Welcome to our community

Be a part of something great, join today!

infinite products defining entire functions

pantboio

Member
Nov 20, 2012
45
Consider the product
$$\displaystyle\prod_{n=1}^{+\infty}(1-e^{-2\pi n}e^{2\pi iz})$$

I've proven that this product converges uniformly on compact subsets of complex plane since the serie $\sum_{n=0}^{+\infty}|\frac{e^{2\pi iz}}{e^{2\pi n}}|$ does.

Now i'm interested to zeros of $F$, the entire function to which the product converges. How can i find them? Can i say that all the zeros of $F$ are those complex numbers $z$ such that $e^{-2\pi n}e^{2\pi i z}=1$? If yes, zeros are of the form
$$z_{n,k}=-in+k$$
$n,k$ integers. Do you think it's correct?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Hi pantboio! :)

Consider the product
$$\displaystyle\prod_{n=1}^{+\infty}(1-e^{-2\pi n}e^{2\pi iz})$$

I've proven that this product converges uniformly on compact subsets of complex plane since the serie $\sum_{n=0}^{+\infty}|\frac{e^{2\pi iz}}{e^{2\pi n}}|$ does.
I don't see that right away, but I'll take your word for it.

Now i'm interested to zeros of $F$, the entire function to which the product converges. How can i find them? Can i say that all the zeros of $F$ are those complex numbers $z$ such that $e^{-2\pi n}e^{2\pi i z}=1$? If yes, zeros are of the form
$$z_{n,k}=-in+k$$
$n,k$ integers. Do you think it's correct?
Yes.
A product of factors is zero if and only if at least 1 of those factors is zero.
This remains true in the limit to infinity.
 

pantboio

Member
Nov 20, 2012
45
Hi pantboio! :)



I don't see that right away, but I'll take your word for it.



Yes.
A product of factors is zero if and only if at least 1 of those factors is zero.
This remains true in the limit to infinity.
Thank you very much.
I used a theorem which tells that if $\{f_j\}$ is a sequence of functions, analytic somewhere, say in a domain $\Omega$, such that the series $\displaystyle\sum_{j=1}^{+\infty}|f_j|$ converges uniformly on compact subsets of $\Omega$, then the infinite product $\prod_{n=1}^{+\infty}(1+f_j)$ also converges , uniformly on compact subsets of $\Omega$, to a holomorphic function.
But my problem is now another. Ok, we have zeros of $F$, namely
$$z_{k,n}=-in+k$$
$n,k$ integers, $n\geq 1$. I have to prove the following:
$$\displaystyle\sum_{k}\frac{1}{|z_k|^{2+\epsilon}}$$
converges for every $\epsilon>0$, while
$$\displaystyle\sum_{k}\frac{1}{|z_k|^{2}}$$ diverges, where $\{z_k\}$ is some enumeration of zeros, but honestly i can't see this. If zeros are $z_{k,n}=-in+k$, then $|z_{k,n}|=\sqrt{n^2+k^2}$ depending both on $n$ and $k$ and i don't know how to proceed....
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Before we start to do anything profound, let's take a look at that statement.

Suppose we pick $z_k = k$, then:

$\displaystyle \sum_k \dfrac 1 {|z_k|^2} = \frac{\pi^2}{6}$​

... but... but... that converges!

If we pick an n other than zero, it will only converge faster.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,705
Thank you very much.
I used a theorem which tells that if $\{f_j\}$ is a sequence of functions, analytic somewhere, say in a domain $\Omega$, such that the series $\displaystyle\sum_{j=1}^{+\infty}|f_j|$ converges uniformly on compact subsets of $\Omega$, then the infinite product $\prod_{n=1}^{+\infty}(1+f_j)$ also converges , uniformly on compact subsets of $\Omega$, to a holomorphic function.
But my problem is now another. Ok, we have zeros of $F$, namely
$$z_{k,n}=-in+k$$
$n,k$ integers, $n\geq 1$. I have to prove the following:
$$\displaystyle\sum_{k}\frac{1}{|z_k|^{2+\epsilon}}$$
converges for every $\epsilon>0$, while
$$\displaystyle\sum_{k}\frac{1}{|z_k|^{2}}$$ diverges, where $\{z_k\}$ is some enumeration of zeros, but honestly i can't see this. If zeros are $z_{k,n}=-in+k$, then $|z_{k,n}|=\sqrt{n^2+k^2}$ depending both on $n$ and $k$ and i don't know how to proceed....
So you want the sum $\displaystyle \sum_{n,k}\frac1{|z_{k,n}|^p} = \sum_{n,k}\frac1{(n^2+k^2)^{p/2}}$ to be divergent when $p=2$, but convergent when $p>2$.

Take the $p=2$ case first. Then the sum is $\displaystyle \sum_{n,k}\frac1{n^2+k^2}$. The sum over $n$ goes from $1$ to $\infty$, and the sum over $k$ goes from $-\infty$ to $\infty$. But if we ignore the negative values of $k$ then the sum of the remaining terms is $$\sum_{n=1}^\infty\sum_{k=0}^\infty \frac1{n^2+k^2}.$$ To estimate this double sum, do a "diagonal count" by considering all the terms for which $k+n=m$ for a fixed $m$. There are $m$ such terms (because $n$ can take any value from $1$ to $m$, and the corresponding value of $k$ goes down from $m-1$ to $0$). Each of these terms satisfies $$\frac1{n^2+k^2} = \frac1{m^2 - 2nk}\geqslant \frac1{m^2},$$ and since there are $m$ of them their sum is at least $\dfrac m{m^2} = \dfrac1m.$ Now when you sum over $m$, you see that the sum diverges.

The case $p>2$ requires a completely different argument. This time we are looking at $\displaystyle \sum_{n,k}\frac1{(n^2+k^2)^{p/2}}$, and my (non-rigorous) suggestion is to use a "two-dimensional" version of the integral test for convergence. If you replace the sum by the integral $$\int^\infty\!\!\int^\infty \frac1{(x^2+y^2)^{p/2}}dx\,dy$$ and transform to polar coordinates, you will be looking at $$\int^\infty r^{1-p}\,dr,$$ which converges when $p>2$. But I do not know of any formal justification for this version of the integral test.
 

pantboio

Member
Nov 20, 2012
45
Two facts we saw
$$\displaystyle\sum_{k}\frac1{|z_k|^{2+\epsilon}}\<\infty\ \forall\epsilon, \ \textrm{and}\ \displaystyle\sum_{k}\frac1{|z_k|^2}=\infty$$

precisely mean that $b=2$, where $b$ is the convergence exponent of the sequence of zeros of $F$, which is defined as
$$b=\inf\{\lambda>0:\displaystyle\sum_{k}\frac1{|z_k|^{\lambda}}<\infty\}$$

Now we define the order of growth of an entire function $f$ as
$$\rho=\{\lambda>0:\displaystyle\sup_{|z|=r}|f(z)|=O(e^{r^{\lambda}}),r\rightarrow\infty\}$$
There's a theorem which ensures
$$b\leq\rho$$
In our case, $\rho\geq 2$. I'm pretty convinced that also the reverse holds, so that $\rho$ is exactly 2, but no idea how to prove this....