0
$\begingroup$

In reference to this question which I asked here couple of days back but didn't get any answer I am posting this question to clarify whether we can able to extend Euler's identity regarding the generating function for partitions to the unit disk from the interval $[0,1).$ I think that I need to take help of analytic continuation. But for that I need to show that the function $F(x)= \prod\limits_{k=1}^{\infty} \frac {1} {1 - x^k}$ is analytic on the unit disk and the function $G(x) = \sum\limits_{k=0}^{\infty} p(k)x^k$ is analytic on $[0,1).$ If they are so then we say that $F$ is an analytic continuation of $G$ on the unit disk. Then only we can say that $G$ can be extended analytically to the unit disk and for all $x \in S^1 \setminus [0,1)$ we assign $G(x)= F(x).$ That will do my job. But why $F$ and $G$ are analytic in their respective domains. Can anybody please help me regarding this?

Thank you very much for your valuable time.

$\endgroup$
2
  • $\begingroup$ Both sides are holomorphic, since they are uniformly convergent on each closed disc $\{z:|z|\le r\}$ for $r<1$. $\endgroup$ Commented May 7, 2019 at 6:05
  • $\begingroup$ Why is it so @Lord Shark the Unknown? Would you please be more explicit about your argument? $\endgroup$ Commented May 7, 2019 at 6:17

0

You must log in to answer this question.