From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Subject: Re: Power Series Question
Date: 15 Aug 2000 05:10:15 -0400
Newsgroups: sci.math
Summary: [missing]

In article <8nar9f$ed1$1@slb7.atl.mindspring.net>,
Daniel Giaimo <dgiaimo@ix.netcom.com> wrote:
:    How does one prove that if a power series, f, centered at the origin
:has radius of convergence equal to R then there is a complex number z of
:absolute value R such that f(z) diverges?
:
 Test your conjecture on the (classical) series

  x + x^2/2 + x^4/4 + x^8/8 + ...

with general term  

  x^(2^n) / 2^n

and radius R=1.

Cheers, ZVK(Slavek)
==============================================================================

From: spamless@Nil.nil
Subject: Re: Power Series Question
Date: 15 Aug 2000 06:20:29 -0400
Newsgroups: sci.math

In article <8nar9f$ed1$1@slb7.atl.mindspring.net>,
Daniel Giaimo <dgiaimo@ix.netcom.com> wrote:
:    How does one prove that if a power series, f, centered at the origin
:has radius of convergence equal to R then there is a complex number z of
:absolute value R such that f(z) diverges?

Well, t'ain't true.

However, there IS a point on the circle at which f is not analytic.

Proof? Suppose f is analytic at each point. Analyticity is a local (not
point) property (there is a neighbourhood of each point in which the
expansion at that point converges). Cover the circle with those
neighbourhoods. FINITELY many cover the circle (since it is compact). In
that case there is a radius R'>R so that f is analytic on the circle
|z|<R'. But then the power series around 0 converges for |z|<R' (larger
than R).

(Proof of the last? 
 Let R<r<r'<R' and prove that the power series around zero for f converges
 for |z|=r. But f^n(0)/n!=(1/(2pi*i))*INTEGRAL[f(t)/t^(n+1)) over |t|=r']
 so, with M(r')=MAX(f(z):|z|=r'), f^n(0)/n!<=M(r')/r'^n and
 SUM[z^n*f^n(0)/n!] has terms dominated (in absolute value) by
 M(r')(r/r')^n, the terms of a absolutely converging geometric series.)
==============================================================================

From: spamless@Nil.nil
Subject: Re: Power Series Question
Date: 15 Aug 2000 14:50:01 -0400
Newsgroups: sci.math

In article <8nar9f$ed1$1@slb7.atl.mindspring.net>,
Daniel Giaimo <dgiaimo@ix.netcom.com> wrote:
:    How does one prove that if a power series, f, centered at the origin
:has radius of convergence equal to R then there is a complex number z of
:absolute value R such that f(z) diverges?

While that isn't true, for *any* complex number z whose modulus is GREATER
than R, it diverges.