2
$\begingroup$

I am sorry beforehand for the length of my post, but I thought I should give some details. I try to figure out where are the singularities of a rather complicated power series.

This series comes from a linear operator that acts on a somehow strange space. I won't try to motivate the problem because I don't think it is important here.

The operator acts on functions of the following form. $$f(\tau,t;s)=\sum_{m\ge0}\sum_{k\ge0}\left( a_{m,k} \frac{t^k s^m}{\tau^{m+k+1}} + b_{m,k}\frac{s^{m+k}}{\tau^m t^{k+1}} \right)$$ Where $\tau$ and $t$ are considered complex coefficients and can be set to be however big is convenient, they can even be considered to be positive. The only variable is considered to be $s$. So the assumption is that given some $\tau$ and $t$, $f(\tau,t;s)$ is an entire function of $s$.

Let $\Gamma(z)=\sum_{n\ge0} \gamma_n z^n$, be a function analytic around 0 with isolated singularities and $\alpha= \partial_s^{-1} \partial_t$. Then we define the formal operator $$\Gamma(\alpha)=\sum_{n\ge0} \gamma_n \alpha^n.$$ This operator is linear and we get $$\Gamma(\alpha)[f(\tau,t;s)]=A(\tau,t;s)+B(\tau,t;s)$$ with $$A(\tau,t;s)=\sum_{m\ge0}\sum_{k\ge0}\sum_{n\ge0} \gamma_n \frac{\binom{k}{n}}{\binom{m+n}{n}}\left( \frac{s}{t} \right)^n a_{m,k} \frac{t^k s^m}{\tau^{m+k+1}} $$ $$B(\tau,t;s)=\sum_{m\ge0}\sum_{k\ge0}\sum_{n\ge0} \gamma_n \frac{\binom{k+n}{n}}{\binom{m+k+n}{n}}\left( -\frac{s}{t} \right)^n b_{m,k}\frac{s^{m+k}}{\tau^m t^{k+1}}$$

While trying to fight with this I stumbled upon the Hadamard product of power series. In short the result is the following. Let $X(z)=\sum_{n\ge0}x_n z^n$ and $Y(z)=\sum_{n\ge0}y_n z^n$, then $$\sum_{n\ge0}x_n y_n z^n=\frac{1}{2\pi i}\int_{C}\frac{X(\zeta)}{\zeta}\;Y\left(\frac{z}{\zeta}\right) d\zeta.$$ The integration is over a circle in the domain of convergence of $X$ and $z$ is such that $z/\zeta$ is in the domain of convergence of $Y$. There is the result that the singularities of the product are on the products of the singularities of the 2 functions on any sheet plus possibly the origin.

We have $$\sum_{n\ge0}\binom{k+n}{n} z^n=\frac{1}{(1-z)^{n+1}}$$ $$\sum_{n\ge0}\binom{k}{n} z^n=(1+z)^k$$ $$\sum_{n\ge0}\frac{z^n}{\binom{m+n}{n}}={}_2 F_1(1,1;m+1;z).$$

Then we have $$\sum_{n\ge0} \gamma_n \frac{\binom{k+n}{n}}{\binom{m+k+n}{n}}\left( -\frac{s}{t} \right)^n = -\frac{1}{4 \pi^2}\int_{C_\xi}\int_{C_\zeta} \frac{{}_2 F_1(1,1;m+k+1;\zeta)}{\xi} \; \frac{\zeta^k}{(\zeta-\xi)^{k+1}}\;\Gamma\left(- \frac{s}{t \, \xi} \right) d\zeta d\xi$$ $$\sum_{n\ge0} \gamma_n \frac{\binom{k}{n}}{\binom{m+n}{n}}\left( \frac{s}{t} \right)^n = -\frac{1}{4 \pi^2}\int_{C_\xi}\int_{C_\zeta} \frac{{}_2 F_1(1,1;m+1;\zeta)}{\xi} \; \frac{(\zeta-\xi)^{k}}{\zeta^{k+1}}\;\Gamma\left( \frac{s}{t \, \xi} \right) d\zeta d\xi. $$

If I am not mistaken, the singularities of these functions are exactly the singularities of the function $\Gamma$ and if $\Gamma$ is entire, then they are entire. My question is whether I can get more singularities when I sum those. So is there a proper pointwise bound of the above functions that will give me convergence of $\Gamma(\alpha)[f(\tau,t;s)]$ assuming that $f$ is entire and $s/t$ is away from the singularities of $\Gamma$?

It goes without saying that if anyone can think of any other way to tackle this, I am more than interested to hear it.

$\endgroup$

Your Answer

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

Browse other questions tagged or ask your own question.