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        general >> complex analysis >> Power series
        
(Message started by: Braincramps on May 6^th, 2004, 6:00pm)

Title: Power series
Post by Braincramps on May 6^th, 2004, 6:00pm

Let f(z) be the power series [smiley=csigma.gif]_n=1[smiley=supinfty.gif] z^(2^(2^n)). Then f is analytic in the unit disk. Show f(re^it) doesn't have a limit as r [smiley=to.gif] 1^- for any t in [-[smiley=cpi.gif], [smiley=cpi.gif]. I guess the case when t is rational with denominator a power of 2 could be easily settled, since each term in the series is 1 for sufficiently large n. Any ideas for the remaining cases?

Title: Re: Power series
Post by Icarus on May 6^th, 2004, 7:26pm

Actually, this holds true for any f that is analytic on all of [bbc] and is not a polynomial. I.e., you might want to see if you can prove it for any f(z) = [sum] a_nzⁿ for which the a_n are not eventually all zero, and which converges for all z.

Oddly, some things are easier to prove in abstraction than they are for specific cases. (I am not guaranteeing that will be the case here, though).

Title: Re: Power series
Post by Braincramps on May 7^th, 2004, 3:01am

I'm afraid it isn't true for any entire f not a polynomial. If f(z)=exp(z), then f(z) [smiley=longrightarrow.gif] 0 as z [smiley=longrightarrow.gif] [smiley=infty.gif] along the negative real axis.

Title: Re: Power series
Post by Icarus on May 7^th, 2004, 7:34pm

You're right. I was making too much of a well-known result (Picard's theorem: The image of any neighborhood of an essential singularity of a meromorphic function under that function will consist of the entire complex plane & [infty], with the possible exception of two points). The theorem does not require that limits along fixed paths diverge, as I was thinking.

I see that I was misreading your problem any way, I thought for some reason that you were letting r go to [infty], not 1. And Hadamard's theorem easily shows that the radius of convergence for this series is 1, not [infty] as I also assumed. :-[

I think the r factor here is not key. If you can show that [sum]e^{it2^(2^n)} does not converge for any t, then when r is near 1, the partial sums of the original series approximate the partial sums of this one, whose non-covergence mean the original series cannot converge either.

So all you really need to do is show that the series does not converge when r = 1. There may be some convenient tool in your toolbag for this, but any I knew have long since faded from memory. The best idea I can offer is by approximations. As you have pointed out, the job is not that hard when 2^kt is an integer for some k. Such numbers are dense in the reals: you can approximate any real number as closely as you like with such a rational. So for an arbitrary real number t, approximate it with a rational s with power-of-2 denominator. Approximate a partial sum of the series for t with a corresponding partial sum for s. Use the divergence of the series for s to show that for any possible value of the series, there are partial sums of arbitrary large number of terms which are bounded away from the value.

It would be a nasty slog to go this way, but I suspect that it will work.