The question is: Is there an entire function $f$ with $f(n)=n/(1+n^2)$ for each $n\in\mathbb N$?

This is how far I came: $g(z)=f(1/z)$ is holomorphic in $z\neq0$ with $g(1/n)=n/(1+n^2)$, so 0 cannot be removable by the Identity Theorem. If 0 is a pole, then $g(z)=h(z)/z^m$ for an entire function $h$ with $h(0)\neq0$ and some $m\in\mathbb N$. But then $h(1/n)=g(1/n)/n^m=1/(n^{m-1}+n^{m+1})\to0$ as $n\to\infty$ and so $h(0)=0$. Contradiction. Thus, 0 must be an essential singularity of $g$ (i.e. $f$ must be transcendental). I don't know how to proceed further. Any hints are highly appreciated. Thanks in advance!


1 Answer 1


The existence of such an entire function follows from the Weierstrass Factorization Theorem, and Mittag-Leffler's Theorem. More generally, if $\bigr(a_n\bigl)_{n\geq1},\bigr(z_n\bigl)_{n\geq1}\in\mathbb{C}$, $a_n\to \infty$ and $a_n$-s are distinct, then there exists an analytic $f$ such that $f(a_n)=z_n$. To see this, construct from WFT an entire function $g$ having simple zeroes at precisely $a_n$-s. From MLT, construct a meromorphic $h$ having no zeroes, but simple poles at precisely $a_n$-s, with residues $\dfrac{z_n}{g'(a_n)}$. Now $f=gh$ has the desired property.

  • $\begingroup$ I had this in mind, but couldn't do the final step. Thank you very much! $\endgroup$
    – sranthrop
    Jul 1, 2016 at 18:19

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