Timeline for Is the sequence of Apéry numbers a Stieltjes moment sequence?
Current License: CC BY-SA 3.0
16 events
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| Oct 15, 2014 at 14:45 | comment | added | Gerald Edgar | I accepted this answer. But anyone reading should see the other answers, too, to get a full discussion. | |
| Oct 15, 2014 at 14:44 | vote | accept | Gerald Edgar | ||
| Sep 1, 2014 at 14:48 | comment | added | Pietro Majer | Well, logarithmic is not bad -- I was scared of stronger singularities that could make $u^2$ non-integrable... | |
| Sep 1, 2014 at 10:48 | comment | added | Gerald Edgar | I am not so sure any longer that the boundary terms all vanish. Maybe we do have to go down to zero. But there are logarithmic singularities there. | |
| Sep 1, 2014 at 9:13 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Sep 1, 2014 at 8:47 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Sep 1, 2014 at 6:38 | comment | added | Pietro Majer | But yes, $(c_0,c)$ would make everything nicer. (btw the 2-order DE can be written in a form of Sturm-Liouville operator, $(gu')'+hu=0$, dividing $2Pu''+P'u'+Qu=0$ by $|P|^{1/2}$ ). | |
| Sep 1, 2014 at 6:15 | comment | added | Pietro Majer | I thought each term $A(x), B(x), C(x)$ should vanish at $x=0$ and $x=c$, and be continuous at $x=c_0$, in order that the boundary terms cancel, because they are coefficients with different powers of $n$ in front. Also, by linearity $2u_1u_2=((u_1+u_2)^2-u_1^2-u_2^2$ is also a solution of the 3rd order DE, so we have $3$ linearly independent solutions ODE. | |
| Aug 31, 2014 at 22:11 | comment | added | Gerald Edgar | My computation says, for a solution of the 3rd-order DE, the boundary terms in the integration by parts will cancel at $c_o$ and at $c$. (The individual terms don't cancel, but taken together they do.) So I'm thinking of using interval $(c_o,c)$ and not trying the tricky thing of going down to $0$. | |
| Aug 31, 2014 at 22:05 | comment | added | Gerald Edgar | I noticed something else. According to the OEIS listing, $A_n = O(c^n/n^{3/2})$. If $w$ has a nonzero limit at $c$, we would only get $O(c^n/n)$ for the moments. So we need a $w$ to have a singularity like $(c-x)^{1/2}$ at $x=c$. Fortunately, that is compatible with the 3rd-order DE (but not with being a square of a solution of the 2nd-order DE). | |
| Aug 31, 2014 at 20:43 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Aug 31, 2014 at 20:19 | comment | added | Pietro Majer | The next step should be, checking if the boundary conditions in 2 are compatible with the solutions $u_1$ and $u_2$. And, of course, if they are in $L^2$. Note that (iii) would be satisfied if $u(c)=0$ because $c$ is then a root of $pw$, $qw$, $rw$ with multiplicity resp. $3,2,2$. Unfortunately this w.e. I'm in a place with no Maple! | |
| Aug 31, 2014 at 19:54 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Aug 31, 2014 at 19:41 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Aug 31, 2014 at 19:35 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Aug 31, 2014 at 19:27 | history | answered | Pietro Majer | CC BY-SA 3.0 |