Author |
Topic: Power series (Read 2585 times) |
|
Braincramps
Newbie
Posts: 9
|
 |
Power series
« on: May 6th, 2004, 6:00pm » |
 Quote  Modify
|
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(reit) 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?
|
|
 IP Logged |
|
|
|
Icarus
wu::riddles Moderator
Uberpuzzler
Boldly going where even angels fear to tread.
Gender: 
Posts: 4863
|
|
Re: Power series
« Reply #1 on: May 6th, 2004, 7:26pm » |
Quote Modify
|
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] anzn for which the an 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).
|
|
IP Logged |
"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
|
|
|
Braincramps
Newbie
Posts: 9
|
|
Re: Power series
« Reply #2 on: May 7th, 2004, 3:01am » |
Quote Modify
|
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.
|
|
IP Logged |
|
|
|
Icarus
wu::riddles Moderator
Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863
|
|
Re: Power series
« Reply #3 on: May 7th, 2004, 7:34pm » |
Quote Modify
|
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]eit2^(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 2kt 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.
|
|
IP Logged |
"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print