Modified Abbasbandy’s method free from second derivative for solving nonlinear equations

1. Introduction

One of the complex problem in science and specially in mathematics is to solve the non-linear equation The solution of such type of equations cannot be find directly except in special cases. Therefore most of the methods for solving such type of equations are iterative methods. In iterative methods, we start with an initial guess \(x_0\) which is improved step by step by means of iterations. In recent years, several iterative methods have been developed by using the different techniques namely: Taylor's series expansion, Adomian decomposition, Quadrature formulae etc. Some basic iterative methods are given in literature [1, 2, 3], and the references therein. Considering (1) and assume that \(\alpha\) is a simple zero of (1) and \(\gamma\) is an initial guess sufficiently close to \(\alpha\) then by using the Taylor's series around \(\gamma\) for (1), we have If \(f^{\prime }(\gamma)\neq 0\), we can evaluate the above expression as follow's: $$ f(x_{k})+(x-x_{k})f^{\prime }(x_{k})=0. $$ If we choose \(x_{k+1}\) the root of equation, then we have This is so-called the Newton's method (NM) [4], for root-finding of nonlinear functions and converges quadratically. From (2), we obtain % This is so-called the Halley's method (HM) [5], for root-finding of nonlinear functions and converges cubically. Simplification of (2) yields another method: This is known as Househ\"older's method [6], for solving nonlinear equations and converges cubically. In this paper, we modified the Abbasbandy's method [7] by making it three step iterative method by taking Newton's method as a pre-predictor and predictor step and Abbasbandy's method as a corrector step. We proved that this new modified methods have twelve, twelve and ten order of convergence and most efficient than existing methods. Some examples are given to show the performance of this method with other famous methods.

2. Iterative methods

Let \(f:X\rightarrow {R},\) \(X\subset{R}\) is a scalar function, then by using Taylor expansion, expanding \(f(x)\) about the point \(x_{k}\), we obtain the Abbasbandy's method as $$ x_{k+1}=x_{k}-\frac{f(x_{k})}{f^{\prime }(x_{k})}-\frac{f^{2}(x_{k})f^{% \prime \prime }(x_{k})}{2f^{\prime 3}(x_{k})}-\frac{f^{3}(x_{k})f^{\prime \prime \prime }(x_{k})}{6f^{\prime 4}(x_{k})}. $$

Algorithm 1. For a given \(x_{0}\), compute the approximate solution \(x_{n+1}\) by the following three step iterative scheme: \begin{eqnarray*}y_{n} &=&x_{n}-\frac{f(x_{n})}{f^{\prime }(x_{n})},n=0,1,2,..., \\ w_{n}&=&y_{n}-\frac{f(y_{n})}{f^{\prime }(y_{n})} \notag\\ x_{n+1}&=&w_{n}-\frac{f(w_{n})}{f^{\prime }(w_{n})}-\frac{f^{2}(w_{n})f^{% \prime \prime }(w_{n})}{2f^{\prime 3}(w_{n})}-\frac{f^{3}(w_{n})f^{\prime \prime \prime }(w_{n})}{6f^{\prime 4}(w_{n})}. \end{eqnarray*}

By following the finite difference scheme, we develop the following algorithms:

Algorithm 2. For a given \(x_{0}\), compute the approximate solution \(x_{n+1}\) by the following iterative schemes: \begin{eqnarray*}y_{n} &=&x_{n}-\frac{f(x_{n})}{f^{\prime }(x_{n})},n=0,1,2,..., \notag\\ w_{n}&=&y_{n}-\frac{f(y_{n})}{f^{\prime }(y_{n})} \notag\\ x_{n+1}&=&w_{n}-\frac{f(w_{n})}{f^{\prime }(w_{n})}-\frac{f^{2}(w_{n})f^{% \prime \prime }(w_{n})}{2f^{\prime 3}(w_{n})}+\frac{f^{3}(w_{n}){f^{\prime }(y_{n})}[f^{\prime \prime }(w_{n})-f^{\prime \prime }(y_{n})]}{6f(y_{n})f^{\prime 4}(w_{n})}.\label{} \end{eqnarray*}

Algorithm 3. For a given \(x_{0}\), compute the approximate solution \(x_{n+1}\) by the following iterative schemes: \begin{eqnarray*}y_{n} &=&x_{n}-\frac{f(x_{n})}{f^{\prime }(x_{n})},n=0,1,2,..., \\ w_{n}&=&y_{n}-\frac{f(y_{n})}{f^{\prime }(y_{n})} \notag\\ x_{n+1}&=&w_{n}-\frac{f(w_{n})}{f^{\prime }(w_{n})}-\frac{f^{\prime }(y_{n})f^{2}(z_{n})}{2f^{\prime 3 }(w_{n})} \left[\frac{f^{\prime }(y_{n})-{f^{\prime }(w_{n})}}{{f(y_{n})}}(1-\frac{f^{\prime }(y_{n})f(w_{n})}{3f(y_{n}) f^{\prime }(w_{n})})+\frac{f^{\prime }(x_{n})f(w_{n})({f^{\prime }(x_{n})}-{f^{\prime }(y_{n})})}{3f(x_{n})f(y_{n}){f^{\prime }(w_{n})}}\right]\label{} \end{eqnarray*}

3. Convergence Analysis

In this section, we prove the convergence of our purposed iterative methods.

Theorem 3.1. Suppose that \(\alpha \) is a root of the equation \(f(x)=0\). If \(f(x)\) is sufficiently smooth in the neighborhood of \(\alpha \), then the convergence order of Algorithm \(1\), Algorithm \(2\) and Algorithm \(3\) is at least twelve, twelve and ten respectively.

Proof. To prove the convergence, suppose that \(\alpha \) is a root of the equation \(f(x)=0\) and \(e_n\) be the error at nth iteration, then \(e_n=x_n-\alpha\) and by using Taylor series expansion, we have \begin{eqnarray*} f(x_n)&=&{f^{\prime }(\alpha)e_n}+\frac{1}{2!}{f^{\prime \prime }(\alpha)e_n^2}+\frac{1}{3!}{f^{\prime \prime \prime }(\alpha)e_n^3}+\frac{1}{4!}{f^{(iv) }(\alpha)e_n^4}+\frac{1}{5!}{f^{(v) }(\alpha)e_n^5}+\frac{1}{6!}{f^{(vi) }(\alpha)e_n^6}+\ldots \end{eqnarray*}

% where $$c_n=\frac{1}{n!}\frac{{f^{(n) }(\alpha)}}{{f^{\prime }(\alpha)}}.$$ With the help of Equation (6) and Equation (7), we get With the help of Equations (8), (9), (10) and (11), we get Using Equations (12), (13), (14), (15) and (16) in Algorithm (1), (2) and (3), we get
\( x_{n+1}=\alpha+(2c_2^{11}-2c_3c_2^9)e_n^{12}+O(e_n^{13}), \)\\ \( x_{n+1}=\alpha+2c_2^{11}e_n^{12}+O(e_n^{13}), \)\\ and \( x_{n+1}=\alpha-\frac{3c_3c_2^7}{2}e_n^{10}+O(e_n^{11}). \)\\ Which implies that Equations \(17\), \(18\) and \(19\) shows that the Algorithms (1), (2) and (3) have convergence of order twelve, twelve and ten respectively.

4. Applications

In this section we solved some nonlinear functions to illustrate the efficiency of our developed algorithms. We compare our developed methods with Newton'smethod (NM), Halley's method (HM and Abbasbanday's method (AM).

Example 1. In this example we solved \(f(x)=x^{3}+4x^{2}-25\) by taking \(x_{0}=-0.8\). It can be observed from Table 1 that NM takes 35 iterations, HM takes 36 iterations, AM takes 13 iterations and our Algorithms (1), (2) and (3) takes 12, 5 and 5 iterations respectively to reach at root of \(f(x)=x^{3}+4x^{2}-25\).

Table 1. Comparison of NM, HM, AM and Algorithms (1), (2) and (3).

Method	\(N\)	\(N_{f}\)	\(|f(x_{n+1})|\)	\(x_{n+1}\)
NM	\(35\)	\(70\)	\(1.105260e-24\)
HM	\(36\)	\(108\)	\(2.995246e-17\)	\(1.365230013414096845760806828980\)
AM	\(13\)	\(52\)	\(6.423767e-20\)
Algorithm 1	\(12\)	\(72\)	\(2.738493e-48\)
Algorithm 2	\(5\)	\(25\)	\(2.812883e-25\)
Algorithm 3	\(5\)	\(20\)	\(3.108248e-83\)

Example 2. In this example we solved \(f(x)=\) \(x^3+x^2-2\) by taking \(x_{0}=-0.1\). It can be observed from Table 2 that NM takes 13 iterations, HM takes 17 iterations, AM takes 19 iterations and our Algorithms (1), (2) and (3) takes 5, 4 and 5 iterations respectively to reach at root of \(f(x)=\) \(x^3+x^2-2\).

Table 2. Comparison of NM, HM, AM and Algorithms (1), (2) and (3).

Method	\(N\)	\(N_{f}\)	\(|f(x_{n+1})|\)	\(x_{n+1}\)
NM	\(13\)	\(26\)	\(2.203086e-19\)
HM	\(17\)	\(51\)	\(4.338982e-22\)	\(1.000000000000000000000000000000\)
AM	\(19\)	\(76\)	\(2.239715e-27\)
Algorithm 1	\(5\)	\(30\)	\(2.338056e-31\)
Algorithm 2	\(4\)	\(20\)	\(5.192250e-45\)
Algorithm 3	\(5\)	\(20\)	\(6.607058e-83\)

Example 3. In this example we solved \(f(x)=\) \(e^{(x^2+7x-30)}-1\) by taking \(x_{0}=4.5\). It can be observed from Table 3 that NM takes 27 iterations, HM takes 14 iterations, AM takes 16 iterations and our Algorithms (1), (2) and (3) takes 8, 7 and 7 iterations respectively to reach at root of \(f(x)=\) \(e^{(x^2+7x-30)}-1\).

Table 3. Comparison of NM, HM, AM and Algorithms (1), (2) and (3)

Method	\(N\)	\(N_{f}\)	\(|f(x_{n+1})|\)	\(x_{n+1}\)
NM	\(27\)	\(54\)	\(6.454129e-23\)
HM	\(14\)	\(42\)	\(1.217550e-25\)	\( 3.000000000000000000000000000000\)
AM	\(16\)	\(64\)	\(1.136732e-17\)
Algorithm 1	\(8\)	\(48\)	\(1.261140e-22\)
Algorithm 2	\(7\)	\(35\)	\(6.546702e-15\)
Algorithm 3	\(7\)	\(28\)	\(9.047215e-71\)

Example 4. In this example we solved \(f(x)=\) \(x^{2}-e^{x}-3x+2\) by taking \(x_{0}=3.5\). It can be observed from Table 4 that NM takes 6 iterations, HM takes 5 iterations, AM takes 5 iterations and our Algorithms (1), (2) and (3) takes 2, 3 and 3 iterations respectively to reach at root of \(f(x)=\) \(x^{2}-e^{x}-3x+2\).

Table 4. Comparison of NM, HM, AM and Algorithms (1), (2) and (3)

Method	\(N\)	\(N_{f}\)	\(|f(x_{n+1})|\)	\(x_{n+1}\)
NM	\(6\)	\(12\)	\(4.925534e-15\)
HM	\(5\)	\(15\)	\(1.463064e-40\)	\(0.257530285439860760455367304937\)
AM	\(5\)	\(20\)	\(1.120893e-28\)
Algorithm 1	\(2\)	\(12\)	\(8.978612e-19\)
Algorithm 2	\(3\)	\(15\)	\(0.000000e +00\)
Algorithm 3	\(3\)	\(12\)	\(4.980111e-66\)

Example 5. In this example we solved \(f(x)=\) \(xe^{x^2}-sin^{2}{x}+3cos{x}+5\) by taking \(x_{0}=1.1\). It can be observed from Table 5 that NM takes 45 iterations, HM takes 44 iterations, AM takes 50 iterations and our Algorithms (1), (2) and (3) takes 14, 12 and 12 iterations respectively to reach at root of \(f(x)=\) \(xe^{x^2}-sin^{2}{x}+3cos{x}+5\).

Table 5. Comparison of NM, HM, AM and Algorithms (1), (2) and (3)

Method	\(N\)	\(N_{f}\)	\(|f(x_{n+1})|\)	\(x_{n+1}\)
NM	45	90	\(1.268546e-15\)
HM	44	132	\(1.169824e-26\)	\( -1.207647827130918927009416758360 \)
AM	\(50\)	\(200\)	\(2.868208e-29\)
Algorithm 1	\(14\)	\(84\)	\(1.935782e-64\)
Algorithm 2	\(12\)	\(60\)	\(4.515078e-97\)
Algorithm 3	\(12\)	\(48\)	\(4.515078e-97\)

Example 6. In this example we solved \(f(x)=\) \(x^{2}+sin(\frac{x}{5})-\frac{1}{4}\) by taking \(x_{0}=2.2\). It can be observed from Table 6 that NM takes 7 iterations, HM takes 5 iterations, AM takes 7 iterations and our Algorithms (1), (2) and (3) takes 2, 2 and 2 iterations respectively to reach at root of \(f(x)=\) \(x^{2}+sin(\frac{x}{5})-\frac{1}{4}\).

Table 6. Comparison of NM, HM, AM and Algorithms (1), (2) and (3)

Method	\(N\)	\(N_{f}\)	\(|f(x_{n+1})|\)	\(x_{n+1}\)
NM	\(7\)	\(14\)	\(7.777907e-23\)
HM	\(5\)	\(15\)	\(1.210132e-42\)	\(0.409992017989137131621258376499\)
AM	\(7\)	\(28\)	\(2.132547e-32\)
Algorithm 1	\(2\)	\(12\)	\(5.800844e-23\)
Algorithm 2	\(2\)	\(10\)	\(5.897018e-23\)
Algorithm 3	\(2\)	\(8\)	\(4.106937e-22\)

Conclusions

Three new algorithms for solving nonlinear functions has been established. We can conclude that the efficiency indexes of algorithms (1), (2) and (3) are 1.5131, 1.6438, and 1.7783 respectively. The convergence orders of algorithms (1), (2) and (3) are twelve, twelve and ten respectively. By solving some examples, the performance of our developed algorithms is discussed. Our developed algorithms are performing well in comparison to Newton's method, Halley's method and Abbasbanday's method.

Author Contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Competing Interests

The author(s) do not have any competing interests in the manuscript.