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In An older question I asked :

( See A Thue-Morse Zeta function (Generalized Riemann Zeta function and new GRH) )

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Consider $t_n$ as the Thue-Morse sequence. Let $m$ be a positive integer and $s$ a complex number, and recall that the Odiuos numbers are the indices of nonzero entries in the Thue-Morse sequence. Now consider the sequence of functions below:

$$f(1,s)=1+2^{-s}+3^{-s}+4^{-s}+\dotsb$$

This is the zeta function valid for $\mathrm{Real}(s)>1$.

$$f(2,s)=1-2^{-s}+3^{-s}-4^{-s}+\dotsb$$

This is the alternating zeta function valid for $\mathrm{Real}(s)>0$.

$$f(3,s)=1-2^{-s}-3^{-s}+4^{-s}+5^{-s}-6^{-s}-7^{-s}+8^{-s}+\dotsb = 4^{-s} (\zeta(s,1/4) - \zeta(s,2/4) - \zeta(s,3/4) + \zeta(s,4/4) ) $$

( $\zeta(s,a)$ is Hurwitz zeta )

I'm not sure if this has an official name yet but it clear that it is valid for $\mathrm{Real}(s)>-1$. This sequence of functions is constructed in the similar way the Thue-Morse sequence is constructed.

$$\begin{align} &\vdots\\ f(\infty,s)&= \sum (-1)^{t_n} n^{-s} \end{align}$$

This is a nice generalization/variant of the Riemann Zeta function and the Dirichlet eta or Dirichlet $L$-functions. It follows that $f(m, s)$ is valid for $\mathrm{Real}(s)>-m+2$. Now there are two logical questions analogue to the questions about the Riemann Zeta function:

  1. What are the functional equations for $f(m,s)$?

  2. Call the $N^\text{th}$ zero $Z_n(m)$. Are all the zero's of $f(m,s)$ for any $m$ with $0<\mathrm{Real}(s)<1$ on the critical line $(\mathrm{Real}(Z_N(m))=1/2)$ ?

  3. Is clearly a generalizations of the Riemann Hypothesis. And I think it might be true! (I made some plots that were convincing but the accuracy was low.)

I wonder if these functions have a name yet and what the answers to the 2 logical questions are. I also invite the readers to make more conjectures and variants with this.

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Some additional questions :

let $T(s) = f(\infty,s) $.

1) Is $T(s)$ meromorphic on The entire complex plane ?

2) how Many poles does $T(s)$ have ? Is it one ?

3) assuming 1) : What is The infinite product representation for $T(s)$ ? ( hadamard type product )

4) assuming 1),2) how fast is this function growing on The complex plane ? As fast as Riemann zeta ?? I assume so.

I think all of these are true. Maybe 2) can Be shown by induction from $f(n,z) $ To $f(n+1,z) $ ?? But infinity is no integer , so maybe not.

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    $\begingroup$ ??? Is this supposed to be a definition of $f(m,s)$ ? Only Euler products lead to non-messy coefficients for both $F(s), \log F(s)$. $\endgroup$ – reuns Nov 29 '18 at 16:28
  • $\begingroup$ @reuns I was talking About hadamard type products. Not euler type. Besides I assume iT had no euler type. $\endgroup$ – mick Nov 29 '18 at 17:49
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    $\begingroup$ If you don't define $f(m,s)$ we won't go far. No Euler product and non-messy coefficients ⟹ no RH. The analytic continuation is obtained by summing by parts $k$-times or looking at the Mellin transform of $\sum_{n=1}^\infty a(n,m) e^{-nx}$. The limit is $h(x)=\prod_{l \ge 1} (1-e^{-2^l x})$. $\log h(x) \approx \sum_l e^{-2^l x} \approx \int_0^\infty e^{-xy} 2^{-2^y} dy= O(...)$ $\endgroup$ – reuns Nov 29 '18 at 21:36
  • $\begingroup$ I started the bounty in order to award it to reuns, I will be able to do that in 23 hours. $\endgroup$ – Klangen Mar 11 at 9:17
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The standard methods for Dirichlet L-functions apply.

  • Let $$h_k(t) = t\prod_{m=0}^{k-1}(1 - t^{2^m}) = \sum_{n=1}^{2^k} a_k(n)t^n$$ $$F_k(s) = \sum_{n=1}^\infty a_k(n \bmod 2^k) n^{-s} $$

$$f_k(x)=\sum_{n=1}^\infty a_k(n \bmod 2^k) e^{- n x}= \frac{h_k(e^{-x})}{1-e^{-2^k x}}$$ Note $h_k(1) = 0$ so $f_k$ is $C^\infty(\mathbb{R})$.

  • For $\Re(s) > 0$ $$\Gamma(s) F_k(s) = \int_0^\infty x^{s-1}\frac{h_k(e^{-x})}{1-e^{-2^k x}}dx$$ For $\Re(s) > -K-1$ $$\Gamma(s) F_k(s) = \sum_{r=0}^K \frac{f_k^{(r)}(0)}{r!} \frac1{s+r}+ \int_0^\infty x^{s-1}(\frac{h_k(e^{-x})}{1-e^{-2^k x}}-1_{x < 1}\sum_{r=0}^K \frac{f_k^{(r)}(0)}{r!} x^r) dx$$ Thus $\Gamma(s) F_k(s)$ is meromorphic everywhere with simple poles at negative integers and $F_k(s)$ is entire.

  • Functional equation : Poisson summation formula, same method as for Dirichlet L-functions and $\sum_n \chi(n) e^{-\pi n^2 x}$.

    Let $\sum_{n=0}^{2^k-1} a_k(n \bmod 2^k) e^{2i \pi mn/2^k}= h_k(e^{2i \pi m/2^k})$ the discrete Fourier transform of $a_k(n \bmod 2^k)$. Then

$$\sum_{n=1}^\infty a_k(n \bmod 2^k) e^{- \pi n^2 x} = (2^k x)^{-1/2} \sum_{m=1}^\infty \frac{h_k(e^{2i \pi m/2^k})}{2^k} e^{- \pi m^2 2^k/ x}$$

$$F_k(s)\Gamma(s/2)\pi^{-s/2}2^{sk/2}= \int_0^\infty x^{s/2-1} \sum_{n=1}^\infty a_k(n \bmod 2^k) e^{- \pi n^2 x/2^k}dx$$ $$= \int_1^\infty (x^{s/2-1} \sum_{n=1}^\infty a_k(n \bmod 2^k) e^{- \pi n^2 x/2^k}+x^{(1-s)/2-1}\sum_{m=1}^\infty \frac{h_k(e^{2i \pi m/2^k})}{2^{k/2}} e^{- \pi m^2 x/2^k}) dx$$

So $F_k(s)$ is a Dirichlet series with functional equation. The standard tools apply, density of zeros, explicit formula for $\log F_k, 1/F_k, F_k'/F_k$ and their Dirichlet series coefficients in term of the non-trivial zeros. But since the $a_k(n \bmod 2^k)$ aren't multiplicative, no Euler product, no Riemann hypothesis.

  • the limit $F_\infty(s) = \lim_{k \to \infty}F_k(s)$. Some properties of the $F_k$ are preserved (the analytic continuation), some are not (functional equation, density of zeros). Asking about a Riemann hypothesis for $F_\infty$ doesn't really make sense.
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  • $\begingroup$ Ok Thank u. Well Maybe a Riemann hypothesis in the sense of Number theory might not make sense. But How about the positions of the zero’s for f(3,s) , f(4,s) or f(Oo,s) in the strip 0 <re(s) < 1 ?? Pictures would be Nice. $\endgroup$ – mick Dec 3 '18 at 4:07
  • $\begingroup$ @mick Forget about it and look at simpler linear combination of Dirichlet L-functions. $\endgroup$ – reuns Dec 3 '18 at 4:15
  • $\begingroup$ 1) no I want to know. Not forget About it. Even if it just Nice pictures. 2) what do you mean simpler lin combinations of dirichlet L-functions ?? Why ?? Which one ? 3) maybe platting the zero’s is better done in the related question 4) but questions 3) and 4) are still unanswered. I assume you actually know the answer. Since the zero’s are requested elsewhere , adding answers to 3),4) might get the accept. $\endgroup$ – mick Dec 3 '18 at 6:34

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