1
$\begingroup$

Let $a,b\in{\mathbb Z}^{\ge0}$ and $h\in{\mathbb R}(d_1,d_2)$ be such that $$a\ge b, \quad h(d_1,d_2)>0~~\forall\,d_1,d_2\in{\mathbb Z}^+, \quad \lim_{d_1,d_2\longrightarrow\infty}\frac{h(d_1,d_2)}{d_1^ad_2^a/(d_1\!+\!d_2)^b}\in{\mathbb R}^+\,.$$ The limit statement means that these fractions are within $\epsilon$ of some number for all $d_1,d_2\!\ge\!d(\epsilon)$.

For $d\!=\!1,2,\ldots$, define the numbers $n_d\!\in\!{\mathbb R}^+$ inductively by $$n_1=1, \qquad n_d=\sum_{\begin{subarray}{c}d_1+d_2=d\\ d_1,d_2\ge1\end{subarray}}\!\!\!\! h(d_1,d_2)n_{d_1}n_{d_2} \quad\forall~d>1.$$ Are the numbers $\sqrt[d]{n_d}$ eventually increasing? In other words, is there $d^*\!\in\!{\mathbb Z}^+$ such that $$\sqrt[d]{n_d}\le\sqrt[d+1]{n_{d+1}} \qquad\forall~ d\ge d^*.$$ This is true if $h(d_1,d_2)=d_1^a d_2^a/(d_1\!+\!d_2)^a$. In general, these roots are bounded above and below away from zero.

$\endgroup$
  • $\begingroup$ Dear Aleksey, Could you say a little more about what this has to do with the recursion relations from Gromov-Witten theory? That might make it easier to put this question in context. $\endgroup$ – Jason Starr Mar 28 '16 at 12:56
  • $\begingroup$ I am sure that you know this already, but you may as well replace the initial condition, $n_1=1$, by the condition that $n_1>0$. The set of solutions is invariant under the transformation $n_d^{\text{new}} = a^d n_d^{\text{old}}$ for $a>0$. Also the conclusion you want to prove is invariant under this transformation. $\endgroup$ – Jason Starr Mar 28 '16 at 15:13
  • $\begingroup$ Okay, here is another silly observation. Assume that $a>0$. By your hypothesis, there exists $d(\epsilon)$ such that for all $d \geq d(\epsilon)$, also $h(d,d) \geq d^{2a-b}/u$ for some $u>0$. Then, for every $v>1$, there is a larger integer $d^*$ such that for all $d\geq d^*$, also $d^{2a-b}/u >v$. Then $\sqrt{2d}{n_{2d}} \geq v\cdot \sqrt{d}{n_d}$. Thus, even if the sequence $(\sqrt{d}{n_d})$ is not increasing, nonetheless it has many increasing subsequences. $\endgroup$ – Jason Starr Mar 28 '16 at 15:51
  • $\begingroup$ Typo correction: The LaTeX was bad for the final 2 sentences. It should have read "Then $\sqrt[2d]{n_{2d}} \geq \sqrt[d]{n_d}$. Thus, even if the sequence $(\sqrt[d]{n_d})_d$ is not increasing, nonetheless it has many increasing subsequences." $\endgroup$ – Jason Starr Mar 28 '16 at 16:27
  • $\begingroup$ In your applications, do you have any a priori upper bounds on $h(d_1,d_2)$ for $d_1+d_2 < 2d(\epsilon)$? If so, you could try to make my previous remark into an induction proof. $\endgroup$ – Jason Starr Mar 28 '16 at 16:57

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.