Timeline for Combinatorial identity involving the square of $\binom{2n}{n}$
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 3, 2018 at 4:09 | answer | added | Max Alekseyev | timeline score: 1 | |
| May 19, 2014 at 21:10 | answer | added | Erwin | timeline score: 3 | |
| May 17, 2014 at 14:40 | comment | added | Suvrit | @Erwin: I saw that in people.fas.harvard.edu/~sfinch/csolve/cbc.pdf who cites several relevant papers that may help! | |
| May 17, 2014 at 14:25 | answer | added | Christian Elsholtz | timeline score: 1 | |
| May 17, 2014 at 3:26 | comment | added | Erwin | Suvrit, that is interesting, because in fact one gets from summation by parts that $$\sum_{k=0}^{n-1}\frac{\sum_{j=0}^k a_j^2}{(2k+1)(2k+3)}=\frac{1}{2}\sum_{k=0}^n\frac{a_k^2}{2k+1}-\frac{1}{2(2n+1)}\sum_{j=0}^na_j^2$$ which allows to write the limit in the form $\lim_{n\to\infty}(2n+1)\left(\frac{\pi}{4}-\frac{1}{2}\sum_{k=0}^n\frac{a_k^2}{2k+1}+\sum_{k=0}^{n-1} \frac{\sum_{j=0}^ka_j^2}{(2k+1)(2k+3)}-\sum_{k=0}^{n-1}\frac{\sum_{j=0}^ka_j^2}{(2k+1)(2k+2)}\right)$ but I don't know if this is better. Where can I see a proof of the identity you cite involving the Catalan constant? | |
| May 17, 2014 at 2:34 | history | edited | Erwin | CC BY-SA 3.0 |
added 16 characters in body
|
| May 17, 2014 at 1:18 | comment | added | Suvrit | also, Z.-W Sun: math.nju.edu.cn/~zwsun seems to be one of the masters of such sums---you may wish to contact him. | |
| May 17, 2014 at 1:03 | comment | added | Suvrit | I guess you must be already aware of related stuff like $\sum_{k=0}^\infty a_k^2/(2k+1) = 4G/\pi$, where $G$ is Catalan's constant... | |
| May 16, 2014 at 22:45 | history | edited | KConrad | CC BY-SA 3.0 |
added 1 character in body
|
| May 16, 2014 at 22:02 | history | edited | Erwin | CC BY-SA 3.0 |
New information added
|
| May 16, 2014 at 21:27 | comment | added | Erwin | I will look into this. I have read the notes on combinatorial identities in H.W. Gould's website without success thus far. But there is a chapter on convolutions so I'll look into that. Thanks! | |
| May 16, 2014 at 18:51 | comment | added | Steven Stadnicki | I don't have my original reference (Melzak's Companion to Concrete Mathematics, IIRC) handy, but isn't there a variant of Hadamard convolution that produces the function $f(x) = \sum_n a_nb_nx^n$ from the two functions $g(x)=\sum_n a_nx^n$ and $h(x)=\sum_n b_nx^n$? | |
| S May 16, 2014 at 17:27 | history | suggested | F. C. |
added one tag for hypergeometric function
|
|
| May 16, 2014 at 17:23 | review | Suggested edits | |||
| S May 16, 2014 at 17:27 | |||||
| May 16, 2014 at 16:39 | history | edited | Emil Jeřábek | CC BY-SA 3.0 |
fix title
|
| May 16, 2014 at 16:38 | answer | added | Igor Rivin | timeline score: 5 | |
| May 16, 2014 at 16:25 | history | asked | Erwin | CC BY-SA 3.0 |