Timeline for Are these two new ways of representing odd zeta values as integrals known?
Current License: CC BY-SA 3.0
13 events
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Sep 27, 2017 at 15:41 | comment | added | Agno | @Paul Enta. Very nice work and thanks for the great update! | |
Sep 26, 2017 at 21:44 | comment | added | Paul Enta | @Agno I poted an new answer which gives an explicit representation of the polynomials which uses Euler polynomials and which should be consistent with your findings in terms of Euler numbers. | |
Jun 6, 2017 at 21:06 | comment | added | Agno | Thanks Paul. I found the formula through "reverse engineering" with the help of this particular Sloane's integer sequence oeis.org/A086646. I discovered that the integers in this triangle correspond to your coefficients as finite sums of Euler numbers weighted by a binomial. One wishes a more direct mathematical proof of course, but I don't have that yet. Note that the closed form becomes even more elegant when you let the index start at 1. It then becomes: $$g_n(y)=-\frac{1}{n\,(2n-1)!}\,\sum_{m=0}^{n}\,y^{2m}\sum_{k=0}^{m}\,E \big(2n-2k \big)\binom{2n}{2k}$$. | |
Jun 6, 2017 at 20:51 | comment | added | Paul Enta | @Agno You are perfectly right. I corrected the typos in $g_5$ and add the expression for $g_6$ which fits your formula too. Congratulations. How did you get this result? | |
Jun 6, 2017 at 20:46 | history | edited | Paul Enta | CC BY-SA 3.0 |
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Jun 5, 2017 at 22:52 | comment | added | Agno | @Paul Enta. I believe the closed form for your polynomials is: $$g_n(y)=-\frac{1}{(n+1)\,(2n+1)!}\,\sum_{m=0}^{n+1}\,y^{2m}\sum_{k=0}^{m}\,E \big( 2\,(n+1-k) \big)\binom{2(n+1)}{2k}$$ Where $E(x)$ is an Euler number. If my formula is indeed correct, I like to conjecture that you made a typo in $g_5$ (that is also indexed wrongly by naming it $g_2$), where the coefficient $95$ should actually be $65$. Grateful if you could confirm. Thanks! | |
Jun 4, 2017 at 18:36 | history | edited | Paul Enta | CC BY-SA 3.0 |
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Jun 3, 2017 at 9:33 | comment | added | Paul Enta | Thank you! I had a lot of fun. I understand the constraint you imposed to the polynomial factor. There can exist many different class of solutions. I just added a starting point at $f_1=x^2p_3(x)$ in your list to show that different expressions can be found. As you noticed in your comment, at least with this method, the structure of the polynomials seems to be very characteristic. I also corrected some typos. | |
Jun 3, 2017 at 9:31 | history | edited | Paul Enta | CC BY-SA 3.0 |
Correct some typos, add few examples
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Jun 3, 2017 at 8:32 | comment | added | Wolfgang | Very nice! When I first looked at it, I obtained experimentally the same polynomials that start with $f_1\equiv1$. By noticing that the coefficients come in pairs (i.e. that $f_{n+1}(x)$ minus the leading $x^{2n}$ term has a factor $(x+1)$), I then figured that there are some degrees of freedom, and if we want to obtain unique polynomials, we must add constraints, e.g. require the last $n$ terms to vanish. So I came up with the factor $x^n$ under the integral. BTW I happened to do all that shortly before Zurab Silagadze posted his answer to the other question, so what was timely. :) | |
Jun 3, 2017 at 7:05 | history | edited | Paul Enta | CC BY-SA 3.0 |
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Jun 2, 2017 at 23:48 | history | edited | Paul Enta | CC BY-SA 3.0 |
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Jun 2, 2017 at 23:43 | history | answered | Paul Enta | CC BY-SA 3.0 |