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Last update as of Jan 27, 2021: I posted this as an article for laymen, here. It is very light mathematically speaking, but section 3 is a little more accurate than my post here.

During some statistical research about iterated convolutions and moving averages, I came up by chance with surprising results for the Dirichlet Eta function related to $\zeta(s)$, and this spiked again my interest about its complex roots in the critical strip.

It started like this. Let $X(t)$ be a time-discrete time series. Apply the following transformation: $$Y(t)=\sum_{k=-N}^N h(k)X(t-k), \mbox{ with } \sum_{k=-N}^N h(k)=1$$

so that $Y = h*X$, where $*$ denotes the discrete convolution operator. Iterate $n$ times: $Y_{n+1} = h * Y_n$ with $Y_0=X$ and $Y_1=Y$. Denote as $h_n$ the $n$-fold self-convolution of $h$, with $h_1=h$, $h_2=h * h$, and so on. In short, $$Y_n(t)=\sum_{k=-N_n}^{N_n} h_n(k)X(t-k), \mbox{ with } \sum_{k=-N_n}^{N_n} h_n(k)=1.$$ Start with $N=N_1=1$ and $h_1(-1) = h_1(0) = h_1(1) = \frac{1}{3}$. Clearly, the properly scaled $h_n(k)$ coefficients have a bell shape, just like the Binomial coefficients. Indeed, they are known as trinomial coefficients, see here. The scaling factor is $O(\sqrt{n})$: this is a direct application of the central limit theorem. Note that $N_n=n$. I wanted to use business data to illustrate the concept, but instead ended up using the real and imaginary part of $\eta(s)$, with $s=0.75 + ti$ a complex number. So in what follows, $X(t)$ represents the real or imaginary part of the complex Dirichlet Eta function $\eta(s)$.

Applying $h_n$ with $n=60$ to $X(t)$ yields the result illustrated in the picture below, after proper scaling.

enter image description here

The blue curve at the bottom is $X(t)$, in this case the real part of $\eta(s)$ for integer values $60\leq t \leq 240$. The red curve is $h_{60} * X$ (convolution) on the same domain. It is very well approximated by a cosine function, so well that you can not visually tell the difference between $h_{60} * X$ and its approximation in the upper part of the picture: the two curves are almost perfectly super-imposed. This is caused by the choice of $h_n$ and its connection to the central limit theorem (my guess); ordinary weighted moving averages don't work. If instead you consider the imaginary part of $\eta(s)$, you get the same kind of graph with same scaling factor, but this time the approximation is with a sine function.

Also, the real part of $\eta(s)$ investigated here is

$$X(t)=\sum_{k=1}^\infty (-1)^{k+1}\frac{\cos(\frac{1}{2} t\log k)}{k^{0.75}}.$$ For the imaginary part, replace cosines by sines. You can not replace $\log k$ by (say) $\sqrt{k}$ or will lose the near-perfect fit. If you replace $\log k$ by a function growing too fast, the behavior of the approximation becomes chaotic. This is illustrated below if you replace $\log k$ by $\sqrt{k}$ in the definition of $X(t)$:

enter image description here

The dips are not a glitch in my computations. I also checked values of $t$ up to $t=600$, with still a perfect fit for the real and imaginary parts of $\eta(s)$, but with additional dips and bumps if you replace $\log k$ by (say) $\sqrt{k}$.

Question

Could this type of formula be of any use to study the zeroes of the Riemann Zeta function? How do you explain such a coincidence (the excellent approximation)? Is my formula also true for $t>600$? Is it peculiar to $\eta(s)$? How to generalize to continuous convolutions?

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  • $\begingroup$ The coefficients $h_n(k)$ are known as triangular coefficients, see oeis.org/… $\endgroup$ Jan 24, 2021 at 10:54

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