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Given an analytic function on some domain. What is the general method to do analytic continuation and the obstruction to do analytic continuation? For example, $g(z)=1+z+z^2+\cdots$ is analytic on $|z|<1$. $f(z)=\frac{1}{1-z}$ is an analytic continuation of $g$ since $f$ is analytic on $\{z \mid z\neq 0\}$ which contains $\{z \mid |z|<1 \}$. Is the general method to find the analytic continuation computing the sum of the series? Is there an analytic function which is analytic on $\mathbb{C}$ which is a analytic continuation of $g(z)$? Why analytic continuation is important?

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I'm confused. First you say $f$ is an analytic continuation of $g$ (correct, except you mean $z\ne1$), then you ask if $g$ has an analytic continuation.... –  anon Oct 1 '11 at 21:24
    
...and then there are functions that you just can't analytically continue, like $$\prod_{k=0}^\infty (1-q^{k+1})$$... –  J. M. Oct 1 '11 at 23:44
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For analytic continuation there is Borel summation for divergent series. No, there is not such analytic continuation of $g$ since the difference $f-g\;$ is equal to zero in the unit circle and therefore its continuation is zero in $\mathbb C$. Analytic continuation is important for many reasons as discussed in complex analysis. For example, the behavior of the analytic continuation of the series $\sum_{n=1}^\infty n^{-z}\;$ in the critical strip is closely related with the distribution of primes.

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