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This question is related to my question Is there an elegant exact formula for the zeta zero counting function? on Math StackExchange.


Question: Does anyone have a Mathematica implementation of the standard $\arg\zeta(s)$ function required to evaluate the function $S(T)=\frac{\Im\left(\log\zeta\left(\frac{1}{2}+i\,T\right)\right)}{\pi}=\frac{\arg\zeta\left(\frac{1}{2}+i\,T\right)}{\pi}$?


I believe Mathematica truncates $\arg\zeta(s)$ to the interval $(-\pi,\pi)$ whereas evaluation of $S(T)$ requires an analytic continuation of $\arg\zeta(s)$.


The answer posted to my question on Math StackExchange implies the analytic continuation of $\arg\zeta(s)$ is derived from the analytic continuation of $\log\zeta(s)$ and the relationship $\arg\zeta(s)=\Im(\log\zeta(s))$. The reason I formulated this question in terms of $\arg\zeta(s)$ instead of $\log\zeta(s)$ is because Mathematica simplifies $\Im(\log\zeta(s))$ to $\arg\zeta(s)$ which initially led to me to believe $\Im(\log\zeta(s))$ is derived from $\arg\zeta(s)$.


The fact that Mathematica does not properly evaluate $\log\zeta(s)$ and $\arg\zeta(s)$ is a bit upsetting and disturbing to me. This also seems to imply Mathematica may not be properly evaluating many of the functions related to $\zeta(s)$ such as those listed at Zeta Functions & Polylogarithms. I've been evaluating many of these functions related to $\zeta(s)$ for about $4\frac{1}{2}$ years now and only recently became aware Mathematica is not evaluating all of these functions correctly.


Assuming the definitions of $N(T)$ and $N_0(T)$ in formulas (1) and (2) below, I've been told $N_0(T)=N(T)\,\forall\,T\ne\Im(\rho)$. This implies one can get a pretty good idea of what $\arg\zeta(s)$ looks like when evaluated along the critical line in the upper half-plane by evaluating formula (3) below.


(1) $\quad N(T)=\sum\limits_{0<\Im(\rho)\le T} 1$

(2) $\quad N_0(T)=1+\frac{\vartheta(T)}{\pi}+\frac{\arg\left(\zeta\left(\frac12+i\,T\right)\right)}{\pi})$

$\qquad\qquad\quad\ =1+\frac{\vartheta(T)}{\pi}+\frac{\Im\left(\log\zeta\left(\frac12+i\,T\right)\right)}{\pi}$

(3) $\quad\arg\zeta\left(\frac{1}{2}+i\,T\right)=\pi\left(N(T)-\left(\frac{\vartheta(T)}{\pi}+1\right)\right),\quad T\ge 0\land T\ne\Im(\rho)$


Figure (1) below illustrates the Mathematica evaluation of $\text{Arg[Zeta[1/2+I T]}$ in orange overlaid on formula (3) for $\arg\zeta\left(\frac{1}{2}+i\,T\right)$ in blue in the interval $0<T<52$. The two dashed horizontal reference lines are at $\pm \pi$. Note the two evaluations are pretty much exactly the same in this interval except where $T=\Im(\rho)$ which I believe is because Mathematica does not truncate $\arg\zeta\left(\frac{1}{2}+i\,T\right)$ in this interval.


Figure (1)

Figure (1): Mathematica evaluation of $\text{Arg[Zeta[1/2+I T]}$ (orange) overlaid on formula (3) for $\arg\zeta\left(\frac{1}{2}+i\,T\right)$ (blue)


Figure (2) below illustrates the Mathematica evaluation of $\text{Arg[Zeta[1/2+I T]}$ in orange overlaid on formula (3) for $\arg\zeta\left(\frac{1}{2}+i\,T\right)$ in blue near Gram point $g_{126}$ and $\Im\left(\rho_{127}\right)$ where Gram's law is violated. The two dashed horizontal reference lines are at $\pm \pi$. Note Mathematica truncates $\arg\zeta\left(\frac{1}{2}+i\,T\right)$ to the interval $(-\pi,\pi)$ which is apparently inconsistent with the analytic continuation of $\log\zeta(s)$ and $\arg\zeta(s)=\Im(\log\zeta(s))$. The orange and red dots represent the discrete evaluation points $(g_{126},\text{Arg[Zeta[1/2+I $g_{126}$])}$ and $\left(\Im\left(\rho_{127}\right),\arg\zeta\left(\frac{1}{2}+i\,\Im\left(\rho_{127}\right)\right)\right)$ respectively.


Figure (2)

Figure (2): Mathematica evaluation of $\text{Arg[Zeta[1/2+I T]}$ (orange) overlaid on formula (3) for $\arg\zeta\left(\frac{1}{2}+i\,T\right)$ (blue)


Figure (3) below illustrates the Mathematica evaluation of $\text{Arg[Zeta[1/2+I T]}$ in orange overlaid on formula (3) for $\arg\zeta\left(\frac{1}{2}+i\,T\right)$ in blue near Gram point $g_{6707}$, $\Im(\rho_{6709})$, and $\Im(\rho_{6710})$ where Lehmer's Phenomenon is exhibited. The two dashed horizontal reference lines are at $\pm \pi$. Note Mathematica once again truncates $\arg\zeta\left(\frac{1}{2}+i\,T\right)$ to the interval $(-\pi,\pi)$. The orange dot represents the discrete evaluation point $(g_{6707},\text{Arg[Zeta[1/2+I $g_{6707}$])}$ and the two red dots represent the evaluation points $\left(\Im\left(\rho_{6709}\right),\arg\zeta\left(\frac{1}{2}+i\,\Im\left(\rho_{6709}\right)\right)\right)$ and $\left(\Im\left(\rho_{6710}\right),\arg\zeta\left(\frac{1}{2}+i\,\Im\left(\rho_{6710}\right)\right)\right)$.


Figure (3)

Figure (3): Mathematica evaluation of $\text{Arg[Zeta[1/2+I T]}$ (orange) overlaid on formula (3) for $\arg\zeta\left(\frac{1}{2}+i\,T\right)$ (blue)

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    $\begingroup$ What is arg supposed to return when $\zeta(s)=0$? I'm unsure what "an analytic continuation of $\arg\zeta(s)$" is exactly. Arg is not a complex-analytic function. I'd guess that you want a smooth function $\Bbb R\rightarrow R$ that maps $T \mapsto \arg\zeta\left({1\over2}+T\right)$. I still think there has to be a discontinuity when the path of ${1\over2}+T$ passes through zero. I also think that if you want an arg that is not path-dependent, a simply connected domain for $s$ that excludes zeros needs to be specified. Am I missing something? $\endgroup$ – Michael E2 Feb 27 at 16:32
  • $\begingroup$ @MichaelE2 I believe the answer posted to my question on Math StackExchange (see math.stackexchange.com/q/4037551) provides more insight. $\endgroup$ – Steven Clark Feb 27 at 16:51
  • $\begingroup$ A related question. $\endgroup$ – J. M.'s torpor Feb 27 at 17:16
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    $\begingroup$ A bit of caution seems in order due to Lehmer's phenomenon of closely-spaced pairs of ζ-zeros, which means that the phase jumps across the zeros are not always of the same sign. For an example, have a look at Plot[Arg[Zeta[1/2+I*t]]/Pi, {t, 7005, 7005.15}] close to the ζ-zero pair ZetaZero[6709] == 0.5 + 7005.06*I and ZetaZero[6710] == 0.5 + 7005.1*I: complementary phase jumps cause trouble. $\endgroup$ – Roman Feb 27 at 19:54
  • $\begingroup$ @Roman The standard definition of $\arg\zeta(s)$ used in the evaluation of the function $S(T)=\frac{\arg\zeta\left(\frac{1}{2}+i\,T\right)}{\pi}$ is not the same as the Mathematica evaluation of "Arg[Zeta[s]]" which is my motivation for this question. $\endgroup$ – Steven Clark Feb 27 at 20:16

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