Math 701: Secant Method

Transcription

1 Math 701: Secant Method The secant method aroximates solutions to f(x = 0 using an iterative scheme similar to Newton s method in which the derivative has been relace by This results in the two-term recurrence f (x n f(x n f(x n 1 x n+1 = x n f(x n( which needs a base of two different aroximations x 0 and x 1 of the solution to get started Let be the exact solution such that f( = 0 and suose f ( 0 Before roving that x 0 and x 1 sufficiently close to imlies x n with order α = (1 + 5/, we first derive this rate of convergence heuristically Intuitive Derivation of the Rate of Convergence Define e n = x n and assume x n as n and further that e n M α for some constants M > 0 and α 1 Let ϵ > 0 be arbitrary By definition Claim that e n+1 = x n+1 = x n f(x n( = e n f(x n(e n = e n f(x n(e n = e ( n f(xn f(x n 1 f(x n (e n = e ( n f(xn f(x n 1 f(x n (e n = f(x n f(x n 1 e n ( f(xn = f(x ( n 1 e n e n 1 e n { } f(xn /e n f(x n 1 /e ( n 1 x n x n 1 = e n 1 f ( and f(x n /e n f(x n 1 / f ( As this is a heuristic derivation of α there is no need to rove the above claims rigorously, but only to justify them from an intuitive oint of view For the first art of the claim note the mean value theorem imlies there is ξ n between x n and x n 1 such that = f (ξ n ( 1

2 It follows that f (ξ n f ( since ξ n as n Therefore 1 f ( as n Consequently, if n is large enough then 1 f ( and the first art of the claim has been justified An intuitive justification of the second art of the claim is similar, but slightly more involved By Taylor s theorem there is η n between x n and such that f(x n = f( + (x n f ( + 1 (x n f ( + 1 3! (x n 3 f (η n Therefore f(x n e n f ( + 1 e nf ( This suggests that f(x n /e n f(x n 1 / f ( f ( + 1 e n f ( f ( = f ( We are now ready to infer a lausible value for α the order of convergence of the secant method Combine the results of the claim with the exression for to obtain e n+1 C e n where C = f ( f ( Substituting the relation e n M α yields Solving for M and α from the relations obtains M e n α M 1+α α CM α M 1+α = CM and α = α + 1 M = C 1/α = f ( f 1/α and α = ( This finishes our heuristic derivation of the rate of convergence of the secant method

3 Rigorous Proof of the Rate of Convergence We assume f is twice continuously differentiable and that is such that f( = 0 and f ( 0 The secant method is x n+1 = x n f(x n First claim if x 0 and x 1 with x 0 x 1 are chosen close enough to that x n as n That is, the secant method converges Let κ(α, β = 1 f (α f (β Since f ( 0 it follows that the limit suremum of κ(α, β is zero as α and β Therefore, there is δ > 0 so small such that κ(α, β γ < 1 whenever α < δ and β < δ Choose x 0 and x 1 such that x 0 < δ and x 1 < δ By the mean-value theorem f(x n f( x n = f (a n and = f (b n for some a n between x n and and for some b n between x n and x n 1 For induction assume x n < δ and x n 1 < δ, in which case a n < δ and b n < δ Denoting e n = x n we obtain e n e n+1 = e n f(x n = e n (f(x n f((e n = e n f (a n e n (e n ( f = (b n (e n 1 f (a n f (b n e n Therefore, e n+1 γ e n and by induction e n+1 γ n e 1 Since γ < 1, it follows that x n as n and the secant method converges Claim there exists C such that e n+1 / e n C as n First note that e n e n+1 = e n f(x n = e n f(x ne n f(x n 1 e n + f(x n 1 e n f(x n = f(x n f(x n 1 e n = f(x n/e n f(x n 1 / f e n (b n ( 3

4 Define A = max{ f (x : x δ } Since f is continuous, it follows that A < By Taylor s theorem we have f(x n = f( + f (e n + Now f (t(x n tdt f(x n e n f(x n 1 = 1 f (t(x n tdt 1 1 f (t(x n 1 tdt e n = J 1 + J where and J 1 = J = 1 ( ( 1 e n 1 f (t(x n tdt f (t(x n tdt 1 f (t(x n 1 tdt = J 3 + J 4 Here J 3 = f (tdt and J 4 = 1 f (t(x n 1 tdt x n 1 Estimate Therefore Also J 1 A 1 1 e n J 1 A A e n e n (x n tdt A e n e n A = 1 f (a n f 0 as n (b n J 3 A e n 0 as n 4

5 The estimate of J 4 will be done more carefully Consider two cases: If x n 1 < x n then the mean-value theorem for integrals imlies ( x n 1 f (t(t x n 1 dt = f (ξ n where ξ n [x n 1, x n ] If x n < x n 1 then xn 1 (x n 1 x n f (t(x n 1 tdt = f (ξ n x n where ξ n [x n, x n 1 ] In either case it holds that as n It follows that J 4 = f (ξ n = f (ξ n ( en + 1 f ( J f ( as n Consequently e n+1 f ( C where C = e n f as n ( Claim that the secant method converges with order α Note that 1 α 1 = α, α 1 = α, and 1 α + 1 α = 1 Define K n = e n+1 / e n and M n = e n+1 / e n α Then It follows that M α n M n 1 = M n = K n M 1/α n 1 Combining the above two inequalities imlies Since K n C as n then K n+1 K 1/α n ( en+1 α ( en ( en+1 α e n α α = = K α e n n and similarly M n+1 = K n+1 M n+1 = K n+1 M 1/α Kn 1/α n 1 M 1/α n C 1 1/α = C α as n 5

6 Since α > 0, the above limit makes sense even when f ( = 0 Let L = 1 + C α By the definition of limit, there exists N large enough such that M n+1 LM 1/α n 1 for all n N The above inequality and the fact that 1/α < 1 imlies that the sequence M n is bounded In articular, suose M n 1 > L α where n N, then M n+1 LM 1/α n 1 < M 1/α+1/α n 1 = M n 1 Therefore M (n+k+1 M n 1 for all k N Similarly if M n > L α for some n N, then M (n+k M n for all k N Having consider both even and odd terms, we conclude in general that M k is bounded Consequently there exists M large enough such that M n M for every n N Thus e n+1 M e n α for every n N and so the secant method converges with order at least α The following references were consulted when rearing the above roof: [1] Burden, Fairs and Burden, Numerical Analysis, Tenth Edition, hint given in for Problem 14 in Section 4 [] Dahlquist and Björck, Numerical Methods in Scientific Comuter, Volume I, roof of Theorem 61 6