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Apr 13, 2017 at 12:58 history edited CommunityBot
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Oct 15, 2014 at 14:44 vote accept Gerald Edgar
Sep 2, 2014 at 2:00 answer added Gerald Edgar timeline score: 3
Aug 31, 2014 at 19:27 answer added Pietro Majer timeline score: 3
Aug 31, 2014 at 15:51 comment added Pietro Majer Good point, $o(c^n)$ just tells $\mu(\{c\})=0$.
Aug 31, 2014 at 13:21 comment added Gerald Edgar It is $o(c^n)$, but that does not make the density vanish at the endpoint. Try density $1$ on $[0,1]$: the moments are $1/(n+1) = o(1)$, but $1$ does not vanish at the endpoint.
Aug 31, 2014 at 12:56 comment added Pietro Majer Another remark: it seems to me that in fact $A(n)=o(c^n)$. Is it true? In this case, $w$ should vanish at $c$.
Aug 30, 2014 at 17:55 comment added Pietro Majer Sure, I'll post it in a couple of days (I'm on travel now).
Aug 30, 2014 at 14:22 answer added Gerald Edgar timeline score: 3
Aug 30, 2014 at 14:07 comment added Gerald Edgar @Pietro: Would you like to add a (partial) answer showing your derivation of the ODE. The set of solutions on $(1/c,c)$ is 2-dimensional, so for this to work we will need some way to pick the right one. I think we cannot get both ends going to zero, however.
Aug 29, 2014 at 21:43 comment added Gerald Edgar This is interesting. Maple solves that ODE in terms of the "Heun G" function. And $1/c$ and $c$ correspond to two of the defining singularities of the HG that we get. Maple's support for HG is rather limited, though, so I have to work to get information.
Aug 29, 2014 at 17:17 comment added Pietro Majer So my guess is: the measure $\mu$ has a.c. part with density $u^2$ for some solution $u$ of the above ODE, maybe singular at $0, 1/c$ or $c$, plus possibly some deltas at these points. And maybe it is a measure on $[1/c,c]$.
Aug 29, 2014 at 17:06 comment added Pietro Majer From the recurrence of the $A_n$ we have $A_n=O(c^n)$ where $c:=17+12\sqrt{2}$ is the larger root of $x^2-34x+1$. Therefore, the measure $\mu$ has support in the interval $[0,c]$ (so we are dealing with the Hausdorff moment problem, and the measure is unique). Moreover, the above linear equation should be studied on $[0,c]$. The series solution with $u(0)=0$ has convergence radius $1/c$ (in fact $1/c$ is the other root).
Aug 29, 2014 at 16:59 comment added Pietro Majer Exact, from the recurrence of $A_n$ I got a third order linear ODE for a density $w=d\mu/dx$. Trying to prove the existence of a positive solution, i.e. $w=u^2$, I noticed that $u$ would satisfy the above second order ODE. I do not know what that means, but seems remarkable. However, I agree that the latter has no nontrivial bounded solutions, but there is another remark:
Aug 29, 2014 at 15:06 comment added Gerald Edgar The series solution for that ODE shows that the only solution with $u(0)=0$ is the identically zero solution. Where did this ODE come from? The recurrence for $A_n$?
Aug 27, 2014 at 19:01 comment added Pietro Majer (I'm just not completely sure that such a solution $u$ does exist)
Aug 27, 2014 at 18:35 comment added Pietro Majer I observed the following fact. Consider the following second order linear ODE with polynomial coefficients: $$(4x^3-136x^2+4x)u''+(8x^2-204x+4)u'+(x-10)u=0.$$ Assume that $u$ is a solution on $(0,+\infty)$, with $u(0)=0$ and $u(x)=o(x^{-n})$ for all natural $n$, and normalized so that $\int_0^\infty u^2dx=1$. Then, the measure $\mu$ with density $d\mu= u^2dx$ has the Apery numbers as a sequence of moments.
Aug 26, 2014 at 19:16 comment added Pietro Majer I think there is some hope to determine a solution, and I'll try some computations.
Aug 23, 2014 at 0:38 history edited Gerald Edgar CC BY-SA 3.0
edited body
Aug 22, 2014 at 23:39 answer added Suvrit timeline score: 11
Aug 22, 2014 at 14:48 comment added Steve Huntsman oeis.org/A228143
Aug 22, 2014 at 13:59 history asked Gerald Edgar CC BY-SA 3.0