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Construct an entire function $f$ whose zero set is $\{im^2: m\in\mathbb N\}\cup\{\sqrt[4]n: n\in\mathbb N\}$, all zeroes being simple. If $g$ is another such function what is the relation between $f$ and $g$? I tried constructing two separate functions one with zero set $A=\{im^2: m\in Z\}$ and another with the zero set $B$, the remainder of what is supplied. The idea is that the product of the functions so constructions would do the job for the former part of the question. I encountered problems however:

I tried condisering the series $\sum\frac1{z-im^2}$. It is a series that converges normally outside $A$. It has simple poles at points in $A$. So when I took reciprocal I got simple zeroes at points in $A$. Sadly the series has infinitely many zeroes on the imaginary axis. So I need to remove them all before taking the reciprocal to ensure that the reciprocal is entire. But I could not get it through.

I considered the function $\cosh(\pi iz^4)$. But its zero set strictly contains $B$ where again I had trouble having to remove infinitely many zeroes.

Any help is greatly appreciated.

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    Checked! Thanks! I stay tuned for the second half of my question2012-12-28

1 Answers 1

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I don't know of a simpler way to get a function for the set B than $ h(z)=\prod_{n=1}^\infty \left(1-\frac{z}{n^{1/4}} \right)\exp\left\{ \frac{z}{n^{1/4}} + \frac{z^2}{2n^{1/2}}+ \frac{z^3}{3n^{3/4}}+ \frac{z^4}{4n}\right\} $ which, admittedly, is taken straight from the proof of the Weierstrass factorization theorem. (If you think this is cheating, you know where the downvote button is.)

The reason for the weird exponentials is that they make the series of logarithms converge. Indeed, the logarithm of the $n$th term involves $ \left(1-\frac{z}{n^{1/4}} \right) = -\frac{z}{n^{1/4}}-\frac{z^2}{2n^{1/2}}-\frac{z^3}{3n^{3/4}}-\frac{z^4}{4n}-\frac{z^5}{5n^{5/4}}-\dots $ but the logarithm of $\exp(\dots)$ cancels the first four summands here. What is left is a convergent series, due to the exponents $5/4>1$ and higher.

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    Oh god really my bad. I forgt to mention that the question demands that all the zeroes are simple indeed. Thanks very much.2012-12-31