The function $f$ below claims to approximate the Riemann zeta function $\zeta(s)=\sum_{n=1}^{\infty} \frac{1}{n^s}$ at all positive integers $s>1$:

$$ f(s) = \frac{\pi^s}{\left\lfloor((2^s - 1)\frac{\pi^s}{2^s}\right\rfloor-1} \approx\zeta(s), $$

where $\lfloor x \rfloor$ represents the greatest integer small than or equal to $x$.

Indeed, tabulating the first few values gives:

\begin{array}{|c|c|c|} \hline s & f(s) & |\zeta(s) - f(s)| \\ \hline 2 & \frac{\pi^2}{6} & 0 \\ \hline 3 & \frac{\pi^3}{26} & 0.00950\ldots \\ \hline 4 & \frac{\pi^4}{90} & 0 \\ \hline 5 & \frac{\pi^5}{295} & 0.00042\ldots \\ \hline 6 & \frac{\pi^6}{945} & 0 \\ \hline 7 & \frac{\pi^7}{2995} & 0.00009\ldots \\ \hline 8 & \frac{\pi^8}{9450} & 0 \\ \hline 9 & \frac{\pi^9}{29749} & 0.000011\ldots \\ \hline \end{array}

I thought there was no function that even came close to "unifying" the values of $\zeta(s)$ at odd and even integers. What is the idea behind the construction of $f$ that allows such behaviour?

  • 1
    $\begingroup$ What about $\zeta(12)$? $\endgroup$ – Angina Seng Apr 3 '18 at 14:33
  • $\begingroup$ Did you calculate the decimal values of the give expressions for f(s), the second column, and compare them to the third column? They are nowhere near! For example, when s= 9, the middle column has $\frac{\pi^9}{29749}= 1.0020202$. That's not at all close to the third column, 0.000011. $\endgroup$ – user247327 Apr 3 '18 at 14:35
  • 1
    $\begingroup$ @user247327 The third column is the difference between $\zeta$ and $f$... $\endgroup$ – Klangen Apr 3 '18 at 14:38
  • $\begingroup$ @LordSharktheUnknown It is $2.302\ldots \times 10^{-7}$ $\endgroup$ – Klangen Apr 3 '18 at 14:39
  • 1
    $\begingroup$ "I thought there was no function" It's an approximation, it's not a closed form. You can always create an approximation (and it's not hard to create an even better approximation). For example the simple expression $\zeta(s) \approx 1 + \frac{1}{2^s}$ is also a very good approximation for large $s$ (say all $s>10$). $\endgroup$ – Winther May 25 '18 at 12:50

For large $s$, $\zeta(s) \approx 1/(1 - 2^{-s})$ (this is the first factor of $\zeta$'s Euler product).
Let us approximate this approximation with something in the form $\pi^s/n(s)$ for some integer $n(s)$.

Then, $n(s)$ has to be close to $\pi^s(1-2^{-s})$. Naturally, picking $n(s) = \lfloor \pi^s(1-2^{-s})\rfloor$, we almost obtain your formula. In fact, if $\epsilon(s) = \pi^s(1-2^{-s}) - n(s)$, then

$\pi^s/n(s) = \pi^s/(\pi^s(1-2^{-s})-\epsilon(s)) = 1/(1 - 2^{-s} - \epsilon(s) \pi^{-s})\\ = 1 + 2^{-s} + \epsilon(s) \pi^{-s} + 4^{-s} + \ldots$

Substracting an extra $1$ from the denominator is the same as increasing $\epsilon(s)$ by $1$, and this makes the term $\epsilon(s) \pi^{-s}$ change into $(1+\epsilon(s))\pi^{-s}$, which I suppose is a bit closer to $3^{-s}$on average (especially for small values of $s$).

In both cases, the error term is of the order of $3^{-s}$ as $s$ gets large.

You can make better formulas (better because they work better for even values of $s$) by using more terms of the Euler product to build $n(s)$.

  • $\begingroup$ Excellent answer, thank you. I can award you the bounty in "19 hours" :) $\endgroup$ – Klangen May 25 '18 at 13:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.