**1. Introduction**

We consider derivative-free methods for finding the multiple root (say, *α*) with multiplicity *m* of a nonlinear equation *f*(*t*) = 0 , i.e., *f* (*j*)(*α*) = 0, *j* = 0, 1, 2, . . . , *m* − 1 and *f* (*m*)(*α*) = 0.

Several higher order methods, with or without the use of modified Newton's method [1]

$$t\_{k+1} = t\_k - m \frac{f(t\_k)}{f'(t\_k)},\tag{1}$$

have been derived and analyzed in literature (see, for example, [2–15] and references cited therein). In such methods, one requires determining the derivatives of either first order or both first and second order. Contrary to this, higher-order derivative-free methods to compute multiple roots are ye<sup>t</sup> to be investigated. These methods are important in the problems where derivative *f* is complicated to process or is costly to evaluate. The basic derivative-free method is the Traub–Steffensen method [16], which uses the approximation

$$f'(t\_k) \simeq \frac{f(t\_k + \beta f(t\_k)) - f(t\_k)}{\beta f(t\_k)}, \quad \beta \in \mathbb{R} - \{0\},$$

or

$$f'(t\_k) \simeq f[\mathbf{s}\_k, t\_k]\_{\prime}$$

for the derivative *f* in the classical Newton method in Equation (1). Here, *sk* = *tk* + *β f*(*tk*) and *f* [*s*, *t*] = *f*(*s*)−*f*(*t*) *s*−*t* is a divided difference of first order. In this way, the modified Newton method in Equation (1) transforms to the modified Traub–Steffensen derivative free method

$$t\_{k+1} = t\_k - m \frac{f(t\_k)}{f[s\_k, t\_k]}.\tag{2}$$

The modified Traub–Steffensen method in Equation (2) is a noticeable improvement over the Newton method, because it preserves the convergence of order two without using any derivative.

In this work, we aim to design derivative-free multiple root methods of high efficient quality, i.e., the methods of higher convergence order that use the computations as small as we please. Proceeding in this way, we introduce a class of derivative-free fourth-order methods that require three new pieces of information of the function *f* per iteration, and hence possess optimal fourth-order convergence in the terminology of Kung–Traub conjecture [17]. This conjecture states that multi-point iterative functions without memory based on *n* function evaluations may attain the convergence order <sup>2</sup>*<sup>n</sup>*−1, which is maximum. The methods achieving this convergence order are usually called optimal methods. The new iterative scheme uses the modified Traub–Steffensen iteration in Equation (2) in the first step and Traub–Steffensen-like iteration in the second step. The methods are examined numerically on many practical problems of different kind. The comparison of performance with existing techniques requiring derivative evaluations verifies the efficient character of the new methods in terms of accuracy and executed CPU time.

The rest of the paper is summarized as follows. In Section 2, the scheme of fourth-order method is proposed and its convergence order is studied for particular cases. The main result for the general case is studied in Section 3. Numerical tests to demonstrate applicability and efficiency of the methods are presented in Section 4. In this section, a comparison of performance with already established methods is also shown. In Section 5, a conclusion of the main points is drawn.

#### **2. Formulation of Method**

To compute a multiple root with multiplicity *m* ≥ 1, consider the following two-step iterative scheme:

$$z\_k = t\_k - m \frac{f(t\_k)}{f[s\_{k'} t\_k]},$$

$$t\_{k+1} = z\_k - H(x\_{k'} y\_k) \frac{f(t\_k)}{f[s\_{k'} t\_k]},\tag{3}$$

where *xk* = *m*0*f*(*zk* ) *f*(*tk* ) , *yk* = *m*0*f*(*zk* ) *f*(*sk* ) and *H* : C<sup>2</sup> → C is analytic in a neighborhood of (0, <sup>0</sup>). Notice that this is a two-step scheme with first step as the Traub–Steffensen iteration in Equation (2) and the next step as the Traub–Steffensen-like iteration. The second step is weighted by the factor *<sup>H</sup>*(*<sup>x</sup>*, *y*), thus we can call it weight factor or more appropriately weight function.

In the sequel, we study the convergence results of proposed iterative scheme in Equation (3). For clarity, the results are obtained separately for different cases based on the multiplicity *m*. Firstly, for the case *m* = 1, the following theorem is proved:

**Theorem 1.** *Assume that f* : C → C *is an analytic function in a domain containing a multiple zero (say, α) with multiplicity m* = 1*. Suppose that the initial point t*0 *is close enough to α, then the convergence order of Equation* (3) *is at least* 4*, provided that H*00 = 0*, H*10 = 1*, H*01 = 0*, H*20 = 2*, H*11 = 11 *and H*02 = 0*, where Hij* = *∂i*+*j <sup>∂</sup>xi∂y<sup>j</sup> <sup>H</sup>*(*xk*, *yk*)|(*xk*=0,*yk*=<sup>0</sup>)*, for* 0 ≤ *i*, *j* ≤ 2*.*

**Proof.** Assume that the error at *k*th stage is *ek* = *tk* − *α*. Using the Taylor's expansion of *f*(*tk*) about *α* and keeping into mind that *f*(*α*) = 0 and *f* (*α*) = 0, we have

$$f(t\_k) = f'(a)e\_k(1 + A\_1e\_k + A\_2e\_k^2 + A\_3e\_k^3 + A\_4e\_k^4 + \cdots),\tag{4}$$

where *An* = 1 (<sup>1</sup>+*n*)! *f*(<sup>1</sup>+*n*)(*α*) *f* (*α*) for *n* ∈ N.

Similarly we have the Taylor's expansion of *f*(*sk*) about *α*

$$f(s\_k) = f'(a)\varepsilon\_{s\_k} \left(1 + A\_1 \varepsilon\_{s\_k} + A\_2 \varepsilon\_{s\_k}^2 + A\_3 \varepsilon\_{s\_k}^3 + A\_4 \varepsilon\_{s\_k}^4 + \dotsb\right),\tag{5}$$

where *esk* = *sk* − *α* = *ek* + *β f* (*α*)*ek*.<sup>1</sup> + *A*1*ek* + *<sup>A</sup>*2*e*2*k* + *<sup>A</sup>*3*e*3*k* + *<sup>A</sup>*4*e*4*k* + ··· /. Then, the first step of Equation (3) yields

$$\begin{split} e\_{\bar{z}\_{k}} &= z\_{k} - a \\ &= (1 + \beta f'(a))A\_{1}e\_{k}^{2} - \left( (2 + 2\beta f'(a) + (\beta f'(a))^{2})A\_{1}^{2} - (2 + 3\beta f'(a) + (\beta f'(a))^{2})A\_{2} \right) e\_{k}^{3} + \left( (4 + 5\beta f'(a))^{2} - (6 + 6\beta f'(a))^{2} \right) e\_{k}^{4} \\ &+ 3(\beta f'(a))^{2} + (\beta f'(a))^{3})A\_{1}^{3} - (7 + 10\beta f'(a) + 7(\beta f'(a))^{2} + 2(\beta f'(a))^{3})A\_{1}A\_{2} + (3 + 6\beta f'(a))e\_{k}^{2} \\ &+ 4(\beta f'(a))^{2} + (\beta f'(a))^{3})A\_{3})e\_{k}^{4} + O(v\_{k}^{5}). \end{split} \tag{6}$$

Expanding *f*(*zk*) about *α*, it follows that

$$f(z\_k) = f'(a)e\_{\overline{z}\_k} \left( 1 + A\_1 e\_{\overline{z}\_k} + A\_2 e\_{\overline{z}\_k}^2 + A\_3 e\_{\overline{z}\_k}^3 + \cdots \right). \tag{7}$$

Using Equations (4), (5) and (7) in *xk* and *yk*, after some simple calculations, we have

$$\begin{aligned} \alpha\_k &= (1 + \beta f'(a))A\_1 e\_k - \left( (3 + 3\beta f'(a) + (\beta f'(a))^2)A\_1^2 - (2 + 3\beta f'(a) + (\beta f'(a))^2)A\_2 \right) e\_k^2 + \left( (8 + 10\beta f'(a))^3 - (\beta f'(a))^3 \right) e\_k^2 \\ &+ 5(\beta f'(a))^2 + (\beta f'(a))^3 (A\_1^3 - 2(5 + 7\beta f'(a) + 4(\beta f'(a))^2 + (\beta f'(a))^3)A\_1 A\_2 + (3 + 6\beta f'(a))^2 e\_k^2 \\ &+ 4(\beta f'(a))^2 + (\beta f'(a))^3 (A\_3) e\_k^3 + O(\varepsilon\_k^4) \end{aligned} \tag{8}$$

and

$$y\_k = A\_1 e\_k - \left( (3 + 2\mathfrak{H}f'(a))A\_1^2 - (2 + \mathfrak{H}f'(a))A\_2 \right) e\_k^2 + \left( (8 + 8\mathfrak{H}f'(a) + 3(\mathfrak{H}f'(a))^2)A\_1^3 \right.\\ \left. \begin{aligned} &(8 + 8\mathfrak{H}f'(a) + 3(\mathfrak{H}f'(a))^2)A\_1^3 \\ &(8 + 3\mathfrak{H}f'(a) + 3(\mathfrak{H}f'(a))^2(A\_1^2 + (3 + 3\mathfrak{H}f'(a) + (\mathfrak{H}f'(a))^2)A\_3)e\_k^3 + O(\varepsilon\_k^4). \end{aligned} \right.$$

Developing *<sup>H</sup>*(*xk*, *yk*) by Taylor series in the neighborhood of origin (0, <sup>0</sup>),

$$H(\mathbf{x}\_k, y\_k) \approx H\_{00} + \mathbf{x}\_k H\_{10} + y\_k H\_{01} + \frac{1}{2} \mathbf{x}\_k^2 H\_{20} + \mathbf{x}\_k y\_k H\_{11} + \frac{1}{2} y\_k^2 H\_{02}.\tag{10}$$

Inserting Equations (4)–(10) into the second step of Equation (3), and then some simple calculations yield

$$\begin{aligned} e\_{k+1} &= -H\_{00}e\_k + \left(H\_{00} - H\_{01} + \beta f'(a)H\_{00} - (-1 + H\_{10})(1 + \beta f'(a))\right)A\_1e\_k^2 - \frac{1}{2}\left((4 + H\_{02} - 8H\_{10} + 2H\_{11} + 2H\_{11} + 2H\_{12}\right)(e\_k - 1) \\ &+ H\_{20} + 4\beta f'(a) - 10\beta f'(a)H\_{10} + 2\beta f'(a)H\_{11} + 2\beta f'(a)H\_{20} + 2(\beta f'(a))^2 - 4(\beta f'(a))^2H\_{10} \\ &+ (\beta f'(a))^2H\_{20} - 2H\_{01}(4 + 3\beta f'(a)) + 2H\_{00}(2 + 2\beta f'(a) + (\beta f'(a))^2))A\_1^2 - 2(2 + \beta f'(a))(H\_{00} \\ &- H\_{01} + \beta f'(a)H\_{00} - (-1 + H\_{10})(1 + \beta f'(a)))A\_2)e\_k^3 + \delta e\_k^4 + O(f\_k^5), \end{aligned}$$

where *δ* = *<sup>δ</sup>*(*β*, *A*1, *A*2, *A*3, *H*00, *H*10, *H*01, *H*20, *H*11, *<sup>H</sup>*02). Here, expression of *δ* is not being produced explicitly since it is very lengthy.

It is clear from Equation (11) that we would obtain at least fourth-order convergence if we set coefficients of *ek*, *e*2*k*and *e*3*k*simultaneously equal to zero. Then, solving the resulting equations, one gets

$$H\_{00} = 0, \ H\_{10} = 1, \ H\_{01} = 0, \ H\_{20} = 2, \ H\_{11} = 1, \ H\_{02} = 0. \tag{12}$$

As a result, the error equation is given by

$$c\_{k+1} = (1 + \beta f'(a))A\_1 \left( (\dots + \dots \oplus f'(a) + (\beta f'(a))^2)A\_1^2 - (1 + \beta f'(a))A\_2 \right) \epsilon\_k^4 + O(\epsilon\_k^5). \tag{13}$$

Thus, the theorem is proved.

Next, we show the conditions for *m* = 2 by the following theorem:

**Theorem 2.** *Using the hypotheses of Theorem 1, the order of convergence of the scheme in Equation* (3) *for the case m* = 2 *is at least* 4*, if H*00 = 0*, H*10 = 1*, H*01 = 1*, and H*20 = 8 − *H*02 − 2*H*11*, wherein* {|*<sup>H</sup>*11|, |*<sup>H</sup>*02|} < ∞*.*

**Proof.** Assume that the error at *k*th stage is *ek* = *tk* − *α*. Using the Taylor's expansion of *f*(*tk*) about *α* and keeping in mind that *f*(*α*) = 0, *f* (*α*) = 0, and *f* (2)(*α*) = 0, we have

$$f(t\_k) = \frac{f^{(2)}(a)}{2!}c\_k^2(1 + B\_1c\_k + B\_2c\_k^2 + B\_3c\_k^3 + B\_4c\_k^4 + \cdots),\tag{14}$$

where *Bn* = 2! (<sup>2</sup>+*n*)! *f*(<sup>2</sup>+*n*)(*α*) *f*(2)(*α*) for *n* ∈ N.

Similarly, we have the Taylor's expansion of *f*(*sk*) about *α*

$$f(\mathbf{s}\_k) = \frac{f^{(2)}(\mathbf{a})}{2!} c\_{\mathbf{s}\_k}^2 \left( 1 + B\_1 \mathbf{e}\_{\mathbf{s}\_k} + B\_2 \mathbf{e}\_{\mathbf{s}\_k}^2 + B\_3 \mathbf{e}\_{\mathbf{s}\_k}^3 + B\_4 \mathbf{e}\_{\mathbf{s}\_k}^4 + \dotsb \right), \tag{15}$$

where *esk* = *sk* − *α* = *ek* + *β f*(2)(*α*) 2! *<sup>e</sup>*2*k*.<sup>1</sup> + *B*1*ek* + *<sup>B</sup>*2*e*2*k* + *<sup>B</sup>*3*e*3*k* + *<sup>B</sup>*4*e*4*k* + ··· /. Then, the first step of Equation (3) yields

$$\begin{aligned} \varepsilon\_{z\_i} &= z\_k - a \\ &= \frac{1}{2} \left( \frac{\mathfrak{F}f^{(2)}(a)}{2} + B\_1 \right) \varepsilon\_k^2 - \frac{1}{16} \left( (\mathfrak{F}f^{(2)}(a))^2 - 8\mathfrak{F}f^{(2)}(a)B\_1 + 12B\_1^2 - 16B\_2 \right) \varepsilon\_k^3 + \frac{1}{64} \left( (\mathfrak{F}f^{(2)}(a))^3 - 3\mathfrak{F}f^{(2)}(a)B\_1 - 6\mathfrak{F}f^{(2)}(a)B\_1 - 3\mathfrak{F}f^{(2)}(a)B\_1 - 3\mathfrak{F}f^{(2)}(a)B\_1 - 2\mathfrak{F}f^{(2)}(a)B\_1 - 2\mathfrak{F}f^{(2)}(a)B\_1 - 3\mathfrak{F}f^{(2)}(a)B\_1 - 2\mathfrak{F}f^{(2)}(a)B\_1 - 2\mathfrak{F}f^{(2)}(a)B\_1 \right) \varepsilon\_k^2 \\ &- 20\mathfrak{F}f^{(2)}(a)B\_1^2 + 72B\_1^3 + 64\mathfrak{F}f^{(2)}(a)B\_2 - 10B\_1 \left( (\mathfrak{F}f^{(2)}(a))^2 + 16B\_2 \right) + 96B\_3 \rfloor \varepsilon\_k^4 + O(\varepsilon\_k^5). \end{aligned}$$

Expanding *f*(*zk*) about *α*, it follows that

$$f(z\_k) = \frac{f^{(2)}(a)}{2!} e\_{z\_k}^2 \left( 1 + B\_1 \varepsilon\_{\overline{z}\_k} + B\_2 \varepsilon\_{\overline{z}\_k}^2 + B\_3 \varepsilon\_{\overline{z}\_k}^3 + B\_4 \varepsilon\_{\overline{z}\_k}^4 + \dotsb \right). \tag{17}$$

Using Equations (14), (15) and (17) in *xk* and *yk*, after some simple calculations, we have

$$\begin{split} \alpha\_{k} &= \frac{1}{2} \left( \frac{\oint f^{(2)}(a)}{2} + B\_{1} \right) e\_{k} - \frac{1}{16} \left( (\beta f^{(2)}(a))^{2} - 6\beta f^{(2)}(a)B\_{1} + 16(B\_{1}^{2} - B\_{2}) \right) e\_{k}^{2} + \frac{1}{64} \left( (\beta f^{(2)}(a))^{3} - 6\beta f^{(2)}(a)B\_{1} + 16(B\_{1}^{2} - B\_{2}) \right) e\_{k}^{3} \\ &- 22\beta f^{(2)}(a)B\_{1}^{2} + 4\left( 29B\_{1}^{3} + 14\beta f^{(2)}(a)B\_{2} \right) - 2B\_{1} \left( 3(\beta f^{(2)}(a))^{2} + 104B\_{2} \right) + 96B\_{3} \beta\_{k}^{3} + O(\varepsilon\_{k}^{4}) \end{split} \tag{18}$$

and

*yk* = 1 2 *β f* (2)(*α*) 2 + *<sup>B</sup>*1*ek* − 116 .3(*β f* (2)(*α*))<sup>2</sup> − 2*β f*(2)(*α*)*<sup>B</sup>*1 + 16(*B*21 − *<sup>B</sup>*2)/*e*2*k* + 164 .7(*β f*(2)(*α*))<sup>3</sup> + 24*β f*(2)(*α*)*<sup>B</sup>*2 − 14*β f* (2)(*α*)*B*21 + 116*B*31 − <sup>2</sup>*B*1.11(*β f*(2)(*α*))<sup>2</sup> + <sup>104</sup>*B*2/ + <sup>96</sup>*B*3/*e*3*k* + *<sup>O</sup>*(*e*4*k* ). (19)

Developing by Taylor series the weight function *<sup>H</sup>*(*xk*, *yk*) in the neighborhood of origin (0, <sup>0</sup>),

$$H(\mathbf{x}\_k, y\_k) \approx H\_{00} + \mathbf{x}\_k H\_{10} + y\_k H\_{01} + \frac{1}{2} \mathbf{x}\_k^2 H\_{20} + \mathbf{x}\_k y\_k H\_{11} + \frac{1}{2} y\_k^2 H\_{02}.\tag{20}$$

Inserting Equations (14)–(20) intothe second step of Equation (3), and then some simple calculations yield

$$e\_{k+1} = -\frac{H\_{00}}{2}e\_k + \frac{1}{4}\left(2 + H\_{00} - H\_{01} - H\_{10}\right)\left(\frac{\beta f^{(2)}(a)}{2} + B\_1\right)e\_k^2 - \frac{1}{64}\left((\beta f^{(2)}(a))^2(4 + 2H\_{00} - 8H\_{01} + H\_{02})\right)e\_k^2 + \frac{1}{2}\left(\frac{\beta^2 f^{(2)}(a)}{2} + H\_{00} - H\_{01}\right)e\_k^2$$

$$-4H\_{10} + 2H\_{11} + H\_{20}) + 4\beta f^{(2)}(a)\left(-8 - 4H\_{00} - H\_{01} + H\_{02} + H\_{10} + 2H\_{11} + H\_{20}\right)B\_1 + 4(12 + 6H\_{00} - H\_{01})e\_k^2 + \frac{1}{2}\left(\frac{\beta^2 f^{(2)}(a)}{2} + H\_{01} - H\_{02}\right)e\_k^2 + \frac{1}{2}\left(\frac{\beta^2 f^{(2)}(a)}{2} + H\_{01} - H\_{02}\right)e\_k^2$$

$$-10H\_{01} + H\_{02} - 10H\_{10} + 2H\_{11} + H\_{20})B\_1^2 - 32(2 + H\_{00} - H\_{01} - H\_{10})B\_2\Big)e\_k^3 + \phi^4e\_k^4 + O(e\_k^5),\tag{21}$$

where *φ* = *φ*(*β*, *B*1, *B*2, *B*3, *H*00, *H*10, *H*01, *H*20, *H*11, *<sup>H</sup>*02). Here, expression of *φ* is not being produced explicitly since it is very lengthy.

It is clear from Equation (21) that we would obtain at least fourth-order convergence if we set coefficients of *ek*, *e*2*k*and *e*3*k*simultaneously equal to zero. Then, solving the resulting equations, one gets

$$H\_{00} = 0, \ H\_{10} = 1, \ H\_{01} = 1, \ H\_{20} = 8 - H\_{02} - 2H\_{11}.\tag{22}$$

As a result, the error equation is given by

$$e\_{k+1} = \frac{1}{32} \left( \frac{\beta f^{(2)}(a)}{2} + B\_1 \right) \left( (2\beta f^{(2)}(a)(3 + H\_{02} + H\_{11})B\_1 + 22B\_1^2 + ((\beta f^{(2)}(a))^2(H\_{02} + H\_{11}) - 8B\_2))e\_k^4 + O(e\_k^5) \right)$$

Thus, the theorem is proved.

Below, we state the theorems (without proof) for the cases *m* = 3, 4, 5 as the proof is similar to the above proved theorems.

**Theorem 3.** *Using the hypotheses of Theorem 1, the order of convergence of scheme in Equation* (3) *for the case m* = 3 *is at least* 4*, if H*00 = 0*, H*10 = 3 − *H*01*, and H*20 = 12 − *H*02 − 2*H*11*, where* {|*<sup>H</sup>*01|, |*<sup>H</sup>*02|, |*<sup>H</sup>*11|} < ∞*. Moreover, the scheme satisfies error equation*

$$\varepsilon\_{k+1} = \frac{1}{54} (\beta f^{(3)}(a)(-3 + H\_{01})\mathbb{C}\_1 + 12\mathbb{C}\_1^3 - 6\mathbb{C}\_1\mathbb{C}\_2)\varepsilon\_k^4 + O(\varepsilon\_k^5),$$

*where Cn* = 3! (<sup>3</sup>+*n*)! *f*(<sup>3</sup>+*n*)(*α*) *f*(3)(*α*) *for n* ∈ N*.*

**Theorem 4.** *Using the hypotheses of Theorem 1, the order of convergence of scheme in Equation* (3) *for the case m* = 4 *is at least* 4*, if H*00 = 0*, H*10 = 4 − *H*01*, and H*20 = 16 − *H*02 − 2*H*11*, where* {|*<sup>H</sup>*01|, |*<sup>H</sup>*02|, |*<sup>H</sup>*11|} < ∞*. Moreover, the scheme satisfies error equation*

$$e\_{k+1} = \frac{1}{128} (13D\_1^3 - 8D\_1D\_2)e\_k^4 + O(e\_k^5),$$

*where Dn* = 4! (<sup>4</sup>+*n*)! *f*(<sup>4</sup>+*n*)(*α*) *f*(4)(*α*) *for n* ∈ N*.* **Theorem 5.** *Using the hypotheses of Theorem 1, the order of convergence of scheme in Equation* (3) *for the case m* = 5 *is at least* 4*, if H*00 = 0*, H*10 = 5 − *H*01*, and H*20 = 20 − *H*02 − 2*H*11*, where* {| *<sup>H</sup>*01|, |*<sup>H</sup>*02|, |*<sup>H</sup>*11|} < ∞*. Moreover, the scheme satisfies error equation*

$$
\epsilon\_{k+1} = \frac{1}{125} \left( \nabla E\_1^3 - 5E\_1 E\_2 \right) \epsilon\_k^4 + O(\epsilon\_k^5),
$$

*where En* = 5! (<sup>5</sup>+*n*)! *f*(<sup>5</sup>+*n*)(*α*) *f*(5)(*α*) *for n* ∈ N*.*

**Remark 1.** *We can observe from the above results that the number of conditions on Hij is* 6, 4, 3, 3, 3 *corresponding to cases m* = 1, 2, 3, 4, 5 *to attain the fourth-order convergence of the method in Equation* (3)*. The cases m* = 3, 4, 5 *satisfy the common conditions, H*00 = 0*, H*10 = *m* − *H*01*, and H*20 = 4*m* − *H*02 − 2*H*11*. Nevertheless, their error equations differ from each other as the parameter β does not appear in the equations for m* = 4, 5*. It has been seen that when m* ≥ 4 *the conditions on Hij are always three in number and the error equation in each such case does not contain β term. This type of symmetry in the results helps us to prove the general result, which is presented in next section.*

## **3. Main Result**

For the multiplicity *m* ≥ 4, we prove the order of convergence of the scheme in Equation (3) by the following theorem:

**Theorem 6.** *Assume that the function f* : C → C *is an analytic in a domain containing zero α having multiplicity m* ≥ 4*. Further, suppose that the initial estimation t*0 *is close enough to α. Then, the convergence of the iteration scheme in Equation* (3) *is of order four, provided that H*00 = 0*, H*10 = *m* − *H*01*, and H*20 = 4*m* − *H*02 − 2*H*11*, wherein* {| *<sup>H</sup>*01|, |*<sup>H</sup>*11|, |*<sup>H</sup>*02|} < ∞*. Moreover, the error in the scheme is given by*

$$
\varepsilon\_{k+1} = \frac{1}{2m^3} \left( (9+m)K\_1^3 - 2mK\_1K\_2 \right) \varepsilon\_k^4 + O(\varepsilon\_k^5).
$$

**Proof.** Taking into account that *f* (*j*)(*α*) = 0, *j* = 0, 1, 2, ... , *m* − 1 and *f <sup>m</sup>*(*α*) = 0, then, developing *f*(*tk*) about *α* in the Taylor's series,

$$f(t\_k) = \frac{f^m(\alpha)}{m!} e\_k^m \left(1 + K\_1 e\_k + K\_2 e\_k^2 + K\_3 e\_k^3 + K\_4 e\_k^4 + \cdots \right),\tag{23}$$

where *Kn* = *m*! (*m*+*n*)! *f*(*m*+*n*)(*α*) *f*(*m*)(*α*) for *n* ∈ N.

> In addition, from the expansion of *f*(*sk*) about *α*, it follows that

$$f(s\_k) = \frac{f^m(a)}{m!}e\_{s\_k}^m(1 + K\_1e\_{s\_k} + K\_2e\_{s\_k}^2 + K\_3e\_{s\_k}^3 + K\_4e\_{s\_k}^4 + \dotsb),\tag{24}$$

where *esk* = *sk* − *α* = *ek* + *β f <sup>m</sup>*(*α*) *m*! *e<sup>m</sup> k* . 1 + *K*1*ek* + *<sup>K</sup>*2*e*<sup>2</sup> *k* + *<sup>K</sup>*3*e*<sup>3</sup> *k* + *<sup>K</sup>*4*e*<sup>4</sup> *k* + ··· /. From the first step of Equation (3),

$$\begin{split} e\_{2\_k} &= z\_k - a \\ &= \frac{K\_1}{m} e\_k^2 + \frac{1}{m^2} \left( 2mK\_2 - (1+m)K\_1^2 \right) e\_k^3 + \frac{1}{m^3} \left( (1+m)^2 K\_1^3 - m(4+3m)K\_1 K\_2 + 3m^2 K\_3 \right) e\_k^4 + O(e\_k^5). \end{split} \tag{25}$$

Expansion of *f*(*zk*) around *α* yields

$$f(z\_k) = \frac{f^m(a)}{m!} e\_{z\_k}^2 \left( 1 + K\_1 e\_{z\_k} + K\_2 e\_{z\_k}^2 + K\_3 e\_{z\_k}^3 + K\_4 e\_{z\_k}^4 + \cdots \right). \tag{26}$$

Using Equations (23), (24) and (26) in the expressions of *xk* and *yk*, we have that

$$\alpha\_k = \frac{K\_1}{m}\varepsilon\_k + \frac{1}{m^2} \left( 2mK\_2 - (2+m)K\_1^2 \right) \varepsilon\_k^2 + \frac{1}{2m^3} \left( (7+7m+2m^2)K\_1^3 - 2m(7+3m)K\_1K\_2 + 6m^2K\_3 \right) \varepsilon\_k^3 + O(\varepsilon\_k^4) \tag{27}$$

and

$$y\_k = \frac{K\_1}{m}e\_k + \frac{1}{m^2} \left( mK\_2 - (2+m)K\_1^2 \right) e\_k^2 + \frac{1}{2m^3} \left( (6+7m+2m^2)K\_1^3 - 2m(6+3m)K\_1K\_2 + 6m^2K\_3 \right) e\_k^3 + O(e\_k^4). \tag{28}$$

Developing *<sup>H</sup>*(*xk*, *yk*) in Taylor's series in the neighborhood of origin (0, <sup>0</sup>),

$$H(\mathbf{x}\_k, y\_k) \approx H\_{00} + \mathbf{x}\_k H\_{10} + y\_k H\_{01} + \frac{1}{2} \mathbf{x}\_k^2 H\_{20} + \mathbf{x}\_k y\_k H\_{11} + \frac{1}{2} y\_k^2 H\_{02}.\tag{29}$$

Inserting Equations (23)–(29) into the second step of Equation (3), it follows that

$$\begin{split} e\_{k+1} &= -\frac{H\_{00}}{m}e\_k + \frac{1}{m^2}(H\_{00} - H\_{01} - H\_{10} + m)K\_1e\_k^2 - \frac{1}{2m^3}((H\_{02} - 6H\_{10} + 2H\_{11} + H\_{20} + 2m - 2mH\_{10} + 2m^2) \\ &+ 2(1+m)H\_{00} - 2(3+m)H\_{01})K\_1^2 - 4m(H\_{00} - H\_{01} - H\_{10} + m)K\_2)e\_k^3 + \frac{1}{2m^4}((5H\_{02} - 13H\_{10} + 10H\_{11} \\ &+ 5H\_{20} + 2m + 2mH\_{02} - 11mH\_{10} + 4mH\_{11} + 2mH\_{20} + 4m^2 - 2m^2H\_{10} + 2m^3 + 2(1+m)^2H\_{00} \\ &- (13+11m+2m^2)H\_{01})K\_1^3 - 2m(2H\_{02} - 11H\_{10} + 4H\_{11} + 2H\_{20} + 4m - 3mH\_{10} + 3m^2 + (4+3m)H\_{00} \\ &- (11+3m)H\_{01})K\_1K\_2 + 6m^2(H\_{00} - H\_{01} - H\_{10} + m)K\_3)e\_k^4 + O(e\_k^5). \end{split}$$

It is clear that we can obtain at least fourth-order convergence if the coefficients of *ek*, *e*2*k* , and *e*3*k* vanish. On solving the resulting equations, we ge<sup>t</sup>

$$H00 = 0, \ H\_{10} = m - H\_{01}, \ H\_{20} = 4m - H\_{02} - 2 \ H\_{11}.\tag{31}$$

Then, error of Equation (30) is given by

$$e\_{k+1} = \frac{1}{2m^3} \left( (9+m)K\_1^3 - 2mK\_1K\_2 \right) e\_k^4 + O(e\_k^5). \tag{32}$$

Thus, the theorem is proved.

**Remark 2.** *The proposed scheme in Equation* (3) *reaches fourth-order convergence provided that the conditions of Theorems 1–3 and 6 are satisfied. This convergence rate is achieved by using only three function evaluations, viz. f*(*tn*)*, f*(*sn*)*, and f*(*zn*)*, per iteration. Therefore, the scheme in Equation* (3) *is optimal by the Kung–Traub hypothesis [17].*

**Remark 3.** *It is important to note that parameter β, which is used in sk, appears only in the error equations of the cases m* = 1, 2, 3 *but not for m* ≥ 4*. For m* ≥ 4*, we have observed that this parameter appears in the coefficients of e*5*kand higher order. However, we do not need such terms to show the required fourth-order convergence.*
