Generalization Of The Secant Method For Nonlinear Equations

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1 Applied Mathematics E-Notes, 8(2008), c ISSN Available free at mirror sites of amen/ Generalization Of The Secant Method For Nonlinear Equations Avram Sidi Received 27 February 2007 Abstract The secant method is a very effective numerical procedure used for solving nonlinear equations of the form f(x) = 0. It is derived via a linear interpolation procedure and employs only values of f(x) at the approximations to the root of f(x) = 0, hence it computes f(x) only once per iteration. In this note, we generalize it by replacing the relevant linear interpolant by a suitable (k + 1)- point polynomial of interpolation, where k is an integer at least 2. Just as the secant method, this generalization too enjoys the property that it computes f(x) only once per iteration. We provide its error in closed form and analyze its order of convergence. We show that this order of convergence is greater than that of the secant method, and it increases as k increases. We also confirm the theory via an illustrative example. 1 Introduction Let α be the solution to the equation f(x) =0. (1) An effective iterative method used for solving (1) that makes direct use of f(x) [but no derivatives of f(x)] is the secant method that is discussed in many books on numerical analysis. See, for example, Atkinson [1], Henrici [2], Ralston and Rabinowitz [3], and Stoer and Bulirsch [5]. See alo the recent note [4] by the author, in which the treatment of the secant method and those of the Newton Raphson, regula falsi, and Steffensen methods are presented in a unified manner. This method is derived by a linear interpolation procedure as follows: Starting with two initial approximations x 0 and x 1 to the solution α of (1), we compute a sequence of approximations {x n } n=0, such that the approximation x n+1 is determined as the point of intersection (in the x-y plane) of the straight line through the points (x n,f(x n )) and (x n 1,f(x n 1 )) with the x-axis. Since the equation of this straight line is y = f(x n )+ f(x n) f(x n 1 ) (x x n ), (2) x n x n 1 Mathematics Subject Classifications: 65H05, 65H99. Computer Science Department, Technion - Israel Institute of Technology, Haifa 32000, Israel 115

2 116 Generalized Secant Method x n+1 is given as f(x n ) x n+1 = x n. (3) f(x n) f(x n 1) x n x n 1 In terms of divided differences, (3) can be written in the form x n+1 = x n f(x n) f[x n,x n 1 ]. (4) We recall that, provided f C 2 (I), where I is an open interval containing α, and f (α) 0, and that x 0 and x 1 are sufficiently close to α, the sequence {x n } n=0 converges to α of order at least (1 + 5)/2 = Another way of obtaining the secant method, of interest to us in the present work, is via a variation of the Newton Raphson method. Recall that in the Newton Raphson method, we start with an initial approximation x 0 and generate a sequence of approximations {x n } n=0 to α through x n+1 = x n f(x n), n =0, 1,.... (5) f (x n ) We also recall that, provided f C 2 (I), where I is an open interval containing α, and f (α) 0, and that x 0 is sufficiently close to α, the sequence {x n } n=0 converges to α of order at least 2. As such, the Newton Raphson method is extremely effective. To avoid computing f (x) [note that f (x) may not always be available or may be costly to compute], and to preserve the excellent convergence properties of the Newton Raphson method, we replace f (x n ) in (5) by the approximation f[x n,x n 1 ]=[f(x n ) f(x n 1 )]/(x n x n 1 ). This results in (4), that is, in the secant method. The justification for this approach is as follows: When convergence takes place, that is, when lim n x n = α, the difference x n x n 1 tends to zero, and this implies that, as n increases, the accuracy of f[x n,x n 1 ] as an approximation to f (x n ) increases as well. Note that the Newton-Raphson iteration requires two function evaluations, one of f(x) and another of f (x), per iteration. The secant method, on the other hand, requires only one function evaluation per iteration, namely, that of f(x). In Section 2 of this note, we consider in detail a generalization of the second of the two approaches described above using polynomial interpolation of degree k with k > 1. The (k + 1)-point iterative method that results from this generalization turns out to be very effective. It is of order higher than that of the secant method and requires only one function evaluation per iteration. In Section 3, we analyze this method and determine its order as well. In Section 4, we confirm our theory via a numerical example. 2 Generalization of Secant Method We start by discussing a known generalization of the secant method (see, for example, Traub [6, Chapters 4, 6, and 10]). In this generalization, we approximate f(x) by the polynomial of interpolation p n,k (x), where p n,k (x i )=f(x i ),i= n, n 1,...,n k, assuming that x 0,x 1,...,x n have all been computed. Following that, we determine

3 A. Sidi 117 x n+1 as a zero of p n,k (x), provided a real solution to p n,k (x) = 0 exists. Thus, x n+1 is the solution to a polynomial equation of degree k. Fork = 1, what we have is nothing but the secant method. For k =2,x n+1 is one of the solutions to a quadratic equation, and the resulting method is known as the method of Müller. Clearly, for k 3, the determination of x n+1 is not easy. This difficulty prompts us to consider the second approach to the secant method we discussed in Section 1, in which we replaced f (x n ) by the slope of the straight line through the points (x n,f(x n )) and (x n 1,f(x n 1 )), that is, by the derivative (at x n ) of the (linear) interpolant to f(x) atx n and x n 1. We generalize this approach by replacing f (x n )byp n,k (x n), the derivative at x n of the polynomial p n,k (x) interpolating f(x) at the points x n i, i =0, 1,...,k, mentioned in the preceding paragraph, with k 2. Because p n,k (x) is a better approximation to f(x) in the neighborhood of x n, p n,k (x n) is a better approximation to f (x n ) when k 2 than f[x n,x n 1 ] used in the secant method. In addition, just as the secant method, the new method computes the function f(x) only once per iteration step, the computation being that of f(x n ). Thus, the new method is described by the following (k + 1)-point iteration: x n+1 = x n f(x n) p n,k (x, n = k, k +1,..., (6) n) with x 0,x 1,...,x k as initial approximations to be provided by the user. Of course, with k fixed, we can start with x 0 and x 1, compute x 2 via the method we have described (with k = 1, namely via the secant method), compute x 3 via the method we have described (with k = 2), and so on, until we have completed the list x 0,x 1,...,x k. We now turn to the computational aspects of this method. What we need is a fast method for computing p n,k (x n). For this, we write p n,k (x) in Newtonian form as follows: p n,k (x) =f(x n )+ k i 1 f[x n,x n 1,...,x n i ] (x x n j ). (7) Here f[x i,x i+1,...,x m ] are divided differences of f(x), and we recall that they can be defined recursively via f[x i ]=f(x i ); f[x i,x j ]= f[x i] f[x j ], x i x j x i x j, (8) and, for m>i+ 1, via f[x i,x i+1,...,x m ]= f[x i,x i+1,...,x m 1 ] f[x i+1,x i+2,...,x m ], x i x m x i x m. (9) We also recall that f[x i,x i+1,...,x m ] is a symmetric function of its arguments, that is, it has the same value under any permutation of {x i,x i+1,...,x m }. Thus, in (7), f[x n,x n 1,...,x n i ]=f[x n i,x n i+1,...,x n ]. In addition, when f C m (I), where I is an open interval containing the points z 0,z 1,...,z m, whether these are distinct or not, there holds f[z 0,z 1,...,z m ]= f (m) (ξ) for some ξ (min{z i }, max{z i }). m! j=0

4 118 Generalized Secant Method x 0 f 0 f 01 x 1 f 1 f 012 f 12 f 0123 x 2 f 2 f 123 f 23 f 1234 x 3 f 3 f 234 f 34 f 2345 x 4 f 4 f 345 f 45 f 3456 x 5 f 5 f 456 f 56 f 4567 x 6 f 6 f 567 f 67 x 7 f 7 Table 1: Table of divided differences over {x 0,x 1,...,x 7} for use to compute x 8 via p 7,3(x) in (6). Note that f i,i+1,...,m stands for f[x i,x i+1,...,x m] throughout. Going back to (7), we note that p n,k (x) there is computed by ordering the x i as x n,x n 1,...,x n k. This ordering enables us to compute p n,k (x n) easily. Indeed, differentiating p n,k (x) in (7), and letting x = x n, we obtain p n,k(x n )=f[x n,x n 1 ]+ k i=2 i 1 f[x n,x n 1,...,x n i ] (x n x n j ). (10) In addition, note that the relevant divided difference table need not be computed anew each iteration; what is needed is adding a new diagonal (from the south-west to north-east) at the bottom of the existing table. To make this point clear, let us look at the following example: Suppose k = 3 and and we have computed x i, i =0, 1,...,7. To compute x 8, we use the divided difference table in Table 1. Letting f i,i+1,...,m stand for f[x i,x i+1,...,x m ], we have j=1 x 8 = x 7 f(x 7) p 7,3 (x 7) = x f 7 7 f 67 + f 567 (x 7 x 6 )+f 4567 (x 7 x 6 )(x 7 x 5 ). To compute x 9, we will need the divided differences f 8,f 78,f 678,f Computing first f 8 = f(x 8 ) with the newly computed x 8, the rest of these divided differences can be computed from the bottom diagonal of Table 1 via the recursion relations f 78 = f 7 f 8 x 7 x 8, f 678 = f 67 f 78 x 6 x 8, f 5678 = f 567 f 678 x 5 x 8, in this order, and appended to the bottom of Table 1. Actually, we can do even better: Since we need only the bottom diagonal of Table 1 to compute x 8, we need to save only this diagonal,

5 A. Sidi 119 namely, only the entries f 7,f 67,f 567,f Once we have computed x 8 and f 8 = f(x 8 ), we can overwrite f 7,f 67,f 567,f 4567 with f 8,f 78,f 678,f Thus, in general, to be able to compute x n+1 via (6), after x n has been determined, we need to store only the entries f n,f n 1,n,...,f n k,n k+1,...,n 1,n along with x n,x n 1,...,x n k. 3 Convergence Analysis We now turn to the analysis of the sequence {x n } n=0 that is generated via (6). Since we already know everything concerning the case k = 1, namely, the secant method, we treat the case k 2. The following theorem gives the main convergence result for the generalized secant method. THEOREM 3.1. Let α be the solution to the equation f(x) = 0. Assume f C k+1 (I), where I in an open interval containing α, and assume also that f (α) 0,in addition to f(α) = 0. Let x 0,x 1,...,x k be distinct initial approximations to α, and generate x n, n = k +1,k+2,..., via x n+1 = x n f(x n) p n,k (x, n = k, k +1,..., (11) n) where p n,k (x) is the polynomial of interpolation to f(x) at the points x n,x n 1,...,x n k. Then, provided x 0,x 1,...,x k are in I and sufficiently close to α, the sequence {x n } n=0 converges to α, and lim n ɛ n+1 i=0 ɛ = ( 1)k+1 n i (k + 1)! f (k+1) (α) f (α) L; ɛ n = x n α n. (12) The order of convergence is s k,1<s k < 2, where s k is the only positive root of the equation s k+1 = k i=0 si and satisfies 2 2 k 1 e<s k < 2 2 k 1 for k 2; s k <s k+1 ; lim s k =2, (13) k where e is the base of natural logarithms, and ɛ n+1 lim = L (sk 1)/k. (14) n ɛ n sk REMARK. Note that, in Theorem 3.1, s 2 =1.839,s 3 =1.928, s 4 =1.966, etc., rounded to three significant figures. For the secant method, we have s 1 =1.618, rounded to three significant figures. PROOF. Below, we shall use the short-hand notation int(a 1,...,a m ) = (min{a 1,...,a m }, max{a 1,...,a m }). We start by deriving a closed-form expression for the error in x n+1. Subtracting α from both sides of (11), and noting that f(x n )=f(x n ) f(α) =f[x n,α](x n α),

6 120 Generalized Secant Method we have x n+1 α = We now note that which, by ( 1 f[x ) n,α] p n,k (x (x n α) = p n,k (x n) f[x n,α] n) p n,k (x (x n α). (15) n) p n,k(x n ) f[x n,α]= { p n,k(x n ) f (x n ) } + { f (x n ) f[x n,α] }, f (x n ) f[x n,α]=f[x n,x n ] f[x n,α]=f[x n,x n,α](x n α) = f (2) (η n ) (x n α) 2! for some η n int(x n,α), and by f (x n ) p n,k(x n ) = f[x n,x n,x n 1,...,x n k ] = f (k+1) (ξ n ) (k + 1)! for some ξ n int(x n,x n 1,...,x n k ), becomes p n,k(x n ) f[x n,α]= f (k+1) (ξ n ) (k + 1)! Substituting (16) and (17) in (15), and letting k (x n x n i ) k (x n x n i ) (16) k (ɛ n ɛ n i )+ f (2) (η n ) ɛ n. (17) 2! we finally obtain D n = f (k+1) (ξ n ) (k + 1)! and Ê n = f (2) (η n ), (18) 2! ɛ n+1 = C n ɛ n ; C n p n,k (x n) f[x n,α] p n,k (x n) = D n (ɛ n ɛ n i )+Ê n ɛ n f (x n )+ D n (ɛ n ɛ n i ). (19) We now prove that convergence takes place. Without loss of generality, we can assume that I =(α T,α + T ) for some T > 0, and that m 1 = min x I f (x) > 0. This is possible since α I and f (α) 0. Let M s = max x I f (s) (x) /s!, s=1, 2,..., and choose the interval J =(α t/2,α+ t/2) sufficiently small to ensure that m 1 > 2M k+1 t k + M 2 t/2. It can now be shown that, provided x n i, i =0, 1,...,k, are all in J, there holds C n M k+1 ɛ n ɛ n i + M 2 ɛ n m 1 M k+1 ɛ n ɛ n i M k+1 ( ɛ n + ɛ n i ) + M 2 ɛ n m 1 M k+1 ( ɛ C, n + ɛ n i )

7 A. Sidi 121 where C M k+1t k + M 2 t/2 m 1 M k+1 t k < 1. Consequently, by (19), ɛ n+1 < ɛ n, which implies that x n+1 J, just like x n i, i =0, 1,...,k. Therefore, if x 0,x 1,...,x k are chosen in J, then C n C < 1 and hence x n+1 J, for all n k. Consequently, by (19), lim n x n = α. As for (12), we proceed as follows: By the fact that lim n x n = α, we first note that lim n p n,k (x n) = f (α) = lim n f[x n,α], and thus lim n C n = 0. This means that lim n (ɛ n+1 /ɛ n ) = 0 and, equivalently, that {x n } n=0 converges of order greater than 1. As a result, lim n (ɛ n /ɛ n i ) = 0, for all i 1, and ɛ n /ɛ n i = o(ɛ n /ɛ n j )asn, for j < i. Consequently, expanding in (19) the product (ɛ n ɛ n i ), we have k k ( [ ]) (ɛ n ɛ n i )= ɛn i 1 ɛn /ɛ n i ( ) [1+O =( 1) k ( )] ɛ n i ɛn /ɛ n 1 as n. (20) Substituting (20) in (19), and defining we obtain D n = D n p n,k (x n), E n = Ê n p n,k (x n), (21) ( ) [1+O ɛ n+1 =( 1) k ( )] D n ɛ n i ɛn /ɛ n 1 + En ɛ 2 n as n. (22) i=0 Dividing both sides of (22) by i=0 ɛ n i, and defining we have Now, σ n = ɛ n+1 i=0 ɛ, (23) n i σ n =( 1) k D n [ 1+O ( ɛn /ɛ n 1 )] + En σ n 1 ɛ n k 1 as n. (24) lim D n = 1 n (k + 1)! f (k+1) (α) f, lim (α) E n = f (2) (α) n 2f (α). (25) Since lim n D n and lim n E n are finite, lim n ɛ n /ɛ n 1 = 0 and lim n ɛ n k 1 = 0, it follows that there exist a positive integer N and positive constants β<1 and D, with E n ɛ n k 1 β when n N, for which (24) gives σ n D + β σ n 1 for all n N. (26)

8 122 Generalized Secant Method Using (26), it is easy to show that σ N+s D 1 βs 1 β + βs σ N, s =1, 2,..., which, by the fact that β<1, implies that {σ n } is a bounded sequence. Making use of this fact, we have lim n E n σ n 1 ɛ n k 1 = 0. Substituting this in (24), and invoking (25), we next obtain lim n σ n =( 1) k lim n D n = L, which is precisely (12). That the order of the method is s k, as defined in the statement of the theorem, follows from [6, Chapter 3]. A weaker version can be proved by letting σ n = L for all n and showing that ɛ n+1 = Q ɛ n sk is possible for s k a solution to the equation s k+1 = k i=0 si and Q = L (sk 1)/k. The proof of this is easy and is left to the reader. This completes our proof. 4 A Numerical Example We apply the method described in Sections 2 and 3 to the solution of the equation f(x) = 0, where f(x) =x 3 8, whose solution is α =2. We take k = 2 in our method. We also choose x 0 = 5 and x 1 = 4, and compute x 2 via one step of the secant method, namely, x 2 = x 1 f(x 1) f[x 0,x 1 ]. (27) Following that, we compute x 3,x 4,..., via x n+1 = x n f(x n ), n =2, 3,.... (28) f[x n,x n 1 ]+f[x n,x n 1,x n 2 ](x n x n 1 ) Our computations were done in quadruple-precision arithmetic (approximately 35- decimal-digit accuracy), and they are given in Table 2. Note that in order to verify the theoretical results concerning iterative methods of order greater that unity, we need to use computer arithmetic of high precision (preferably, of variable precision, if available) because the number of correct significant decimal digits increases dramatically from one iteration to the next as we are approaching the solution. From Theorem 3.1, and lim n ɛ n+1 = ( 1)3 ɛ n ɛ n 1 ɛ n 2 3! f (3) (2) f (2) = 1 = log ɛ n+1 /ɛ n lim n log ɛ n /ɛ n 1 = s 2 = , and these seem to be confirmed in Table 2. Also, x 9 should have a little under 50 correct significant figures, even though we do not see this in Table 2 due to the fact that the arithmetic we have used to generate Table 2 can provide an accuracy of at most 35 digits approximately.

9 A. Sidi 123 n x n ɛ n L n Q n Table 2: Results obtained by applying the generalized secant method with k = 2, as shown in (27) and (28), to the equation x 3 8 = 0. Here, L n = ɛ n+1 /(ɛ n ɛ n 1 ɛ n 2 ) and Q n = ( log ɛ n+1 /ɛ n ) / ( log ɛ n /ɛ n 1 ). References [1] K. E. Atkinson, An Introduction to Numerical Analysis, Wiley, New York, second edition, [2] P. Henrici, Elements of Numerical Analysis, Wiley, New York, [3] A. Ralston and P. Rabinowitz, A First Course in Numerical Analysis, McGraw-Hill, New York, second edition, [4] A. Sidi, Unified treatment of regula falsi, Newton Raphson, secant, and Steffensen methods for nonlinear equations, J. Online Math. Appl., 6, (Available from the URL library/4/vol6/sidi/sidi.pdf). [5] J. Stoer and R. Bulirsch, Introduction to Numerical Analysis, Springer-Verlag, New York, third edition, [6] J. F. Traub, Iterative Methods for the Solution of Equations, Prentice Hall, Englewood Cliffs, N.J., 1964.