1
$\begingroup$

Consider

$$f(z)=\sum_{w\in C}\frac{1}{z-w}$$ Where $C$ is the set of complex integers. What I would like to know is where can I find any information about this function (name perhaps). For instance, I don't even know how the summation would be carried out considering there are so many different ways to sum over the lattice. Does it converge? How fast does it grow? Etc... I think it must be related to elliptic functions because it is doubly periodic, but knowing nothing else of EF, I am sortof stuck. Any thoughts would be great.

$\endgroup$
  • $\begingroup$ Why do you think that function is defined? $\endgroup$ – Thomas Andrews Jun 1 '14 at 14:36
  • $\begingroup$ Well it seems like the simplest function periodic in two directions. I'm hoping it exists (converges, that is) because it is related to how many integer lattice points lie within a circle of certain radius (Gauss' circle problem), and is just generally interesting IMO. $\endgroup$ – Elie Bergman Jun 1 '14 at 14:38
  • $\begingroup$ It doesn't work, unfortunately. The basic examples of doubly-periodic functions involve squaring the $\frac{1}{z-w}$ because otherwise there is no limit for this sum. See en.wikipedia.org/wiki/Weierstrass's_elliptic_functions $\endgroup$ – Thomas Andrews Jun 1 '14 at 14:42
2
$\begingroup$

The following is taken pretty much verbatim from Apostol. In general, consider the function $f_a:\mathbb{C}-\Lambda\to \mathbb{C}$ (with $\Lambda =\{n+im|(n,m) \in \mathbb{Z}^2\}$) $$f_a(z)=\sum_{(m,n )\in \Lambda}\frac{1}{(z-(n+im))^a}.$$ Consider the unit square with centre $C$ that $z$ is in. The minimum and maximum distance from $z$ to the vertices of the cell are $R,r>0$ respectively. Then consider the square with the same centre with side-length $3$. It has $4 \cdot 2$ points, all of which are at a distance between $2\cdot R$ and $2\cdot r$ from $z$. Repeating this procedure to the all squares with centre $C$ and edd side length, we have

$$\sum_{k \ge1}\frac{4k}{(kr)^a}\le \sum_{(m,n) \in \Lambda}\left|\frac{1}{(z-(n+im))^a}\right|\le \sum_{k \ge 1}\frac{4k}{(kR)^a}$$ $$\frac{4}{r^a}\zeta(a-1)\le \sum_{(m,n) \in \Lambda}\left|\frac{1}{(z-(n+im))^a}\right|\le \frac{4}{R^a}\zeta(a-1).$$

Thus $f_a(z)$ is convergent iff $\zeta(a-1)$ is convergent (i.e. $a>2$).

$\endgroup$
  • $\begingroup$ Pretty sweet. Which book is this from? $\endgroup$ – Elie Bergman Jun 1 '14 at 16:28
  • $\begingroup$ This one. $\endgroup$ – Meow Jun 1 '14 at 16:30
  • $\begingroup$ Although this sum won't converge, the fact that the function you listed does for a>2 -> by analytic continuation, our sum should exist. A valid argument? $\endgroup$ – Elie Bergman Jun 1 '14 at 16:32
  • $\begingroup$ It's foreseeable that you could analytically continue $f_a(z)$ for all complex $a \ne 2$. $\endgroup$ – Meow Jun 1 '14 at 16:37
  • $\begingroup$ That's my question on the sum of all complex numbers. $\endgroup$ – Elie Bergman Jun 1 '14 at 16:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.