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I was fooling around, trying to come up with a pseudo-rapid way to compute $\pi$. Then I remembered that for the Riemann zeta function ($\zeta$), we always have: \begin{equation} \zeta(2n)=c\pi^{2n}, \end{equation} where $n$ is a positive integer and where $c$ is some rational number. Thus, I thought that an efficient way of computing $\pi$ would be to be simply to invert this relation (denoting $s=2n$): \begin{equation} \pi=(\zeta(s)/c)^{1/s} \end{equation} Then I thought that to make this even faster, one could find a power $s$ such that $c$ takes the form $c=1/d$, with $d$ an integer (since, numerically, multiplication is faster than division). Then we would have: \begin{equation} \pi=(d\zeta(s))^{1/s} \end{equation} I think the largest the $s$ power is, the faster one can compute $\pi$ up to some precision (by summing a finite number of terms of the zeta-function).

  • However, I wasn't able to find $s>10$ such that $c=1/d$... Using Mathematica I searched up to $s=1000$.

This inspired me to ask the following question:

Is there a $s>10$ such that $c=1/d$ with $d\in\mathbb{N}$ ?

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    $\begingroup$ Two things: (1) I think it'd be easier to just write $\zeta(2n) = c\pi^{2n}$ instead of explicitly requiring the number $s$ to be even - unless you want to open the question up to all zeta-values, in which case (I believe) the answer is that nobody even knows whether zeta functions at odd values are rational multiples of the corresponding power of pi; there are good reasons to believe they're not, but we don't have enough info to say... $\endgroup$ – Steven Stadnicki Apr 1 '14 at 0:44
  • $\begingroup$ Sure, I'll update the question. Indeed my question is only for even powers. $\endgroup$ – VanillaSpinIce Apr 1 '14 at 0:48
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    $\begingroup$ (2) While this is a great problem in its own right and well worth asking about, it's worth pointing out that there are much faster ways of computing $\pi$ than even this; any direct (that is, based on summing the zeta series directly) approach will yield a number of digits that increases linearly at each iteration, but there are other methods (for instance, those based on the arithmetico-geometric mean) that yield much faster growth in accuracy at each iteration; see en.wikipedia.org/wiki/AGM_method for a basic introduction. $\endgroup$ – Steven Stadnicki Apr 1 '14 at 0:48
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    $\begingroup$ Fun fact: That approach has been turned around slightly, and used by Plouffe and Fee to compute high-indexed Bernoulli numbers. $\endgroup$ – ccorn Apr 1 '14 at 0:53
  • $\begingroup$ Considering valuation with respect to $p=2$ may narrow down the search. Apart from that, indices $2n$ where the numerator of the Bernoulli number $B_{2n}$ contains prime factors $p>2n$ can be shown not to fulfill the required condition. But numerators aren't easy... $\endgroup$ – ccorn Apr 1 '14 at 1:48
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I rant too much to post comments!

Let me begin by saying that this is equivalent to proving the existence, of $2n>10$, of $$\dfrac{\pi^{2n}}{\zeta(2n)}\in\mathbb{N}$$ We know that, $B_n$ are the Bernoulli numbers, $$\dfrac{\pi^{2n}}{\zeta(2n)}=\dfrac{(2n)!}{(-1)^{n+1}B_{2n}2^{2n-1}}$$ Note that $\dfrac{\pi^{2n}}{\zeta(2n)}=\dfrac{B_n}{A_n}$, where $A_n,B_n$ are the sequences $A002432$ and  $A046988$, respectively, in OEIS.

Now, if we write $\xi_n=\dfrac{B_n}{A_n}$, we see that we can define inductively, $$\xi_1=\dfrac{1}{6}\\ \xi_n=\sum_{i=1}^{n-1}(-1)^{i-1}\frac{\xi_{n-i}}{(2i+1)!}+(-1)^{n+1}\frac{n}{(2n+1)!}\\ \implies \xi_n=\dfrac{\pi^{2n}}{\sum^\infty_{k=1}\dfrac{1}{k^{2n}}}$$ Using mathematica, one gets the values of $\xi_n$. I am not sure if for all finite values of $n>5$, $\xi_n\in\mathbb{N}$. From this MO post, you can see that $\zeta(2n)$ is a rational multiple of $\pi^{2n}$, and for a proof, see here.

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    $\begingroup$ How is this a rant...? =D $\endgroup$ – Pedro Tamaroff Apr 1 '14 at 1:33
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    $\begingroup$ Not an answer. $\xi_5$ corresponds to index $10$, and for $\xi_6$ I get a denominator of $691$. $\endgroup$ – ccorn Apr 1 '14 at 1:34
  • $\begingroup$ @PedroTamaroff :-) This was just too long to be a comment! $\endgroup$ – user122283 Apr 1 '14 at 1:34
  • $\begingroup$ @ccorn How is this not an answer? $\endgroup$ – user122283 Apr 1 '14 at 1:35
  • $\begingroup$ @SanathDevalapurkar: An answer would be to find some $\xi_n\in\mathbb{N}$ for $n>5$, or to show that there aren't any. $n=6$ does not work. $\endgroup$ – ccorn Apr 1 '14 at 1:36

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