# 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.

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 $$X(t)$$:

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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