Is this zeta-type function meromorphic?












1














In An older question I asked :



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



——



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.



——



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.










share|cite|improve this question




















  • 2




    ??? Is this supposed to be a definition of $f(m,s)$ ? Only Euler products lead to non-messy coefficients for both $F(s), log F(s)$.
    – reuns
    Nov 29 '18 at 16:28












  • @reuns I was talking About hadamard type products. Not euler type. Besides I assume iT had no euler type.
    – mick
    Nov 29 '18 at 17:49






  • 1




    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 $h(x)=prod_{l ge 1} (1-e^{-2^l x})$. $log h(x) approx sum_l e^{-2^l x} approx int_0^infty e^{-xy} 2^{-2^y} dy= O(...)$
    – reuns
    Nov 29 '18 at 21:36
















1














In An older question I asked :



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



——



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.



——



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.










share|cite|improve this question




















  • 2




    ??? Is this supposed to be a definition of $f(m,s)$ ? Only Euler products lead to non-messy coefficients for both $F(s), log F(s)$.
    – reuns
    Nov 29 '18 at 16:28












  • @reuns I was talking About hadamard type products. Not euler type. Besides I assume iT had no euler type.
    – mick
    Nov 29 '18 at 17:49






  • 1




    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 $h(x)=prod_{l ge 1} (1-e^{-2^l x})$. $log h(x) approx sum_l e^{-2^l x} approx int_0^infty e^{-xy} 2^{-2^y} dy= O(...)$
    – reuns
    Nov 29 '18 at 21:36














1












1








1


3





In An older question I asked :



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



——



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.



——



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.










share|cite|improve this question















In An older question I asked :



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



——



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.



——



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.







complex-analysis analytic-number-theory infinite-product zeta-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 3 '18 at 3:04







mick

















asked Nov 29 '18 at 14:47









mickmick

5,09422064




5,09422064








  • 2




    ??? Is this supposed to be a definition of $f(m,s)$ ? Only Euler products lead to non-messy coefficients for both $F(s), log F(s)$.
    – reuns
    Nov 29 '18 at 16:28












  • @reuns I was talking About hadamard type products. Not euler type. Besides I assume iT had no euler type.
    – mick
    Nov 29 '18 at 17:49






  • 1




    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 $h(x)=prod_{l ge 1} (1-e^{-2^l x})$. $log h(x) approx sum_l e^{-2^l x} approx int_0^infty e^{-xy} 2^{-2^y} dy= O(...)$
    – reuns
    Nov 29 '18 at 21:36














  • 2




    ??? Is this supposed to be a definition of $f(m,s)$ ? Only Euler products lead to non-messy coefficients for both $F(s), log F(s)$.
    – reuns
    Nov 29 '18 at 16:28












  • @reuns I was talking About hadamard type products. Not euler type. Besides I assume iT had no euler type.
    – mick
    Nov 29 '18 at 17:49






  • 1




    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 $h(x)=prod_{l ge 1} (1-e^{-2^l x})$. $log h(x) approx sum_l e^{-2^l x} approx int_0^infty e^{-xy} 2^{-2^y} dy= O(...)$
    – reuns
    Nov 29 '18 at 21:36








2




2




??? Is this supposed to be a definition of $f(m,s)$ ? Only Euler products lead to non-messy coefficients for both $F(s), log F(s)$.
– reuns
Nov 29 '18 at 16:28






??? Is this supposed to be a definition of $f(m,s)$ ? Only Euler products lead to non-messy coefficients for both $F(s), log F(s)$.
– reuns
Nov 29 '18 at 16:28














@reuns I was talking About hadamard type products. Not euler type. Besides I assume iT had no euler type.
– mick
Nov 29 '18 at 17:49




@reuns I was talking About hadamard type products. Not euler type. Besides I assume iT had no euler type.
– mick
Nov 29 '18 at 17:49




1




1




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 $h(x)=prod_{l ge 1} (1-e^{-2^l x})$. $log h(x) approx sum_l e^{-2^l x} approx int_0^infty e^{-xy} 2^{-2^y} dy= O(...)$
– reuns
Nov 29 '18 at 21:36




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 $h(x)=prod_{l ge 1} (1-e^{-2^l x})$. $log h(x) approx sum_l e^{-2^l x} approx int_0^infty e^{-xy} 2^{-2^y} dy= O(...)$
– reuns
Nov 29 '18 at 21:36










1 Answer
1






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oldest

votes


















1














The standard methods for Dirichlet L-functions apply.




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


$$f_k(x)=sum_{n=1}^infty a_k(n bmod 2^k) e^{- n x}= frac{h_k(e^{-x})}{1-e^{-2^k x}}$$
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.






share|cite|improve this answer























  • 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.
    – mick
    Dec 3 '18 at 4:07










  • @mick Forget about it and look at simpler linear combination of Dirichlet L-functions.
    – reuns
    Dec 3 '18 at 4:15










  • 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.
    – mick
    Dec 3 '18 at 6:34











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1 Answer
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active

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active

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1














The standard methods for Dirichlet L-functions apply.




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


$$f_k(x)=sum_{n=1}^infty a_k(n bmod 2^k) e^{- n x}= frac{h_k(e^{-x})}{1-e^{-2^k x}}$$
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.






share|cite|improve this answer























  • 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.
    – mick
    Dec 3 '18 at 4:07










  • @mick Forget about it and look at simpler linear combination of Dirichlet L-functions.
    – reuns
    Dec 3 '18 at 4:15










  • 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.
    – mick
    Dec 3 '18 at 6:34
















1














The standard methods for Dirichlet L-functions apply.




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


$$f_k(x)=sum_{n=1}^infty a_k(n bmod 2^k) e^{- n x}= frac{h_k(e^{-x})}{1-e^{-2^k x}}$$
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.






share|cite|improve this answer























  • 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.
    – mick
    Dec 3 '18 at 4:07










  • @mick Forget about it and look at simpler linear combination of Dirichlet L-functions.
    – reuns
    Dec 3 '18 at 4:15










  • 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.
    – mick
    Dec 3 '18 at 6:34














1












1








1






The standard methods for Dirichlet L-functions apply.




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


$$f_k(x)=sum_{n=1}^infty a_k(n bmod 2^k) e^{- n x}= frac{h_k(e^{-x})}{1-e^{-2^k x}}$$
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.






share|cite|improve this answer














The standard methods for Dirichlet L-functions apply.




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


$$f_k(x)=sum_{n=1}^infty a_k(n bmod 2^k) e^{- n x}= frac{h_k(e^{-x})}{1-e^{-2^k x}}$$
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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 3 '18 at 3:57

























answered Dec 3 '18 at 3:46









reunsreuns

19.7k21046




19.7k21046












  • 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.
    – mick
    Dec 3 '18 at 4:07










  • @mick Forget about it and look at simpler linear combination of Dirichlet L-functions.
    – reuns
    Dec 3 '18 at 4:15










  • 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.
    – mick
    Dec 3 '18 at 6:34


















  • 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.
    – mick
    Dec 3 '18 at 4:07










  • @mick Forget about it and look at simpler linear combination of Dirichlet L-functions.
    – reuns
    Dec 3 '18 at 4:15










  • 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.
    – mick
    Dec 3 '18 at 6:34
















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.
– mick
Dec 3 '18 at 4:07




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.
– mick
Dec 3 '18 at 4:07












@mick Forget about it and look at simpler linear combination of Dirichlet L-functions.
– reuns
Dec 3 '18 at 4:15




@mick Forget about it and look at simpler linear combination of Dirichlet L-functions.
– reuns
Dec 3 '18 at 4:15












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.
– mick
Dec 3 '18 at 6:34




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.
– mick
Dec 3 '18 at 6:34


















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