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        riddles >> putnam exam (pure math) >> Power Series
        
(Message started by: Braincramps on May 16^th, 2004, 2:04am)

Title: Power Series
Post by Braincramps on May 16^th, 2004, 2:04am

Suppose f(r)= [smiley=csigma.gif]^{[smiley=infty.gif]} _n=1a_nrⁿ has (-1,1) as its interval of convergence (f may or may not converge at endpoints) and f(r) has a finite limit as r [smiley=longrightarrow.gif]1^-. Let N(r) be a positive integer-valued function such that N(r)[smiley=longrightarrow.gif][smiley=infty.gif] as r[smiley=longrightarrow.gif]1^-. Does [smiley=csigma.gif]^{[smiley=infty.gif]}_n=N(r) a_nrⁿ have a finite limit as r[smiley=longrightarrow.gif]1^-?

Title: Re: Power Series
Post by Barukh on May 16^th, 2004, 9:17am

I just wanted to clarify:

Is it the case, that when n runs through all the possible values of N(r), r also runs through the whole convergence interval, so that the last series is no longer a power series?

Also, it seems that N(r) will take the same value more than once (in fact, infinitely often) - what terms of the series are taken then?

::)

Title: Re: Power Series
Post by TenaliRaman on May 16^th, 2004, 9:34am

My initial thoughts :
sum(n=N(r) to inf)a_nrⁿ
= sum(n=0 to inf)a_nrⁿ - sum(n=0 to N(r))a_nrⁿ ... (*)

Now i know that convergence is not affected in a power series as long as we remove finite number of terms from the start of the power series. So as long as N(r) is finite , the * converges.

now that i read Barukhs post, i am forced to make some more thinking over it , whether or not taking limit r->1- on (*) would be proper since N(r) isn't continuous (???)

Title: Re: Power Series
Post by Braincramps on May 16^th, 2004, 10:28am

on 05/16/04 at 09:17:05, Barukh wrote:

Is it the case, that when n runs through all the possible values of N(r), r also runs through the whole convergence interval, so that the last series is no longer a power series?

r doesn`t vary as n varies - but the number of terms of the series vary as r varies. As an example we could take N(r), the lower limit of the sum at r, to be an increasing step function satisfying 1/ [smiley=surd.gif](1-r) < N(r) < 1/(1-r).