Let $\{z_j\}_{j \geq 1}$ and $\{p_j\}_{j \geq 1}$ be sequences of distinct complex numbers. Set $$M(z)=1+\sum_{j=1}^{+\infty} \frac{r_j}{z-p_j}, \quad z \in \mathbb{C}.$$ Under which conditions we can find $\{r_j\}_{j \geq 1} \subset \mathbb{C}$ such that $M$ is meromorphic and $$M(z_k)=0, \quad \forall k \geq 1.$$ ? I guess it is linked to the Mittag-Leffler theorem but I don't see how to use it, nor if it gives sharp conditions.


The Mittag-Leffler theorem says that there exists a meromorphic function with principal part $r_j/(z-p_j)$ at $p_j$; it does not say that the funtion is given by that sum. (And in fact the conclusion of ML still holds when that sum does not converge.)

It's clear that for any sequence $p_j$ there exists a sequence $r_j>0$ such that the sum converges to a meromorphic function.

  • $\begingroup$ I don't see why it is meaningless ? I have a function given in this form ($p_j$ fixed, $r_j$ to be chosen) and I would like to know if I can prescribe arbitrary zeros of this function by adjusting the $r_j$'s. If not, what are the conditions on $z_j$. $\endgroup$ – perturbation May 28 '18 at 17:33
  • $\begingroup$ You'd like to know if you can prescribe arbitrary zeroes, fine. That's not what you asked. $\endgroup$ – David C. Ullrich May 28 '18 at 17:41
  • $\begingroup$ How do you suggest that I rewrite my question then ? For me it was clear as I wrote "Let $z_j$ and $p_j$..." (so they are fixed) and then "we can find $r_j$...?" $\endgroup$ – perturbation May 28 '18 at 18:28
  • $\begingroup$ Sorry. I missed the first sentence - yes, it is perfectly clear. $\endgroup$ – David C. Ullrich May 28 '18 at 19:21
  • $\begingroup$ No problem. Do you have any hint then ? I thought it was linked to the Mittag-Leffler expansion, as in this post: math.stackexchange.com/questions/784659/… $\endgroup$ – perturbation May 28 '18 at 19:45

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.