Xi }}_2\left(\overline{⋋}\right)-\overline{⋋}{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}.$$

(13)

This fixed-point formulation is used to suggest the following Algorithms 1–4.

Algorithm 1 A new q-iterative scheme of second order convergence

For a given ${\widetilde{⌀}}_0$ (initial guess), an approximate solution ${\widetilde{⌀}}_{n+1}$ is computed by the following iterative scheme $${\widetilde{⌀}}_{n+1}=\frac{{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)-{\widetilde{⌀}}_n{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)},$$ (14) where $n=0,1,2,\dots .$

From (4) and (7), we have

$$H\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)={\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)-{\mathrm{\Xi }}_2\left(\overline{⋋}\right)-\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1-\overline{⋋}\right){\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{{\widetilde{⌀}}_0+{\widetilde{⌀}}_1+\overline{⋋}}{2}\right).$$

Thus,

$$\begin{array}{cc}\hfill {\widetilde{⌀}}_1+{\widetilde{⌀}}_2& =M_{\mathrm{q}}\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)=\frac{H\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{{\widetilde{⌀}}_0+{\widetilde{⌀}}_1+\overline{⋋}}{2}\right)}+\frac{{\mathrm{\Xi }}_2\left(\overline{⋋}\right)-\overline{⋋}}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{{\widetilde{⌀}}_0+{\widetilde{⌀}}_1+\overline{⋋}}{2}\right)}\hfill \\ \hfill & =\frac{{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)-{\mathrm{\Xi }}_2\left(\overline{⋋}\right)-\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1-\overline{⋋}\right){\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{{\widetilde{⌀}}_0+{\widetilde{⌀}}_1+\overline{⋋}}{2}\right)+{\mathrm{\Xi }}_2\left(\overline{⋋}\right)-\overline{⋋}}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{{\widetilde{⌀}}_0+{\widetilde{⌀}}_1+\overline{⋋}}{2}\right)}\hfill \\ \hfill & =\frac{{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)-\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1-\overline{⋋}\right){\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{{\widetilde{⌀}}_0+{\widetilde{⌀}}_1+\overline{⋋}}{2}\right)-\overline{⋋}}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{{\widetilde{⌀}}_0+{\widetilde{⌀}}_1+\overline{⋋}}{2}\right)}.\hfill \end{array}$$

For $n=2,$

$$\begin{array}{cc}\hfill \widetilde{⌀}\approx & U_2={\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2=\theta +M_{\mathrm{q}}\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)\hfill \\ \hfill & =\overline{⋋}+\frac{{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)-\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1-\overline{⋋}\right){\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{{\widetilde{⌀}}_0+{\widetilde{⌀}}_1+\overline{⋋}}{2}\right)-\overline{⋋}}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{{\widetilde{⌀}}_0+{\widetilde{⌀}}_1+\overline{⋋}}{2}\right)}\hfill \\ \hfill & =\frac{{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)-\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right){\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{{\widetilde{⌀}}_0+{\widetilde{⌀}}_1+\overline{⋋}}{2}\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{{\widetilde{⌀}}_0+{\widetilde{⌀}}_1+\overline{⋋}}{2}\right)}.\hfill \end{array}$$

Take,

$$\begin{array}{cc}\hfill {\widetilde{⌀}}_0+{\widetilde{⌀}}_1& =v=\frac{{\mathrm{\Xi }}_2\left(\overline{⋋}\right)-\overline{⋋}{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}\hfill \\ \hfill & =\frac{{\mathrm{\Xi }}_2\left(v\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{v+\overline{⋋}}{2}\right)}-\frac{v{D}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{v+\overline{⋋}}{2}\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{v+\overline{⋋}}{2}\right)}.\hfill \end{array}$$

This relation yields the following two-step method for solving nonlinear Equation (1).

Algorithm 2 A new q-iterative scheme of third order convergence

For a given initial guess ${\widetilde{⌀}}_0,$ the approximated solution ${\widetilde{⌀}}_{{}_{n+1}}$ can be computed by the following iterative schemes. $$v_n=\frac{{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)-{\widetilde{⌀}}_n{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)}$$ (15) and $${\widetilde{⌀}}_{n+1}=\frac{{\mathrm{\Xi }}_2\left(v_n\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{v_n+{\widetilde{⌀}}_n}{2}\right)}-\frac{v_n{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{v_n+{\widetilde{⌀}}_n}{2}\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{v_n+{\widetilde{⌀}}_n}{2}\right)},$$ (16) where $n=0,1,2,\dots .$

It is noted that

$${\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2=w=\frac{{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{{\widetilde{⌀}}_0+{\widetilde{⌀}}_1+\overline{⋋}}{2}\right)}-\frac{\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right){\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{{\widetilde{⌀}}_0+{\widetilde{⌀}}_1+\overline{⋋}}{2}\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{{\widetilde{⌀}}_0+{\widetilde{⌀}}_1+\overline{⋋}}{2}\right)}.$$

(17)

From (4) and (7), we can write

$$H\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2\right)={\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2\right)-{\mathrm{\Xi }}_2\left(\overline{⋋}\right)-\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2-\overline{⋋}\right){\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{{\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2+\overline{⋋}}{2}\right)$$

and

$$\begin{array}{cc}\hfill {\widetilde{⌀}}_1+{\widetilde{⌀}}_2+{\widetilde{⌀}}_3& =M_{\mathrm{q}}\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2\right)=\frac{H\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{{\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2+\overline{⋋}}{2}\right)}+\frac{{\mathrm{\Xi }}_2\left(\overline{⋋}\right)-\overline{⋋}}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{{\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2+\overline{⋋}}{2}\right)}\hfill \\ \hfill & =\frac{{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2\right)-{\mathrm{\Xi }}_2\left(\overline{⋋}\right)-\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2-\overline{⋋}\right){\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{{\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2+\overline{⋋}}{2}\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{{\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2+\overline{⋋}}{2}\right)}\hfill \\ \hfill & +\frac{{\mathrm{\Xi }}_2\left(\overline{⋋}\right)-\overline{⋋}}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{{\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2+\overline{⋋}}{2}\right)}\hfill \\ \hfill & =\frac{{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{{\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2+\overline{⋋}}{2}\right)}-\frac{\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2-\overline{⋋}\right){\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{{\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2+\overline{⋋}}{2}\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{{\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2+\overline{⋋}}{2}\right)}\hfill \\ \hfill & -\frac{\overline{⋋}}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{{\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2+\overline{⋋}}{2}\right)}.\hfill \end{array}$$

For $n=3,$

$$\begin{array}{cc}\hfill \widetilde{⌀}\approx U_3& ={\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2+{\widetilde{⌀}}_3={\widetilde{⌀}}_0+M_{\mathrm{q}}\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2\right).\hfill \\ \hfill & =\overline{⋋}-\frac{{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{{\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2+\overline{⋋}}{2}\right)}-\frac{\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2-\overline{⋋}\right){\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{{\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2+\overline{⋋}}{2}\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{{\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2+\overline{⋋}}{2}\right)}\hfill \\ \hfill & -\frac{\overline{⋋}}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{{\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2+\overline{⋋}}{2}\right)}.\hfill \end{array}$$

Using (17), we have

$$\begin{array}{cc}\hfill U_3& =\overline{⋋}-\frac{{\mathrm{\Xi }}_2\left(w\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{w+\overline{⋋}}{2}\right)}-\frac{\left(w-\overline{⋋}\right){\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{w+\overline{⋋}}{2}\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{w+\overline{⋋}}{2}\right)}-\frac{\overline{⋋}}{1-{\mathrm{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{w+\overline{⋋}}{2}\right)}\hfill \\ \hfill & =\frac{{\mathrm{\Xi }}_2\left(w\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{w+\overline{⋋}}{2}\right)}-\frac{w{D}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{w+\overline{⋋}}{2}\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{w+\overline{⋋}}{2}\right)}.\hfill \end{array}$$

Using this relation, we have the following three-step method for solving nonlinear Equation (1).

Algorithm 3 A new q-iterative scheme of fourth order convergence

For a given initial guess ${\widetilde{⌀}}_0$, compute the approximate solution ${\widetilde{⌀}}_{n+1}$ by the following iterative scheme $$v_n=\frac{{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)-{\widetilde{⌀}}_n{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)},$$ $$w_n=\frac{{\mathrm{\Xi }}_2\left(v_n\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{v_n+{\widetilde{⌀}}_n}{2}\right)}-\frac{v_n{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{v_n+{\widetilde{⌀}}_n}{2}\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{v_n+{\widetilde{⌀}}_n}{2}\right)}$$ (18) and $${\widetilde{⌀}}_{n+1}=\frac{{\mathrm{\Xi }}_2\left(w_n\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{w_n+{\widetilde{⌀}}_n}{2}\right)}-\frac{w_n{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{w_n+{\widetilde{⌀}}_n}{2}\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{w_n+{\widetilde{⌀}}_n}{2}\right)}.$$ (19) where $n=0,1,2,\dots .$

2.2. Trapezoidal Rule

Again using the technique of He [23] and fundamental law of calculus along with Trapezoidal rule in the $\mathrm{q}$-calculus, we can obtain

$$\widetilde{⌀}={\mathrm{\Xi }}_2\left(\overline{⋋}\right)+\left(\widetilde{⌀}-\overline{⋋}\right)\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\widetilde{⌀}\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)+H\left(\widetilde{⌀}\right)$$

and

$$\begin{array}{cc}\hfill H\left(\widetilde{⌀}\right)& ={\mathrm{\Xi }}_2\left(\widetilde{⌀}\right)-{\mathrm{\Xi }}_2\left(\overline{⋋}\right)-\left(\widetilde{⌀}-\overline{⋋}\right)\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\widetilde{⌀}\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)\hfill \\ \hfill & =\widetilde{⌀}\left(1-\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\widetilde{⌀}\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)-{\mathrm{\Xi }}_2\left(\overline{⋋}\right)+\overline{⋋}\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\widetilde{⌀}\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right),\hfill \end{array}$$

(20)

from which it follows that

$$\begin{array}{cc}\hfill \widetilde{⌀}& =\frac{H\left(\widetilde{⌀}\right)}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\widetilde{⌀}\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}+\frac{{\mathrm{\Xi }}_2\left(\overline{⋋}\right)-\overline{⋋}\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\widetilde{⌀}\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\widetilde{⌀}\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}\hfill \\ \hfill & =\theta +M_{\mathrm{q}}\left(\widetilde{⌀}\right),\hfill \end{array}$$

(21)

where

$$\theta =\overline{⋋}$$

(22)

and

$$M_{\mathrm{q}}\left(\widetilde{⌀}\right)=\frac{H\left(\widetilde{⌀}\right)}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\widetilde{⌀}\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}+\frac{{\mathrm{\Xi }}_2\left(\overline{⋋}\right)-\overline{⋋}}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\widetilde{⌀}\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}.$$

(23)

Now, applying the decomposition technique of Daftardar-Gejji and Jafari [17], we have

For $n=0,$

$$\widetilde{⌀}\approx U_0={\widetilde{⌀}}_0=\theta =\overline{⋋}.$$

(24)

From (20), it can easily be computed as

$$H\left({\widetilde{⌀}}_0\right)=0.$$

Using (23), we obtain

$$\begin{array}{cc}\hfill {\widetilde{⌀}}_1=M_{\mathrm{q}}\left({\widetilde{⌀}}_0\right)& =\frac{H\left({\widetilde{⌀}}_0\right)}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}+\frac{{\mathrm{\Xi }}_2\left(\overline{⋋}\right)-\overline{⋋}}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}\hfill \\ \hfill & =\frac{{\mathrm{\Xi }}_2\left(\overline{⋋}\right)-\overline{⋋}}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}.\hfill \end{array}$$

For $n=1,$

$$\begin{array}{cc}\hfill \widetilde{⌀}\approx U_1& ={\widetilde{⌀}}_0+{\widetilde{⌀}}_1={\widetilde{⌀}}_0+M_{\mathrm{q}}\left({\widetilde{⌀}}_0\right)\hfill \\ \hfill & =\overline{⋋}+\frac{{\mathrm{\Xi }}_2\left(\overline{⋋}\right)-\overline{⋋}}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}.\hfill \end{array}$$

(25)

Using (25), we have

$$\widetilde{⌀}=\frac{{\mathrm{\Xi }}_2\left(\overline{⋋}\right)-\overline{⋋}{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}.$$

(26)

This formulation determines the Algorithm 1.

From (20) and (23), we have

$$H\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)={\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)-{\mathrm{\Xi }}_2\left(\overline{⋋}\right)-\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1-\overline{⋋}\right)\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)$$

and

$$\begin{array}{cc}\hfill {\widetilde{⌀}}_1+{\widetilde{⌀}}_2& =M_{\mathrm{q}}\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)\hfill \\ \hfill & =\frac{H\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}+\frac{{\mathrm{\Xi }}_2\left(\overline{⋋}\right)-\overline{⋋}}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}\hfill \\ \hfill & =\frac{{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)-{\mathrm{\Xi }}_2\left(\overline{⋋}\right)-\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1-\overline{⋋}\right)\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}+\frac{{\mathrm{\Xi }}_2\left(\overline{⋋}\right)-\overline{⋋}}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}\hfill \\ \hfill & =\frac{{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}-\frac{\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1-\overline{⋋}\right)\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}-\frac{\overline{⋋}}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}.\hfill \end{array}$$

For $n=2,$

$$\begin{array}{cc}\hfill \widetilde{⌀}\approx U_2& ={\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2={\widetilde{⌀}}_0+M_{\mathrm{q}}\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)\hfill \\ \hfill & =\overline{⋋}+\frac{{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}-\frac{\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1-\overline{⋋}\right)\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}-\frac{\overline{⋋}}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}.\hfill \end{array}$$

Take

$$\begin{array}{cc}\hfill {\widetilde{⌀}}_0+{\widetilde{⌀}}_1=v& =\frac{{\mathrm{\Xi }}_2\left(\overline{⋋}\right)-\overline{⋋}{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}\hfill \\ \hfill & =\overline{⋋}+\frac{{\mathrm{\Xi }}_2\left(v\right)}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(v\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}-\frac{\left(v-\overline{⋋}\right)\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(v\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(v\right)+{\mathrm{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}-\frac{\overline{⋋}}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(v\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}\hfill \\ \hfill & =\frac{{\mathrm{\Xi }}_2\left(v\right)}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(v\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}-\frac{v\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(v\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(v\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}.\hfill \end{array}$$

This relation yields the following two-step method for solving nonlinear Equation (1).

Algorithm 4 A new q-iterative scheme of third order convergence

For a given initial guess ${\widetilde{⌀}}_0$, the approximate solution ${\widetilde{⌀}}_{n+1}$ can be computed by the following iterative scheme $$v_n=\frac{{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)-{\widetilde{⌀}}_n{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)}$$ (27) and $${\widetilde{⌀}}_{n+1}=\frac{{\mathrm{\Xi }}_2\left(v_n\right)}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(v_n\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)}{2}\right)}-\frac{v_n\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(v_n\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)}{2}\right)}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(v_n\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)}{2}\right)}.n=0,1,2,\dots .$$ (28)

It is noted that

$${\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2=w=\frac{{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}-\frac{v\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}.$$

(29)

From (20) and (23), we can write

$$H\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2\right)={\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2\right)-{\mathrm{\Xi }}_2\left(\overline{⋋}\right)-\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2-\overline{⋋}\right)\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)$$

and

$$\begin{array}{cc}\hfill {\widetilde{⌀}}_1+{\widetilde{⌀}}_2+{\widetilde{⌀}}_3& =M_{\mathrm{q}}\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2\right)\hfill \\ \hfill & =\frac{H\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2\right)}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}+\frac{{\mathrm{\Xi }}_2\left(\overline{⋋}\right)-\overline{⋋}}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}\hfill \\ \hfill & =\frac{{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2\right)-{\mathrm{\Xi }}_2\left(\overline{⋋}\right)-\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2-\overline{⋋}\right)\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}\hfill \\ \hfill & +\frac{{\mathrm{\Xi }}_2\left(\overline{⋋}\right)-\overline{⋋}}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}\hfill \\ \hfill & =\frac{{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2\right)}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}-\frac{\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2-\overline{⋋}\right)\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}\hfill \\ \hfill & -\frac{\overline{⋋}}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}.\hfill \end{array}$$

For $n=3,$

$$\begin{array}{cc}\hfill \widetilde{⌀}\approx U_3& ={\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2+{\widetilde{⌀}}_3={\widetilde{⌀}}_0+M_{\mathrm{q}}\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2\right)\hfill \\ \hfill & =\overline{⋋}+\frac{{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2\right)}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}-\frac{\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2-\overline{⋋}\right)\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}\hfill \\ \hfill & -\frac{\overline{⋋}}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_0+{\widetilde{⌀}}_1+{\widetilde{⌀}}_2\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}.\hfill \end{array}$$

Using (28), we have

$$\begin{array}{cc}\hfill U_3& =\overline{⋋}+\frac{{\mathrm{\Xi }}_2\left(w\right)}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(w\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}-\frac{\left(w-\overline{⋋}\right)\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(w\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(w\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}-\frac{\overline{⋋}}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(w\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}\hfill \\ \hfill & =\frac{{\mathrm{\Xi }}_2\left(w\right)}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(w\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}-\frac{w\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(w\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(w\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\overline{⋋}\right)}{2}\right)}.\hfill \end{array}$$

This formulation yields the following three-step method for solving nonlinear Equation (1).

3. Order of Convergence

This section comprises the convergence analysis of the $\mathrm{q}$-iterative methods determined by Algorithms 1–5 in the previous section.

In the following theorem, we have found the order of convergence of the Algorithm 1, and it is a quadratic order of convergence.

Theorem 1.

Let $I\subset \mathbb{R}$ be an open interval, and ${\mathrm{\Xi }}_1:I\to \mathbb{R}$ is differential function. If $\beta \in I$ be a simple root of ${\mathrm{\Xi }}_1\left(\widetilde{⌀}\right)=0$ and ${\widetilde{⌀}}_0$ is sufficiently close to β, then multi-step method defined by Algorithm 1 has quadratic convergence.

Proof. 

Let $\beta$ be the root of nonlinear equation ${\mathrm{\Xi }}_1\left(\widetilde{⌀}\right)=0$, or equivalently $\widetilde{⌀}={\mathrm{\Xi }}_2\left(\widetilde{⌀}\right)$. Let $e_n$ and $e_{n+1}$ be the errors at $nth$ and $\left(n+1\right)$ iterations, respectively.

Algorithm 5 A new q-iterative scheme of fourth order convergence

For a given initial guess ${\widetilde{⌀}}_0$, the approximated solution ${\widetilde{⌀}}_{n+1}$ is computed by the following iterative schemes. $$v_n=\frac{{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)-{\widetilde{⌀}}_n{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)},$$ $$w_n=\frac{{\mathrm{\Xi }}_2\left(v_n\right)}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(v_n\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)}{2}\right)}-\frac{v_n\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(v_n\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)}{2}\right)}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(v_n\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)}{2}\right)},$$ (30) and $${\widetilde{⌀}}_{n+1}=\frac{{\mathrm{\Xi }}_2\left(w_n\right)}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(w_n\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)}{2}\right)}-\frac{w_n\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(w_n\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)}{2}\right)}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(w_n\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)}{2}\right)}.n=0,1,2,\dots .$$ (31)

Now, expanding ${\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)$ and ${\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)$ in terms of $\mathrm{q}$-Taylor’s series about $\beta$, to obtain

$$\begin{array}{cc}\hfill {\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)& ={\mathfrak{D}}_{\mathrm{q}}^0{\mathrm{\Xi }}_2\left(\beta \right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\beta \right)e_n+\frac{1}{\left[2\right]!}{\mathfrak{D}}_{\mathrm{q}}^2{\mathrm{\Xi }}_2\left(\beta \right)e_n^2+\frac{1}{\left[3\right]!}{\mathfrak{D}}_{\mathrm{q}}^3{\mathrm{\Xi }}_2\left(\beta \right)e_n^3+\frac{1}{\left[4\right]!}{\mathfrak{D}}_{\mathrm{q}}^4{\mathrm{\Xi }}_2\left(\beta \right)e_n^4+O\left(e_n^5\right)\hfill \\ \hfill & =\beta +{\theta }_1{e}_n+{\theta }_2{e}_n^2+{\theta }_3{e}_n^3+{\theta }_4{e}_n^4+O\left(e_n^5\right).\hfill \end{array}$$

(32)

and

$$\begin{array}{cc}\hfill {\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)& ={\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\beta \right)+{\mathfrak{D}}_{\mathrm{q}}^2{\mathrm{\Xi }}_2\left(\beta \right)e_n+\frac{1}{\left[2\right]!}{\mathfrak{D}}_{\mathrm{q}}^3{\mathrm{\Xi }}_2\left(\beta \right)e_n^2+\frac{1}{\left[3\right]!}{\mathfrak{D}}_{\mathrm{q}}^4{\mathrm{\Xi }}_2\left(\beta \right)e_n^3+\frac{1}{\left[4\right]!}{\mathfrak{D}}_{\mathrm{q}}^5{\mathrm{\Xi }}_2\left(\beta \right)e_n^4+O\left(e_n^5\right)\hfill \\ \hfill & ={\theta }_1+2{\theta }_2{e}_n+3{\theta }_3{e}_n^2+4{\theta }_4{e}_n^3+5{\theta }_5{e}_n^4+O\left(e_n^5\right),\hfill \end{array}$$

(33)

where

$${\theta }_m=\frac{{\mathfrak{D}}_{\mathrm{q}}^m{\mathrm{\Xi }}_2\left(\beta \right)}{\left[m\right]!}\mathrm{for}m=1,2,3,\dots$$

and

$$e_n={\widetilde{⌀}}_n-\beta$$

$$\begin{array}{cc}\hfill {\widetilde{⌀}}_n{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)& =\beta {\theta }_1+2\beta {\theta }_2{e}_n+3\beta {\theta }_3{e}_n^2+4\beta {\theta }_4{e}_n^3+5\beta {\theta }_5{e}_n^4\hfill \\ \hfill & +{\theta }_1{e}_n+2{\theta }_2{e}_n^2+3{\theta }_3{e}_n^3+4{\theta }_4{e}_n^4+O\left(e_n^5\right).\hfill \end{array}$$

(34)

Subtracting (32) from (34), we have

$$\begin{array}{cc}\hfill {\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)-{\widetilde{⌀}}_n{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)& =\beta -\beta {\theta }_1-2\beta {\theta }_2{e}_n-3\beta {\theta }_3{e}_n^2-4\beta {\theta }_4{e}_n^3-5\beta {\theta }_5{e}_n^4\hfill \\ \hfill & -{\theta }_2{e}_n^2-2{\theta }_3{e}_n^3-3{\theta }_4{e}_n^4+O\left(e_n^5\right).\hfill \end{array}$$

(35)

and

$$1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)=1-{\theta }_1-2{\theta }_2{e}_n-3{\theta }_3{e}_n^2-4{\theta }_4{e}_n^3-5{\theta }_5{e}_n^4+O\left(e_n^5\right).$$

(36)

Dividing (35) and (36), we have

$$\begin{array}{cc}\hfill \frac{{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)-{\widetilde{⌀}}_n{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)}& =\left[\beta -\beta {\theta }_1-2\beta {\theta }_2{e}_n-3\beta {\theta }_3{e}_n^2-4\beta {\theta }_4{e}_n^3-5\beta {\theta }_5{e}_n^4-{\theta }_2{e}_n^2-2{\theta }_3{e}_n^3\right.\hfill \\ \hfill & \left.-3{\theta }_4{e}_n^4+O\left(e_n^5\right)\right]{\left[1-\left({\theta }_1+2{\theta }_2{e}_n+3{\theta }_3{e}_n^2+4{\theta }_4{e}_n^3+O\left(e_n^4\right)\right)\right]}^{-1}.\hfill \end{array}$$

Using (14), we have

$$\begin{array}{cc}\hfill {\widetilde{⌀}}_{n+1}& =\left[\beta -\beta {\theta }_1-2\beta {\theta }_2{e}_n-3\beta {\theta }_3{e}_n^2-4\beta {\theta }_4{e}_n^3-5\beta {\theta }_5{e}_n^4-{\theta }_2{e}_n^2\right.\hfill \\ \hfill & \left.-2{\theta }_3{e}_n^3-3{\theta }_4{e}_n^4+O\left(e_n^5\right)\right]\left[\frac{1}{\left(1-{\theta }_1\right)}+\frac{1}{{\left(1-{\theta }_1\right)}^2}2{\theta }_2{e}_n\right.\hfill \\ \hfill & \left.+\frac{1}{{\left(1-{\theta }_1\right)}^3}\left(3{\theta }_3-3{\theta }_1{\theta }_3+4{\theta }_2^2\right)e_n^2+O\left(e_n^3\right)\right]\hfill \\ \hfill & =\beta +\frac{1}{\left(-1+{\theta }_1\right)}{\theta }_2{e}_n^2+\frac{1}{{\left(-1+{\theta }_1\right)}^2}\left(2{\theta }_1{\theta }_3-2{\theta }_3-2{\theta }_2^2\right)e_n^3\hfill \\ \hfill & +\frac{1}{{\left(-1+{\theta }_1\right)}^3}\left(3{\theta }_4+4{\theta }_2^3-6{\theta }_1{\theta }_4+3{\theta }_1^2{\theta }_4+7{\theta }_2{\theta }_3-7{\theta }_1{\theta }_2{\theta }_3\right)e_n^4+O\left(e_n^5\right).\hfill \end{array}$$

(37)

Hence

$$e_{n+1}=\frac{1}{\left(-1+{\theta }_1\right)}{\theta }_2{e}_n^2+O\left(e_n^3\right).$$

□

In the following theorem, we have found the order of convergence of the Algorithm 2, and it is the third order of convergence.

Theorem 2.

Let $I\subset \mathbb{R}$ be an open interval, and ${\mathrm{\Xi }}_1:I\to \mathbb{R}$ is differential function. If $\beta \in I$ be a simple root of ${\mathrm{\Xi }}_1\left(\widetilde{⌀}\right)=0$ and ${\widetilde{⌀}}_0$ is sufficiently close to β, then multi-step method defined by Algorithm 2 has a third order of convergence.

Proof. 

From (37), we obtain

$$\begin{array}{cc}\hfill {\widetilde{⌀}}_{n+1}& =\beta +\frac{1}{\left(-1+{\theta }_1\right)}{\theta }_2{e}_n^2+\frac{1}{{\left(-1+{\theta }_1\right)}^2}\left(2{\theta }_1{\theta }_3-2{\theta }_3-2{\theta }_2^2\right)e_n^3\hfill \\ \hfill & +\frac{1}{{\left(-1+{\theta }_1\right)}^3}\left(3{\theta }_4+4{\theta }_2^3-6{\theta }_1{\theta }_4+3{\theta }_1^2{\theta }_4+7{\theta }_2{\theta }_3-7{\theta }_1{\theta }_2{\theta }_3\right)e_n^4+O\left(e_n^5\right).\hfill \end{array}$$

Using (15), we have

$$\begin{array}{cc}\hfill v_n& =\beta +\frac{1}{\left(-1+{\theta }_1\right)}{\theta }_2{e}_n^2+\frac{1}{{\left(-1+{\theta }_1\right)}^2}\left(2{\theta }_1{\theta }_3-2{\theta }_3-2{\theta }_2^2\right)e_n^3\hfill \\ \hfill & +\frac{1}{{\left(-1+{\theta }_1\right)}^3}\left(3{\theta }_4+4{\theta }_2^3-6{\theta }_1{\theta }_4+3{\theta }_1^2{\theta }_4+7{\theta }_2{\theta }_3-7{\theta }_1{\theta }_2{\theta }_3\right)e_n^4+O\left(e_n^5\right).\hfill \end{array}$$

(38)

Expanding ${\mathrm{\Xi }}_2\left(v_n\right)$ in terms of $\mathrm{q}$-Taylor’s series about $\beta ,$ we obtain

$$\begin{array}{cc}\hfill {\mathrm{\Xi }}_2\left(v_n\right)& =\beta +\frac{1}{\left(-1+{\theta }_1\right)}{\theta }_1{\theta }_2{e}_n^2+\frac{1}{{\left(-1+{\theta }_1\right)}^2}{\theta }_1\left(2{\theta }_1{\theta }_3-2{\theta }_3-2{\theta }_2^2\right)e_n^3\hfill \\ \hfill & +\frac{1}{{\left(-1+{\theta }_1\right)}^3}{\theta }_1\left(3{\theta }_4+4{\theta }_2^3-6{\theta }_1{\theta }_4+3{\theta }_1^2{\theta }_4+7{\theta }_2{\theta }_3-7{\theta }_1{\theta }_2{\theta }_3\right)e_n^4+O\left(e_n^5\right).\hfill \end{array}$$

(39)

Expanding ${\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{v_{n+{\widetilde{⌀}}_n}}{2}\right)$ in terms of $\mathrm{q}$-Taylor’s series about $\beta ,$ we obtain

$$\begin{array}{cc}\hfill {\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{v_{n+{\widetilde{⌀}}_n}}{2}\right)& ={\theta }_1+{\theta }_2{e}_n+\frac{1}{\left(-1+{\theta }_1\right)}2{\theta }_2^2{e}_n^2+\frac{1}{{\left(-1+{\theta }_1\right)}^2}2{\theta }_2\left(2{\theta }_1{\theta }_3-2{\theta }_3-2{\theta }_2^2\right)e_n^3\hfill \\ \hfill & +\frac{1}{{\left(-1+{\theta }_1\right)}^3}2{\theta }_2\left(3{\theta }_4+4{\theta }_2^3-6{\theta }_1{\theta }_4+3{\theta }_1^2{\theta }_4+7{\theta }_2{\theta }_3-7{\theta }_1{\theta }_2{\theta }_3\right)e_n^4+O\left(e_n^5\right).\hfill \end{array}$$

(40)

and

$$\begin{array}{cc}\hfill v_n{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{v_{n+{\widetilde{⌀}}_n}}{2}\right)& =\beta {\theta }_1+\beta {\theta }_2{e}_n+\frac{1}{\left(-1+{\theta }_1\right)}2\beta {\theta }_2^2{e}_n^2+\frac{1}{{\left(-1+{\theta }_1\right)}^2}2\beta {\theta }_2\left(2{\theta }_1{\theta }_3-2{\theta }_3-2{\theta }_2^2\right)e_n^3\hfill \\ \hfill & +\frac{1}{{\left(-1+{\theta }_1\right)}^3}2\beta {\theta }_2\left(3{\theta }_4+4{\theta }_2^3-6{\theta }_1{\theta }_4+3{\theta }_1^2{\theta }_4+7{\theta }_2{\theta }_3-7{\theta }_1{\theta }_2{\theta }_3\right)e_n^4\hfill \\ \hfill & +\frac{1}{\left(-1+{\theta }_1\right)}{\theta }_1{\theta }_2{e}_n^2+\frac{1}{{\left(-1+{\theta }_1\right)}^2}{\theta }_1\left(2{\theta }_1{\theta }_3-2{\theta }_3-2{\theta }_2^2\right)e_n^3\hfill \\ \hfill & +\frac{1}{{\left(-1+{\theta }_1\right)}^3}{\theta }_1\left(3{\theta }_4+4{\theta }_2^3-6{\theta }_1{\theta }_4+3{\theta }_1^2{\theta }_4+7{\theta }_2{\theta }_3-7{\theta }_1{\theta }_2{\theta }_3\right)e_n^4\hfill \\ \hfill & +\frac{1}{\left(-1+{\theta }_1\right)}{\theta }_2^2{e}_n^3+\frac{1}{{\left(-1+{\theta }_1\right)}^2}{\theta }_2\left(2{\theta }_1{\theta }_3-2{\theta }_3-2{\theta }_2^2\right)e_n^4\hfill \\ \hfill & +\frac{1}{{\left(-1+{\theta }_1\right)}^2}2{\theta }_2^3{e}_n^4+O\left(e_n^5\right).\hfill \end{array}$$

(41)

Subtracting (39) from (41), we have

$$\begin{array}{cc}\hfill {\mathrm{\Xi }}_2\left(v_n\right)-v_n{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{v_{n+{\widetilde{⌀}}_n}}{2}\right)& =-\beta \left(-1+{\theta }_1\right)-\beta {\theta }_2{e}_n+\frac{1}{\left(-1+{\theta }_1\right)}2\beta {\theta }_2^2{e}_n^2\hfill \\ \hfill & -\frac{1}{{\left(-1+{\theta }_1\right)}^2}2\beta {\theta }_2\left(2{\theta }_1{\theta }_3-2{\theta }_3-2{\theta }_2^2\right)e_n^3\hfill \\ \hfill & -\frac{1}{{\left(-1+{\theta }_1\right)}^3}2\beta {\theta }_2\left(3{\theta }_4+4{\theta }_2^3-6{\theta }_1{\theta }_4+3{\theta }_1^2{\theta }_4+7{\theta }_2{\theta }_3-7{\theta }_1{\theta }_2{\theta }_3\right)e_n^4\hfill \\ \hfill & -\frac{1}{\left(-1+{\theta }_1\right)}{\theta }_2^2{e}_n^3-\frac{1}{{\left(-1+{\theta }_1\right)}^2}{\theta }_2\left(2{\theta }_1{\theta }_3-2{\theta }_3-2{\theta }_2^2\right)e_n^4\hfill \end{array}$$

(42)

$$\begin{array}{cc}\hfill & -\frac{1}{{\left(-1+{\theta }_1\right)}^2}2{\theta }_2^3{e}_n^4+O\left(e_n^5\right).\hfill \end{array}$$

(43)

Now,

$$\begin{array}{cc}\hfill 1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{v_{n+{\widetilde{⌀}}_n}}{2}\right)& =1-{\theta }_1-{\theta }_2{e}_n-\frac{1}{\left(-1+{\theta }_1\right)}2{\theta }_2^2{e}_n^2-\frac{1}{{\left(-1+{\theta }_1\right)}^2}2{\theta }_2\left(2{\theta }_1{\theta }_3-2{\theta }_3-2{\theta }_2^2\right)e_n^3\hfill \\ \hfill & -\frac{1}{{\left(-1+{\theta }_1\right)}^3}2{\theta }_2\left(3{\theta }_4+4{\theta }_2^3-6{\theta }_1{\theta }_4+3{\theta }_1^2{\theta }_4+7{\theta }_2{\theta }_3-7{\theta }_1{\theta }_2{\theta }_3\right)e_n^4+O\left(e_n^5\right).\hfill \end{array}$$

(44)

Dividing (43) and (44), we obtain

$$\begin{array}{cc}\hfill \frac{{\mathrm{\Xi }}_2\left(v_n\right)-v_n{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{v_{n+{\widetilde{⌀}}_n}}{2}\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{v_{n+{\widetilde{⌀}}_n}}{2}\right)}& =\beta +\frac{1}{{\left(-1+{\theta }_1\right)}^2}{\theta }_2^2{e}_n^3-\frac{1}{{\left(-1+{\theta }_1\right)}^3}{\theta }_2^3{e}_n^4+\frac{1}{{\left(-1+{\theta }_1\right)}^3}{\theta }_2\left(2{\theta }_1{\theta }_3-2{\theta }_3-2{\theta }_2^2\right)e_n^4\hfill \\ \hfill & +\frac{1}{{\left(-1+{\theta }_1\right)}^3}2{\theta }_2^3{e}_n^4+O\left(e_n^5\right).\hfill \end{array}$$

From (16), we obtain

$$\begin{array}{cc}\hfill {\widetilde{⌀}}_{n+1}& =\beta +\frac{1}{{\left(-1+{\theta }_1\right)}^2}{\theta }_2^2{e}_n^3-\frac{1}{{\left(-1+{\theta }_1\right)}^3}{\theta }_2^3{e}_n^4+\frac{1}{{\left(-1+{\theta }_1\right)}^3}{\theta }_2\left(2{\theta }_1{\theta }_3-2{\theta }_3-2{\theta }_2^2\right)e_n^4\hfill \\ \hfill & +\frac{1}{{\left(-1+{\theta }_1\right)}^3}2{\theta }_2^3{e}_n^4+O\left(e_n^5\right).\hfill \end{array}$$

(45)

Therefore,

$$e_{n+1}=\frac{1}{{\left(-1+{\theta }_1\right)}^2}{\theta }_2^2{e}_n^3+O\left(e_n^4\right).$$

□

In the following theorem, we have found the order of convergence of the Algorithm 3, and it is the fourth order of convergence.

Theorem 3.

Let $I\subset \mathbb{R}$ be an open interval, and ${\mathrm{\Xi }}_1:I\to \mathbb{R}$ is differential function. If $\beta \in I$ be a simple root of ${\mathrm{\Xi }}_1\left(\widetilde{⌀}\right)=0$ and ${\widetilde{⌀}}_0$ is sufficiently close to β, then the multi-step method defined by Algorithm 3 has the fourth order of convergence.

Proof. 

From (38), we obtain

$$\begin{array}{cc}\hfill v_n& =\beta +\frac{1}{\left(-1+{\theta }_1\right)}{\theta }_2{e}_n^2+\frac{1}{{\left(-1+{\theta }_1\right)}^2}\left(2{\theta }_1{\theta }_3-2{\theta }_3-2{\theta }_2^2\right)e_n^3\hfill \\ \hfill & +\frac{1}{{\left(-1+{\theta }_1\right)}^3}\left(3{\theta }_4+4{\theta }_2^3-6{\theta }_1{\theta }_4+3{\theta }_1^2{\theta }_4+7{\theta }_2{\theta }_3-7{\theta }_1{\theta }_2{\theta }_3\right)e_n^4+O\left(e_n^5\right).\hfill \end{array}$$

From (45), we obtain

$$\begin{array}{cc}\hfill w_n& =\beta +\frac{1}{{\left(-1+{\theta }_1\right)}^2}{\theta }_2^2{e}_n^3-\frac{1}{{\left(-1+{\theta }_1\right)}^3}{\theta }_2^3{e}_n^4\hfill \\ \hfill & +\frac{1}{{\left(-1+{\theta }_1\right)}^3}{\theta }_2\left(2{\theta }_1{\theta }_3-2{\theta }_3-2{\theta }_2^2\right)e_n^4+\frac{1}{{\left(-1+{\theta }_1\right)}^3}2{\theta }_2^3{e}_n^4+O\left(e_n^5\right)\hfill \\ \hfill & =\beta +\frac{1}{{\left(-1+{\theta }_1\right)}^2}{\theta }_2^2{e}_n^3+\frac{1}{{\left(-1+{\theta }_1\right)}^3}\left(2{\theta }_1{\theta }_2{\theta }_3-2{\theta }_2{\theta }_3-{\theta }_2^3\right)e_n^4+O\left(e_n^5\right).\hfill \end{array}$$

Expanding ${\mathrm{\Xi }}_2\left(w_n\right)$ in terms of $\mathrm{q}$-Taylor’s series about $\beta ,$ we have

$${\mathrm{\Xi }}_2\left(w_n\right)=\beta +\frac{1}{{\left(-1+{\theta }_1\right)}^2}{\theta }_1{\theta }_2^2{e}_n^3+\frac{1}{{\left(-1+{\theta }_1\right)}^3}{\theta }_1\left(2{\theta }_1{\theta }_2{\theta }_3-2{\theta }_2{\theta }_3-{\theta }_2^3\right)e_n^4+O\left(e_n^5\right).$$

(46)

Expanding ${\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{w_n+{\widetilde{⌀}}_n}{2}\right)$ in terms of $\mathrm{q}$-Taylor’s series about $\beta ,$ we have

$${\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{w_n+{\widetilde{⌀}}_n}{2}\right)={\theta }_1+{\theta }_2{e}_n+\frac{1}{{\left(-1+{\theta }_1\right)}^2}2{\theta }_2^3{e}_n^3+\frac{1}{{\left(-1+{\theta }_1\right)}^3}2{\theta }_2\left(2{\theta }_1{\theta }_2{\theta }_3-2{\theta }_2{\theta }_3-{\theta }_2^3\right)e_n^4+O\left(e_n^5\right).$$

(47)

Now,

$$\begin{array}{cc}\hfill w_n{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{w_n+{\widetilde{⌀}}_n}{2}\right)& =\beta {\theta }_1+\beta {\theta }_2{e}_n+\frac{1}{{\left(-1+{\theta }_1\right)}^2}2\beta {\theta }_2^3{e}_n^3+\frac{1}{{\left(-1+{\theta }_1\right)}^3}2\beta {\theta }_2\left(2{\theta }_1{\theta }_2{\theta }_3-2{\theta }_2{\theta }_3-{\theta }_2^3\right)e_n^4\hfill \\ \hfill & +\frac{1}{{\left(-1+{\theta }_1\right)}^2}{\theta }_1{\theta }_2^2{e}_n^3+\frac{1}{{\left(-1+{\theta }_1\right)}^3}{\theta }_1\left(2{\theta }_1{\theta }_2{\theta }_3-2{\theta }_2{\theta }_3-{\theta }_2^3\right)e_n^4\hfill \\ \hfill & +\frac{1}{{\left(-1+{\theta }_1\right)}^2}{\theta }_2^3{e}_n^4+O\left(e_n^5\right).\hfill \end{array}$$

(48)

Subtracting (46) and (48), we obtain

$$\begin{array}{cc}\hfill {\mathrm{\Xi }}_2\left(w_n\right)-w_n{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{w_n+{\widetilde{⌀}}_n}{2}\right)& =-\beta \left(-1+{\theta }_1\right)-\beta {\theta }_2{e}_n-\frac{1}{{\left(-1+{\theta }_1\right)}^2}2\beta {\theta }_2^3{e}_n^3-\frac{1}{{\left(-1+{\theta }_1\right)}^3}2\beta {\theta }_2\left(2{\theta }_1{\theta }_2{\theta }_3\right.\hfill \\ \hfill & \left.-2{\theta }_2{\theta }_3-{\theta }_2^3\right)e_n^4-\frac{1}{{\left(-1+{\theta }_1\right)}^2}{\theta }_2^3{e}_n^4+O\left(e_n^5\right).\hfill \end{array}$$

(49)

Now,

$$\begin{array}{cc}\hfill 1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{w_n+{\widetilde{⌀}}_n}{2}\right)& =1-{\theta }_1-{\theta }_2{e}_n-\frac{1}{{\left(-1+{\theta }_1\right)}^2}2{\theta }_2^3{e}_n^3\hfill \\ \hfill & -\frac{1}{{\left(-1+{\theta }_1\right)}^3}2{\theta }_2\left(2{\theta }_1{\theta }_2{\theta }_3-2{\theta }_2{\theta }_3-{\theta }_2^3\right)e_n^4+O\left(e_n^5\right).\hfill \end{array}$$

(50)

Dividing (49) and (50), we obtain

$$\frac{{\mathrm{\Xi }}_2\left(w_n\right)-w_n{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{w_n+{\widetilde{⌀}}_n}{2}\right)}{1-{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(\frac{w_n+{\widetilde{⌀}}_n}{2}\right)}=\beta +\frac{1}{{\left(-1+{\theta }_1\right)}^3}{\theta }_2^3{e}_n^4+O\left(e_n^5\right).$$

Using (19), we obtain

$${\widetilde{⌀}}_{n+1}=\beta +\frac{1}{{\left(-1+{\theta }_1\right)}^3}{\theta }_2^3{e}_n^4+O\left(e_n^5\right).$$

Therefore,

$$e_{n+1}=\frac{1}{{\left(-1+{\theta }_1\right)}^3}{\theta }_2^3{e}_n^4+O\left(e_n^5\right).$$

□

In the following theorem, we have found the order of convergence of the Algorithm 4 and it is third order of convergence.

Theorem 4.

Let $I\subset \mathbb{R}$ be an open interval, and ${\mathrm{\Xi }}_1:I\to \mathbb{R}$ is differential function. If $\beta \in I$ be a simple root of ${\mathrm{\Xi }}_1\left(\widetilde{⌀}\right)=0$ and ${\widetilde{⌀}}_0$ is sufficiently close to β, then multi-step method defined by Algorithm 4 has the third order of convergence.

Proof. 

From (38), we obtain

$$\begin{array}{cc}\hfill v_n& =\beta +\frac{1}{\left(-1+{\theta }_1\right)}{\theta }_2{e}_n^2+\frac{1}{{\left(-1+{\theta }_1\right)}^2}\left(2{\theta }_1{\theta }_3-2{\theta }_3-2{\theta }_2^2\right)e_n^3\hfill \\ \hfill & +\frac{1}{{\left(-1+{\theta }_1\right)}^3}\left(3{\theta }_4+4{\theta }_2^3-6{\theta }_1{\theta }_4+3{\theta }_1^2{\theta }_4+7{\theta }_2{\theta }_3-7{\theta }_1{\theta }_2{\theta }_3\right)e_n^4+O\left(e_n^5\right).\hfill \end{array}$$

Expanding ${\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(v_n\right),$ in terms of $\mathrm{q}$-Taylor’s series, we obtain

$$\begin{array}{cc}\hfill {\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(v_n\right)& ={\theta }_1+\frac{1}{\left(-1+{\theta }_1\right)}2{\theta }_2^2{e}_n^2+\frac{1}{{\left(-1+{\theta }_1\right)}^2}2{\theta }_2\left(2{\theta }_1{\theta }_3-2{\theta }_3-2{\theta }_2^2\right)e_n^3\hfill \\ \hfill & +\frac{1}{{\left(-1+{\theta }_1\right)}^3}2{\theta }_2\left(3{\theta }_4+4{\theta }_2^3-6{\theta }_1{\theta }_4+3{\theta }_1^2{\theta }_4+7{\theta }_2{\theta }_3-7{\theta }_1{\theta }_2{\theta }_3\right)e_n^4+O\left(e_n^5\right).\hfill \end{array}$$

(51)

Using (33) and (51), we obtain

$$\begin{array}{cc}\hfill & \frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(v_n\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)}{2}\hfill \\ \hfill & ={\theta }_1+{\theta }_2{e}_n+\frac{3}{2}{\theta }_3{e}_n^2+\frac{1}{\left(-1+{\theta }_1\right)}{\theta }_2^2{e}_n^2+2{\theta }_4{e}_n^3+\frac{1}{{\left(-1+{\theta }_1\right)}^2}{\theta }_2\left(2{\theta }_1{\theta }_3-2{\theta }_3-2{\theta }_2^2\right)e_n^3\hfill \\ \hfill & +\frac{5}{2}{\theta }_5{e}_n^4+\frac{1}{{\left(-1+{\theta }_1\right)}^3}{\theta }_2\left(3{\theta }_4+4{\theta }_2^3-6{\theta }_1{\theta }_4+3{\theta }_1^2{\theta }_4+7{\theta }_2{\theta }_3-7{\theta }_1{\theta }_2{\theta }_3\right)e_n^4+O\left(e_n^5\right).\hfill \end{array}$$

(52)

Now,

$$\begin{array}{cc}\hfill & v_n\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(v_n\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)}{2}\right)\hfill \\ \hfill & =\beta {\theta }_1+\beta {\theta }_2{e}_n+\frac{3}{2}\beta {\theta }_3{e}_n^2+\frac{1}{\left(-1+{\theta }_1\right)}\beta {\theta }_2^2{e}_n^2+2\beta {\theta }_4{e}_n^3+\frac{1}{{\left(-1+{\theta }_1\right)}^2}\beta {\theta }_2\left(2{\theta }_1{\theta }_3-2{\theta }_3-2{\theta }_2^2\right)e_n^3\hfill \\ \hfill & +\frac{5}{2}\beta {\theta }_5{e}_n^4+\frac{1}{{\left(-1+{\theta }_1\right)}^3}\beta {\theta }_2\left(3{\theta }_4+4{\theta }_2^3-6{\theta }_1{\theta }_4+3{\theta }_1^2{\theta }_4+7{\theta }_2{\theta }_3-7{\theta }_1{\theta }_2{\theta }_3\right)e_n^4\hfill \\ \hfill & +\frac{1}{\left(-1+{\theta }_1\right)}{\theta }_1{\theta }_2{e}_n^2+\frac{1}{\left(-1+{\theta }_1\right)}{\theta }_2^2{e}_n^3+\frac{1}{\left(-1+{\theta }_1\right)}3{\theta }_2{\theta }_3{e}_n^4+\frac{1}{{\left(-1+{\theta }_1\right)}^2}{\theta }_2^3{e}_n^4\hfill \\ \hfill & +\frac{1}{{\left(-1+{\theta }_1\right)}^2}{\theta }_1\left(2{\theta }_1{\theta }_3-2{\theta }_3-2{\theta }_2^2\right)e_n^4+\frac{1}{{\left(-1+{\theta }_1\right)}^2}{\theta }_2\left(2{\theta }_1{\theta }_3-2{\theta }_3-2{\theta }_2^2\right)e_n^4\hfill \\ \hfill & +\frac{1}{{\left(-1+{\theta }_1\right)}^3}{\theta }_1\left(3{\theta }_4+4{\theta }_2^3-6{\theta }_1{\theta }_4+3{\theta }_1^2{\theta }_4+7{\theta }_2{\theta }_3-7{\theta }_1{\theta }_2{\theta }_3\right)e_n^4+O\left(e_n^5\right).\hfill \end{array}$$

(53)

Subtracting (39), from (53), we have

$$\begin{array}{cc}\hfill & {\mathrm{\Xi }}_2\left(v_n\right)-v_n\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(v_n\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)}{2}\right)\hfill \\ \hfill & =-\beta \left(-1+{\theta }_1\right)-\beta {\theta }_2{e}_n-\frac{3}{2}\beta {\theta }_3{e}_n^2-\frac{1}{\left(-1+{\theta }_1\right)}\beta {\theta }_2^2{e}_n^2-2\beta {\theta }_4{e}_n^3-\frac{1}{{\left(-1+{\theta }_1\right)}^2}\beta {\theta }_2\left(2{\theta }_1{\theta }_3-2{\theta }_3\right.\hfill \\ \hfill & \left.-2{\theta }_2^2\right)e_n^3-\frac{5}{2}\beta {\theta }_5{e}_n^4-\frac{1}{{\left(-1+{\theta }_1\right)}^3}\beta {\theta }_2\left(3{\theta }_4+4{\theta }_2^3-6{\theta }_1{\theta }_4+3{\theta }_1^2{\theta }_4+7{\theta }_2{\theta }_3-7{\theta }_1{\theta }_2{\theta }_3\right)e_n^4\hfill \\ \hfill & -\frac{1}{\left(-1+{\theta }_1\right)}{\theta }_2^2{e}_n^3-\frac{1}{\left(-1+{\theta }_1\right)}3{\theta }_2{\theta }_3{e}_n^4-\frac{1}{{\left(-1+{\theta }_1\right)}^2}{\theta }_2^3{e}_n^4-\frac{1}{{\left(-1+{\theta }_1\right)}^2}{\theta }_2\left(2{\theta }_1{\theta }_3\right.\hfill \\ \hfill & \left.-2{\theta }_3-\right)2{\theta }_2^2{e}_n^4+O\left(e_n^5\right).\hfill \end{array}$$

(54)

Now,

$$\begin{array}{cc}\hfill & 1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(v_n\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)}{2}\right)\hfill \\ \hfill & =1-{\theta }_1-{\theta }_2{e}_n-\frac{3}{2}{\theta }_3{e}_n^2-\frac{1}{\left(-1+{\theta }_1\right)}{\theta }_2^2{e}_n^2-2{\theta }_4{e}_n^3-\frac{1}{{\left(-1+{\theta }_1\right)}^2}{\theta }_2\left(2{\theta }_1{\theta }_3-2{\theta }_3-2{\theta }_2^2\right)e_n^3\hfill \\ \hfill & -\frac{5}{2}{\theta }_5{e}_n^4-\frac{1}{{\left(-1+{\theta }_1\right)}^3}{\theta }_2\left(3{\theta }_4+4{\theta }_2^3-6{\theta }_1{\theta }_4+3{\theta }_1^2{\theta }_4+7{\theta }_2{\theta }_3-7{\theta }_1{\theta }_2{\theta }_3\right)e_n^4+O\left(e_n^5\right).\hfill \end{array}$$

(55)

Dividing (54) and (55), we obtain

$$\frac{{\mathrm{\Xi }}_2\left(v_n\right)-v_n\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(v_n\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)}{2}\right)}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(v_n\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)}{2}\right)}=\beta +\frac{1}{{\left(-1+{\theta }_1\right)}^2}{\theta }_2^2{e}_n^3+\frac{1}{2{\left(-1+{\theta }_1\right)}^2}\left(4{\theta }_1{\theta }_2{\theta }_3-{\theta }_2{\theta }_3-4{\theta }_2^3\right)e_n^4+O\left(e_n^5\right).$$

Using (28), we have

$$\begin{array}{cc}\hfill {\widetilde{⌀}}_{n+1}& =\beta +\frac{1}{{\left(-1+{\theta }_1\right)}^2}{\theta }_2^2{e}_n^3\hfill \\ \hfill & +\frac{1}{2{\left(-1+{\theta }_1\right)}^2}\left(4{\theta }_1{\theta }_2{\theta }_3-{\theta }_2{\theta }_3-4{\theta }_2^3\right)e_n^4+O\left(e_n^5\right).\hfill \end{array}$$

(56)

Therefore,

$$e_{n+1}=\frac{1}{{\left(-1+{\theta }_1\right)}^2}{\theta }_2^2{e}_n^3+\frac{1}{2{\left(-1+{\theta }_1\right)}^2}\left(4{\theta }_1{\theta }_2{\theta }_3-{\theta }_2{\theta }_3-4{\theta }_2^3\right)e_n^4+O\left(e_n^5\right).$$

□

In the following theorem we have found the order of convergence of the Algorithm 5, and it is the fourth order of convergence.

Theorem 5.

Let $I\subset \mathbb{R}$ be an open interval, and ${\mathrm{\Xi }}_1:I\to \mathbb{R}$ be a differential function. If $\beta \in I$ be a simple root of ${\mathrm{\Xi }}_1\left(\widetilde{⌀}\right)=0$ and ${\widetilde{⌀}}_0$ is sufficiently close to β, then the multi-step method defined by Algorithm 5 has the fourth order of convergence.

Proof. 

From (38), we obtain

$$\begin{array}{cc}\hfill v_n& =\beta +\frac{1}{\left(-1+{\theta }_1\right)}{\theta }_2{e}_n^2+\frac{1}{{\left(-1+{\theta }_1\right)}^2}\left(2{\theta }_1{\theta }_3-2{\theta }_3-2{\theta }_2^2\right)e_n^3\hfill \\ \hfill & +\frac{1}{{\left(-1+{\theta }_1\right)}^3}\left(3{\theta }_4+4{\theta }_2^3-6{\theta }_1{\theta }_4+3{\theta }_1^2{\theta }_4+7{\theta }_2{\theta }_3-7{\theta }_1{\theta }_2{\theta }_3\right)e_n^4+O\left(e_n^5\right).\hfill \end{array}$$

From (56), we have

$$w_n=\beta +\frac{1}{{\left(-1+{\theta }_1\right)}^2}{\theta }_2^2{e}_n^3+\frac{1}{2{\left(-1+{\theta }_1\right)}^2}\left(4{\theta }_1{\theta }_2{\theta }_3-{\theta }_2{\theta }_3-4{\theta }_2^3\right)e_n^4+O\left(e_n^5\right).$$

Expanding ${\mathrm{\Xi }}_2\left(w_n\right)$ in terms of $\mathrm{q}$-Taylor’s series, we have

$$\begin{array}{cc}\hfill {\mathrm{\Xi }}_2\left(w_n\right)& =\beta +\frac{1}{{\left(-1+{\theta }_1\right)}^2}{\theta }_1{\theta }_2^2{e}_n^3\hfill \\ \hfill & +\frac{1}{2{\left(-1+{\theta }_1\right)}^2}{\theta }_1\left(4{\theta }_1{\theta }_2{\theta }_3-{\theta }_2{\theta }_3-4{\theta }_2^3\right)e_n^4+O\left(e_n^5\right).\hfill \end{array}$$

(57)

Expanding ${\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(w_n\right)$ in terms of $\mathrm{q}$-Taylor’s series, we have

$$\begin{array}{cc}\hfill {\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(w_n\right)& ={\theta }_1+\frac{1}{{\left(-1+{\theta }_1\right)}^2}2{\theta }_2^3{e}_n^3\hfill \\ \hfill & +\frac{1}{2{\left(-1+{\theta }_1\right)}^2}{\theta }_2\left(4{\theta }_1{\theta }_2{\theta }_3-{\theta }_2{\theta }_3-4{\theta }_2^3\right)e_n^4+O\left(e_n^5\right).\hfill \end{array}$$

(58)

Using (33) and (58), we have

$$\begin{array}{cc}\hfill \frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(w_n\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)}{2}& ={\theta }_1+{\theta }_2{e}_n+\frac{3}{2}{\theta }_3{e}_n^2+\frac{1}{{\left(-1+{\theta }_1\right)}^2}{\theta }_2^3{e}_n^3+2{\theta }_4{e}_n^3\hfill \\ \hfill & +\frac{1}{{\left(-1+{\theta }_1\right)}^2}{\theta }_2\left(2{\theta }_1{\theta }_3-2{\theta }_3-2{\theta }_2^2\right)e_n^4+\frac{5}{2}{\theta }_5{e}_n^4+O\left(e_n^5\right).\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill w_n\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(w_n\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)}{2}\right)& =\beta {\theta }_1+\beta {\theta }_2{e}_n+\frac{3}{2}\beta {\theta }_3{e}_n^2+\frac{1}{{\left(-1+{\theta }_1\right)}^2}\beta {\theta }_2^3{e}_n^3+2\beta {\theta }_4{e}_n^3\hfill \\ \hfill & +\frac{1}{{\left(-1+{\theta }_1\right)}^2}\beta {\theta }_2\left(2{\theta }_1{\theta }_3-2{\theta }_3-2{\theta }_2^2\right)e_n^4+\frac{5}{2}\beta {\theta }_5{e}_n^4+\frac{1}{{\left(-1+{\theta }_1\right)}^2}{\theta }_1{\theta }_2^2{e}_n^3\hfill \\ \hfill & +\frac{1}{{\left(-1+{\theta }_1\right)}^2}{\theta }_2^3{e}_n^4+\frac{1}{2{\left(-1+{\theta }_1\right)}^2}{\theta }_1\left(4{\theta }_1{\theta }_2{\theta }_3-{\theta }_2{\theta }_3-4{\theta }_2^3\right)e_n^4+O\left(e_n^5\right).\hfill \end{array}$$

(59)

Subtracting (57) from (59), we have

$$\begin{array}{cc}\hfill {\mathrm{\Xi }}_2\left(w_n\right)-w_n\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(w_n\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)}{2}\right)& =-\beta \left(-1+{\theta }_1\right)-\beta {\theta }_2{e}_n-\frac{3}{2}\beta {\theta }_3{e}_n^2-\frac{1}{{\left(-1+{\theta }_1\right)}^2}\beta {\theta }_2^3{e}_n^3+2\beta {\theta }_4{e}_n^3\hfill \\ \hfill & -\frac{1}{{\left(-1+{\theta }_1\right)}^2}\beta {\theta }_2\left(2{\theta }_1{\theta }_3-2{\theta }_3-2{\theta }_2^2\right)e_n^4-\frac{5}{2}\beta {\theta }_5{e}_n^4\hfill \\ \hfill & -\frac{1}{{\left(-1+{\theta }_1\right)}^2}{\theta }_2^3{e}_n^4+O\left(e_n^5\right).\hfill \end{array}$$

(60)

Now,

$$\begin{array}{cc}\hfill 1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(w_n\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)}{2}\right)& =1-{\theta }_1-{\theta }_2{e}_n-\frac{3}{2}{\theta }_3{e}_n^2-\frac{1}{{\left(-1+{\theta }_1\right)}^2}{\theta }_2^3{e}_n^3-2{\theta }_4{e}_n^3-\frac{1}{{\left(-1+{\theta }_1\right)}^2}{\theta }_2\left(2{\theta }_1{\theta }_3\right.\hfill \\ \hfill & \left.-2{\theta }_3-2{\theta }_2^2\right)e_n^4-\frac{5}{2}{\theta }_5{e}_n^4+O\left(e_n^5\right).\hfill \end{array}$$

(61)

Dividing (60) and (61), we have

$$\frac{{\mathrm{\Xi }}_2\left(w_n\right)-w_n\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(w_n\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)}{2}\right)}{1-\left(\frac{{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left(w_n\right)+{\mathfrak{D}}_{\mathrm{q}}{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)}{2}\right)}=\beta +\frac{1}{{\left(-1+{\theta }_1\right)}^3}{\theta }_2^3{e}_n^4+O\left(e_n^5\right).$$

Using (31), we have

$${\widetilde{⌀}}_{n+1}=\beta +\frac{1}{{\left(-1+{\theta }_1\right)}^3}{\theta }_2^3{e}_n^4+O\left(e_n^5\right).$$

Therefore,

$$e_{n+1}=\frac{1}{{\left(-1+{\theta }_1\right)}^3}{\theta }_2^3{e}_n^4+O\left(e_n^5\right).$$

□

4. Numerical Examples and Comparison Results

This section elaborates on the efficacy of algorithms introduced in this chapter with the support of examples. All the numerical experiments are performed with Intel (R) core [TM] $2\times 2.1$ GHz, 12 GB of RAM, and all the codes are written in maple. We use $\epsilon ={10}^{-5}$ and obtain an approximated simple root rather than the exact based on the exactness $\epsilon$ of the computer.

The abbreviation CAG is used for the classical iterative method and QAG for the $\mathrm{q}$-analogue of the classical iterative method.

Recall the classical algorithm $2A$ in [8] $\left(CAG1\right),$ defined by

$${\widetilde{⌀}}_{n+1}=\frac{{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)-{\widetilde{⌀}}_n{\mathrm{\Xi }}_2^{\prime }\left({\widetilde{⌀}}_n\right)}{1-{\mathrm{\Xi }}_2^{\prime }\left({\widetilde{⌀}}_n\right)}$$

and classical algorithm $2B$ in [8] $\left(CAG2\right),$ defined by

$$\begin{array}{cc}\hfill v_n& =\frac{{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)-{\widetilde{⌀}}_n{\mathrm{\Xi }}_2^{\prime }\left({\widetilde{⌀}}_n\right)}{1-{\mathrm{\Xi }}_2^{\prime }\left({\widetilde{⌀}}_n\right)}\hfill \\ \hfill {\widetilde{⌀}}_{n+1}& =\frac{{\mathrm{\Xi }}_2\left(v_n\right)-v_n{\mathrm{\Xi }}_2^{\prime }\left(\frac{{\widetilde{⌀}}_n+v_n}{2}\right)}{1-{\mathrm{\Xi }}_2^{\prime }\left(\frac{{\widetilde{⌀}}_n+v_n}{2}\right)},n=0,1,2,\dots \hfill \end{array}$$

and classical algorithm $2C$ in [8] $\left(CAG3\right),$ defined by

$$\begin{array}{cc}\hfill v_n& =\frac{{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)-{\widetilde{⌀}}_n{\mathrm{\Xi }}_2^{\prime }\left({\widetilde{⌀}}_n\right)}{1-{\mathrm{\Xi }}_2^{\prime }\left({\widetilde{⌀}}_n\right)}\hfill \\ \hfill w_n& =\frac{{\mathrm{\Xi }}_2\left(v_n\right)-v_n{\mathrm{\Xi }}_2^{\prime }\left(\frac{{\widetilde{⌀}}_n+v_n}{2}\right)}{1-{\mathrm{\Xi }}_2^{\prime }\left(\frac{{\widetilde{⌀}}_n+v_n}{2}\right)}\hfill \\ \hfill {\widetilde{⌀}}_{n+1}& =\frac{{\mathrm{\Xi }}_2\left(w_n\right)-w_n{\mathrm{\Xi }}_2^{\prime }\left(\frac{w_n+v_n}{2}\right)}{1-{\mathrm{\Xi }}_2^{\prime }\left(\frac{w_n+v_n}{2}\right)},n=0,1,2,\dots \hfill \end{array}$$

and classical algorithm $2\mathfrak{D}$ in [8] $\left(CAG4\right),$ defined by

$$\begin{array}{cc}\hfill v_n& =\frac{{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)-{\widetilde{⌀}}_n{\mathrm{\Xi }}_2^{\prime }\left({\widetilde{⌀}}_n\right)}{1-{\mathrm{\Xi }}_2^{\prime }\left({\widetilde{⌀}}_n\right)}\hfill \\ \hfill {\widetilde{⌀}}_{n+1}& =\frac{{\mathrm{\Xi }}_2\left(v_n\right)-v_n{\mathrm{\Xi }}_2^{\prime }\left(\frac{{\mathrm{\Xi }}_2^{\prime }\left(v_n\right)+{\mathrm{\Xi }}_2^{\prime }\left({\widetilde{⌀}}_n\right)}{2}\right)}{1-{\mathrm{\Xi }}_2^{\prime }\left(\frac{{\mathrm{\Xi }}_2^{\prime }\left(v_n\right)+{\mathrm{\Xi }}_2^{\prime }\left({\widetilde{⌀}}_n\right)}{2}\right)},n=0,1,2,\dots \hfill \end{array}$$

and classical algorithm $2E$ in [8] $\left(CAG5\right),$ defined by

$$\begin{array}{cc}\hfill v_n& =\frac{{\mathrm{\Xi }}_2\left({\widetilde{⌀}}_n\right)-{\widetilde{⌀}}_n{\mathrm{\Xi }}_2^{\prime }\left({\widetilde{⌀}}_n\right)}{1-{\mathrm{\Xi }}_2^{\prime }\left({\widetilde{⌀}}_n\right)}\hfill \\ \hfill w_{n+1}& =\frac{{\mathrm{\Xi }}_2\left(v_n\right)-v_n{\mathrm{\Xi }}_2^{\prime }\left(\frac{{\mathrm{\Xi }}_2^{\prime }\left(v_n\right)+{\mathrm{\Xi }}_2^{\prime }\left({\widetilde{⌀}}_n\right)}{2}\right)}{1-{\mathrm{\Xi }}_2^{\prime }\left(\frac{{\mathrm{\Xi }}_2^{\prime }\left(v_n\right)+{\mathrm{\Xi }}_2^{\prime }\left({\widetilde{⌀}}_n\right)}{2}\right)}\hfill \\ \hfill {\widetilde{⌀}}_{n+1}& =\frac{{\mathrm{\Xi }}_2\left(w_n\right)-w_n{\mathrm{\Xi }}_2^{\prime }\left(\frac{{\mathrm{\Xi }}_2^{\prime }\left(w_n\right)+{\mathrm{\Xi }}_2^{\prime }\left({\widetilde{⌀}}_n\right)}{2}\right)}{1-{\mathrm{\Xi }}_2^{\prime }\left(\frac{{\mathrm{\Xi }}_2^{\prime }\left(w_n\right)+{\mathrm{\Xi }}_2^{\prime }\left({\widetilde{⌀}}_n\right)}{2}\right)},n=0,1,2,\dots .\hfill \end{array}$$

For simplicity, we denote the Algorithms 1–5 by QAG1, QAG2, QAG3, QAG4 and QAG5, respectively. The computational results are presented in tables to elaborate on the performance and efficacy of our $\mathrm{q}$-iterative methods, which is the main motivation for transforming the classical methods toward the $\mathrm{q}$-iterative methods.

The first two examples are concerned with the performance of $\mathrm{q}$-iterative methods with the classical methods for different values of $\mathrm{q}$ up to three iterations. Similarly, we can check the performance of the $\mathrm{q}$-iterative methods for different values of $\mathrm{q}$ for the rest of the iterations until we achieve the desired accuracy.

Example 1.

For ${\mathrm{\Xi }}_1\left(\widetilde{⌀}\right)={\widetilde{⌀}}^3-10,{\mathrm{\Xi }}_2\left(\widetilde{⌀}\right)=\sqrt{\frac{10}{\widetilde{⌀}}}$ and ${\widetilde{⌀}}_0=1.5.$

Table 1 shows the computation of ${\widetilde{⌀}}_i$ and ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_i\right)$ for $i=1,2,3$ and different values of $\mathrm{q}$ by using QAG1.

Table 1. Calculation of ${\widetilde{⌀}}_i$ and ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_i\right)$ for $i=1,2,3$ and different values of $\mathrm{q}$ by using QAG1.

q	${\widetilde{⌀}}_1$	${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right)$	${\widetilde{⌀}}_2$	${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right)$	${\widetilde{⌀}}_3$	${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)$
$1.02$	$2.0855038$	$-0.9294639$	$2.1542111$	$-0.31131\times {10}^{-2}$	$2.1544357$	$0.14\times {10}^{-4}$
$1.01$	$2.0835147$	$-0.9553929$	$2.1540089$	$-0.59279\times {10}^{-2}$	$2.1544358$	$0.15\times {10}^{-4}$
$0.9999$	$2.0812097$	$-0.9853779$	$2.1537933$	$-0.89285\times {10}^{-2}$	$2.1544345$	$-0.26\times {10}^{-5}$
$0.98$	$2.0774328$	$-1.0343671$	$2.1533170$	$-0.155555\times {10}^{-2}$	$2.1544288$	$-0.820\times {10}^{-4}$
$0.97$	$2.0753646$	$-1.0611178$	$2.1530570$	$-0.191717\times {10}^{-2}$	$2.1544241$	$-0.1475\times {10}^{-3}$
$0.99$	$2.0794806$	$-1.0078277$	$2.1535621$	$-0.121457\times {10}^{-2}$	$2.1544325$	$-0.305\times {10}^{-4}$
$0.96$	$2.0732769$	$-1.0880667$	$2.1527815$	$-0.230026\times {10}^{-2}$	$2.1544175$	$-0.2394\times {10}^{-3}$
$0.95$	$2.0711676$	$-1.1152394$	$2.1524896$	$-0.270605\times {10}^{-2}$	$2.1544091$	$-0.3563\times {10}^{-3}$
$0.9$	$2.0602939$	$-1.2544419$	$2.1507718$	$-0.509182\times {10}^{-2}$	$2.1543358$	$-0.13770\times {10}^{-2}$
$0.8$	$2.0367340$	$-1.5510461$	$2.1457907$	$-0.1198833$	$2.1539345$	$-0.69634\times {10}^{-2}$
$0.7$	$2.0103303$	$-1.8753950$	$2.1380819$	$-0.2259848$	$2.1529026$	$-0.213188\times {10}^{-1}$
$0.6$	$1.9804056$	$-2.2328367$	$2.1265044$	$-0.3839025$	$2.1506354$	$-0.528110\times {10}^{-1}$
$0.5$	$1.9459983$	$-2.6306808$	$2.1092951$	$-0.6154807$	$2.1459875$	$-0.1171646$
$0.4$	$1.9056623$	$-3.0794944$	$2.0835889$	$-0.9544265$	$2.1366846$	$-0.2451351$

We can observe from Table 1 that we can obtain more accurate values of ${\widetilde{⌀}}_1, {\widetilde{⌀}}_2, {\widetilde{⌀}}_3$ when $\mathrm{q}\to 1^{-}$ and for which ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right), {\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right), {\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)$ tend towards zero. The values of ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right)=-0.9853779,$ ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right)=-0.89285\times {10}^{-2},$ $X{i}_1\left({\widetilde{⌀}}_3\right)=-0.26\times {10}^{-5}$ calculated by QAG1 at $\mathrm{q}=0.9999$ are closer to zero as compared to the values ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right)=-0.9815116,{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right)=-0.89397\times {10}^{-2},$ ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)=-0.13\times {10}^{-5}$ calculated by CAG1.

Table 2 shows the computation of ${\widetilde{⌀}}_i$ and ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_i\right)$ for $i=1,2,3$ and different values of $\mathrm{q}$ by using QAG2.

Table 2. Calculation of ${\widetilde{⌀}}_i$ and ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_i\right)$ for $i=1,2,3$ and different values of $\mathrm{q}$ by using QAG2.

q	${\widetilde{⌀}}_1$	${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right)$	${\widetilde{⌀}}_2$	${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right)$	${\widetilde{⌀}}_3$	${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)$
$1.02$	$2.1487224$	$-0.793316\times {10}^{-1}$	$2.1544346$	$-0.13\times {10}^{-5}$	$2.1544348$	$0.2\times {10}^{-5}$
$1.01$	$2.1483621$	$-0.843213\times {10}^{-1}$	$2.1544347$	$0.0\times {10}^{-20}$	$2.1544348$	$0.2\times {10}^{-5}$
$0.9999$	$2.1479640$	$-0.898325\times {10}^{-1}$	$2.1544348$	$0.2\times {10}^{-5}$	$2.1544348$	$0.2\times {10}^{-5}$
$0.98$	$2.1471908$	$-0.1005307$	$2.1544345$	$-0.26\times {10}^{-5}$	$2.1544348$	$0.2\times {10}^{-5}$
$0.97$	$2.1467681$	$-0.1063760$	$2.1544342$	$-0.68\times {10}^{-5}$	$2.1544348$	$0.2\times {10}^{-5}$
$0.99$	$2.1475969$	$-0.949128\times {10}^{-1}$	$2.1544346$	$-0.13\times {10}^{-5}$	$2.1544345$	$-0.26\times {10}^{-5}$
$0.96$	$2.1463291$	$-0.1124443$	$2.1544337$	$-0.138\times {10}^{-5}$	$2.1544347$	0
$0.95$	$2.1458727$	$-0.1187505$	$2.1544331$	$-0.221\times {10}^{-4}$	$2.1544346$	$-0.13\times {10}^{-5}$
$0.9$	$2.1433145$	$-0.1540482$	$2.1544260$	$-0.1210\times {10}^{-3}$	$2.1544346$	$-0.13\times {10}^{-5}$
$0.8$	$2.1365749$	$-0.2466375$	$2.1543726$	$-0.8646\times {10}^{-3}$	$2.1544344$	$-0.40\times {10}^{-5}$
$0.7$	$2.1270331$	$-0.3767284$	$2.1541870$	$-0.34486\times {10}^{-2}$	$2.1544328$	$-0.263\times {10}^{-4}$
$0.6$	$2.1136122$	$-0.5577407$	$2.1536621$	$-0.107543\times {10}^{-1}$	$2.1544211$	$-0.1892\times {10}^{-3}$
$0.5$	$2.0946713$	$-0.8093200$	$2.1523059$	$-0.296136\times {10}^{-1}$	$2.1543658$	$-0.9592\times {10}^{-3}$
$0.4$	$2.0675721$	$-1.1614303$	$2.1489197$	$-0.765985\times {10}^{-1}$	$2.1541200$	$-0.43813\times {10}^{-2}$

We can observe from Table 2 that we can obtain more accurate values of ${\widetilde{⌀}}_1,{\widetilde{⌀}}_2,{\widetilde{⌀}}_3$ when $\mathrm{q}\to 1^{-}$ and for which ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right),{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right),{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)$ tend towards zero. The values of ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right)=-0.898325\times {10}^{-1},{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right)=0.2\times {10}^{-5},{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)=0.2\times {10}^{-5}$ calculated by QAG2 at $\mathrm{q}=0.9999$ are closer to zero as compared to the values ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right)=-0.895114\times {10}^{-1},{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right)=0.0\times {10}^{-5},$ ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)=0$ calculated by CAG2.

Table 3 shows the computation of ${\widetilde{⌀}}_i$ and ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_i\right)$ for $i=1,2,3$ and different values of $\mathrm{q}$ by using QAG3.

Table 3. Calculation of ${\widetilde{⌀}}_i$ and ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_i\right)$ for $i=1,2,3$ and different values of $\mathrm{q}$ by using QAG3.

q	${\widetilde{⌀}}_1$	${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right)$	${\widetilde{⌀}}_2$	${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right)$	${\widetilde{⌀}}_3$	${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)$
$1.02$	$2.1539757$	$-0.63900\times {10}^{-2}$	$2.1544347$	0	$2.1544346$	$-0.13\times {10}^{-5}$
$1.01$	$2.1539305$	$-0.70191\times {10}^{-2}$	$2.1544346$	$-0.13\times {10}^{-5}$	$2.1544347$	0
$0.9999$	$2.1538776$	$-0.77553\times {10}^{-2}$	$2.1544348$	$0.2\times {10}^{-5}$	$2.1544346$	$-0.13\times {10}^{-5}$
$0.98$	$2.1537743$	$-0.91930\times {10}^{-2}$	$2.1544347$	0	$2.1544348$	$0.2\times {10}^{-5}$
$0.97$	$2.1537146$	$-0.100237\times {10}^{-1}$	$2.1544347$	0	$2.1544347$	0
$0.99$	$2.1538300$	$-0.84178\times {10}^{-1}$	$2.1544345$	$-0.26\times {10}^{-5}$	$2.1544347$	0
$0.96$	$2.1536507$	$-0.109129\times {10}^{-1}$	$2.1544347$	0	$2.1544346$	$-0.13\times {10}^{-5}$
$0.95$	$2.1535820$	$-0.118688\times {10}^{-1}$	$2.1544347$	0	$2.1544345$	$-0.26\times {10}^{-5}$
$0.9$	$2.1531639$	$-0.176850\times {10}^{-1}$	$2.1544347$	0	$2.1544347$	0
$0.8$	$2.1518179$	$-0.363940\times {10}^{-1}$	$2.1544342$	$-0.68\times {10}^{-5}$	$2.1544348$	$0.2\times {10}^{-5}$
$0.7$	$2.1494112$	$-0.697879\times {10}^{-1}$	$2.1544308$	$-0.542\times {10}^{-4}$	$2.1544346$	$-0.13\times {10}^{-5}$
$0.6$	$2.1452102$	$-0.1278997$	$2.1544128$	$-0.3048\times {10}^{-3}$	$2.1544346$	$-0.13\times {10}^{-5}$
$0.5$	$2.1379260$	$-0.2281227$	$2.1543352$	$-0.13853\times {10}^{-2}$	$2.1544342$	$-0.68\times {10}^{-5}$
$0.4$	$2.1251725$	$-0.4019598$	$2.1540180$	$-0.58012\times {10}^{-2}$	$2.1544291$	$-0.778\times {10}^{-4}$

We can observe from Table 4 that we can obtain more accurate values of ${\widetilde{⌀}}_1,{\widetilde{⌀}}_2,{\widetilde{⌀}}_3$ when $\mathrm{q}\to 1^{-}$ and for which ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right),{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right),{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)$ tend towards zero. The values of ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right)=-1.8947724,$ ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right)=-0.797291\times {10}^{-1},$ ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)=-0.1154\times {10}^{-3}$ calculated by QAG4 at $\mathrm{q}=0.9999$ are closer to zero as compared to the values ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right)=-0.1078526,$ ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right)=0,{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)=0$ calculated by CAG4.

Table 4. Calculation of ${\widetilde{⌀}}_i$ and ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_i\right)$ for $i=1,2,3$ and different values of $\mathrm{q}$ by using QAG4.

q	${\widetilde{⌀}}_1$	${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right)$	${\widetilde{⌀}}_2$	${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right)$	${\widetilde{⌀}}_3$	${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)$
$1.02$	$2.0155906$	$-1.8114506$	$2.1507447$	$-0.512943\times {10}^{-1}$	$2.1544725$	$0.000526$
$1.01$	$2.0124687$	$-1.8494409$	$2.1497784$	$-0.646977\times {10}^{-1}$	$2.1544547$	$0.000279$
$0.9999$	$2.0087308$	$-1.8947724$	$2.1486937$	$-0.797291\times {10}^{-1}$	$2.1544264$	$-0.1154\times {10}^{-3}$
$0.98$	$2.0031070$	$-1.9626580$	$2.1466791$	$-0.1076065$	$2.1543339$	$-0.14034\times {10}^{-2}$
$0.97$	$1.9999832$	$-2.0002016$	$2.1455786$	$-0.1228128$	$2.1542693$	$-0.23028\times {10}^{-2}$
$0.99$	$2.0062296$	$-1.9250117$	$2.1477472$	$-0.928330\times {10}^{-1}$	$2.1543865$	$-0.6710\times {10}^{-3}$
$0.96$	$1.9968590$	$-2.0376328$	$2.1444435$	$-0.1384808$	$2.1541914$	$-0.33874\times {10}^{-2}$
$0.95$	$1.9937319$	$-2.0749817$	$2.1432740$	$-0.1546063$	$2.1541000$	$-0.46598\times {10}^{-2}$
$0.9$	$1.9780388$	$-2.2606512$	$2.1368988$	$-0.2422011$	$2.1534184$	$-0.141449\times {10}^{-1}$
$0.8$	$1.9461316$	$-2.6291663$	$2.1213124$	$-0.4541657$	$2.1506768$	$-0.522365\times {10}^{-1}$
$0.7$	$1.9130363$	$-2.9988460$	$2.1013440$	$-0.7212075$	$2.1454002$	$-0.1252764$
$0.6$	$1.8780908$	$-3.3755511$	$2.0758939$	$-1.0542768$	$2.1363381$	$-0.2498801$
$0.5$	$1.8404890$	$-3.7655280$	$2.0432788$	$-1.4693350$	$2.1214338$	$-0.4525267$
$0.4$	$1.7991591$	$-4.1761697$	$2.0009064$	$-1.9891183$	$2.0971373$	$-0.7768219$

Table 5 shows the computation of ${\widetilde{⌀}}_i$ and ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_i\right)$ for $i=1,2,3$ and different values of $\mathrm{q}$ by using QAG5.

Table 5. Calculation of ${\widetilde{⌀}}_i$ and ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_i\right)$ for $i=1,2,3$ and different values of $\mathrm{q}$ by using QAG5.

q	${\widetilde{⌀}}_1$	${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right)$	${\widetilde{⌀}}_2$	${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right)$	${\widetilde{⌀}}_3$	${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)$
$1.02$	$1.8525042$	$-3.6426283$	$2.1056498$	$-0.6640519$	$2.1542450$	$-0.26412\times {10}^{-2}$
$1.01$	$1.8475243$	$-3.6937602$	$2.1008763$	$-0.7274017$	$2.1534289$	$-0.139989\times {10}^{-1}$
$0.9999$	$1.8410817$	$-3.7595029$	$2.0950306$	$-0.8045897$	$2.1523801$	$-0.285824\times {10}^{-1}$
$0.98$	$1.8330136$	$-3.8411864$	$2.0863755$	$-0.9180852$	$2.1502738$	$-0.578276\times {10}^{-1}$
$0.97$	$1.8283072$	$-3.8885043$	$2.0814894$	$-0.9817429$	$2.1489927$	$-0.755872\times {10}^{-1}$
$0.99$	$1.8377864$	$-3.7929521$	$2.0912292$	$-0.8545537$	$2.1514391$	$-0.416549\times {10}^{-1}$
$0.96$	$1.8236674$	$-3.9349148$	$2.0765881$	$-1.0452991$	$2.1476055$	$-0.947938\times {10}^{-1}$
$0.95$	$1.8190888$	$-3.9804822$	$2.0716658$	$-1.1088264$	$2.1461121$	$-0.1154430$
$0.9$	$1.7970567$	$-4.1965621$	$2.0467795$	$-1.4254137$	$2.1371176$	$-0.2392034$
$0.8$	$1.7566430$	$-4.5793605$	$1.9958259$	$-2.0499847$	$2.1120648$	$-0.5784639$
$0.7$	$1.7199113$	$-4.9123392$	$1.9432472$	$-2.6618911$	$2.0779929$	$-1.0271134$
$0.6$	$1.6856517$	$-5.2103528$	$1.8884984$	$-3.2648098$	$2.0342962$	$-1.5813479$
$0.5$	$1.6528545$	$-5.4845205$	$1.8306772$	$-3.8647069$	$1.9792413$	$-2.2465278$
$0.4$	$1.6206639$	$-5.7432428$	$1.7686734$	$-4.4672260$	$1.9100080$	$-3.0320414$

We can observe from Table 3 that we can obtain more accurate values of ${\widetilde{⌀}}_1,{\widetilde{⌀}}_2,{\widetilde{⌀}}_3$ when $\mathrm{q}\to 1^{-}$ and for which ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right),{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right),{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)$ tend towards zero. The values of ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right)=-0.77553\times {10}^{-2},$ ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right)=0.2\times {10}^{-5},{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)=-0.13\times {10}^{-5}$ calculated by QAG3 at $\mathrm{q}=0.9999$ are closer to zero as compared to the values ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right)=-0.76913\times {10}^{-2},{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right)=0.2\times {10}^{-5},{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)=0$ calculated by CAG3.

Table 4 shows the computation of ${\widetilde{⌀}}_i$ and ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_i\right)$ for $i=1,2,3$ and different values of $\mathrm{q}$ by using QAG4.

We can observe from Table 5 that we can obtain more accurate values of ${\widetilde{⌀}}_1,{\widetilde{⌀}}_2,{\widetilde{⌀}}_3$ when $\mathrm{q}\to 1^{-}$ and for which ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right),{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right),{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)$ tend towards zero. The values of ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right)=-3.7595029,$ ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right)=-0.8045897,{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)=-0.285824\times {10}^{-1}$ calculated by QAG5 at $\mathrm{q}=0.9999$ are closer to zero as compared to the values ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right)=-0.116031\times {10}^{-1},{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right)=0,{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)=0$ calculated by CAG5.

Example 2.

For ${\mathrm{\Xi }}_1\left(\widetilde{⌀}\right)=cos\widetilde{⌀}-\widetilde{⌀},{\mathrm{\Xi }}_2\left(\widetilde{⌀}\right)=cos\widetilde{⌀}$ and ${\widetilde{⌀}}_0=1.5.$

Table 6 shows the computation of ${\widetilde{⌀}}_i$ and ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_i\right)$ for $i=1,2,3$ and different values of $\mathrm{q}$ by using QAG1.

Table 6. Calculation of ${\widetilde{⌀}}_i$ and ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_i\right)$ for $i=1,2,3$ and different values of $\mathrm{q}$ by using QAG1.

q	${\widetilde{⌀}}_1$	${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right)$	${\widetilde{⌀}}_2$	${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right)$	${\widetilde{⌀}}_3$	${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)$
$1.02$	$0.78065181$	$-0.7019683\times {10}^{-1}$	$0.73958325$	$-0.83375\times {10}^{-3}$	$0.73908681$	$-0.281\times {10}^{-5}$
$1.01$	$0.78122381$	$-0.7117148\times {10}^{-1}$	$0.73952786$	$-0.74103\times {10}^{-3}$	$0.73908586$	$-0.122\times {10}^{-5}$
$0.9999$	$0.78174266$	$-0.720\times {10}^{-1}$	$0.73946761$	$-0.64017\times {10}^{-3}$	$0.73908513$	$1.0\times {10}^{-8}$
$0.98$	$0.78267158$	$-0.7363944\times {10}^{-1}$	$0.73934401$	$-0.43328\times {10}^{-3}$	$0.73908429$	$0.141\times {10}^{-5}$
$0.97$	$0.78306522$	$-0.7431072\times {10}^{-1}$	$0.73927746$	$-0.32189\times {10}^{-3}$	$0.73908415$	$0.165\times {10}^{-5}$
$0.99$	$0.78223329$	$-0.7289215\times {10}^{-1}$	$0.73940801$	$-0.54041\times {10}^{-3}$	$0.73908461$	$8.8\times {10}^{-7}$
$0.96$	$0.78341443$	$-0.7490633\times {10}^{-1}$	$0.73920858$	$-0.20661\times {10}^{-3}$	$0.73908431$	$0.138\times {10}^{-5}$
$0.95$	$0.78371939$	$-0.7542653\times {10}^{-1}$	$0.73913744$	$-0.8754\times {10}^{-4}$	$0.73908469$	$7.4\times {10}^{-7}$
$0.9$	$0.78458021$	$-0.7689529\times {10}^{-1}$	$0.73875617$	$0.00055052$	$0.73909074$	$-0.938\times {10}^{-5}$
$0.8$	$0.78299283$	$-0.7418727\times {10}^{-1}$	$0.73795273$	$0.00189473$	$0.73912535$	$-0.6731\times {10}^{-4}$
$0.7$	$0.77698500$	$-0.6395431\times {10}^{-1}$	$0.73731293$	$0.00296482$	$0.73918318$	$-0.16410\times {10}^{-3}$
$0.6$	$0.76650309$	$-0.4616249\times {10}^{-1}$	$0.73716075$	$0.00321930$	$0.73923216$	$-0.24607\times {10}^{-3}$
$0.5$	$0.75144522$	$-0.2074223\times {10}^{-1}$	$0.73789775$	$0.00198670$	$0.73920251$	$-0.19645\times {10}^{-3}$
$0.4$	$0.73166283$	$0.01240165$	$0.74000770$	$-0.154433\times {10}^{-2}$	$0.73897200$	$0.18934\times {10}^{-3}$

We can observe from Table 6 that we can obtain more accurate values of ${\widetilde{⌀}}_1,{\widetilde{⌀}}_2,{\widetilde{⌀}}_3$ when $\mathrm{q}\to 1^{-}$ and for which ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right),{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right),{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)$ tend towards zero. The values of ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right)=-0.720\times {10}^{-1},{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right)=-0.64017\times {10}^{-3},{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)=1.0\times {10}^{-8}$ calculated by QAG1 at $\mathrm{q}=0.9999$ are closer to zero as compared to the values ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right)=-0.7206977\times {10}^{-1},{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right)=-0.64310\times {10}^{-3},{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)=-4.0\times {10}^{-8}$ calculated by CAG1.

Table 7 shows the computation of ${\widetilde{⌀}}_i$ and ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_i\right)$ for $i=1,2,3$ and different values of $\mathrm{q}$ by using QAG2.

Table 7. Calculation of ${\widetilde{⌀}}_i$ and ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_i\right)$ for $i=1,2,3$ and different values of $\mathrm{q}$ by using QAG2.

q	${\widetilde{⌀}}_1$	${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right)$	${\widetilde{⌀}}_2$	${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right)$	${\widetilde{⌀}}_3$	${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)$
$1.02$	$0.74465007$	$-0.932497\times {10}^{-2}$	$0.73908526$	$-2.1\times {10}^{-7}$	$0.73908513$	$1.0\times {10}^{-8}$
$1.01$	$0.74468755$	$-0.938785\times {10}^{-2}$	$0.73908516$	$-4.0\times {10}^{-8}$	$0.73908512$	$2.0\times {10}^{-8}$
$0.9999$	$0.74471713$	$-0.943748\times {10}^{-2}$	$0.73908516$	$-4.0\times {10}^{-8}$	$0.73908511$	$4.0\times {10}^{-8}$
$0.98$	$0.74475421$	$-0.949969\times {10}^{-2}$	$0.73908514$	$-1.0\times {10}^{-8}$	$0.73908514$	$-1.0\times {10}^{-8}$
$0.97$	$0.74476156$	$-0.951202\times {10}^{-2}$	$0.73908520$	$-1.1\times {10}^{-7}$	$0.73908513$	$1.0\times {10}^{-8}$
$0.99$	$0.74473951$	$-0.947503\times {10}^{-2}$	$0.73908515$	$-3.0\times {10}^{-8}$	$0.73908514$	$-1.0\times {10}^{-8}$
$0.96$	$0.74476156$	$-0.951202\times {10}^{-2}$	$0.73908533$	$-3.3\times {10}^{-7}$	$0.73908516$	$-4.0\times {10}^{-8}$
$0.95$	$0.74475442$	$-0.950004\times {10}^{-2}$	$0.73908540$	$-4.5\times {10}^{-7}$	$0.73908514$	$-1.0\times {10}^{-8}$
$0.9$	$0.74461545$	$-0.926689\times {10}^{-2}$	$0.73908654$	$-0.235\times {10}^{-5}$	$0.73908513$	$1.0\times {10}^{-8}$
$0.8$	$0.74387765$	$-0.802929\times {10}^{-2}$	$0.73909070$	$-0.932\times {10}^{-5}$	$0.73908511$	$4.0\times {10}^{-8}$
$0.7$	$0.74267074$	$-0.600566\times {10}^{-2}$	$0.73909560$	$-0.1752\times {10}^{-4}$	$0.73908516$	$-4.0\times {10}^{-8}$
$0.6$	$0.74121862$	$-0.357231\times {10}^{-2}$	$0.73909734$	$-0.2043\times {10}^{-4}$	$0.73908517$	$-6.0\times {10}^{-8}$
$0.5$	$0.73981044$	$-0.121408\times {10}^{-2}$	$0.73909217$	$-0.1178\times {10}^{-4}$	$0.73908522$	$-1.5\times {10}^{-7}$
$0.4$	$0.73880901$	$0.46209\times {10}^{-3}$	$0.73908096$	$0.698\times {10}^{-5}$	$0.73908509$	$7.0\times {10}^{-8}$

We can observe from Table 7 that we can obtain more accurate values of ${\widetilde{⌀}}_1,{\widetilde{⌀}}_2,{\widetilde{⌀}}_3$ when $\mathrm{q}\to 1^{-}$ and for which ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right),{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right),{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)$ tend towards zero. The values of ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right)=-0.943748\times {10}^{-2},$ ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right)=-4.0\times {10}^{-8},{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)=4.0\times {10}^{-8}$ calculated by QAG2 at $\mathrm{q}=0.9999$ are closer to zero as compared to the values ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right)=-0.943785\times {10}^{-2},{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right)=1.0\times {10}^{-8},{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)=-4.0\times {10}^{-8}$ calculated by CAG2.

Table 8 shows the computation of ${\widetilde{⌀}}_i$ and ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_i\right)$ for $i=1,2,3$ and different values of $\mathrm{q}$ by using QAG3.

Table 8. Calculation of ${\widetilde{⌀}}_i$ and ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_i\right)$ for $i=1,2,3$ and different values of $\mathrm{q}$ by using QAG3.

q	${\widetilde{⌀}}_1$	${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right)$	${\widetilde{⌀}}_2$	${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right)$	${\widetilde{⌀}}_3$	${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)$
$1.02$	$0.73985322$	$-0.128570\times {10}^{-2}$	$0.73908511$	$4.0\times {10}^{-8}$	$0.73908518$	$-8.0\times {10}^{-8}$
$1.01$	$0.73985332$	$-0.128586\times {10}^{-2}$	$0.73908513$	$1.0\times {10}^{-8}$	$0.73908518$	$-8.0\times {10}^{-8}$
$0.9999$	$0.73985157$	$-0.128293\times {10}^{-2}$	$0.73908514$	$-1.0\times {10}^{-8}$	$0.73908511$	$4.0\times {10}^{-8}$
$0.98$	$0.73984646$	$-0.127438\times {10}^{-2}$	$0.73908513$	$1.0\times {10}^{-8}$	$0.73908514$	$-1.0\times {10}^{-8}$
$0.97$	$0.73984183$	$-0.126663\times {10}^{-2}$	$0.73908511$	$4.0\times {10}^{-8}$	$0.73908509$	$7.0\times {10}^{-8}$
$0.99$	$0.73984995$	$-0.128022\times {10}^{-2}$	$0.73908512$	$2.0\times {10}^{-8}$	$0.73908511$	$4.0\times {10}^{-8}$
$0.96$	$0.73983612$	$-0.125707\times {10}^{-2}$	$0.73908516$	$-4.0\times {10}^{-8}$	$0.73908513$	$1.0\times {10}^{-8}$
$0.95$	$0.73982930$	$-0.124565\times {10}^{-2}$	$0.73908511$	$4.0\times {10}^{-8}$	$0.73908511$	$4.0\times {10}^{-8}$
$0.9$	$0.73978066$	$-0.116422\times {10}^{-2}$	$0.73908512$	$2.0\times {10}^{-8}$	$0.73908509$	$7.0\times {10}^{-8}$
$0.8$	$0.73962677$	$-0.90660\times {10}^{-3}$	$0.73908508$	$9.0\times {10}^{-8}$	$0.73908510$	$6.0\times {10}^{-8}$
$0.7$	$0.73943607$	$-0.58738\times {10}^{-3}$	$0.73908510$	$6.0\times {10}^{-8}$	$0.73908511$	$4.0\times {10}^{-8}$
$0.6$	$0.73925621$	$-0.28633\times {10}^{-3}$	$0.73908509$	$7.0\times {10}^{-8}$	$0.73908513$	$1.0\times {10}^{-8}$
$0.5$	$0.73912846$	$-0.7251\times {10}^{-4}$	$0.73908512$	$2.0\times {10}^{-8}$	$0.73908514$	$-1.0\times {10}^{-8}$
$0.4$	$0.73907500$	$0.1696\times {10}^{-4}$	$0.73908514$	$-1.0\times {10}^{-8}$	$0.73908515$	$-3.0\times {10}^{-8}$

We can observe from Table 8 that we can obtain more accurate values of ${\widetilde{⌀}}_1,{\widetilde{⌀}}_2,{\widetilde{⌀}}_3$ when $\mathrm{q}\to 1^{-}$ and for which ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right),{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right),{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)$ tend towards zero. The values of ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right)=-0.128293\times {10}^{-2},$ ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right)=-1.0\times {10}^{-8},{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)=4.0\times {10}^{-8}$ calculated by QAG3 at $\mathrm{q}=0.9999$ are closer to zero as compared to the values ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right)=-0.128409\times {10}^{-2},{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right)=-3.0\times {10}^{-8},{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)=-4.0\times {10}^{-8}$ calculated by CAG3.

Table 9 shows the computation of ${\widetilde{⌀}}_i$ and ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_i\right)$ for $i=1,2,3$ and different values of $\mathrm{q}$ by using QAG4.

Table 9. Calculation of ${\widetilde{⌀}}_i$ and ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_i\right)$ for $i=1,2,3$ and different values of $\mathrm{q}$ by using QAG4.

q	${\widetilde{⌀}}_1$	${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right)$	${\widetilde{⌀}}_2$	${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right)$	${\widetilde{⌀}}_3$	${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)$
$1.02$	$0.72875514$	$0.01724884$	$0.73908597$	$-0.140\times {10}^{-5}$	$0.73908512$	$2.0\times {10}^{-8}$
$1.01$	$0.72804826$	$0.01842628$	$0.73908432$	$0.136\times {10}^{-5}$	$0.73908514$	$-1.0\times {10}^{-8}$
$0.9999$	$0.72731877$	$0.01964098$	$0.73908229$	$0.476\times {10}^{-5}$	$0.73908518$	$-8.0\times {10}^{-8}$
$0.98$	$0.72582276$	$0.02213081$	$0.73907685$	$0.1386\times {10}^{-4}$	$0.73908508$	$9.0\times {10}^{-8}$
$0.97$	$0.72504916$	$0.02341766$	$0.73907340$	$0.1964\times {10}^{-4}$	$0.73908516$	$-4.0\times {10}^{-8}$
$0.99$	$0.72658119$	$0.02086875$	$0.73907980$	$0.893\times {10}^{-5}$	$0.73908515$	$-3.0\times {10}^{-8}$
$0.96$	$0.72426152$	$0.02472741$	$0.73906940$	$0.2633\times {10}^{-4}$	$0.73908518$	$-8.0\times {10}^{-8}$
$0.95$	$0.72346100$	$0.02605810$	$0.73906479$	$0.3405\times {10}^{-4}$	$0.73908516$	$-4.0\times {10}^{-8}$
$0.9$	$0.71931432$	$0.03294336$	$0.73903128$	$0.9013\times {10}^{-4}$	$0.73908507$	$1.1\times {10}^{-7}$
$0.8$	$0.71102869$	$0.04666225$	$0.73889767$	$0.31373\times {10}^{-3}$	$0.73908399$	$0.191\times {10}^{-5}$
$0.7$	$0.70499831$	$0.05661434$	$0.73867172$	$0.69183\times {10}^{-3}$	$0.73908049$	$0.777\times {10}^{-5}$
$0.6$	$0.70549106$	$0.05580218$	$0.73844540$	$0.107051\times {10}^{-2}$	$0.73907346$	$0.1954\times {10}^{-4}$
$0.5$	$0.71907370$	$0.03334249$	$0.73853449$	$0.92145\times {10}^{-3}$	$0.73907013$	$0.2511\times {10}^{-4}$
$0.4$	$0.75440547$	$-0.2572663\times {10}^{-1}$	$0.73966849$	$-0.97644\times {10}^{-3}$	$0.73910731$	$-0.3712\times {10}^{-4}$

We can observe from Table 9 that we can obtain more accurate values of ${\widetilde{⌀}}_1,{\widetilde{⌀}}_2,{\widetilde{⌀}}_3$ when $\mathrm{q}\to 1^{-}$ and for which ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right),{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right),{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)$ tend towards zero. The values of ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right)=0.1964098\times {10}^{-1},{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right)=0.476\times {10}^{-5},{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)=-8.0\times {10}^{-8}$ calculated by QAG4 at $\mathrm{q}=0.9999$ are closer to zero as compared to the values ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right)=-0.128409\times {10}^{-2},{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right)=-3.0\times {10}^{-8},{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)=-4.0\times {10}^{-8}$ calculated by CAG4.

Table 10 shows the computation of ${\widetilde{⌀}}_i$ and ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_i\right)$ for $i=1,2,3$ and different values of $\mathrm{q}$ by using QAG5.

Table 10. Calculation of ${\widetilde{⌀}}_i$ and ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_i\right)$ for $i=1,2,3$ and different values of $\mathrm{q}$ by using QAG5.

q	${\widetilde{⌀}}_1$	${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right)$	${\widetilde{⌀}}_2$	${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right)$	${\widetilde{⌀}}_3$	${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)$
$1.02$	$0.74177078$	$-0.449739\times {10}^{-2}$	$0.73908522$	$-1.5\times {10}^{-7}$	$0.73908510$	$6.0\times {10}^{-8}$
$1.01$	$0.74210039$	$-0.504973\times {10}^{-2}$	$0.73908519$	$-1.0\times {10}^{-7}$	$0.73908514$	$-1.0\times {10}^{-8}$
$0.9999$	$0.74245983$	$-0.565214\times {10}^{-2}$	$0.73908514$	$-1.0\times {10}^{-8}$	$0.73908510$	$6.0\times {10}^{-8}$
$0.98$	$0.74325814$	$-0.699042\times {10}^{-2}$	$0.73908500$	$2.2\times {10}^{-7}$	$0.73908513$	$1.0\times {10}^{-8}$
$0.97$	$0.74370437$	$-0.773868\times {10}^{-2}$	$0.73908490$	$3.9\times {10}^{-7}$	$0.73908509$	$7.0\times {10}^{-8}$
$0.99$	$0.74284286$	$-0.629419\times {10}^{-2}$	$0.73908507$	$1.1\times {10}^{-7}$	$0.73908513$	$1.0\times {10}^{-8}$
$0.96$	$0.74418248$	$-0.854057\times {10}^{-2}$	$0.73908481$	$5.4\times {10}^{-7}$	$0.73908516$	$-4.0\times {10}^{-8}$
$0.95$	$0.74469363$	$-0.939805\times {10}^{-2}$	$0.73908464$	$8.3\times {10}^{-7}$	$0.73908514$	$-1.0\times {10}^{-8}$
$0.9$	$0.74777644$	$-0.1457372\times {10}^{-1}$	$0.73908285$	$0.382\times {10}^{-5}$	$0.73908512$	$2.0\times {10}^{-8}$
$0.8$	$0.75674474$	$-0.2966996\times {10}^{-1}$	$0.73906877$	$0.2739\times {10}^{-4}$	$0.73908516$	$-4.0\times {10}^{-8}$
$0.7$	$0.76856553$	$-0.4965701\times {10}^{-1}$	$0.73901927$	$0.11023\times {10}^{-3}$	$0.73908528$	$-2.5\times {10}^{-7}$
$0.6$	$0.77810379$	$-0.6585797\times {10}^{-1}$	$0.73890963$	$0.29371\times {10}^{-3}$	$0.73908586$	$-0.122\times {10}^{-5}$
$0.5$	$0.77021498$	$-0.5245398\times {10}^{-1}$	$0.73884034$	$0.40967\times {10}^{-3}$	$0.73908693$	$0.73908693$
$0.4$	$0.70625933$	$0.05453553$	$0.73944871$	$-0.60854\times {10}^{-3}$	$0.73908086$	$0.715\times {10}^{-5}$

We can observe from Table 10 that we can obtain more accurate values of ${\widetilde{⌀}}_1,{\widetilde{⌀}}_2,{\widetilde{⌀}}_3$ when $\mathrm{q}\to 1^{-}$ and for which ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right),{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right),{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)$ tend towards zero. The values of ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right)=-0.565214\times {10}^{-2},{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right)=-1.0\times {10}^{-8},{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)=6.0\times {10}^{-8}$ calculated by QAG5 at $\mathrm{q}=0.9999$ are closer to zero as compared to the values ${\mathrm{\Xi }}_1\left({\widetilde{⌀}}_1\right)=-0.614538\times {10}^{-2},{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_2\right)=-1.0\times {10}^{-8},{\mathrm{\Xi }}_1\left({\widetilde{⌀}}_3\right)=-4.0\times {10}^{-8}$ calculated by CAG5.

Comparison of the Classical and $\mathrm{q}$-Analogue of Iterative Methods

Here, we check the efficiency of our new iterative methods by considering some of the nonlinear equations. Furthermore, we compare the standard Newton’s method (NM), Halley method (HM), Algorithm 2B [8] and Algorithm 2C [8] with our new iterative Algorithms 1–5. In the tables, we display the number of iterations (IT), the approximate root $x_n$, the value ${\mathrm{\Xi }}_1\left(x_n\right)$ and $\delta$ be the distance between two successive estimations. It is important to mention that in order to obtain better computational results of the $\mathrm{q}$-iterative methods, we take the value of $\mathrm{q}=0.9999.$

Example 3.

For ${\mathrm{\Xi }}_1\left(\widetilde{⌀}\right)={\widetilde{⌀}}^3-10,{\mathrm{\Xi }}_2\left(\widetilde{⌀}\right)=\sqrt{\frac{10}{\widetilde{⌀}}}$ and ${\widetilde{⌀}}_0=1.5.$

Methods	IT	${\widetilde{\mathbf{⌀}}}_{\mathbf{n}}$	${\mathbf{\Xi }}_{\mathbf{1}}\left({\widetilde{\mathbf{⌀}}}_{\mathbf{n}}\right)$	$\mathbf{\delta }=\mid {\widetilde{\mathbf{⌀}}}_{\mathbf{n}}-{\widetilde{\mathbf{⌀}}}_{\mathbf{n}-\mathbf{1}}\mid$	CUP-Time
NM [8]	5	$2.1544346900319185890$	$4.85518\times {10}^{-13}$	$2.740789522304\times {10}^{-7}$	1.203
HM [8]	5	$2.1544346900319185890$	$4.85518\times {10}^{-13}$	$2.740789522304\times {10}^{-7}$	1.171
Algorithm 2B [8]	3	$2.1544346900318837218$	$1.0\times {10}^{-13}$	$3.6257326037\times {10}^{-9}$	1.156
Algorithm 2C [8]	3	$2.1544346900318837217$	$-8.0\times {10}^{-19}$	$1.457\times {10}^{-16}$	0.828
QAG1	4	$2.1544346900302749510$	$-2.24017572\times {10}^{-11}$	$6.43262623793\times {10}^{-8}$	0.140
QAG2	3	$2.1544346900318837193$	$-3.42\times {10}^{-17}$	$3.8786590757\times {10}^{-9}$	0.156
QAG3	3	$2.1544346900318837217$	$-8.0\times {10}^{-19}$	$3.926\times {10}^{-16}$	0.140
QAG4	4	$2.1544346895391400096$	$-6.8613411350\times {10}^{-9}$	$0.85676129438478\times {10}^{-5}$	0.109
QAG5	5	$2.1544346897346676286$	$-4.1386646970\times {10}^{-9}$	$0.24019786061687\times {10}^{-5}$	0.171

Example 4.

For ${\mathrm{\Xi }}_1\left(\widetilde{⌀}\right)=cos\left(\widetilde{⌀}\right)-\widetilde{⌀},{\mathrm{\Xi }}_2\left(\widetilde{⌀}\right)=cos\left(\widetilde{⌀}\right)$ and ${\widetilde{⌀}}_0=1.7.$

Methods	IT	${\widetilde{\mathbf{⌀}}}_{\mathbf{n}}$	${\mathbf{\Xi }}_{\mathbf{1}}\left({\widetilde{\mathbf{⌀}}}_{\mathbf{n}}\right)$	$\mathbf{\delta }=\mid {\widetilde{\mathbf{⌀}}}_{\mathbf{n}}-{\widetilde{\mathbf{⌀}}}_{\mathbf{n}-\mathbf{1}}\mid$	CUP-Time
HM [8]	4	$0.73908513321516087614$	$3.9244\times {10}^{-16}$	$3.258805388731\times {10}^{-8}$	1.062
Algorithm 2B [8]	3	$0.73908513321516064166$	$-1.0\times {10}^{-20}$	$8.63747112426\times {10}^{-9}$	1.406
Algorithm 2C [8]	3	$0.73908513321516064166$	$-1.0\times {10}^{-20}$	$3.72234\times {10}^{-15}$	1.109
QAG1	4	$0.73908513321473278667$	$7.1606325\times {10}^{-13}$	$2.622628874279\times {10}^{-8}$	0.390
QAG2	3	$0.73908513321516064390$	$-3.76\times {10}^{-18}$	$8.41276738939\times {10}^{-9}$	0.375
QAG3	3	$0.73908513321516064390$	$3.0\times {10}^{-20}$	$2.74617\times {10}^{-15}$	0.390
QAG4	3	$0.73908513321054913628$	$7.71787087\times {10}^{-12}$	$0.288824123785448\times {10}^{-5}$	0.390
QAG5	3	$0.73908513321515746853$	$5.31058\times {10}^{-15}$	$2.193899796764\times {10}^{-8}$	0.390

Example 5.

For ${\mathrm{\Xi }}_1\left(\widetilde{⌀}\right)={sin}^2\left(\widetilde{⌀}\right)-{\widetilde{⌀}}^2+1,{\mathrm{\Xi }}_2\left(\widetilde{⌀}\right)=sin\left(\widetilde{⌀}\right)+\frac{1}{sin\left(\widetilde{⌀}\right)+\widetilde{⌀}}$ and ${\widetilde{⌀}}_0=1.$

Methods	IT	${\widetilde{\mathbf{⌀}}}_{\mathbf{n}}$	${\mathbf{\Xi }}_{\mathbf{1}}\left({\widetilde{\mathbf{⌀}}}_{\mathbf{n}}\right)$	$\mathbf{\delta }=\mid {\widetilde{\mathbf{⌀}}}_{\mathbf{n}}-{\widetilde{\mathbf{⌀}}}_{\mathbf{n}-\mathbf{1}}\mid$	CUP-Time
NM [8]	5	$1.4044916482156470349$	$7.591622\times {10}^{-13}$	$6.247205954873\times {10}^{-7}$	1.421
HM [8]	5	$1.4044916482156470349$	$7.591622\times {10}^{-13}$	$6.247205954873\times {10}^{-7}$	1.406
Algorithm 2B [8]	3	$1.4044916482153412261$	$-2.0\times {10}^{-19}$	$4.830530998\times {10}^{-10}$	1.640
Algorithm 2C [8]	3	$1.4044916482153412261$	$-2.0\times {10}^{-19}$	$3.27\times {10}^{-17}$	1.328
QAG1	4	$1.4044916482154012263$	$-1.489491\times {10}^{-13}$	$1.4431460426\times {10}^{-9}$	0.609
QAG2	3	$1.4044916482153412250$	$2.58\times {10}^{-18}$	$5.627896370\times {10}^{-10}$	0.625
QAG3	3	$1.4044916482153412260$	$1.0\times {10}^{-19}$	$6.0\times {10}^{-19}$	0.593
QAG4	4	$1.4044916482140525197$	$3.19917839\times {10}^{-12}$	$2.37987556765\times {10}^{-8}$	0.203
QAG5	4	$1.4044916482133742657$	$4.88292555\times {10}^{-12}$	$2.79102149666\times {10}^{-8}$	0.250

Example 6.

For ${\mathrm{\Xi }}_1\left(\widetilde{⌀}\right)=\times {10}^{\widetilde{⌀}}-3{\widetilde{⌀}}^2,{\mathrm{\Xi }}_2\left(\widetilde{⌀}\right)=\sqrt{\frac{\times {10}^{\widetilde{⌀}}}{3}}$ and ${\widetilde{⌀}}_0=0.8.$

Methods	IT	${\widetilde{\mathbf{⌀}}}_{\mathbf{n}}$	${\mathbf{\Xi }}_{\mathbf{1}}\left({\widetilde{\mathbf{⌀}}}_{\mathbf{n}}\right)$	$\mathbf{\delta }=\mid {\widetilde{\mathbf{⌀}}}_{\mathbf{n}}-{\widetilde{\mathbf{⌀}}}_{\mathbf{n}-\mathbf{1}}\mid$	CUP-Time
NM [8]	4	$0.91000757248870906142$	$-2.3\times {10}^{-18}$	$1.13400774851\times {10}^{-9}$	0.390
HM [8]	4	$0.91000757248870906142$	$-2.3\times {10}^{-18}$	$1.13400774848\times {10}^{-9}$	0.281
Algorithm 2B [8]	3	$0.91000757248870906069$	$-1.0\times {10}^{-19}$	$5.97372\times {10}^{-15}$	0.203
Algorithm 2C [8]	2	$0.91000757248870906058$	$2.0\times {10}^{-19}$	$0.114201199652787\times {10}^{-5}$	0.218
QAG1	3	$0.91000757246605133264$	$6.74226939\times {10}^{-11}$	$0.117769471538602\times {10}^{-5}$	0.187
QAG2	3	$0.91000757248870906066$	$-1.0\times {10}^{-19}$	$4.576774\times {10}^{-14}$	0.171
QAG3	2	$0.91000757248870906062$	$1.0\times {10}^{-19}$	$0.114452189637435\times {10}^{-5}$	0.156
QAG4	3	$0.91000757248830221190$	$1.2106615\times {10}^{-12}$	$5.910997202999\times {10}^{-8}$	0.187
QAG5	3	$0.91000757248870097044$	$2.40741\times {10}^{-14}$	$3.24759020938\times {10}^{-9}$	0.171

Tables compare the solutions obtained by using the classical and our $\mathrm{q}$-iterative methods. The results show that our $\mathrm{q}$-analogue iterative methods QAG1, QAG2, QAG3, QAG4 and QAG5 give the same results as the classical methods NM, HM, Algorithm 2B [8] and Algorithm 2C [8].

Remark 1.

The efficiency index is considered as $E=p^{\frac{1}{m}}$, where p represents the order of the method and m is the total number of function evaluations per iteration necessary by the method.

Methods	Efficiency Index	Methods	Efficiency Index
Algoritm 2A [8]	$1.414213$	Algorithm 1	$1.414213$
Algoritm 2B [8]	$1.316074$	Algorithm 2	$1.316074$
Algoritm 2C [8]	$1.259921$	Algorithm 3	$1.259921$
Algoritm 2D [8]	$1.316074$	Algorithm 4	$1.316074$
Algoritm 2E [8]	$1.259921$	Algorithm 5	$1.259921$

We conclude that the efficiency indexes calculated by Algoritms 1, Algoritm 2, Algoritm 3, Algoritm 4 and Algoritm 5 are the same as those calculated by Algoritm 2A [8], Algoritm 2B [8], Algoritm 2C [8], Algoritm 2D [8] and Algoritm 2E [8], respectively.

5. Conclusions

In this paper, we have introduced some new multi-step algorithms using the Daftardar–Jafari decomposition technique. The comparison of these newly established algorithms with the classical methods reflects that the proposed $\mathrm{q}$-iterative methods are reliable and the best alternatives to the already known algorithms. The computational results conclude that the $\mathrm{q}$-analogue of the iterative methods for solving the nonlinear equations generates the same results as the classical methods, but the convergence rate towards approaching the root is higher than the convergence rate suggested by the classical methods. It is worth mentioning here that one can extend the results obtained in this paper by using the post quantum calculus techniques. This will be an interesting problem for future research.