my book wrote this:

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It says that the function must fail to be analytic somewhere on the circle of convergence. Is this a correct way of proving that this must hold, and is there also a simpler reason why they may say this?

Assume for contradiction that the function is analytic on the entire circle. By definition(?) that means that on every point on the circle of convergence there is a small open circle where the function is analytic. Since the circle of convergence is a compact set, we look at we must have a minimum radius of these small covering circles which is larger than zero(call it $r_{small}$, by the extreme value theorem. If we now cover each point on the circle of convergence with these small circles, then we see that that we can extend the original radius R with $r_{small}/2$.

Is this proof correct? The reason I chose to divide with 2 was that when we move along the circle of convergence and cover each point by a circle with radius $r_{small}$ is it really that obvious that the the area we cover have a radius $R+r_{small}$? Or could some points inside the circle of radius $+r_{small}$ be missing?

  • $\begingroup$ Your argument shows that $f$ would be analytic on a larger disk. This does not immediatly show that the Taylor series converges in this larger disk (but it is true and possibly contained in what your quote summarizes) $\endgroup$ – Hagen von Eitzen Oct 9 '14 at 17:33
  • $\begingroup$ @HagenvonEitzen If I use Cauchys integral formula, and inside the integral change the denominator with an appropritae geometric series I get a power series. And then if I use that this power series must be the Taylor series, am I done then? $\endgroup$ – user119615 Oct 9 '14 at 17:43

This is not true!

Take the function $$ f(z)=\sum_{n=3}^\infty \frac{z^n}{n(n-1)(n-2)}. $$ Then,

a. Its radius of convergence is $r=1$.

b. The power series converges if and only if $\,\lvert z\rvert\le 1$.

c. It defines a continuous function in the closed unit disk.

d. It is differentiable even on the boundary of the unit disk!


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