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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.

Karthik C
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    For $A$ the function $z \prod_{m>0}(1-\tfrac{z}{im^2})$ should do. – WimC Dec 28 '12 at 13:48
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    Well, I think the product $z\prod (1-z/im^2)$ is convergent... and this should solve the problem for the subset $A$. For the subset $B$, i guess you don't want an answer involving the Weiestrass factorization theorem, do you? – wisefool Dec 28 '12 at 13:48
  • I have edited. I question excluded 0. Only $im^2$ with $m$ nonzero. I regret for the error. The Weierstrass factorisation theorem is new to me. I would however appreciate any solution. – Karthik C Dec 28 '12 at 14:30
  • The summation of logs of the terms in the product are just big Oh of $1/m^2$. I mean as $m\rightarrow\infty$ the limit of $\frac{\log (1-\frac z{m^2})}{1/m^2}$ is -z. By comparison test the series converges for each z. So the product converges. Thanks for the useful input! Need to still check normal convergence. – Karthik C Dec 28 '12 at 14:37
  • Checked! Thanks! I stay tuned for the second half of my question – Karthik C Dec 28 '12 at 15:09

1 Answers1

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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.

  • If $g$ is another such function, then what is the relation between $f$ and $g$? At least $\frac fg$ is non vanishing and entire. That apart, is there any relation more specific? For instance, can one find an entire h such that $\frac fg=\exp h$ in this case? – Karthik C Dec 31 '12 at 06:17
  • @Aneesh If the functions $f$ and $g$ have zeroes of the same order at every point of your set (for example, if all zeroes are simple), then $f/g$ is indeed nonvanishing and entire. Therefore, $\log(f/g)$ is well defined and holomorphic in $\mathbb C$. It follows that $f=ge^h$ as you suggested. // However, if the orders of zeroes of $f$ and $g$ are not the same, the ratio $f/g$ could vanish and could have poles. For example, $g$ could be equal to $f^2$. –  Dec 31 '12 at 06:21
  • Oh god really my bad. I forgt to mention that the question demands that all the zeroes are simple indeed. Thanks very much. – Karthik C Dec 31 '12 at 06:28