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Where can I find a proof of this theorem: For each $r \in \mathbb{Z}_{+}$, there exists a complex entire function $f(z)$ such that $f(r) \neq 0$ but $f(r+1)=f(r+2)=\cdots =0$, i.e. $f(z) \in I_{r+1}$ but $f(z) \neq I_{r}$, where $I_{r}= \{ f \in R \ | f(r)=f(r+1)= \cdots =0\}$ where $R$ is the ring of complex entire functions

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closed as too localized by Robin Chapman, Felipe Voloch, S. Carnahan Jul 31 '10 at 19:21

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2 Answers 2

up vote 5 down vote accepted

Why don't you start with a function with zeros at the integers, for instance $\sin\pi z$, and then somehow eliminate the zero at $r$?

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Since your question title has Weierstrass, presumably you know about Weierstrass products. You should use them! They let you build an entire function having zeros at any discrete set. (The more refined Hadamard product construction can be used when the chosen zeros are spread out enough, which is the case for your situation.)

There is a direct answer to your question using the reciprocal of the $\Gamma$-function with a linear change of variables. You really should look at the Weierstrass product of the $\Gamma$-function to understand what is going on here so that you see what its Weierstrass product achieves in terms of zeros/poles.

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