From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) Subject: Re: taylor polynomials and dy/dx Date: 31 Oct 2000 12:55:55 -0500 Newsgroups: sci.math Summary: Functions which are nowhere smooth or smooth but nowhere analytic In article , G.E. Ivey wrote: :jolt64@my-deja.com wrote: : :> kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) wrote: :>> In article <8tajdt$frg$1@nnrp1.deja.com>, wrote: :>> :If a function f(x) is differentiable across all numbers in its :>> :domain, does there necessarily exist a Taylor polynomial :>> :convergent to the function? :>> :>> No. :>> :>> If you ask a more specific question, you can get a more specific :>> answer. (Is the domain real or complex? What is the shape of the :>> domain? And what kind of differentiability - real or complex?) :>> :>> Cheers, ZVK(Slavek). : :>If the answer really is no, can you give me an example that does :>not have ths convergence, and explain why? Thanks. :>Jolt : The classic example is: y(x)= exp(-x^2) if x is not 0, y(0)=0. :All derivatives exist even at 0 (all derivatives are a polynomial :times exp(-x^2) and the exponential dominates) and all derivatives :are 0 at x=0. The "MacLaurin" series (Taylor's series about x=0) is :0+ 0x+ 0x^2+ ... which trivially converges for all x but just as :trivially does NOT converge to y(x) except at x=0. You can have other complications: (1) The series f(x) = sum[n=0 to infinity] (20)^(-n) * sin(10^n * x) results in a function which is once continuously differentiable everywhere but nowhere twice differentiable. That's why its Taylor expansion cannot have more terms. Proof of this failure requires some tedious manipulations and discussions, but the example is sort of classical, too. (2) The series g(x) = sum[n=0 to infinity] (1/n!) * cos(2^n * x) is infinitely differentiable but its formal Taylor expansion diverges everywhere except for the centre, no matter where the centre is located. The proof uses the fact that the (magnitudes of the even-indexed) coefficients of the series grow too fast to pass a convergence test. (But you wanted to prove facts about even and odd functions, and Taylor expansions are not required for the proofs.) Cheers, ZVK(Slavek). ============================================================================== From: ullrich@math.okstate.edu (David C. Ullrich) Subject: Re: analytic function question Date: Wed, 01 Nov 2000 14:02:13 GMT Newsgroups: sci.math On Tue, 31 Oct 2000 20:43:31 -0500, "TTL" wrote: >Let Sum_k=0 to infinity a_k z^k be a power series with radius of convergence >0I believe that there exists an open neighborhood N of a such that the >analytic function defined by the power series can be >extended analytically to N. >Where can I find a proof? Or maybe I am wrong? You're wrong. In fact there exist examples where the power series converges at _every_ point of the boundary, the boundary function is infinitely differentiable (infinitely differentiable in the real-variables sense), but the function cannot be analytically continued past any boundary point. Now, this leaves open the question of whether there are any such examples at 7:30 in the morning. Just offhand I _think_ that a modification of ZK's "weirder" example does it: Let f(x) = sum[m=0 to infinity] x^(m!)/(m!)^m , |x| <= 1. It's clear that if you differentiate the series N times you get something that converges absolutely on the unit circle, so f has infinitely differentiable boundary values. (Or rather "So f has radius of convergence at least 1, and also has infinitely differentiable boundary values".) And the radius of convergence is no larger than 1; this is because if x > 1 the terms tend to infinity. (The logarithm is m! log(x) - m log(m!) > m! log(x) - m^2 log(m), which certainly blows up if x > 1.) So the radius of convergence is 1 and the function is infinitely differentiable on the boundary. Why can it be analytically continued past no boundary point? Roughly speaking, the fact that each exponent is a factor of the next exponent shows that all boundary points are the same - if it could be continued past one boundary point it could be continued to a larger disk centered at 0, contradicting the fact that the radius of convergence is 1. Why are all boundary points the same? Well of course that's not a very precise way to put it. Note that if you delete the first N terms then all the remaining exponents are divisible by N!, hence what remains is a function of z^(N!). That is, for N = 1, 2, ... there exist a polynomial P = P_N and a function g = g_N such that f(x) = P(x) + g(x^(N!)) . Now g satisfies g(ax) = x, where a = exp(2Pi i/N!). So if g can be continued across an arc on the boundary of length greater than 2Pi/N! then g can be continued past the entire boundary. That's impossible because the radius of convergence is 1. So g_N cannot be continued past any arc of length greater than 2Pi/N!. Hence f cannot be continued past any arc of length greater than 2Pi/N!, and hence f cannot be continued past any arc on the boundary at all.