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Does there exist a non-constant entire function $f : \mathbb{C}\to\mathbb{C}$ such that $f(n+\dfrac{1}{n})=0$ for all $n\in \Bbb N$?

Let $f$ be a non-constant entire function such that $f(n+\dfrac{1}{n})=0\forall n\in \Bbb N$.

Then $f(2)=0;f(3+\frac{1}{3})=0$ and so on.But the problem is the set of zeros of $f$ does not have a limit point.

How can I conclude whether such a function exists or not?Please help

dtldarek
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    The Weierstrass factorization theorem allow you to find an entire function such that has zeros exactly at a sequence ${z_n}_{n\geq 1}$ such that $z_n \to \infty$. – Sorombo Oct 14 '16 at 05:12
  • $\prod_{n=1}^\infty (1-\frac{z}{n+1/n})e^{z/(n+1/n)}$ – reuns Oct 14 '16 at 05:13
  • Otherwise, you should be able to modify $\displaystyle\frac{1}{\Gamma(\frac{z+(z^2-4)^{1/2}}{-2})}$ such that it is entire – reuns Oct 14 '16 at 05:15
  • Related: http://math.stackexchange.com/questions/1161375/entire-function-vanishing-at-n-frac1n-for-n-geq-1/1162183#1162183 – Micah Oct 14 '16 at 05:20

1 Answers1

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There exists such a function. An infinite product such as

$$f(z) = \prod_{n =1}^\infty \left(1-\frac{z^2}{(n+1/n)^2}\right)$$

determines a nonconstant entire function of $z$ with zeroes at $z= \pm (n+1/n)$.

Bruno Joyal
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