For $t \in \mathbb{R}$ define $$ F(t) = \sum_{n=1}^{[t]} \frac{(-1)^{(n-1)}}{n^{\frac12 + it}}$$

Let $\operatorname{Arg}(t)$ be $\operatorname{atan2}(\Im t , \Re t)$ - basically this is $\arctan$, but the sign depends on the quadrant.

Observation $\operatorname{Arg}(F(t))$ jumps (usually from negative to positive) very near all nontrivial zeta zeros on the critical line and in seemingly rare occasions without zeros. Computing $F(t)$ the naiive way is not efficient for me.

$F(t)$ is truncated Dirichlet eta function on the critical line, but it is not $0$ at zeros of zeta, though $|F(t)|$ has local minima near zeros.

Added Wolfram Alpha found closed form for $F(t)$ in terms of Hurwitz zeta and zeta:

\begin{align} & \sum_{n=1}^k\frac{(-1)^{n-1}}{n^{\frac12 + i t}} = \\ & 2^{-1/2-i t}(-(-1)^k \zeta(i t+1/2, (k+1)/2)+ \\ & (-1)^k \zeta(i t+1/2, (k+2)/2)+2^{1/2+i t} \zeta(1/2 i (2 t-i))-2 \zeta(1/2 i (2 t-i))) \end{align}

Setting $k=[t]$ gives $F(t)$. Numerical evidence supports the closed form.

In comments Greg Martin suggested $F(t)$ might not be correlated to higher zeros, though numerical evidence suggests it is correlated at height $10^6$, including closely spaced zeros.

Another observation is $|F(t)|$ appears to have local minima close to zeta zeros on the critical line.

Setting $$ G(t) = \sum_{n=1}^{[t]} \frac{(-1)^{(n-1)}}{n^{1 + it}}$$

the jumps of $G(t)$ appear zeros of $\eta(1+i t)$ and looks like $|G(t)| \sim |\eta(1 + i t)|$

Can this be explained?



Closely spaced zeros and modulus


  • $\begingroup$ I imagine $F(t)$ will be correlated with the first several zeros (the larger $t$ is, the more zeros will be involved) but that eventually it will be more or less completely agnostic about high-up zeros. One could probably make this rigorous using a trace formula of some kind. $\endgroup$ – Greg Martin Jul 8 '13 at 17:01
  • $\begingroup$ @GregMartin There is closed form for $F(t)$ in terms of Hurwitz zeta and zeta, edited the question. The modulus appears related to zeros too and it works for $t=10^6$ and closely spaced zeros. $\endgroup$ – joro Jul 9 '13 at 6:45

I'm not sure why this is mysterious. The function $F(t)$ is a partial sum for the "$\eta$-function" $$ \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^{1/2+i t}}=(1-2^{1-s})\zeta(s) $$ a function which converges conditionally, and uniformly for $t$ in a compact set. Now $F(t)$ has discontinuities due to the truncation $[t]$. However, for large $t$ where the density of zeros of $\eta$ is $2\pi/\log(t)$ we will see many zeros in between each discontinuity.

The function $\eta(1/2+it)$ will trace out a curve in the complex plane, passing repeatedly through the origin. It's argument will have a discontinuity of $+\pi$ each time (I'm assuming RH here.) Since $F(t)$ tracks this function closely, it's argument will change rapidly, even when $F(t)$ does not pass through the origin.

The places where $\arg F(t)$ seems to jump away from a Riemann zero are more mysterious, but I would first consider the possibility the $\arg$ code is lying to you.

| cite | improve this answer | |
  • $\begingroup$ Thank you Stopple. I wrote it is truncated eta too. How do you explain local minima of |F(t)| are approximations of zeros without known counterexamples so far? And for "how large $t$" do you expect counterexamples (so far it works for t=10^6 for closely spaced zeros). More mysterious to me are closely spaced zeros. $\endgroup$ – joro Jul 17 '13 at 5:58
  • $\begingroup$ @joro: The answer to your question about the local minima of $|F(t)|$ is exactly the same: $|F(t)|$ is a good approximation to $\eta(1/2+i t)$. In contrast to what Greg Martin argues, I'm claiming this does not fail for large $t$; in fact the approximation gets better. $\endgroup$ – Stopple Jul 17 '13 at 14:38
  • $\begingroup$ The arg jump far away from zero on the first plot might be explained by $\Im \eta(1/2+it)=0$ $\endgroup$ – joro Jul 17 '13 at 14:58
  • $\begingroup$ The plot jumps from (about) $-\pi$ to (about) $+\pi$ near $1006.5$. This happens when the function crosses the negative $x$ axis, it's the defined behavior of atan2:en.wikipedia.org/wiki/Atan2 $\endgroup$ – Stopple Jul 17 '13 at 17:42
  • $\begingroup$ And the reason such things are rare is related to this answer:mathoverflow.net/questions/73098/… $\endgroup$ – Stopple Jul 18 '13 at 1:40

About the local minima approximating zeros, have you considered Hurwitz theorem? if the limit of the sequence of partial sums is 0 (not identically), then, in a disk close neighbourhood of the concerned zero, each nth partial sum with n greater than an index k (k depending on the size of said disk) would feature a zero. Perhaps, your minima correspond indeed to said zeroes ... of course, in general, to exactly hit zero, instead of a minimum just close to it, you will have to move within said disk also a bit away from the critical line.

| cite | improve this answer | |

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.