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. \(\square\)