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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


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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. – Greg Martin Jul 8 '13 at 17:01
@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. – joro Jul 9 '13 at 6:45
up vote 4 down vote accepted

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.

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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. – joro Jul 17 '13 at 5:58
@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. – Stopple Jul 17 '13 at 14:38
The arg jump far away from zero on the first plot might be explained by $\Im \eta(1/2+it)=0$ – joro Jul 17 '13 at 14:58
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 – Stopple Jul 17 '13 at 17:42
And the reason such things are rare is related to this… – Stopple Jul 18 '13 at 1:40

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