Timeline for $\zeta(0)$ and the cotangent function
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Jan 19, 2018 at 18:26 | answer | added | echinodermata | timeline score: 7 | |
| Dec 20, 2014 at 18:04 | vote | accept | GH from MO | ||
| Dec 6, 2014 at 21:11 | history | edited | GH from MO |
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| Dec 6, 2014 at 11:23 | answer | added | juan | timeline score: 19 | |
| Dec 4, 2014 at 17:23 | comment | added | GH from MO | @FredKline: Edwards does not emphasize the validity of his formula for $n=0$, in fact the previous display suggests that he meant it only for $n\geq 1$. At any rate, what you say is fine, but it does not answer my question. | |
| Dec 4, 2014 at 17:15 | comment | added | Fred Daniel Kline | From Edwards, p 12, (1): for $n=0$,$$\zeta (2 n)=\frac{(-1)^{n+1} 2^{2 n-1} \pi ^{2 n} B_{2 n}}{(2 n)!}=\frac{(-1)^{n+1} 2^{2 n-1} \pi ^{2 n}}{(2 n)!}=-\frac{1}{2},$$ with and without the Bernoulli number. | |
| Dec 4, 2014 at 14:37 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Dec 4, 2014 at 8:09 | comment | added | GH from MO | @TomCopeland: I am not so sure about that. Of course the Taylor series for $\psi(1+x)$ together with the last identity you mention yields an alternate proof of the 2nd identity in my post. But I don't see how this could answer my question, given that $\zeta(0)$ does not pop up naturally in the mentioned series. On the other hand, the $1/x$ term in the digamma identity has a clear and simple source, namely the identity $\Gamma(x+1)=x\Gamma(x)$. Once again, the issue is not why in my 2nd identity the constant term is $-1/2$ but why it equals $\zeta(0)$ as one would expect from that identity. | |
| Dec 2, 2014 at 18:07 | comment | added | GH from MO | @echinodermata: Excellent point! | |
| Dec 2, 2014 at 17:53 | comment | added | echinodermata | Don't forget the series also predicts that $\zeta(-2k)=0$ for negative $k$. | |
| Nov 29, 2014 at 17:25 | comment | added | GH from MO | @Agno: Your first formula is really about the relation of $\Gamma$ to $\sin$ and $\cos$, since by the functional equation for $\zeta$, the right hand side is a product of two ratios of $\Gamma$-values. So I think it is unrelated to my question, no matter how pretty the formula is. | |
| Nov 29, 2014 at 17:00 | comment | added | Agno | Just one additional thought. When equating both $\cot$ functions (by multiplying mine by $\frac{z^2}{2}$) and starting the infinite sum at $k=0$, a quite beautiful relation emerges between a (weighted) sum and a product of zetas that is valid in the domain $0<|z|<1$: $$\frac{\zeta(1+2\,z) \, \zeta(1-2\,z)}{\zeta(2\,z) \, \zeta(-2\,z)} = 2 \, \sum_{k=0}^\infty\zeta(2k)\,z^{2k-2}$$ | |
| Nov 29, 2014 at 15:48 | comment | added | Agno | I guess this is probably unrelated to your question, but just wanted to share another nice link between $\cot$ and $\zeta$ that I recently found. It is easy to derive from the reflection formula (just multiply $\zeta(s)$ and $\zeta(-s)$). $$-\frac{\pi}{z}\cot\left(\pi\, z\right)=\frac{\zeta(1+2\,z) \, \zeta(1-2\,z)}{\zeta(2\,z) \, \zeta(-2\,z)}$$ | |
| Nov 29, 2014 at 13:23 | comment | added | Liviu Nicolaescu | I do not have a convincing answer to your question. | |
| Nov 29, 2014 at 13:19 | comment | added | GH from MO | @LiviuNicolaescu: Thank you for your valuable comments. I looked at both sources, and indeed they do discuss Bernoulli numbers/polynomials, the analytic continuation of $\zeta(s)$, and the evaluation of $\zeta(0)$ and $\zeta(2k)$. Cartier also discusses a variant of my initial identity in Exercise 17, expressing $\coth(z/2)$ by a similar sum. Still, it is not clear to me how these discussions answer my question, perhaps I should read them in detail. | |
| Nov 29, 2014 at 12:39 | comment | added | Liviu Nicolaescu | The proof I know use Euler's summation method. Hardy's less known gem, Divergent series, discusses this, I think in the last chapter. Another place I think I saw this is P. Cartier's contribution in the volume From Number Theory to Physics, Springer 1992. (I don't have any of these sources in front of me.) | |
| Nov 29, 2014 at 12:24 | history | asked | GH from MO | CC BY-SA 3.0 |