Is there an example of an analytic function in the unit disc whose zeros are only the points $z_n=1-1/n$?
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$\begingroup$ If there is such a function it should be $\prod_{n=1}^\infty (z-z_n)$. $\endgroup$– azarelApr 16, 2012 at 20:06
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$\begingroup$ @azarel: that doesn't converge. $\endgroup$– GEdgarApr 16, 2012 at 20:14
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$\begingroup$ @GEdgar Thanks for pointing that out. $\endgroup$– azarelApr 16, 2012 at 20:25
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3 Answers
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Another one: $$ f(z) = \frac{1}{\Gamma\left(\frac{1}{z-1}\right)} $$ This is analytic in the plane, except one point $z=1$, and has zeros exactly $1-1/n$, $n=1,2,3\dots$. Unlike Henning's, which also has zeros ${} \gt 1$.
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$\begingroup$ Thanks for the examples given, but how did you come to it? Maybe Henning's could have come to mind, but how did you think about this in general? $\endgroup$ Apr 16, 2012 at 21:34
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$\begingroup$ I think it's just a matter of already knowing $1/\Gamma(z)$ as a standard example of a holomorphic function whose zeroes form a singly infinite arithmetic sequence. $\endgroup$ Apr 16, 2012 at 22:03
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How about $f(z) = \sin(\frac{\pi}{1-z})$?
answered Apr 16, 2012 at 20:11
hmakholm left over Monicahmakholm left over Monica
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For the general question of functions with prescribed zeros, consider Weierstraß' factorization theorem.
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$\begingroup$ Weierstrass factorization theorem is usually stated for entire functions, though. $\endgroup$– mrfApr 16, 2012 at 20:15
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$\begingroup$ True - I refered to this sentence from the Wikipedia page: "The theorem generalizes to the following: sequences in open subsets (and hence regions) of the Riemann sphere have associated functions that are holomorphic in those subsets and have zeroes at the points of the sequence." $\endgroup$– DirkApr 16, 2012 at 20:18
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1$\begingroup$ Just use a fractional linear transformation to go from a question about a sequence of prescribed zeros with a finite limit $\alpha$ to a sequence of prescribed zeros with a limit of $\infty$: get an entire function, and transform it back (obtaining a function analytic on ${\mathbb C} \backslash \{ \alpha \}$). $\endgroup$ Apr 16, 2012 at 20:47