Timeline for Sum of squared hypergeometric polynomials
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Dec 6 at 9:00 | history | bounty ended | CommunityBot | ||
| Dec 2 at 5:47 | comment | added | Hjalmar Rosengren | @DimaPasechnik There is no formal power series that satisfies $uL'(u)=-1/2$. To expand the left-hand side as a series in $1-u$ is a very natural idea but of course it would be a different proof. You run into similar convergence problems. Note that the identity we want to prove is only true analytically for $0<u<1$. Something bad happens at both end-points. | |
| Dec 1 at 19:30 | comment | added | Dima Pasechnik | I'd have started with replacing $u$ with $1-u$ at the very beginning. | |
| Dec 1 at 19:24 | comment | added | Dima Pasechnik | technically speaking, it's not a formal power series. Rather one should talk about $uL'(u)=-1/2$. | |
| Dec 1 at 15:52 | comment | added | Hjalmar Rosengren | @DimaPasechnik You can also ponder how it is possible to start with a power series $L(u)$ in $u$, differentiate it and obtain $L'(u)=-1/2u$. | |
| Nov 30 at 13:48 | comment | added | Hjalmar Rosengren | @DimaPasechnik To me, this seems very far from a "usual formal power series argument". The whole issue is how to make sense of the sum over $m$ in (5). This series does not involve $u$, so it is irrelevant whether you consider $u$ as a complex or a formal variable. You could insert another formal power series variable $t^m$, but then you first need to explain how this variable appears in (6) and then how to make sense of letting $t=1$. | |
| Nov 30 at 12:06 | comment | added | Dima Pasechnik | It seems it's just a usual formal power series argument; worries about convergence are somewhat overblown here. | |
| Nov 30 at 7:57 | comment | added | Carlo Beenakker | I would think that Fred deserves the bounty ... | |
| Nov 30 at 6:49 | comment | added | Hjalmar Rosengren | @FredHucht I now realized how to make one of my own attempted proofs work. I post this as a separate answer. One could try to similarly insert a convergence factor $t^m$ in your computation, but it is not clear to me how to make this work. I hope someone can sort this out so I can award the bounty. | |
| Nov 30 at 4:11 | comment | added | Hjalmar Rosengren | @FredHucht Wow, that's amazing! My experience is that anything that wild usually gives wrong results. The terms in the series are $\mathcal O(m^{2k+2l-1})$, so they don't even tend to zero. If $k$ and $l$ were not integers, the sum in $m$ is a multiple of ${}_3F_2(k+1,l+1,2;2-k,2-l;1)$. This is still divergent for large $k$ and $l$, but when it converges it can be summed by Dixon's formula, which it seems that Mathematica knows. I don't know how to convert what you did to a valid proof. | |
| Nov 29 at 23:00 | comment | added | Fred Hucht | @HjalmarRosengren I used Mathematica for all calculations. The infinite sum over $m$ in \eqref{eq:5} is in fact divergent for integer $k,l$, but can be summed to \eqref{eq:6} without this condition. I guess that some regularization mechanism is doing the magic, and consistently shifts all contributions to the $k=l=0$ term. Maybe you have a better explanation, as you have much more experience with hypergeometric sums. | |
| Nov 29 at 21:51 | comment | added | Hjalmar Rosengren | How do you compute the sum over $m$ in the step from (5) to (6)? It looks to me like it is divergent. | |
| Nov 29 at 20:28 | comment | added | Carlo Beenakker | impressive; this question was unanswered for more than 7 years ... | |
| Nov 29 at 19:31 | history | edited | Fred Hucht | CC BY-SA 4.0 |
fixed some typos
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| Nov 29 at 19:23 | history | answered | Fred Hucht | CC BY-SA 4.0 |