Connective constant of a hexagonal lattice

pr.likelihood – Incredibly correct recursions for the Riemann Zeta duty Answer

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pr.likelihood – Incredibly correct recursions for the Riemann Zeta duty

During some statistical analysis about iterated convolutions and stirring averages, I got here up by random with shocking outcomes for the Dirichlet Eta duty associated to $zeta(s)$, and this spiked once more my curiosity about its complicated roots within the captious divest.

It began love this. Let $X(t)$ breathe a time-discrete time succession. Apply the next transformation:
$$Y(t)=sum_{ok=-N}^N h(ok)X(t-k), mbox{ with } sum_{ok=-N}^N h(ok)=1$$

in order that $Y = h*X$, the place $*$ denotes the discrete convolution operator. Iterate $n$ instances: $Y_{n+1} = h * Y_n$ 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 forth. In quick,
$$Y_n(t)=sum_{ok=-N_n}^{N_n} h_n(ok)X(t-k), mbox{ with } sum_{ok=-N_n}^{N_n} h_n(ok)=1.$$
Start with $N=N_1=1$ and $h_1(-1) = h_1(0) = h_1(1) = frac{1}{3}$. Clearly, the correctly scaled $h_n(ok)$ coefficients have a bell configuration, simply love the Binomial coefficients. The scaling issue is $O(sqrt{n})$: it is a direct utility of the central restrict theorem. Note that $N_n=n$. I wished to employ industry information as an example the conception, however as an alternative ended up utilizing the actual and imaginary sever of $eta(s)$, with $s=0.75 + ti$ a complicated quantity. So in what follows, $X(t)$ represents the actual or imaginary sever of the complicated Dirichlet Eta duty $eta(s)$.

Applying $h_n$ with $n=60$ to $X(t)$ yields the outcome illustrated within the portray under, after capable scaling.

enter image description here

The blue round on the backside is $X(t)$, on this illustration the actual sever of $eta(s)$ for integer values $60leq t leq 240$. The purple round is $h_{60} * X$ (convolution) on the identical province. It could be very properly approximated by a cosine duty, so properly that you could not visually inform the dissimilarity between $h_{60} * X$ and its approximation within the higher sever of the portray: the 2 curves are virtually completely super-imposed. This is brought on by the altenative of $h_n$ and its connection to the central restrict theorem (my speculate); peculiar weighted stirring averages do not labor. If as an alternative you deem the imaginary sever of $eta(s)$, you get the identical kindly of graph with identical scaling issue, however this time the approximation is with a sine duty.

Also, the actual sever of $eta(s)$ investigated right here is

$$X(t)=sum_{ok=1}^infty (-1)^{ok+1}frac{cos(tlog ok)}{ok^{0.75}}.$$
For the imaginary sever, supplant cosines by sines. You cannot supplant $log ok$ by (say) $sqrt{ok}$ or will lose the near-perfect reconcile. If you supplant $log ok$ by a duty rising too steadfast, the conduct of the approximation turns into chaotic. This is illustrated under when you supplant $log ok$ by $sqrt{ok}$ within the definition of $X(t)$:

enter image description here

The dips should not a glitch in my computations. I too checked values of $t$ as much as $t=600$, with quiet a excellent reconcile for the actual and imaginary components of $eta(s)$, however with further dips and bumps when you supplant $log ok$ by (say) $sqrt{ok}$.

Question

Could this kind of system breathe of any employ to seek the zeroes of the Riemann Zeta duty? How do you elucidate such a coincidence (the superb approximation)? Is my system too undoubted for $t>600$? Is it unusual to $eta(s)$? How to generalize to steady convolutions?

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