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Basically what the title says.

I saw this through another Math Platform but did not get any response to it. The original question was to find distinct integers of $ x $ and $y$ such that $ \zeta(x) = \zeta(y) $.

I'm not too familiar with zeta functions but I'm sure that there must be a simple explanation for this.

Can anyone help me? Thank you.

GohP.iHan
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  • Can someone fix my question? I'm not sure how to amend it to Henrik's and whacka's comment. I don't know the exact words to use to describe my question. Yes, I'm talking about extending this to the non-conventional sense. – GohP.iHan Sep 08 '15 at 21:47
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    @whacka, hasnt the zeta function zeroes over all negative even integers (trivial zeroes)? – Nikos M. Sep 08 '15 at 21:51
  • Yup. (Did you mean to address that to OP?) – anon Sep 08 '15 at 21:52
  • @whacka, no i wanted to say negative integers, but then i looked it up (again), it is negative even integers! :) – Nikos M. Sep 08 '15 at 21:53
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    See https://en.wikipedia.org/wiki/Particular_values_of_Riemann_zeta_function#Negative_integers (and click through to the entry on Bernoulli numbers). – Barry Cipra Sep 08 '15 at 21:56

3 Answers3

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The general formula for the Riemann zeta function at negative integers, taken from the Wikipedia entry, is

$$\zeta(-n)=-{B_{n+1}\over n+1},$$

where $B_{n+1}$ is a Bernoulli number. As it happens, $B_2=1/6$ and $B_{14}=7/6$, so

$$\zeta(-1)=\zeta(-13)=-\frac{1}{12}.$$

As far as I know, the equality can be chalked up to coincidence, and any proof will require a fair amount of analytic infrastructure, such as the functional equation

$$\zeta(s)=2^s\pi^{s-1}\sin\left(\pi s\over2\right)\Gamma(1-s)\zeta(1-s).$$

Benedict W. J. Irwin
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Barry Cipra
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  • Thank you for your answer. By the way, it's $2^s$, not $2^2$. Follow up question: Prove that the only distinct integer solution for $\zeta(x) = \zeta(y)$ are $(x,y)=(-1,-13),(-13,-1)$. Is it something easy/reasonable to prove? – GohP.iHan Sep 08 '15 at 22:55
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    By the Von-Staudt-Clausen theorem the denominator of $B_{14}$ is the product of the primes $p$ such that $p-1$ divides $14$, so the equality of denominators (between $B_2$ and $B_{14}$) is not a mere coincidence. – Jack D'Aurizio Sep 08 '15 at 23:18
  • @GohP.iHan, you might try asking that as a separate question. – Barry Cipra Sep 08 '15 at 23:41
  • @Jack: I'm not familiar with any of these. Should I (still) ask a separate question? Or does your answer suffice? – GohP.iHan Sep 09 '15 at 13:46
  • @GohP.iHan: just have a look at the Wikipedia page about the Von-Staudt-Clausen theorem, there is no need of asking another question (about what?) – Jack D'Aurizio Sep 09 '15 at 14:10
  • @JackD'Aurizio: Pardon me, but it looks to me that we're only matching the denominators. How do we know the numerators are the same as well? Or have I interpreted it wrongly? – GohP.iHan Sep 09 '15 at 14:24
  • @JackD'Aurizio, are you saying the Von-Staudt-Clausen theorem proves that $mB_{2n}=nB_{2m}$ only for $(m,n)=(1,7)$? – Barry Cipra Sep 09 '15 at 14:26
  • @BarryCipra: no, I am just saying that the equality of the denominators of $B_2$ and $B_{14}$ is not a mere coincidence, but a consequence of such a theorem. The equality of numerators between $\zeta(-1)$ and $\zeta(-13)$, however, can be regarded as a mere coincidence. – Jack D'Aurizio Sep 09 '15 at 14:33
  • @JackD'Aurizio, Ok, thanks. I think the OP's follow-up (at least what I was suggesting as a separate question) might be best stated in the form of my last comment: Is $(m,n)=(1,7)$ the only solution to $mB_{2n}=nB_{2m}$ with $1\le m\lt n$? (Formulating it this way would avoid the issue of whether $\zeta(2n+1)$ is ever rational for positive $n$.) – Barry Cipra Sep 09 '15 at 14:38
  • @JackD'Aurizio and BarryCipra: Okay. I've posted another problem here. Thank you! – GohP.iHan Sep 09 '15 at 21:34
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Continuing the functions described in this answer and this answer, which are differences of finite sums of reciprocals of powers of integers and their Euler-Maclaurin Sum Formula approximations, set $$ \begin{align} \zeta_n(z) &=\sum_{k=1}^n\small\frac1{k^z}-\frac1{1-z}n^{1-z}-\frac12n^{-z}+\frac{z}{12}n^{-1-z}-\frac{z(z+1)(z+2)}{120\cdot3!}n^{-3-z}\\ &\small+\frac{z(z+1)(z+2)(z+3)(z+4)}{252\cdot5!}n^{-5-z}\\ &\small-\frac{z(z+1)(z+2)(z+3)(z+4)(z+5)(z+6)}{240\cdot7!}n^{-7-z}\\ &\small+\frac{z(z+1)(z+2)(z+3)(z+4)(z+5)(z+6)(z+7)(z+8)}{132\cdot9!}n^{-9-z}\\ &-\tfrac{691z(z+1)(z+2)(z+3)(z+4)(z+5)(z+6)(z+7)(z+8)(z+9)(z+10)}{32760\cdot11!}n^{-11-z}\\ &+\tfrac{z(z+1)(z+2)(z+3)(z+4)(z+5)(z+6)(z+7)(z+8)(z+9)(z+10)(z+11)(z+12)}{12\cdot13!}n^{-13-z} \end{align} $$ then for $\mathrm{Re}(z)\gt-15$, $$ \lim_{n\to\infty}\zeta_n(z)=\zeta(z) $$ For $z=-1$, $$ \zeta_n(-1)=\underbrace{\sum_{k=1}^nk-\left(\frac{n^2}2+\frac{n}2\right)}_{\text{these must cancel}}\underbrace{\vphantom{\sum_{k=1}^n}\ -\frac1{12}\ \ }_{\zeta(-1)}\underbrace{\vphantom{\sum_{k=1}^n}+O\left(n^{-2}\right)}_{\text{vanishes}} $$ For $z=-13$, $$ \zeta_n(-13)=\underbrace{\sum_{k=1}^nk^{13}-P_{13}(n)}_{\text{these must cancel}}\underbrace{\vphantom{\sum_{k=1}^n}\ -\frac1{12}\ \ }_{\zeta(-13)} $$ The parts that "must cancel" must cancel because, when $z$ is a non-positive integer, they are the difference of two polynomials with no constant terms. If they did not match, there would be no way for the limit to exist. Other than those terms are the constant term and the terms that vanish as $n\to\infty$. Thus, the constant term is $\zeta(z)$.

As Barry Cipra says, I don't know that there is any reason that these are the same other than coincidence.

Community
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robjohn
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$$\zeta(1-2k)=-\frac{B_{2k}}{2k}:k\gt 0$$ $$\displaystyle \sum_{k=0}^{n-1}k^{m}=B_{m}n+A_{2}n^{2}+...+A_{m+1}n^{m+1}$$ $$B_{m}=\displaystyle \frac{-1}{m+1}\sum_{k=0}^{m-1}\binom{m+1}{k}B_{k}$$ $$B_{0}=1:B_{1}=-\frac{1}{2},B_{2}=\frac{1}{6},B_{14}=\frac{7}{6}$$ $$k=1,\zeta(-1)=-\frac{B_{2}}{2\times 1}=-\frac{\frac{1}{6}}{2}=-\frac{1}{12}$$ $$k=7,\zeta(-13)=-\frac{B_{14}}{2\times 7}=-\frac{\frac{7}{6}}{2\times 7}=-\frac{1}{12}$$

Bouzari Abdelkader
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