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Say we have a function $\displaystyle f(z)=\sum_{n=0}^\infty a_n z^n$ with radius of convergence $R>0$. Why is the radius of convergence only $R$? Can we conclude that there must be a pole, branch cut or discontinuity for some $z_0$ with $|z_0|=R$? What does that mean for functions like
$$f(z)=\begin{cases} 0 & \text{for $z=0$} \\\ e^{-\frac{1}{z^2}} & \text{for $z \neq 0$} \end{cases}$$
that have a radius of convergence $0$?

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    @KH: Please rectify your question. I am not able to see it clearly.2010-08-12
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    @Chandru1: Fixed. Thanks for catching it.2010-08-12

1 Answers 1

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If the radius of convergence is $R$, that means there is a singular point on the circle $|z| = R$. In other words, there is a point $\xi$ on the circle of radius $R$ such that the function cannot be extended via "analytic continuation" in a neighborhood of $\xi$. This is a straightforward application of compactness of the circle and can be found in books on complex analysis, e.g. Rudin's.

However, it does not mean that there is a pole, branch cut, or discontinuity, though those would cause singular values. Indeed, a "pole" on the boundary would only make sense if you can analytically continue the power series to some proper domain containing the disk $D_R(0)$, and this is generally impossible. For instance, the power series $\sum z^{2^j}$ cannot be continued in any way outside the unit disk, because it is unbounded along any ray whose angle is a dyadic fraction. The unit circle is its natural boundary, though it does not make sense to say that the function has a branch point or pole there. (More generally, one can show that given any domain in the plane, there is a holomorphic function in that domain which cannot be extended any further, essentially using variations on the same theme.)

The function $\sum_j \frac{z^j}{j^2}$, incidentally, is continuous on the closed unit disk, but even though there is a singular point there. So continuity may happen at singular points.

The last function you mention does not have a power series expansion in a neighborhood of zero. In fact, it is not continuous at zero, because it blows up if you approach zero along the imaginary axis.

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    Could you please tell me why the function $\sum_ \frac{z^j}{j^2}$ has singular point on closed unit disk?at which point it is?2013-04-27
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    @Tsotsi: There's a singular point at $z = 1$; this is a special case of a theorem of Pringsheim. See http://books.google.com/books?id=uP8SF4jf7GEC&pg=PA235&lpg=PA235&dq=vivanti+pringsheim&source=bl&ots=yMqVs6tz3B&sig=jzagbZHNuI4WpQQZouONAjxACVw&hl=en&sa=X&ei=5n58Ub7QEI6u0AHZ5oG4BA&ved=0CDsQ6AEwAg#v=onepage&q=vivanti%20pringsheim&f=false2013-04-28