A New Application of Gauss Quadrature Method for Solving Systems of Nonlinear Equations

S S symmetry

Article A New Application of Gauss Quadrature Method for Solving Systems of Nonlinear Equations

Hari M. Srivastava 1,2,3 , Javed Iqbal 4, Muhammad Arif 4 , Alamgir Khan 4, Yusif S. Gasimov 3 and Ronnason Chinram 5,*

1 Department of Mathematics and Statistics, University of Victoria, Victoria, BC V8W 3R4, Canada; [email protected] 2 Department of Medical Research, China Medical University Hospital, China Medical University, Taichung 40402, Taiwan 3 Department of Mathematics and Informatics, Azerbaijan University, 71 Jeyhun Hajibeyli Street, Baku AZ1007, Azerbaijan; [email protected] 4 Department of Mathematics, Abdul Wali Khan University, Mardan 23200, KPK, Pakistan; [email protected] (J.I.); [email protected] (M.A.); [email protected] (A.K.) 5 Division of Computational Science, Faculty of Science, Prince of Songkla University, Hat Yai, Songkhla 90110, Thailand * Correspondence: [email protected]

Abstract: In this paper, we introduce a new three-step Newton method for solving a system of nonlinear equations. This new method based on Gauss quadrature rule has sixth order of convergence (with n = 3). The proposed method solves nonlinear boundary-value problems and integral equations in few iterations with good accuracy. Numerical comparison shows that the new method is remarkably effective for solving systems of nonlinear equations.

Citation: Srivastava, H.M.; Iqbal, J.; Keywords: nonlinear equations; gauss quadrature formula; ordinary differential equation (ODE); Arif, M.; Khan, A.; Gasimov, Y.S.; error equations; sixth-order convergence; numerical examples Chinram, R. A New Application of Gauss Quadrature Method for Solving Systems of Nonlinear Equations. Symmetry 2021, 13, 432. 1. Introduction https://doi.org/10.3390/sym13030432 In numerical analysis and other branches of scientiﬁc interests, solving a system of nonlinear equations by means of computational methods has always been very well Academic Editor: Sun Young Cho motivated and convincing for researchers. For a system of nonlinear equations:

Received: 26 January 2021 T P(S) = p (S) p (S) ··· p (S) = Accepted: 1 March 2021 1 , 2 , , n 0, (1) Published: 7 March 2021 T n n where S = (s1, s2, ··· , sn) and P : D ⊆ R −→ R is a nonlinear system, and pi, i = 1, 2, ··· , n : D ⊆ n −→ is a nonlinear mapping. The solution of the nonlinear system Publisher’s Note: MDPI stays neutral R R S∗ = (s∗ s∗ ··· s∗)T with regard to jurisdictional claims in of equations in (1) may be defined as the process of finding a vector 1, 2, , n ( ∗) = published maps and institutional affil- such that P S 0. The classical Newton method is one of the most commonly used iations. iterative methods: −1 S(k+1) = Sk − P0S(k) PS(k) (k = 0, 1, 2, ··· ),

where P0(S(k)) is the Jacobian matrix of the nonlinear function P(S) in the kth iteration at Copyright: c 2021 by the authors. the point S(k) (see [1–3]). Newton’s method quadratically converges to the solution S∗ if Licensee MDPI, Basel, Switzerland. the function P is continuous and differentiable. In recent years, several methods have been This article is an open access article developed to analyze the solution of systems of nonlinear equations to improve interaction distributed under the terms and by using the quadrature formulas and fractional iterative method in the literature (see [4–12]). conditions of the Creative Commons Attribution (CC BY) license (https:// creativecommons.org/licenses/by/ 4.0/).

Symmetry 2021, 13, 432. https://doi.org/10.3390/sym13030432 https://www.mdpi.com/journal/symmetry Symmetry 2021, 13, 432 2 of 12

In particular, Codero and Torregrosa [9] developed the third-order Newton–Simpson method as follows: " ! #−1 S(k) + Y(k) S(k+1) = S(k) − 6 P0(S(k)) + 4P0 + P0(Y(k)) P(S(k)), (2) 2

and the Open Newton method:

S(k) + 3Z(k) S(k) + Z(k) S(k+1) = S(k) − 3 2P0 − P0 4 2 − (3) 3S(k) + Z(k) 1 + 2P0 P(S(k)), 4

where Z(k) represents the Newton approximation. Khirallah and Haﬁz [13] suggested a cubically convergent method using the four-point Newton–Cotes formula for solving systems of nonlinear equations as follows:

2S(k) + Z(k) S(k) + 2Z(k) S(k+1) = S(k) − 8 P0(S(k)) + 3P0 + 3P0 2 2 (4) −1 + P0(Z(k)) P(S(k)).

The quadrature rule is used to approximate the deﬁnite integral of a function. The general form of a quadrature rule is given by [14]

Z b m Q = v(r)s(r)dr ' Qm = ∑ wis(ri) a i=0

where v(r) is a weight function, wi, i = 0, 1, 2, ... m are coefﬁcients (weights). ri are points of the rule and s is a given function integrable on the interval [a, b] with the weight function v. Motivated and inspired by the research going on in this area, we have introduced a new iterative for solving nonlinear equations. Several numerical examples are considered to show the effectiveness of the proposed method. The new iterative method shows the compatibility of numerical results with the scheme’s theoretical analysis. We have solved nonlinear boundary-value problems by using the proposed method. Our method gives better results than the other methods and converges more rapidly to the solution. Section5 concludes the paper.

2. Three-Step Newton Method Let P : D ⊆ Rn −→ Rn, be s-times Fréchet differentiable function on a convex set ( ) D ⊆ Rn. Using the Mean-Value Theorem of multi-variable vectors function P(S k ) (see [1]), we have Z 1 P(S) − P(S(k)) = P0S(k) + r(S − S(k))(S − S(k))dr. (5) 0 Using the left rectangular rule, the right-hand side of (5) can be written as

Z 1 P0S(k) + r(S − S(k))(S − S(k))dr =∼ P0(S(k))(S − S(k)). (6) 0 From (5) and (6), we get

S = S(k) − P0(S(k))−1P(S(k))(k = 0, 1, 2, ··· ). (7)

Replacing S by S(k+1) in (7), we get the Newton method. Using (5) and different numerical integration formulas, one can obtain different iterative methods such as (2), (3), Symmetry 2021, 13, 432 3 of 12

and (4). To develop the new iterative method, we approximate the integral in (5) by the following three-point Gauss Legendre integration formula: r ! ! Z y y − x y + x 5 3 y + x f (t)dt ≈ 4 f + f (y − x) − + x 9 2 2 5 2 r ! ! (8) 5 3 y + x + f (y − x) + , 2 5 2

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

Z 1 P0S(k) + r(S − S(k))(S − S(k))dr 0 (k) (k) r (k) S − S 0 S + S 5 0 ( ) 3 S + S ≈ 4P + P (S − S k ) − + (9) 9 2 2 5 2 r 5 3 S + S(k) + P0 (S − S(k)) + . 2 5 2

Moreover, from (1), (5), and (9), we get ! S − S(k) S + S(k) 0 ≈ P S(k) + 4P0 9 2 r 5 3 S + S(k) + P0 S − S(k) − + (10) 2 5 2 r 5 3 S + S(k) + P0 S − S(k) + . 2 5 2

From (10), the iterative scheme is given by

S + S(k) S ≈ S(k) − 9 4P0 2 r 5 3 S + S(k) + P0 S − S(k) − + (11) 2 5 2 r −1 5 3 S + S(k) + P0 S − S(k) + PS(k). 2 5 2

Subsequently, we use the kth iteration T(k) and Z(k) of Newton’s method to replace S and S(k) respectively on the right-hand side of (11) and obtain a new iterative scheme as follows: Symmetry 2021, 13, 432 4 of 12

Algorithm 1: Three-Step Newton Method ( ) Step 1: Select an initial guess S 0 ∈ Rn and start k from 0. Step 2: Compute 5 S(k+1) = T(k) − 9 4P0(H(k)) + P0(H(k) + W(k)) 2 − 5 1 + P0(H(k) + J(k)) P(T(k)). 2 Step 3: Set for the next step: T(k) + Z(k) H(k) = , 2 r 3 J(k) = (T(k) − Z(k)) 5

and r 3 W(k) = (T(k) − Z(k)) − . 5 Step 4: Compute Z(k) = S(k) − P0(S(k))−1P(S(k))(k = 0, 1, 2, ··· ) T(k) = Z(k) − P0(Z(k))−1P(Z(k))(k = 0, 1, 2, ··· ).

Step 5: If ||S(k+1) − S(k)|| < e, then stop; otherwise. put k = k + 1 and go to Step 2.

In the next section, we discuss the convergence of the proposed method.

3. Convergence Analysis In the following theorem, we prove the convergence of the proposed method.

Theorem 1. Suppose that the function P : U ⊆ Rn −→ Rn is sufﬁciently Fréchet differentiable ∗ at each point of an open convex neighborhood U of the solution S ∈ Rn of (1). Assume also that P0(S) is continuous and nonsingular at S = S∗. Then, the sequence {S(k)} generated by Algorithm 5 converges to S∗ with the sixth order of convergence and the error equation is given by

5 6 7 ek+1 = 3C2ek + O(ek ), (12)

where (k) ∗ ek = S − S .

Proof. Let P : D ⊆ Rn −→ Rn be s-times Fréchet differentiable in U. Then, by using the ( ) usual notation for the mth derivative of P at v ∈ Rn, the m-linear function P m (v) : Rn × n n (m) n ∗ n · · · × R −→ R is such that P (v)(u1, u2, ··· , un) ∈ R . Suppose now that S + h ∈ R lies in the neighborhood of S∗. The Taylor polynomial for P(S∗ + h) can be of the form:

f −1 ! ∗ 0 ∗ m f P(S + h) = P (S ) h + ∑ Cmh + O(h ), (13) m=2

where 1 C = [P0(S∗)]−1P(m)(S∗)(m ≥ 2). m m! m n We observe that Cmh ∈ R , since

( ) ∗ 0 ∗ − P m (S ) ∈ L ∈ (Rn × Rn, ··· , Rn) and [P (X )] 1 ∈ L ∈ Rn. Symmetry 2021, 13, 432 5 of 12

In addition, we can express P0 as follows:

f −1 ! 0 ∗ 0 ∗ m−1 f P (S + h) = P (S ) I + ∑ mCmh + O(h ), (14) m=2

n×n m−1 n where I ∈ R is the identity matrix. We note that mCmh ∈ L ∈ R . From (13) and (14), we get

(k) 0 ∗ 2 3 4 5 6 P(S ) = P (S ) ek + C2ek + C3ek + C4ek + C5ek + C6ek + ··· . (15)

0 (k) 0 ∗ 2 3 4 5 P (S ) = P (S ) I + 2C2ek + 3C3ek + 4C4ek + 5C5ek + 6C5ek + ··· , (16) where 1 C = ( )[P0(S∗)]−1P(k)(S∗)(k = 2, 3, 4, ··· ) k k! and (k) ∗ ek = S − S . From (15), we have

0 (k) −1 2 2 P (S ) = I − 2C2ek + (4C2 − 3C3)ek + (−4C4 + 6C2C3 + 6C3C2 3 3 4 2 2 4 − 8C2)ek + (16C2 − 36C2C3 + 16C2C4 + 9C3 − 5C5)ek (17) −1 + ··· P0(S∗) .

−1 By multiplying P0S(k) and PS(k), we obtain

0 (k) −1 (k) 2 2 3 3 P (S ) P(S ) = ek − C2ek + 2(C2 − C3)ek + (−4C2 + 4C2C3 + 3C3C2 4 4 2 2 5 − 3C4)e + (8C − 20C C3 + 6C + 10C2C4 − 4C5)e k 2 2 3 k (18) 5 3 2 2 + (−16C2 + 52C2C3 − 33C2C3 − 28C2C4 + 17C3C4+ 6 13C2C5 − 5C6)ek + ··· .

Taylor’s series expansion of PZ(k) is given by

(k) 0 ∗ (k) ∗ (k) ∗2 (k) ∗ 3 (k) ∗4 P Z = P (S ) Z − S + C2 Z − S + C3(Z − S ) + C4 Z − S (19) (k) ∗5 (k) ∗6 − C5 Z − S + C6 Z − S + ··· ,

where 1 −1 C = P0(S∗) P(k)(S∗)(k = 2, 3, 4, ··· ). k k! Moreover, Z(k) can be written as follows:

(k) ∗ 2 2 3 3 4 Z = S + C2ek − 2(C2 − C3)ek − − 4C2 + 7C2C3 − 3C4 ek 4 2 2 5 5 + (4C5 − 12C + 24C C3 − 10C2C4 − 6C )e + 16C 2 2 3 k 2 (20) 3 2 2 − 52C2C3 + 28C2C4 + 33C2C3 − 13C2C5 − 17C3C4 6 + 5C6 ek + ··· .

Taylor’s series expansion of P0Z(k) is given by

0 (k) 0 ∗ (k) ∗ (k) ∗ 2 (k) ∗ 3 P (Z ) = P (S ) I + 2C2(Z − S ) + 3C3(Z − S ) + 4C4)(Z − S ) (21) (k) ∗ 4 (k) ∗ 5 + 5C5(Z − S ) + O(Z − S ) . Symmetry 2021, 13, 432 6 of 12

Putting (20) in (21), we have

0 (k) 0 ∗ 2 2 2 3 3 4 P (S ) = P (S ) I + 2C2ek − 4C2(C2 − C3)ek − C2(−8C2 + 11C2C3 − 6C4)ek + (−16C5 + 28C3C − 20C2C + 8C C )e5 + (32C6 − 68C4C 2 2 3 2 4 2 5 k 2 2 3 (22) 3 2 3 6 + 60C2C4 − 26C2C5 − 16C2C − 3C4 + 12C3 + 10C2 C6)ek + ··· .

−1 Upon multiplying P0Z(k) by PZ(k), we have

0 (k) −1 (k) 2 2 3 3 4 P (Z ) P(Z ) = C2ek + 2(−C2 + C3)ek + (3C2 − 7C2C3 + 3C4)ek 4 2 2 5 + (−4C + 16C C3 − 10C2C4 − 6C + 4C5)e 2 2 3 k (23) 5 3 2 2 + (6C2 − 32C2C3 + 22C2C4 + 29C2C3 − 13C2C5 6 − 17C3C4 + 5C6)ek + ··· .

The expression for T(k) is given below:

(k) ∗ 3 4 4 2 2 5 T = S + 3C2ek − 16C2 − 20C2C3 + 6C3 + 10C2C4C5 ek 5 3 2 2 5 − − 24C2 + 96C2C3 − 41C2C3 − 40C2C4 − 12C2 + 17C3C4 (24) 6 6 + 13C2C5 + 8C2 − 5C6 ek + ··· .

Similarly, PT(k) can be written as follows:

(k) 0 ∗ 3 4 4 2 2 5 P T = P (S ) 3C2ek − 16C2 − 20C2C3 + 6C3 + 10C2C4 − 4C5 ek 5 3 2 2 5 − − 24C2 + 96C2C3 − 41C2C3 − 40C2C4 − 12C2 (25) 6 6 + 17C3C4 + 13C2C5 + 8C2 − 5C6 ek + ··· .

Furthermore, we have

1 7 3 5 H(k) = C e2 + − C2 + C e3 + − C C + C + C3 e4 + − 6C4 2 2 k 2 3 k 2 2 3 2 4 2 2 k 2 69 + 12C2C − 5C C − 3C2 + 2C + C3C − 10C2C e5 (26) 2 3 2 4 3 5 5 2 3 2 4 k 37 13 5 + C C2 + 17C2C − C C − 36C3C + C e6 + ··· 2 2 3 2 4 2 2 5 2 3 2 6 k

and 1 1 √ 2 √ 2 √ J(k) = C + 15C e2 + − C2 + C − 15C2 + 15C e3 2 2 5 2 k 2 3 5 2 5 3 k 3 5 3 √ 3 √ 7 √ + C + C3 + 15C3 + 15C − 15C C (27) 2 4 2 2 5 2 5 4 5 2 3 7 − C C e4 + ··· . 2 2 3 k Similarly, we have

1 1 √ 2 √ 2 √ W(k) = C + 15C e2 + − C2 + C − 15C2 + 15C e3 2 2 5 2 k 2 3 5 2 5 3 k 3 5 3 √ 3 √ 7 √ + C + C3 + 15C3 + 15C − 15C C (28) 2 4 2 2 5 2 5 4 5 2 3 7 − C C e4 + ··· . 2 2 3 k Symmetry 2021, 13, 432 7 of 12

The expression P0H(k) + J(k) can be written as follows:

2 √ 4 √ P0H(k) + J(k) = I + C2 − 15C2 e2 + 2C3 − C 15C + 2C C 2 5 2 k 2 5 2 3 2 3 4 √ 11 √ 6 √ + 15C3 e3 + 5C4 + 15C2C − C 15C (29) 5 2 k 2 5 2 3 5 2 4 89 6 √ − C2C + 3C C − 15C4 e4 + ··· 20 2 3 2 4 5 2 k

and 2 √ 4 √ P0H(k) + W(k) = I + C2 − 15C2 e2 + − 2C3 + C 15C + 2C C 2 5 2 k 2 5 2 3 2 3 4 √ 11 √ 6 √ − 15C3 e3 + 5C4 − 15C2C + C 15C (30) 5 2 k 2 5 2 3 5 2 4 89 6 √ − C2C + 3C C + 15C4 e4 + ··· . 20 2 3 2 4 5 2 k

Furthermore, we have

25 P0H(k) = I + C2e2 + − 2C3 + 2C C e3 + 5C4 − C2C + 3C C e4 2 k 2 2 3 k 2 4 2 3 2 4 k (31) 5 3 2 2 5 + − 12C2 + 21C2C3 − 10C2C4 − 3C2C3 + 4C2C5 ek + ··· .

From (29)–(31), we have

− 5 5 1 4P0H(k) + P0H(k) + W(k) + P0H(k) + J(k) 2 2 1 2 = − C2e2 + C3 3 2 k 3 2 (32) 2 2 7 5 10 − C C e3 + − C4 + C2C − C C e4 + − C3C + C2C 3 2 3 k 3 2 4 2 3 2 4 k 3 2 3 3 2 4 1 4 − C C32 − C C e5 + ··· . 3 2 3 2 5 k

From (18)–(32), we obtain

5 6 6 4 3 2 2 ek+1 = 3C e + − 38C + 78C C3 + −32C C4 − 28C C 2 k 2 2 2 2 3 (33) 2 7 + 8C2C5 + 12C2C3C4 ek + ··· .

From (33), we conclude that the proposed method yields convergence of order 6.

4. Numerical Results In this section, we consider some problems to show the performance and efﬁciency of the newly developed method. We compare Newton’s method (NM) (see [6]) and methods (4), (5), (23), (25) and (27) in [15] with Algorithm 1. The stopping criterion is

(k+1) (k) −15 Error = ||S − S ||∞ < 10 ,

and k denotes the number of iterations. The computational order of convergence q (see [16]) is approximated by Symmetry 2021, 13, 432 8 of 12

ln(||S(k+1) − S(k)||/||S(k) − S(k−1)||) q ≈ . (34) ln ||S(k) − S(k−1)||/||S(k−1) − S(k−2)|| Consider the following systems of nonlinear equations (see [16]).

y2 Problem 1. −x2 + + y − 17 = 0, 6 x2 − y − 19 = 0,

Problem 2. x2 + y2 + z2 − 1 = 0, 2x2 + y2 − 4z = 0, 3x2 − 4y2 + z2 = 0.

2 Problem 3. ex + 8x sin(y) = 0, x + 2y − 1 = 0.

Problem 4. cos(x) − sin(y) = 0, zx + 1/y = 0, ex − z2 = 0.

Problem 5. x2 + y2 + z2 − 9 = 0, xyz = 1, x + y − z2 = 0.

Problem 6. xy + xz + yz − 1 = 0, yz + w(y + z) = 0, xz + w(x + z) = 0, xy + w(x + y) = 0.

Numerical results are given in Table1 below.

Table 1. Numerical results for the Problems 1 to 6.

Method Initial Guess k Approximate Solution q Problem 1 NM 6 (5.0000000000000, 6.0000000000000)T 2.0 (4) 4 (5.0000000000000, 6.0000000000000)T 2.9 (5) 4 (5.0000000000000, 6.0000000000000)T 2.9 (5.5, 6.8)T (23) 4 (5.0000000000000, 6.0000000000000)T 3.9 (25) 4 (5.0000000000000, 6.0000000000000)T 3.9 (27) 4 (5.0000000000000, 6.0000000000000)T 3.9 Algorithm 1 3 (5.0000000000000, 6.0000000000000)T 5.9 Symmetry 2021, 13, 432 9 of 12

Table 1. Cont.

Method Initial Guess k Approximate Solution q Problem 2 NM 6 (0.698288610, 0.628524230, 0.342564189)T 2.0 (4) 4 (0.698288610, 0.628524230, 0.342564189)T 3.0 (5) 4 (0.698288610, 0.628524230, 0.342564189)T 3.0 (0.5, 0.5, 0.5)T (23) 4 (0.698288610, 0.628524230, 0.342564189)T 4.0 (25) 4 (0.698288610, 0.628524230, 0.342564189)T 4.0 (27) 4 (0.698288610, 0.628524230, 0.342564189)T 4.0 Algorithm 1 3 (0.698288610, 0.628524230, 0.342564189)T 5.9 Problem 3 NM 7 (−0.22850805121143, 0.61425402560572)T 2.0 (4) 4 (−0.22850805121143, 0.61425402560572)T 3.0 (5) 4 (−0.22850805121143, 0.61425402560572)T 3.0 (0.5, 1.0)T (23) 4 (−0.22850805121143, 0.61425402560572)T 3.8 (25) 4 (−0.22850805121143, 0.61425402560572)T 4.2 (27) 4 (−0.22850805121143, 0.61425402560572)T 4.2 Algorithm 1 3 (−0.22850805121143, 0.61425402560572)T 6.1 Problem 4 NM 7 (0.90956949, 0.66122683, 1.57583414)T 2.0 (4) 5 (0.90956949, 0.66122683, 1.57583414)T 3.0 (5) 5 (0.90956949, 0.66122683, 1.57583414)T 3.0 (1.0, 0.5, 1.5)T (23) 4 (0.90956949, 0.66122683, 1.57583414)T 4.7 (25) 4 (0.90956949, 0.66122683, 1.57583414)T 4.1 (27) 4 (0.90956949, 0.66122683, 1.57583414)T 4.1 Algorithm 1 3 (0.90956949, 0.66122683, 1.57583414)T 6.3 Problem 5 NM 7 (2.49137571, 0.2427458788, 1.653517941)T 2.0 (4) 5 (2.49137571, 0.2427458788, 1.653517941)T 3.0 (5) 5 (2.49137571, 0.2427458788, 1.653517941)T 3.0 (2.5, 0.5, 1.5)T (23) 4 (2.49137571, 0.2427458788, 1.653517941)T 4.5 (25) 4 (2.49137571, 0.2427458788, 1.653517941)T 4.5 (27) 4 (2.49137571, 0.2427458788, 1.653517941)T 4.5 Algorithm 1 3 (2.49137571, 0.2427458788, 1.653517941)T 6.1 Problem 6 NM 5 (0.577350, 0.577350, 0.577350,-0.288680)T 2.1 (4) 4 (0.577350, 0.577350, 0.577350,-0.288680)T 3.3 (5) 4 (0.577350, 0.577350, 0.577350,-0.288680)T 3.3 (0.6, 0.6, 0.6, −0.2)T (23) 3 (0.577350, 0.577359, 0.577350,-0.288680)T 5.5 (25) 3 (0.577350, 0.577350, 0.577350,-0.288680)T 5.5 (27) 3 (0.577350, 0.577350, 0.577350,-0.288680)T 5.5 Algorithm 1 3 (0.577350, 0.577350, 0.577350,-0.288680)T 6.0

Problem 7 ([15]). Consider a nonlinear boundary-value problem of the following form:

y00(t) + y1+b(t) = 0 (t ∈ [0, 1]; b > 0) (35) y(0) = 0, y(1) = 1.

Here we have discretized the above nonlinear ODE (35) by using the finite difference method. Symmetry 2021, 13, 432 10 of 12

By taking b = 2.5 and n = 10, we obtain the following system of nonlinear equations.

3.5 100y2 − 200y1 + y1 = 0, 3.5 100y3 − 200y2 + 100y1 + y2 = 0, 3.5 100y4 − 200y3 + 100y2 + y3 = 0, 3.5 100y5 − 200y4 + 100y3 + y4 = 0, 3.5 100y6 − 200y5 + 100y4 + y5 = 0, (36) 3.5 100y7 − 200y6 + 100y5 + y6 = 0, 3.5 100y8 − 200y7 + 100y6 + y7 = 0, 3.5 100y9 − 200y8 + 100y7 + y8 = 0, 3.5 −200y9 + 100y8 + 100 + y9 = 0,

where y(0) = (1, 1, 1, 1, 1, 1, 1, 1, 1)T is the initial guess. We obtain the approximate solution as follows:

y∗ = (0.1039574502033 ··· , 0.2079117381290 ··· , 0.3118302489670 ··· , 0.4156008747144 ··· , 0.5189667289214 ··· , 0.6214486996519 ··· , (37) 0.7222575415768 ··· , 0.8201966396108 ··· , 0.9135562704267 ··· )T.

We compare Algorithm 1 with the Newton–Simpson method (NS-M) and the Open Newton method (ON-M) (see [9]), the four-point method (KH-M) (see [13]), the New- ton–Gauss method (NG-M), and the ﬁfth-order scheme (M 14) (see [15]). The numerical results are shown in Table2 below.

Table 2. Numerical results for Problem 7.

Method k 1 2 3 4 5 −2 −7 −21 −65 ||Sk+1 − Sk||2 1.61121 4.7021 × 10 5.7614 × 10 1.2500 × 10 1.3253 × 10 NS-M q 3.20022 2.98531 2.99899 2.99999 3.00000 −3 −8 −22 −66 −198 ||F(Sk)||2 4.3531 × 10 5.6034 × 10 1.2677 × 10 1.3581 × 10 1.6345 × 10 −2 −7 −21 −65 ||Sk+1 − Sk||2 1.61111 46671 × 10 5.6464 × 10 1.17711 × 10 1.10665 × 10 ON-M q 3.19700 2.98555 2.99891 2.99999 3.00000 −3 −8 −22 −66 −199 ||F(Sk)||2 4.3121 × 10 5.4934 × 10 1.1944 × 10 1.13333 × 10 9.4989 × 10 −2 −7 −21 −65 ||Sk+1 − Sk||2 1 .61111 4.6911 × 10 5.7277 × 10 1.2288 × 10 1.2577 × 10 KH-M q 3.19921 2.98555 2.99800 2.99999 3.00000 −3 −8 −22 −66 −198 ||F(Sk)||2 4.3411 × 10 5.5731 × 10 1.2455 × 10 1.2888 × 10 1.3951 × 10 −1 −6 −21 −65 ||Sk+1 − Sk||2 6.44555 1.8681 × 10 2.2633 × 10 4.7391 × 10 4.5099 × 10 NG-M q 3.19711 2.98555 2.99899 2.99999 3.00000 −3 −8 −22 −66 −198 ||F(Sk)||2 4.3175 × 10 5.5051 × 10 1.2011 × 10 1.1544 × 10 1.0044 × 10 −3 −18 −92 −463 ||Sk+1 − Sk||2 1.64888 2.0166 × 10 2.8188 × 10 1.6500 × 10 1.1465 × 10 M(14) q 5.10032 4.99732 4.99999 5.00000 5.00000 −4 −19 −93 −464 −2320 ||F(Sk)||2 1.9300 × 10 2.8741 × 10 1.6941 × 10 1.1788 × 10 1.9077 × 10 −48 ||Sk+1 − Sk||2 0.1622 × 10 0 - - - Alg. 1 q 6.10040 6 - - - −47 ||F(Sk)||2 0.7544 × 10 0 - - -

From Table2, we see that the proposed method converges to the solution in just two iterations. To illustrate the performance of the new method, we plot the approximate solution against the Maple solution in Figure1. Symmetry 2021, 13, 432 11 of 12

Figure 1. Comparison between the exact solution ( Maple solution) and the approximate solution

In the next problem, we compare Algorithm 1 with M6 [17] of order 6.

Problem 8. Consider the following integral equation:

y(r) Z 1 y(r) − 1 − µy(r) k(r, u)y(u)du = 0. (38) 4 0 Solving (38), we have the following system of nonlinear equations:

8 1 ui βj y ≈ 1 + y y , i = 1, 2...8. (39) i i ∑ + j 8 j=1 ui uj

For more detail see [17]. We compare Algorithm 1 with M6 [17] in Table3.

Table 3. Numerical results and comparison for Problem 8.

Method Number of Iterations Error Newton 5 3.1408 × 10−16 M6 3 2.2204 × 10−16 Algorithm 1 3 1.0000 × 10−19

From the last column of Table3, we conclude that the new method is more accurate than M6 [17].

5. Conclusions In this article, we have implemented a new three-step Newton method for solving a system of nonlinear equations. The order of convergence of the proposed method is six. To show the effectiveness of the new method, we have provided some numerical tests. The graphical illustration shows the accuracy of the proposed method. Numerical results conﬁrmed that the suggested method converges to the solution in fewer iterations with high accuracy, which justiﬁes the advantage of this method.

Author Contributions: Conceptualization, H.M.S., J.I. and A.K.; methodology, J.I., M.A. and A.K.; formal analysis, Y.S.G., J.I. and R.C.; investigation, review and editing, H.M.S., J.I. and M.A.; writing, Symmetry 2021, 13, 432 12 of 12

J.I. and A.K.; funding acquisition, R.C. All authors have read and agreed to the published version of the manuscript. Funding: This work has been supported by the grant provided by Division of Computational Science, Faculty of Science, Prince of Songkla University, Hat Yai, Songkhla 90110, Thailand. Institutional Review Board Statement: Not applicable. Informed Consent Statement: Not applicable. Data Availability Statement: Not applicable. Conﬂicts of Interest: The authors declare no conﬂict of interest.

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