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I would like to know the convergence radius of the following two double power series of $(x,y) \in \mathbb{C}^2$:

\begin{align} \sum_{m,n=0}^\infty \frac{(d-a)_{n+m}(d+b)_{n+m}(d+a)_n(d-b)_n}{n!m!(2d-c)_n(d)_{2n+m}}x^ny^m \end{align}

\begin{align} \sum_{m,n=0}^\infty \frac{(n+m)!(n+m)!}{(2n+m)!m!}x^ny^m \end{align}

where $(u)_v:=\Gamma(u+v)/\Gamma(u)$ is the Pochhammer symbol, and $a,b,c,d$ are complex numbers.

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  • $\begingroup$ The radius of convergence of a multivariate series is not a number. The region of convergence is a convex domain in the $(\log|x|,\log|y|)$ plane. $\endgroup$ Commented Feb 22, 2021 at 12:12
  • $\begingroup$ Exactly. My question is about the convergence region. Thanks for the clarification. $\endgroup$
    – H1son0
    Commented Feb 22, 2021 at 12:18

2 Answers 2

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This is to give an explicit description of the domain of convergence of your second series provided in the answer by Alexandre Eremenko.

In the $(s,t)$-plane, the domain is the interior of the set \begin{equation} \{(s,t)\in\mathbb R^2\colon s<g(t)\}, \end{equation} where \begin{equation} g(t):=\inf_{a\in(0,1)}h(a,t) \end{equation} and \begin{equation} h(a,t):=\frac{(1-a) \ln (1-a)+(1+a) \ln (1+a)-(1-a) t}a. \end{equation} The partial derivative of $h(a,t)$ in $a$ is \begin{equation} \frac{t-\ln(1-a^2)}{a^2}. \end{equation} So, if $t>0$, then $h(a,t)$ is increasing in $a\in(0,1)$ from $h(0+,t)=-\infty$ and hence $g(t)=-\infty$. If $t<0$, then $h(a,t)$ attains its minimum in $a\in(0,1)$ when $a$ is the root of the equation $t-\ln(1-a^2)=0$, that is, when $a=\sqrt{1-e^t}$. Thus, the domain of convergence for your second series is \begin{equation} \{(s,t)\in\mathbb R^2\colon t<0\ \&\ s<h(\sqrt{1-e^t},t)\}. \end{equation}

Part of this region (for $t>-8$) is shown here:

enter image description here

Note that $h(\sqrt{1-e^t},t)\to\ln4=1.386\dots$ as $t\to-\infty$.

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  • $\begingroup$ Thanks. I could only make a rough sketch by hand. $\endgroup$ Commented Feb 22, 2021 at 16:09
  • $\begingroup$ @AlexandreEremenko : I was working on an answer, but it did not occur to me to first consider the implicit description of the domain that you provided. After seeing your answer, the life got much easier. :-) $\endgroup$ Commented Feb 22, 2021 at 16:15
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The region of convergence is the union of polydisks $\{(x,y):|x|<r_1,|y|<r_2\}$ such that $$\limsup_{m+n\to\infty}|c_{m,n}r_1^nr_2^m|^{1/(m+n)}\leq 1,$$ which is the generalization of the Cauchy-Hadamard formula. Computation for your second series leads to the following description of this region. In the plane $(s,t)=(\log r_1,\log r_2)$ the region is the intersection of half-planes $H_\alpha,\; 0\leq\alpha\leq 1$, where $$H_\alpha=\{(s,t):\alpha s+(1-\alpha)t<(1+\alpha)\log(1+\alpha)+(1-\alpha)\log(1-\alpha)\}.$$ I do not see how to describe this intersection more explicitly. One can do a similar computation for the first series.

The reference for the Cauchy-Hadamard formula is B. A. Fuks, Theory of analytic functions of several complex variables, AMS, Providence, R.I. 1963, Ch. I, Sect. 3, Theorem 3.7.

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  • $\begingroup$ Thanks a lot for the answer! But, I am afraid it is not clear for me how the half-plane $H_\alpha$ is derived from the formula. I would be grateful if you could kindly explain it. $\endgroup$
    – H1son0
    Commented Feb 25, 2021 at 5:40
  • $\begingroup$ I used Stirling's formula, and the Cauchy-Hadamard that I wrote. Notation $\alpha=\lim n/(m+n), (m+n)\to\infty$. Take log of the expression in CH, apply Stirling, and do arithmetic. $\endgroup$ Commented Feb 25, 2021 at 12:29

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