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See https://aimath.org/news/gl3/zfunction.html

Actually, this question doesnt make sense, the Davenport-Heilbronn "zeta function" isn't even an L-series so it isn't right to call it a "zeta function" it's only a function which shares one particular symmetry of a subset of its zeros ... since its not an L-series there is no Z-function corresponding to it

An L-function is a Dirichlet series L(s) = ∑n an n-s with certain special properties. For the purpose of this discussion, the relevant property is the functional equation Λ(s) = Qs ∏j Γ(λj s +μj) L(s) = ε $\Lambda (1-s)$ .

where

  • L(s) The L-function
  • s=$\sigma$ +i t a complex number
  • an The Dirichlet series coefficients of the L-function, normalized so that a1=1. The (very deep!) Ramanujan conjecture asserts that an = O(n&epsilon). This implies (but is stronger than) the Dirichlet series converges for σ>1.
  • Γ The Gamma function
  • λj Positive real numbers. It is conjectured that one can take λj = ½ for all j . The number 2 ∑ &lambdaj is called the degree of the L-function, which is conjectured to be an integer.
  • μj Numbers with non-negative real parts.
  • Q A positive number. ε A number with absolute value 1, called the sign of the functional equation.
  • Complex conjugate, operating on functions by f*(z) := f(z*)*, where the * on the right side is conjugation in the usual sense.

where it states that the Z function can be constructed for an arbitrary L functions such that its real-valued on the real-line and shares the same zeros as the L function (after transforming)

The functional equation can also be written in an asymmetric form L(s) = ε X(s) L*(1-s), where X(s) can be obtained by rearranging the functional equation written above.

Finally, we have the Z-function associated to L, defined by $Z(t) = (ε X(½ + i t))^½ L(½+i t)$ .

In https://arxiv.org/pdf/1602.06328.pdf they demonstrate that the Davenport-Heilbron "zeta function" is not even an L function, but a linear combination of two. so they only satisfy a "Riemann zeta"-like functional equation, so the conclusions drawn from it aren't even applicable to all L functions.

"We are reluctant to call these zeros counterexamples to the Riemann Hypothesis ( RH ) since the function f  s  is not a Dirichlet L - function for which the generalized RH has been stated , neither does it belong to the Selberg class for which the Grand RH is expected to be true . They are simply an illustration of the fact that a Riemann - type of functional equation implies this symmetry of some non trivial zeros with respect to the critical line . Functions of this type can be constructed for an arbitrary modulus q possessing complex conjugate characters "

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Let $$F(s) = \sum_{n=1}^\infty a_n n ^{-s}, \qquad \Lambda(s) = \varepsilon Q^{s} \prod_{m=1}^M \Gamma(\omega_m s+\mu_m) F(s), \qquad \Lambda(s) = \overline{\Lambda(1-\overline{s})}$$ an L-function or a Dirichlet series with a functional equation.

Then $\Lambda(1/2+it) = \overline{\Lambda(1-(1/2-it))}$ so that $Z(t) = \xi(1/2+it)$ is real for $t$ real

The basic examples are

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  • $\begingroup$ math.stackexchange.com/questions/2170824/… $\endgroup$ – crow Mar 3 '17 at 22:26
  • $\begingroup$ Actually, this question doesnt make sense, the Davenport-Heilbronn "zeta function" isn't even an L-series so it isn't right to call it a "zeta function" it's only a function which shares one particular symmetry of a subset of its zeros ... since its not an L-series there is no Z-function corresponding to it $\endgroup$ – crow Mar 3 '17 at 22:27
  • $\begingroup$ That's why I mentioned Dirichlet series with a functional equation (i.e. it can have no Euler product) @crow $\endgroup$ – reuns Mar 3 '17 at 22:31
  • $\begingroup$ well fuggit, the RH must be true, no one can even construct an example of one with the zeros off the line $\endgroup$ – crow Aug 11 '18 at 17:03

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