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How to find an entire function $f(z)$ such that $f(n) = \sqrt {|n|}$ for every integer $n$?

Now my thinking is to create a series $\displaystyle\sum_{k=-\infty}^\infty f_{k}(z)$ such that $f_{k}(z)=\sqrt{|k|}$ and the series is convergent for every $z$.

Thanks for any help

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What have you tried so far? What don't you understand? Where are you stuck? – fdart17 Apr 19 '11 at 18:45
I thought of $$f(z)=\sum_{n=-\infty}^{\infty}\sqrt{|n|} \frac{\sin2\pi x}{x-n}$$ because for any interger $m$ $$f(z)=\sum_{n=-m}^{m}\sqrt{|n|} \frac{\sin2\pi x}{x-n}$$ is entire and has the values you want on all the integers between $-m$ and $m$. Unfortunately the infinite series has some serious convergence issues, and doesn't work at all. – Eric Naslund Apr 19 '11 at 21:48

the existence of such thing is standard, see Rudin under 'interpolation'. To actually construct it , you are multiplying by $\sum \sqrt{\vert n \vert} \frac 1 {x-n}$ which has the correct poles at n but doesn't converge. Try $\sum \sqrt{\vert n \vert} (\frac 1 {x-n} - \frac 1 n)$ which has the same poles & residues but speeds up the convergence

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