For solving nonlinear equations, we drive the derivative-free iterative technique by using the approximate version of the first derivative of \(g'(x_{m})\) by
$$\begin{aligned} g'(x_{m})\approx \big (g(x_{m}+\theta g(x_{m}))-g(x_{m}-\theta g(x_{m}))\big )\big /2 \theta g(x_{m}), \end{aligned}$$
(1)
where \(\theta \in {\mathbb {R}}\) and not equal zero. Let us consider the method in [15]:
$$\begin{aligned} y_{m}& = x_{m}-\frac{g(x_{m})}{g'(x_{m})},\nonumber \\ z_{m}& = x_{m}-\bigg (1+\frac{g(y_{m})}{g(x_{m})-2g(y_{m})}\bigg )\frac{g(x_{m})}{g'(x_{m})},\nonumber \\ x_{m+1}& = z_{m}-\bigg (1+\frac{2g(y_{m})}{g(x_{m})-2g(y_{m})}\bigg )\frac{g(z_{m})}{g'(x_{m})}. \end{aligned}$$
(2)
By using Eq. (1), we obtain the following new eighth order algorithm in the absence of the derivative which using for solving a nonlinear equation as follows.
Eighth order derivative free iteration algorithm (8th BM): Further, we substitute the approximation of the derivative \(g'(x)\) in Eq. (2) by Eq. (1), we get the proposed algorithm free from derivatives, as follows:
8th BM: Given an initial approximation \(x_{0}\) (close to \(\alpha\) ) the root of \(g(x)=0\). We find the approximate solution
$$\begin{aligned} y_{m}& = x_{m}-\frac{2 \theta g^{2}(x_{m})}{g(x_{m}+\theta g(x_{m})) -g(x_{m}-\theta g(x_{m}))},\nonumber \\ z_{m}& = x_{m}-\bigg (\frac{g^{2}(x_{m})-g(x_{m})g(y_{m})+g^{2}(y_{m})}{g^{2}(x_{m}) -2g(x_{m})g(y_{m})+g^{2}(y_{m})}\bigg )\frac{2 \theta g^{2}(x_{m})}{g(x_{m}+\theta g(x_{m}))-g(x_{m}-\theta g(x_{m}))},\nonumber \\ x_{m+1}& = z_{m}-\frac{2 \theta g^{2}(z_{m})}{g(z_{m}+\theta g(z_{m})) -g(z_{m}-\theta g(z_{m}))}. \end{aligned}$$
(3)
Steps for calculating root using 8th BM
Step 1: Define the function g(x).
Step 2: Nominate an approximation guess \(x_0\).
Step 3: By using the formula (3), we calculate the next approximation of the root \(x_{i+1}, (i=0,1,2,\cdots ).\)
Step 4: We use a specific accuracy \(\epsilon\) as \(|x_{i}-\alpha |<\epsilon\), and repeat Step 3 until we get desired approximate root which satisfy the condition. In order to prove the convergence of 8th BM, we establish the following theorem with the help of Maple software.
Theorem
Suppose that \(g(x):{\mathbb {R}}\longrightarrow {\mathbb {R}}\) for the interval (a, b). Assume that g(x) has sufficiently continuous derivatives in (a, b). If \(\alpha\) has a simple root of g(x) and if \(x_{0}\) is closed to \(\alpha\) then 8th BM satisfies the following error equation:
$$\begin{aligned} e_{m+1}=(\theta ^4 F^4 c^2_{3}+2 \theta ^2 F^2 c^2_{3}-4 \theta ^2 F^2 c^2_{2} c_{3}+c^2_{3}-4 c_{3} c^2_{2}+4 c^4_{2}) c^3_{2} e^8_{m}+O (e^9_{m}). \end{aligned}$$
(4)
Proof
Let the error at step m be denoted by \(e_{m} = x_{m} -\alpha\) and \(F= g'(\alpha )\) and \(c_{k}=\frac{1}{k!}\frac{g^{k}(\alpha )}{g'(\alpha )} ,\quad k=2,3,.....\quad\). If we expand \(g(x_{m})\) around the root \(\alpha\) and express it in terms of powers of error \(e_{m}\) we obtain
$$\begin{aligned} g(x_{m})& = g(\alpha )+(x_{m}-\alpha )g'(\alpha )+\frac{(x_{m}-\alpha )^2}{2!}g^{(2)}(\alpha )+\frac{(x_{m}-\alpha )^3}{3!} g^{(3)}(\alpha )\nonumber \\&\quad+\,\frac{(x_{m}-\alpha )^4}{4!}g^{(4)}(\alpha )+ \frac{(x_{m}-\alpha )^5}{5!}g^{(5)}(\alpha )+\frac{(x_{m} -\alpha )^6}{6!}g^{(6)}(\alpha )\nonumber \\&\quad+\,\frac{(x_{m}-\alpha )^7}{7!}g^{(7)}(\alpha )+\frac{(x_{m}-\alpha )^8}{8!}g^{(8)}(\alpha )+\ldots \nonumber \\& = F\big (e_{m}+c_{2}e^{2}_{m}+c_{3}e^{3}_{m}+ c_{4}e^{4}_{m}+c_{5}e^{5}_{m}+c_{6} e^{6}_{m}+c_{7}e^{7}_{m}+c_{8}e^{8}_{m}+\ldots \big ). \end{aligned}$$
(5)
Computing \(g^{2}(x_{m})\) using Eq. (5), then multiply by \(2\theta\) we get
$$\begin{aligned} 2 \theta g^{2}(x_{m})& = 2 \theta F^2 e^2_{m}+4 \theta F^2 c_{2} e^3_{m}+2 \theta F^2 (c^2_{2}+2 c_{3}) e^4_{m}+4 \theta F^2 (c_{2} c_{3}+c_{4}) e^5_{m}\nonumber \\&\quad+\,2 \theta F^2 (2 c_{2} c_{4}+2 c_{5}+c^2_{3}) e^6_{m}+... \end{aligned}$$
(6)
Expand \(g(x_{m}+\theta g(x_{m}))\) and \(g(x_{m}-\theta g(x_{m}))\) around the root \(\alpha\) and express it in terms of powers of error \(e_{m}\) we get
$$\begin{aligned} g(x_{m}+\theta g(x_{m}))& = F (1+\theta F) e_{m}+F c_{2} (3 \theta F+1+\theta ^2 F^2) e^2_{m}+F (2 \theta F c^2_{2}\nonumber \\&\quad+\,2 \theta ^2 F^2 c^2_{2}+c_{3}+4 \theta F c_{3}+3 c_{3} \theta ^2 F^2+\theta ^3 F^3 c_{3}) e^3_{m}\nonumber \\ &\quad+\,F (5 \theta F c_{2} c_{3}+8 \theta ^2 F^2 c_{2} c_{3}+3 \theta ^3 F^3 c_{2} c_{3}+c_{4}\nonumber \\&\quad+\,5 \theta F c_{4}+6 c_{4} \theta ^2 F^2+4 c_{4} \theta ^3 F^3+c_{4} \theta ^4 F^4+\theta ^2 F^2 c^3_{2}) e^4_{m}+..., \end{aligned}$$
(7)
$$\begin{aligned} g(x_{m}-\theta g(x_{m}))& = -F (-1+\theta F) e_{m}+F c_{2} (-3 \theta F+1+\theta ^2 F^2) e^2_{m}-F (2 \theta F c^2_{2}\nonumber \\&\quad-\,2 \theta ^2 F^2 c^2_{2}-c_{3}+4 \theta F c_{3}-3 c_{3} \theta ^2 F^2+\theta ^3 F^3 c_{3}) e^3_{m}\nonumber \\&\quad+\,F (-5 \theta F c_{2} c_{3}+8 \theta ^2 F^2 c_{2} c_{3}-3 \theta ^3 F^3 c_{2} c_{3}+c_{4}\nonumber \\&\quad-\,5 \theta F c_{4}+6 c_{4} \theta ^2 F^2-4 c_{4} \theta ^3 F^3+c_{4} \theta ^4 F^4+\theta ^2 F^2 c^3_{2}) e^4_{m}-... \end{aligned}$$
(8)
Using Eqs. (7) and (8), we have
$$\begin{aligned} g(x_{m}+\theta g(x_{m})) -g(x_{m}-\theta g(x_{m}))&=2 \theta F^2 e_{m}+6 \theta F^2 c_{2} e^2_{m}+(4 c^2_{2} \theta F^2+8 \theta F^2 c_{3}+2 c_{3} \theta ^3 F^4) e^3_{m}\nonumber \\&\quad +(10 c_{3} \theta F^2 c_{2}+6 c_{3} \theta ^3 F^4 c_{2}+10 \theta F^2 c_{4}+8 c_{4} \theta ^3 F^4) e^4_{m}+... \end{aligned}$$
(9)
Combining Eqs. (6) and (9), we get
$$\begin{aligned} \frac{2 \theta g^{2}(x_{m})}{g(x_{m}+\theta g(x_{m})) -g(x_{m}-\theta g(x_{m}))}& = e_{m}-c_{2} e^2_{m}+(2 c^2_{m}-2 c_{3}-c_{3} \theta ^2 F^2) e^3_{m}+(7 c_{2} c_{3}\nonumber \\&\quad+\,\theta ^2 F^2 c_{2} c_{3}-3 c_{4}-4 c_{4} \theta ^2 F^2-4 c^3_{2}) e^4_{m}+... \end{aligned}$$
(10)
By considering these relations and \(y_m\) in Eq. (3), we get
$$\begin{aligned} y_{m}& = \alpha +c_{2} e^2_{m}+(-2 c^2_{2}+2 c_{3}+c_{3} \theta ^2 F^2) e^3_{m}+(-7 c_{2} c_{3}\nonumber \\&\quad-\,\theta ^2 F^2 c_{2} c_{3}+3 c_{4}+4 c_{4} \theta ^2 F^2+4 c^3_{2}) e^4_{m}+... \end{aligned}$$
(11)
At this time, we expand \(g(y_m)\) around \(\alpha\) by using the result in Eq. (11), as accordingly, we get
$$\begin{aligned} g(y_{m})& = F c_{2} e^2_{m}+F (-2 c^2_{2}+2 c_{3}+c_{3} \theta ^2 F^2) e^3_{m}-F (7 c_{2} c_{3}\nonumber \\&\quad+\,\theta ^2 F^2 c_{2} c_{3}-3 c_{4}-4 c_{4} \theta ^2 F^2-5 c^3_{2}) e^4_{m}-... \end{aligned}$$
(12)
By considering these relations and \(z_m\) in Eq. (3), we get
$$\begin{aligned} z_{m}& = \alpha +(2 c^3_{2}-c_{2} c_{3}-\theta ^2 F^2 c_{2} c_{3}) e^4_{m}+(-10 c^4_{2}+14 c_{3} c^2_{2}\nonumber \\&\quad+\,5 \theta ^2 F^2 c^2_{2} c_{3}-2 c^2_{3}-3 \theta ^2 F^2 c^2_{3}-\theta ^4 F^4 c^2_{3}-2 c_{2} c_{4}-4 \theta ^2 F^2 c_{2} c_{4}) e^5_{m}+... \end{aligned}$$
(13)
Expanding \(g(z_{m})\) and about \(\alpha\) and using Eq. (13), we obtain
$$\begin{aligned} g(z_{m})& = -F c_{2} (-2 c^2_{2}+c_{3}+c_{3} \theta ^2 F^2) e^4_{m}-F (10 c^4_{2}-14 c_{3} c^2_{2}\nonumber \\&\quad-\,5 \theta ^2 F^2 c^2_{2} c_{3}+2 c^2_{3}+3 \theta ^2 F^2 c^2_{3}+\theta ^4 F^4 c^2_{3}+2 c_{2} c_{4}+4 \theta ^2 F^2 c_{2} c_{4}) e^5_{m}+... \end{aligned}$$
(14)
Combining Eqs. (13) and (14) we get
$$\begin{aligned} \frac{2 \theta g^{2}(z_{m})}{g(z_{m}+\theta g(z_{m})) -g(z_{m}-\theta g(z_{m}))}& = -(-2 c^2_{2}+c_{3}+c_{3} \theta ^2 F^2) c_{2} e^4_{m}+(-10 c^4_{2}+14 c_{3} c^2_{2}+5 \theta ^2 F^2 c^2_{2} c_{3}\nonumber \\&\quad-\,2 c^2_{3}-3 \theta ^2 F^2 c^2_{3}-\theta ^4 F^4 c^2_{3}-2 c_{2} c_{4}-4 \theta ^2 F^2 c_{2} c_{4}) e^5_{m}+... \end{aligned}$$
(15)
By using Eqs. (13) and (15) in the last expression of Eq. (3), we obtain
$$\begin{aligned} x_{m+1}& = \alpha +(\theta ^4 F^4 c^2_{3}+2 \theta ^2 F^2 c^2_{3}-4 \theta ^2 F^2 c^2_{2} c_{3}+c^2_{3}-4 c_{3} c^2_{2}+4 c^4_{2}) c^3_{2} e^8_{m}+O (e^9_{m}). \end{aligned}$$
(16)
From Eq. (16) and \(e_{m+1} = x_{m+1}-\alpha\) finally we have
$$\begin{aligned} e_{m+1}=(\theta ^4 F^4 c^2_{3}+2 \theta ^2 F^2 c^2_{3}-4 \theta ^2 F^2 c^2_{2} c_{3}+c^2_{3}-4 c_{3} c^2_{2}+4 c^4_{2}) c^3_{2} e^8_{m}+O (e^9_{m}). \end{aligned}$$
(17)
The last equation shows that 8th BM is eight order of convergence. This completes the proof. \(\square\)