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A cubically convergent Newton-type method under weak conditions
Journal of Computational and Applied Mathematics 220 (2008) 409 – 412 www.elsevier.com/locate/cam
A cubically convergent Newton-type method under weak conditions Liang Fanga, b,∗ , Guoping Hec , Zhongyong Hub a Department of Mathematics, Shanghai Jiao Tong University, 200240, Shanghai, China b Department of Mathematics and System Science, Tai Shan University, 271021, Tai’an, China c College of Information Science and Engineering, Shandong University of Science and Technology, 266510, Qingdao, China
Received 22 January 2007; received in revised form 9 September 2007
Abstract Under weak conditions, we present an iteration formula to improve Newton’s method for solving nonlinear equations. The method is free from second derivatives, permitting f (x) = 0 in some points and per iteration it requires two evaluations of the given function and one evaluation of its derivative. Analysis of convergence demonstrates that the new method is cubically convergent. Some numerical examples illustrate that the algorithm is more efficient and performs better than classical Newton’s method. © 2007 Elsevier B.V. All rights reserved. MSC: 47H17; 65J15; 41A25; 65D99 Keywords: Newton’s method; Nonlinear equations; Iterative method; Third-order convergence
1. Introduction In this paper, we consider iterative methods to find a simple root x ∗ of a nonlinear equation f (x) = 0, i.e., f (x ∗ ) = 0, where f : D ⊆ R → R for an open interval D is a scalar function. It is well known that Newton’s method is one of the best iterative methods for solving a single nonlinear equation by using xn+1 = xn −
f (xn ) , f (xn )
(1)
which converges quadratically in some neighborhood of x ∗ . We know that Newton’s method has a problem: condition f (x) = 0 in a neighborhood of root x ∗ is severe indeed for its convergence and its applications are restricted. In recent years, many iterative methods with cubic convergence for solving nonlinear equations have been developed [1–10]. However, besides Newton’s method, these modified Newton’s methods also have such a problem which restricts their applications. Thus, the research of developing third-order methods which are free from the restriction of derivative of f (x) is required for practical application and becomes very active now. ∗ Corresponding author.
E-mail address: [email protected]. 0377-0427/$ - see front matter © 2007 Elsevier B.V. All rights reserved. doi:10.1016/j.cam.2007.08.013
410
L. Fang et al. / Journal of Computational and Applied Mathematics 220 (2008) 409 – 412
Motivated and inspired by the recent activities in this direction, we present a new modified Newton-type method with order of convergence three, in which f (x) = 0 is permitted in some points, for solving the above problem. The method is free from second derivatives and per iteration it only requires two evaluations of the given function and one evaluation of its derivative. By numerical examples, the method is efficient in some cases where some other modifications of Newton’s method have failed, in particular, when the derivative of the function f (x) at an iterate is singular or almost singular. In the following sections we construct the new iteration formula and prove that it is third-order convergent. Lastly, with the help of Matlab, several numerical examples are given to compare the efficiency and performance of it with classical Newton’s method. 2. The method and analysis of its convergence Some modifications of Newton’s method with cubic convergence have been developed in [2,3,5,6,9,7], by considering different quadrature formulae for the computation of the integral arising from Newton’s theorem: f (x) = f (yn ) +
x
f (t) dt,
(2)
yn
where yn = xn + f (xn )/f (xn ). Using midpoint rule to approximate right-hand integral, we obtain
x yn
f (t) dt ≈ (x − yn )f
x + yn 2
.
(3)
Thus we obtain an implicit method xn+1 = yn −
f ((x
f (yn ) , n+1 + yn )/2)
(4)
which requires the (n + 1)th iterate xn+1 to calculate the (n + 1)th iterate itself. To obtain the explicit form, we make use of the Newton iterative step to compute the (n + 1)th iterate xn+1 on the right-hand side of (4), namely replacing ∗ ∗ f ((xn+1 + yn )/2) with f ((xn+1 + yn )/2), where xn+1 = xn − f (xn )/f (xn ) is the Newton iterate, thus we obtain a method xn+1 = xn −
f (xn + f (xn )/f (xn )) − f (xn ) . f (xn )
(5)
Now, replacing f (xn ) with n f (xn ) + f (xn ), we obtain xn+1 = xn −
f (xn + f (xn )/(n f (xn ) + f (xn ))) − f (xn ) , n f (xn ) + f (xn )
(6)
where n ∈ R, 0 |n | 1 with n = 0, 1, 2, . . . are parameters. These parameters are chosen such that sign(n f (xn )) = sign(f (xn )), where sign(x) is the sign function, i.e., sign(x) =
1 if x 0, −1 if x < 0.
In practical application, the iterative formula (6) may be also expressed as xn+1 = yn −
f (yn ) , n f (xn ) + f (xn )
(7)
L. Fang et al. / Journal of Computational and Applied Mathematics 220 (2008) 409 – 412
411
where yn = xn +
f (xn ) . n f (xn ) + f (xn )
(8)
For (6), we have: Theorem 1. Assume that the function f : D ⊆ R → R has a single root x ∗ ∈ D, where D is an open interval. If f (x) has first, second and third derivatives in the interval D, then the method defined by (6) converges cubically to x ∗ in a neighborhood of x ∗ and satisfies the following error equation: en+1 = (−2n + 3c2 n − 2c3 )en3 + O(en4 ),
(9)
where en = xn − x ∗ , ck = f (k) (x ∗ )/(k!f (x ∗ )), k = 2, 3. Proof. Let en = xn − x ∗ and dn = yn − xn , where yn is defined by (8). Using Taylor expansion and taking into account f (x ∗ ) = 0, we have f (xn ) = f (x ∗ )[en + c2 en2 + c3 en3 + O(en4 )].
(10)
Furthermore, we have f (xn ) = f (x ∗ )[1 + 2c2 en + 3c3 en2 + O(en3 )],
(11)
f (xn ) = f (x ∗ )[2c2 + 6c3 en + O(en2 )],
(12)
f (3) (xn ) = f (x ∗ )[6c3 + O(en )].
(13)
Substituting (10) and (11) into (8), we obtain dn =
f (xn ) = en − (c2 + n )en2 + (2c22 + 2c2 n + 2n − 2c3 )en3 + O(en4 ). n f (xn ) + f (xn )
(14)
Again expanding f (yn ) about xn , we obtain f (yn ) = f (xn ) + f (xn )dn + 21 f (xn )dn2 + 16 f (3) (xn )dn3 + O(en4 ).
(15)
Substituting (11)–(14) into (15), we have f (yn ) = f (x ∗ )[2en + (3c2 − n )en2 + (2n − 2c22 + 6c3 − 2c2 n )en3 + O(en4 )].
(16)
From (6), we have en+1 = en −
f (yn ) − f (xn ) . n f (xn ) + f (xn )
(17)
Substituting (10), (11) and (16) into (17), we obtain en+1 = (−2n + 3c2 n − 2c3 )en3 + O(en4 ).
(18)
This means that the method defined by (6) is cubically convergent. The proof is completed. 3. Numerical example Iterative formulas (1), (6) are called the classical Newton’s method (NT) and modified Newton’s method (MDNT(6)), respectively. Now, we employ MDNT(6) to find the simple roots of some nonlinear equations and compare it with NT. For simplicity, in the example, we choose n , such that |n | = 0.5. Displayed in Table 1 are the number of iteration (IT) required such that |f (xn )| < 1E − 15. The computational results in Table 1 show that MDNT(6) requires less IT than NT. Therefore, it works better than classical Newton’s method.
412
L. Fang et al. / Journal of Computational and Applied Mathematics 220 (2008) 409 – 412
Table 1 Comparison between MDNT(6) and NT
f1 f2 f3 f4
MDNT (6) IT
MDNT (6) xn
MDNT (6) |f (xn )|
NT
13 13 17 12
−2.5462 1.7549 −0.5044 −3
2.22E-16 8.88E-16 4.44E-16 0
Failure Failure Failure Divergent
In Table 1, we use the following functions and initial points: f1 (x) = x sin x + cos x − 0.6, f2 (x) = x − 2x + x − 1, 3
2
x0 = 0.
x0 = 1.
f3 (x) = x − e − 3x + x + 2, 3
x
2
f4 (x) = 3x 2 − 6x − 45,
x0 = 0.
x0 = 1.
4. Conclusions We have obtained a new modification of Newton’s method for solving nonlinear equations. From Theorem 1, we prove that the order of convergence of the new algorithm is three. Analysis of efficiency shows that the new method is comparable to the well-known Newton’s method. From numerical examples, we show that the new algorithm is efficient and performs well, in particular, when the derivatives of the function f (x) at some iterates are singular or almost singular. References [1] C. Chun, Construction of Newton-like iteration methods for solving nonlinear equations, Numer. Math. 104 (3) (2006) 297–315. [2] M. Frontini, E. Sormani, Some variants of Newtons method with third-order convergence, J. Comput. Appl. Math. 140 (2004) 419–426. [3] M. Frontini, E. Sormani, Third-order methods from quadrature formulae for solving systems of nonlinear equations, Appl. Math. Comput. 149 (2004) 771–782. [4] A. Galantai, The theory of Newton’s method, J. Comput. Appl. Math. 124 (2000) 25–44. [5] H.H.H. Homeier, A modified Newton method for rootfinding with cubic convergence, J. Comput. Appl. Math. 157 (2003) 227–230. [6] H.H.H. Homeier, A modified Newton method with cubic convergence: the multivariate case, J. Comput. Appl. Math. 169 (2004) 161–169. [7] H.H.H. Homeier, On Newton-type methods with cubic convergence, J. Comput. Appl. Math. 176 (2005) 425–432. [8] A.Y. Özban, Some new variants of Newton’s method, Appl. Math. Lett. 17 (2004) 677–682. [9] J.S. Weerakoon, T.G.I. Fernando, A variant of Newtons method with accelerated third-order convergence, Appl. Math. Lett. 13 (2000) 87–93. [10] X.Y. Wu, A new continuation Newton-like method and its deformation, Appl. Math. Comput. 112 (2000) 75–78.
A cubically convergent Newton-type method under weak conditions Liang Fanga, b,∗ , Guoping Hec , Zhongyong Hub a Department of Mathematics, Shanghai Jiao Tong University, 200240, Shanghai, China b Department of Mathematics and System Science, Tai Shan University, 271021, Tai’an, China c College of Information Science and Engineering, Shandong University of Science and Technology, 266510, Qingdao, China
Received 22 January 2007; received in revised form 9 September 2007
Abstract Under weak conditions, we present an iteration formula to improve Newton’s method for solving nonlinear equations. The method is free from second derivatives, permitting f (x) = 0 in some points and per iteration it requires two evaluations of the given function and one evaluation of its derivative. Analysis of convergence demonstrates that the new method is cubically convergent. Some numerical examples illustrate that the algorithm is more efficient and performs better than classical Newton’s method. © 2007 Elsevier B.V. All rights reserved. MSC: 47H17; 65J15; 41A25; 65D99 Keywords: Newton’s method; Nonlinear equations; Iterative method; Third-order convergence
1. Introduction In this paper, we consider iterative methods to find a simple root x ∗ of a nonlinear equation f (x) = 0, i.e., f (x ∗ ) = 0, where f : D ⊆ R → R for an open interval D is a scalar function. It is well known that Newton’s method is one of the best iterative methods for solving a single nonlinear equation by using xn+1 = xn −
f (xn ) , f (xn )
(1)
which converges quadratically in some neighborhood of x ∗ . We know that Newton’s method has a problem: condition f (x) = 0 in a neighborhood of root x ∗ is severe indeed for its convergence and its applications are restricted. In recent years, many iterative methods with cubic convergence for solving nonlinear equations have been developed [1–10]. However, besides Newton’s method, these modified Newton’s methods also have such a problem which restricts their applications. Thus, the research of developing third-order methods which are free from the restriction of derivative of f (x) is required for practical application and becomes very active now. ∗ Corresponding author.
E-mail address: [email protected]. 0377-0427/$ - see front matter © 2007 Elsevier B.V. All rights reserved. doi:10.1016/j.cam.2007.08.013
410
L. Fang et al. / Journal of Computational and Applied Mathematics 220 (2008) 409 – 412
Motivated and inspired by the recent activities in this direction, we present a new modified Newton-type method with order of convergence three, in which f (x) = 0 is permitted in some points, for solving the above problem. The method is free from second derivatives and per iteration it only requires two evaluations of the given function and one evaluation of its derivative. By numerical examples, the method is efficient in some cases where some other modifications of Newton’s method have failed, in particular, when the derivative of the function f (x) at an iterate is singular or almost singular. In the following sections we construct the new iteration formula and prove that it is third-order convergent. Lastly, with the help of Matlab, several numerical examples are given to compare the efficiency and performance of it with classical Newton’s method. 2. The method and analysis of its convergence Some modifications of Newton’s method with cubic convergence have been developed in [2,3,5,6,9,7], by considering different quadrature formulae for the computation of the integral arising from Newton’s theorem: f (x) = f (yn ) +
x
f (t) dt,
(2)
yn
where yn = xn + f (xn )/f (xn ). Using midpoint rule to approximate right-hand integral, we obtain
x yn
f (t) dt ≈ (x − yn )f
x + yn 2
.
(3)
Thus we obtain an implicit method xn+1 = yn −
f ((x
f (yn ) , n+1 + yn )/2)
(4)
which requires the (n + 1)th iterate xn+1 to calculate the (n + 1)th iterate itself. To obtain the explicit form, we make use of the Newton iterative step to compute the (n + 1)th iterate xn+1 on the right-hand side of (4), namely replacing ∗ ∗ f ((xn+1 + yn )/2) with f ((xn+1 + yn )/2), where xn+1 = xn − f (xn )/f (xn ) is the Newton iterate, thus we obtain a method xn+1 = xn −
f (xn + f (xn )/f (xn )) − f (xn ) . f (xn )
(5)
Now, replacing f (xn ) with n f (xn ) + f (xn ), we obtain xn+1 = xn −
f (xn + f (xn )/(n f (xn ) + f (xn ))) − f (xn ) , n f (xn ) + f (xn )
(6)
where n ∈ R, 0 |n | 1 with n = 0, 1, 2, . . . are parameters. These parameters are chosen such that sign(n f (xn )) = sign(f (xn )), where sign(x) is the sign function, i.e., sign(x) =
1 if x 0, −1 if x < 0.
In practical application, the iterative formula (6) may be also expressed as xn+1 = yn −
f (yn ) , n f (xn ) + f (xn )
(7)
L. Fang et al. / Journal of Computational and Applied Mathematics 220 (2008) 409 – 412
411
where yn = xn +
f (xn ) . n f (xn ) + f (xn )
(8)
For (6), we have: Theorem 1. Assume that the function f : D ⊆ R → R has a single root x ∗ ∈ D, where D is an open interval. If f (x) has first, second and third derivatives in the interval D, then the method defined by (6) converges cubically to x ∗ in a neighborhood of x ∗ and satisfies the following error equation: en+1 = (−2n + 3c2 n − 2c3 )en3 + O(en4 ),
(9)
where en = xn − x ∗ , ck = f (k) (x ∗ )/(k!f (x ∗ )), k = 2, 3. Proof. Let en = xn − x ∗ and dn = yn − xn , where yn is defined by (8). Using Taylor expansion and taking into account f (x ∗ ) = 0, we have f (xn ) = f (x ∗ )[en + c2 en2 + c3 en3 + O(en4 )].
(10)
Furthermore, we have f (xn ) = f (x ∗ )[1 + 2c2 en + 3c3 en2 + O(en3 )],
(11)
f (xn ) = f (x ∗ )[2c2 + 6c3 en + O(en2 )],
(12)
f (3) (xn ) = f (x ∗ )[6c3 + O(en )].
(13)
Substituting (10) and (11) into (8), we obtain dn =
f (xn ) = en − (c2 + n )en2 + (2c22 + 2c2 n + 2n − 2c3 )en3 + O(en4 ). n f (xn ) + f (xn )
(14)
Again expanding f (yn ) about xn , we obtain f (yn ) = f (xn ) + f (xn )dn + 21 f (xn )dn2 + 16 f (3) (xn )dn3 + O(en4 ).
(15)
Substituting (11)–(14) into (15), we have f (yn ) = f (x ∗ )[2en + (3c2 − n )en2 + (2n − 2c22 + 6c3 − 2c2 n )en3 + O(en4 )].
(16)
From (6), we have en+1 = en −
f (yn ) − f (xn ) . n f (xn ) + f (xn )
(17)
Substituting (10), (11) and (16) into (17), we obtain en+1 = (−2n + 3c2 n − 2c3 )en3 + O(en4 ).
(18)
This means that the method defined by (6) is cubically convergent. The proof is completed. 3. Numerical example Iterative formulas (1), (6) are called the classical Newton’s method (NT) and modified Newton’s method (MDNT(6)), respectively. Now, we employ MDNT(6) to find the simple roots of some nonlinear equations and compare it with NT. For simplicity, in the example, we choose n , such that |n | = 0.5. Displayed in Table 1 are the number of iteration (IT) required such that |f (xn )| < 1E − 15. The computational results in Table 1 show that MDNT(6) requires less IT than NT. Therefore, it works better than classical Newton’s method.
412
L. Fang et al. / Journal of Computational and Applied Mathematics 220 (2008) 409 – 412
Table 1 Comparison between MDNT(6) and NT
f1 f2 f3 f4
MDNT (6) IT
MDNT (6) xn
MDNT (6) |f (xn )|
NT
13 13 17 12
−2.5462 1.7549 −0.5044 −3
2.22E-16 8.88E-16 4.44E-16 0
Failure Failure Failure Divergent
In Table 1, we use the following functions and initial points: f1 (x) = x sin x + cos x − 0.6, f2 (x) = x − 2x + x − 1, 3
2
x0 = 0.
x0 = 1.
f3 (x) = x − e − 3x + x + 2, 3
x
2
f4 (x) = 3x 2 − 6x − 45,
x0 = 0.
x0 = 1.
4. Conclusions We have obtained a new modification of Newton’s method for solving nonlinear equations. From Theorem 1, we prove that the order of convergence of the new algorithm is three. Analysis of efficiency shows that the new method is comparable to the well-known Newton’s method. From numerical examples, we show that the new algorithm is efficient and performs well, in particular, when the derivatives of the function f (x) at some iterates are singular or almost singular. References [1] C. Chun, Construction of Newton-like iteration methods for solving nonlinear equations, Numer. Math. 104 (3) (2006) 297–315. [2] M. Frontini, E. Sormani, Some variants of Newtons method with third-order convergence, J. Comput. Appl. Math. 140 (2004) 419–426. [3] M. Frontini, E. Sormani, Third-order methods from quadrature formulae for solving systems of nonlinear equations, Appl. Math. Comput. 149 (2004) 771–782. [4] A. Galantai, The theory of Newton’s method, J. Comput. Appl. Math. 124 (2000) 25–44. [5] H.H.H. Homeier, A modified Newton method for rootfinding with cubic convergence, J. Comput. Appl. Math. 157 (2003) 227–230. [6] H.H.H. Homeier, A modified Newton method with cubic convergence: the multivariate case, J. Comput. Appl. Math. 169 (2004) 161–169. [7] H.H.H. Homeier, On Newton-type methods with cubic convergence, J. Comput. Appl. Math. 176 (2005) 425–432. [8] A.Y. Özban, Some new variants of Newton’s method, Appl. Math. Lett. 17 (2004) 677–682. [9] J.S. Weerakoon, T.G.I. Fernando, A variant of Newtons method with accelerated third-order convergence, Appl. Math. Lett. 13 (2000) 87–93. [10] X.Y. Wu, A new continuation Newton-like method and its deformation, Appl. Math. Comput. 112 (2000) 75–78.