4
$\begingroup$

This observation is based on the numerical calculation of the exponential sum: $$\sum_{n=1}^{[\sqrt{t/2\pi}]} n^{-\frac{1}{2}+it}$$ It is known that this sum is related to the famous Riemann–Siegel formula.

Now if we denote this sum as $S_1$, and define sum $S_2$ as follows: $$\sum_{n=1}^{[\sqrt[4]{t/2\pi}]} n^{-\frac{1}{2}+it}$$

Numerical calculation shows that there exists a strong similarity between $|S_1|$ and $|S_2|$, more precisely, it seems that we may have $|S_2|\ll |S_1| \ll |S_2|$.

Here is a plot of $|S_1|$ (blue) and $|S_2|$ (green), with $t$ varies in $[10^4-1000,10^4]$

and another with $t$ varies in $[10^6-1000,10^6]$, more calculations have affirmed this phenomenon.

So there goes the question: Does there exist a good mathematical explanation of this phenomenon?

$\endgroup$

2 Answers 2

4
$\begingroup$

The assertion is generally believed to be true. In fact, much more is conjectured: For every $\epsilon>0$ we have $\left|\zeta(\frac{1}{2}+it)-\sum_{n\leq t^\epsilon} n^{-\frac{1}{2}-it}\right|= \mathcal{O}(t^\epsilon)$. This statement is equivalent to the Lindelöff hypothesis. One direction is obvious: If $\zeta$ can be approximated by short sums, then $\zeta$ cannot be too large. The other direction can be found in Titchmarsh's book.

$\endgroup$
7
  • 2
    $\begingroup$ Sorry but this is not correct. It is true that to the right of the critical line the Zeta function can be approximated well by a short Dirichlet polynomial. But this is no longer true on the critical line. For example the mean square of the Zeta function is of size log t but the mean square of the short Dirichlet polynomial is only epsilon log t. $\endgroup$
    – Lucia
    Sep 15, 2017 at 16:52
  • $\begingroup$ Sorry, I don't have Titchmarsh with me, so I cannot check the correct statement. $\endgroup$ Sep 15, 2017 at 17:24
  • $\begingroup$ @Jan-ChristophSchlage-Puchta there chapter V he looks at short sums $\sum_{n \le a} n^{-1/2-it}$ $\endgroup$
    – reuns
    Sep 15, 2017 at 17:45
  • 3
    $\begingroup$ Lindelöf hypothesis is equivalent to $|\zeta(\frac{1}{2}+it)-\sum_{n\leq t^\epsilon} n^{-\frac{1}{2}-it}|= O(t^\delta)$ $\endgroup$
    – reuns
    Sep 15, 2017 at 18:09
  • $\begingroup$ Right. edited the answer accordingly. $\endgroup$ Sep 15, 2017 at 19:51
0
$\begingroup$

The explanation is quite trivial as you can think of the sum of length $t^{1/4}$ as approximately a smoothing of the sum of length $t^{1/2}$ in an interval of length approximately $4 / \log t$. So of course typically the two sums are not too far away. Moreover if the sum of length $t^{1/4}$ is large then there is a nearby large value of the sum of length $t^{1/2}$.

In particular the phenomenon you're observing has nothing to do with analytic number theory, it's a general fact of harmonic analysis applied to Dirichlet polynomials.

P.S: I don't understand why the other answer keeps getting upvoted while in fact not explaining anything. The equivalent statement of Lindelof stated there is trivial because the "approximating" Dirichlet polynomial is trivially bounded by $t^{\varepsilon}$. So the said equivalence is by the triangle inequality equivalent to $|\zeta(\tfrac 12 + it)| \ll t^{\varepsilon}$ and as such the "equivalence" doesn't involve anything about short Dirichlet polynomials and is not explaining anything... Note that Lindelof does not imply that the polynomial of length t^{1/4} is close to $\zeta(\tfrac 12 +it)$, this statement is simply false point-wise, even though one could conjecture that the ratios of the two Dirichlet polynomials are typically not too far apart. In any case the point-wise approximation can't be because the zeros of the short polynomial are in the first place not on the half-line...

$\endgroup$
4
  • $\begingroup$ Yes, the other answer is not great. But neither is yours! These Dirichlet polynomials oscillate a lot on the scale of $1/\log t$, so smoothing them out over such an interval destroys the behavior at any given point. In any case the phenomenon noted will not exist always for large $t$, and the question is not clear about what actually is intended. $\endgroup$
    – Lucia
    Sep 16, 2017 at 16:25
  • $\begingroup$ Given the Shanon interpolation theorem one can think of the short sum as being an average of roughly O(1) estimations of the long sum at points that are 1/log t spaced. Of course massive cancellations within these O(1) points are possible, but heuristically one would expect them to be rare and this should explain the phenomena observed by OP. For instance I would conjecture that the log of the ratio of the two Dirichlet polynomials (the long one over the short one) converges to some limit law.In particular since the first poly is typically either very large or very small so would be the other $\endgroup$
    – trivial
    Sep 16, 2017 at 19:25
  • $\begingroup$ What you say makes sense to me as far as averages of $\log \zeta(1/2+it)$ on such short intervals. It is unclear to me if the same should hold for $\zeta$ on such intervals. For example, the functional equation already tells you the argument of $\zeta$ (mod $\pi$), and this fluctuates a lot on the scale of $1/\log$. Would that not suggest more cancellation than what you say on this scale? $\endgroup$
    – Lucia
    Sep 16, 2017 at 22:51
  • $\begingroup$ So your point is $f_N(t) = \sum_{n < N} n^{-1/2+it}$ is band-limited on $[0,\log N)$ so $f_N(t) = \sum_k \frac{\sin(\pi \frac{t-k}{2 \log N})}{\pi (t-k)} f_N(\frac{k}{2 \log N})$ and $f_{N^{1/2}}(t) =\sum_k \frac{\sin(\pi \frac{t-k}{2 \log N^{1/2}})}{\pi (t-k)} f_N(\frac{k}{2 \log N^{1/2}})$. $\endgroup$
    – reuns
    Dec 23, 2017 at 1:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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