NEWTON S METHOD FOR EQUATIONS RELATED TO EXPONENTIAL FUNCTION. Moonja Jeong

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Kangweon-Kyungki Math. Jour. 9 2001), No. 1, pp. 67 73 NEWTON S METHOD FOR EQUATIONS RELATED TO EXPONENTIAL FUNCTION Moonja Jeong Abstract. For some equation related with exponential function, we seek roots and find the properties of the roots. By using the relation of the roots and attractors, we find a region in the basin of attraction of the attractor at infinity for Newton s method for solving given equation. 1. Introduction We often need to find the roots of an equation fz) = 0. To be sure, if f is a linear or quadratic polynomial, formulas for writing exact solutions exist and are well known. For general f, Newton s method is a clever numerical procedure for solving equations fz) = 0. For Newton s method we begin with a guess z 0, for a root to the equation fz) = 0. We use the Newton iteration function, given by N f z) = z fz)/f z) to find a better approximation z 1 = N f z 0 ). Newton s method is to iterate the function N f, by successively computing z 2 = N f z 1 ) = N f N f z 0 )), z 3 = N f z 2 ) = N f N f N f z 0 ))) and so on. It is not always true that Newton s method yields approximations that converge to the root see [7; p. 497]). The simple roots of fz) = 0 are fixed points of N f z) satisfying N f z) = z and we call them attracting fixed points for N f z) or simply attractors for N f z) since the derivative N f z) = fz) f z) f z) 2 ) = 0 there see [1; p.214]). Hence, a Newton sequence {z k } given by Newton s method converges to a root of fz) = 0 if z 0 is a proper initial guess. For a simple root z satisfying fz ) = 0, the basin of attraction of z is the set of all points whose orbits tend to z. Received January 30, 2001. 1991 Mathematics Subject Classification: 30D05. Key words and phrases: Newton s method, Newton iteration function, attractor, basin of attraction.

68 Moonja Jeong The equation z 3 1 = 0 has three roots and each of the roots becomes an attractor for Newton s method. The graphical image of dynamics of Newton s method for it is fractal in [5; p. 334] and one of the interesting features of the figure is the boundaries between the three regions and every boundary point is actually next to all three regions. We can conjecture similar behavior for the equation which has infinitely many roots. For example, the equation e z 1 = 0 has infinitely many roots z = 2kπi, where k is an integer. These roots become attractors of Newton s method for e z 1 = 0. Let M ζ,α z) = exp α ζ + z ) ζ z where α > 0 and ζ = 1. The equation M ζ,α z) 1 = 0 has all its infinite number of roots on the unit circle see [4]). More properties including Denjoy-Wolff points of M ζ,α can be found in [2, 3]. In this note, a family of functions E A,w0 z) = exp i z w ) 0 Aw 0 where A = 1, w 0 C with Imw 0 > 0 is considered. The mapping property of the function E A,w0 can be found in [6; p. 60] and it has singularity at z = A. E A,w0 is a generalization of the form of M ζ,α. In section 2, we check the behavior of the roots by comparing the distances of the adjacent roots and by comparing the distances between the roots and A. By using the relation of the roots and attractors, we find a region in the basin of attraction of the attractor at for Newton s method for solving given equation. This note has much importance in the sense that we characterize the dynamical behavior of Newton s method for solving E A,w0 z) 1 = 0 with given condition and provide the basis for the graphical image. 2. Newton s method for E A,w0 z) 1 = 0 The roots of the equation E A,w0 z) 1 = 0 are the attractors of N EA,w0 1z). First, we will check the location of them. If w 0 = αi where α > 0, then E A,w0 equals M A,α and then we locate the roots

Newton s method for equations related to exponential function 69 of E A,w0 as in [4]. Through this paper we assume that A = 1 and w 0 C with Imw 0 > 0 without loss of generality. The equation E A,w0 z) 1 = 0 has the roots ζ k where ζ k w 0 Aw 0 ζ k A = 2kπ with an integer k. Hence, ζ k = A2kπ w 0 )/2kπ w 0 ). Since ζ k = 1, the roots ζ k are located on the unit circle. Therefore, the equation E A,w0 z) 1 = 0 has infinitely many roots ζ k and all of them lie on the unit circle. Theorem 1. As k, the roots ζ k of the equation E A,w0 z) 1 = 0 approach to A in a counterclockwise direction and the roots ζ k approach to A in a clockwise direction. For any positive integer k, the followings hold: i) the roots ζ k and ζ k are symmetric about the ray of angle φ iff Rew 0 = 0 ii) the root ζ k is closer to the singular point A than the root ζ k is to A iff Rew 0 > 0. Proof. As k, ζ k A = 2Imw 0 / 2kπ w 0 approaches to 0. Since ζ k /A = 2kπ w 0 2kπ w 0, we see that the roots ζ k approach to A in a counterclockwise direction and the roots ζ k approach to A in a clockwise direction as k by considering the Argζ k /A). Since ζ k A = 2Imw 0 / 2kπ w 0, we see that ζ k A > ζ k A if 2kπ w 0 < 2kπ w 0 if krew 0 ) > 0. Hence the inequality ζ k A > ζ k A holds for any positive integer k iff Rew 0 > 0 and the equality ζ k A = ζ k A holds for any integer k iff Rew 0 = 0. It means that for any positive integer k, the roots ζ k and ζ k are symmetric about the ray of angle φ where A = e iφ iff Rew 0 = 0 and the root ζ k is closer to the singular point A than the root ζ k is to A iff Rew 0 > 0. Next, we study the behavior of the roots of the equation E A,w0 z) 1 = 0 more carefully. To do this we need the following two lemmas.

70 Moonja Jeong Lemma 2. Let ζ k be the roots of the equation E A,w0 z) 1 = 0. If 2k 1)π < Rew 0 < 2kπ, then ζ k ζ k 1 is the biggest among the distance between the adjacent roots. If Rew 0 = 2kπ, then ζ k ζ k 1 = ζ k+1 ζ k is the biggest among the distance between the adjacent roots. Proof. Since ζ k ζ k 1 = 2kπ w 0 2k 1)π w 0, 2kπ w 0 2k 1)π w 0 we can compare the distances between two adjacent roots by calculating ζ k ζ k 1 2 ζ k+1 ζ k 2 = 4πImw 0 ) 2 8π2kπ Rew 0 ) 2k 1)π w 0 )2kπ w 0 )2k + 1)π w 0 ) 2. Hence, ζ k ζ k 1 > ζ k+1 ζ k iff 2kπ > Rew 0. Therefore, if 2k 1)π < Rew 0 2kπ, then < ζ k 2 ζ k 3 < ζ k 1 ζ k 2 < ζ k ζ k 1 and < ζ k+2 ζ k+1 < ζ k+1 ζ k ζ k ζ k 1 hold and it induces the desired conclusion. Lemma 3. Let ζ k be the roots of the equation E A,w0 z) 1 = 0. If 2k 1)π < Rew 0 < 2k + 1)π, then ζ k is the farthest root from A. If Rew 0 = 2k 1)π, then ζ k and ζ k 1 are the farthest roots from A. Proof. Since ζ k A = 2Imw 0 2kπ w 0, we can compare the distances between the roots and A by calculating ζ k A 2 ζ k 1 A 2 = 2Imw 0 ) 2 4π2k 1)π Rew0 ) 2kπ w 0 2 2k 1)π w 0 2 Hence, ζ k A > ζ k 1 A iff 2k 1)π < Rew 0. It means that if 2k 1)π Rew 0 < 2k + 1)π, then ζ k A ζ k 1 A > ζ k 2 A > and ζ k A > ζ k+1 A > ζ k+2 A > hold and it leads to the conclusion. By combining Lemma 2 and Lemma 3, we conclude ).

Newton s method for equations related to exponential function 71 Theorem 4. Let ζ k be the roots of the equation E A,w0 z) 1 = 0. If 2k 1)π < Rew 0 < 2k 1)π, then ζ k ζ k 1 is the biggest among the distance between the adjacent roots and ζ k 1 is the farthest root from A. If 2k 1)π < Rew 0 < 2kπ, then ζ k ζ k 1 is the biggest among the distance between the adjacent roots and ζ k is the farthest root from A. If Rew 0 = 2k 1)π, then ζ k ζ k 1 is the biggest among the distance between the adjacent roots and ζ k and ζ k 1 are the farthest roots from A. If Rew 0 = 2kπ, then ζ k ζ k 1 = ζ k+1 ζ k is the biggest among the distance between the adjacent roots and ζ k is the farthest root from A. To understand the location of the roots, we consider the following two examples. Example 5. If w 0 = i and A = 1, then ζ 0 = 1 and ζ k = 2kπ i 2kπ+i = 2kπ) 2 1 4kπi 2kπ) 2 +1. Hence π/2 < Arg ζ k < 0 and 0 < Arg ζ k < π/2 for any positive integer k. Also, Arg ζ k 0 and Arg ζ k 0 + as k. Note that ζ k = ζ k in this case. By Theorem 4, ζ 0 ζ 1 = ζ 1 ζ 0 is the biggest among the distance between the adjacent roots and ζ 0 is the farthest root from A. Example 6. If w 0 = 51 + i) and A = 1, then ζ 0 = i and ζ k = 2kπ 5) 2 5 2 102kπ 5)i 2kπ 5) 2 +5. Hence for k 2, π/2 < Arg ζ 2 k < 0 and for any k 1, 0 < Arg ζ k < π/2. Also Arg ζ k 0 and Arg ζ k 0 + as k. By Theorem 1, for any positive integer k ζ k 1 < ζ k 1. By Theorem 4, ζ 1 ζ 0 is the biggest among the distance between the adjacent roots and ζ 1 is the farthest root from A. In Theorem 7 we check the behavior of the Newton sequence near. Theorem 7. The region satisfying 1) z > 2 + 4Imw 0 1 e Imw 0/3 is in the basin of attraction of the attractor at infinity for Newton s method for the equation E A,w0 z) 1 = 0.

72 Moonja Jeong Proof. Newton iteration function is N EA,w0 1z) = z E A,w 0 z) 1 E A,w 0 z) = z + )2 2AImw 0 [ 1 exp i z w )] 0 Aw 0. By letting F z) = 1/N EA,w0 11/z), we check that F z) 0 and F z) 0 as z 0. Hence z = 0 is an attracting fixed point for F z) and so z = is an attracting fixed point for N EA,w0 1z). Now, we prove the region satisfying 1) is in the basin of attraction of the attractor at infinity for N EA,w0 1z). If 4Imw 0 z > 2 + 1 e, Imw 0/3 then z > 2. Since [ Re i z w ] [ 0 Aw 0 z w0 Aw 0 ) z = Im Ā) ] 2 = Imw 0 z 2 1) 2, it holds exp i z w 0 Aw 0 ) [ = exp Re i z w 0 Aw 0 exp < exp Imw 0 z 2 ) 1) z + 1) 2 Imw ) 0 3 ]) when z > 2. Note that we assumed Imw 0 > 0 through this paper.

Newton s method for equations related to exponential function 73 Therefore if 1) holds, then N EA,w0 1z) 2 2 Imw 0 1 exp i z w ) 0 Aw 0 z z 1)2 > 1 e Imw0/3 ) z 2 Imw 0 1 e Imw 0 /3 > z z 1 ) e Imw0/3 + Imw 0 2 Imw 0 Imw 0 > z. It implies that Newton sequence generated by an initial value z 0 in the region satisfying 1) diverges to. So the region satisfying 1) is contained in the basin of attraction of the attractor at. References [1] R. Devaney, A first course in chaotic dynamical systems: theory and experiment, Addison-Wesley Publishing Company, Reading, MA, 1992. [2] H. O. Kim and H. S. Kim, Iterations, Denjoy-Wolff points and Ergodic properties of point-mass singular inner functions, J. of Math. Anal. and Appl. 175 1993), 105 117. [3], Dynamics of two parameter family of point-mass singular inner functions, J. KMS 33 1996), 39 46. [4] Y. B. Kim, H. O. Kim, H. S. Kim, and S. Y. Shin, Infinite-corner-point fractal image generation by Newton s method for solving exp α w+z w z ) 1 = 0, Comput. & Graphics 17 1993), 705 711. [5] E. Scheinerman, Dynamical systems, Prentice Hall, New Jersey, 1996. [6] H. Silverman, Complex Variables, Houghton Mifflin Company, Boston, MA, 1975. [7] D. Varberg, E. Purcell, and S. Rigdon, Calculus, Prentice Hall, New Jersey, 2000. Department of Mathematics The University of Suwon Suwon P.O.Box 77, Kyungkido, 440-600 E-mail: mjeong@mail.suwon.ac.kr