The non-trivial zeros of $\zeta^{k}(s)$, with $k=k^{th}$ derivative, do not lie on a line and seem to be distributed randomly in the region $\sigma > \frac12$. However the non-real zeros in the critical strip of:

$$\zeta^{k}(s) \pm \zeta^{k}(1-s)$$

all appear to reside on the critical line (with maybe a finite number of exceptions lying outside the critical strip). Could this be proven with similar techniques as outlined here $\zeta(s)-\zeta(1-s)$ ?

The reason I ask is that Speiser(1934), Levinson & Montgomery (1974) and recently Yildirim have proven that assuming RH, $\zeta^{1}(s)$, $\zeta^{2}(s)$ and $\zeta^{3}(s)$ have no zeros in $0 < \Re(s) < \frac12$, but also that the number of zeros of $\zeta^{k}(s)$ residing in the region $\Re(s) < \frac12$, must be finite (there is actually only one pair found for $\zeta^{2}(s)$ and $\zeta^{3}(s)$ in $\Re(s) < 0$).

Now suppose $k=1..3$ and it can indeed be proven that all zeros of $\zeta^{k}(s) \pm \zeta^{k}(1-s)$ must lie on the critical line, then the only possibility for a zero of $\zeta^{k}(s)$ to hide in $0 < \Re(s) < \frac12$ (and thereby falsifying the RH), is when $\zeta^{k}(s)=\zeta^{k}(1-s)=0$. This then immediately raises the second question on whether contrary to $\zeta(s)$ and the absence of a reflexive functional equation for its derivatives, it could be shown that when $\zeta^{k}(s)$ is a zero, $\zeta^{k}(1-s)$ cannot be one?

  • $\begingroup$ As a comment: It is known that all proper derivatives of $\zeta$ must have infinetely many zeros in $\Re s > 1/2$, even $\{ \Re \rho : \zeta^{(k)}(\rho) = 0 \}$ are dense in $1/2 < \Re s <1$. It is a consequence of $RH$ that $\zeta'$ does not have any zero in $0< \Re s <1/2$. $\endgroup$
    – Marc Palm
    Commented Apr 6, 2012 at 15:13
  • $\begingroup$ Jus commenting, since GH's answer has used that $\zeta$ does not vanish off the critcal line in the critical region, whereas $\zeta'$ does... $\endgroup$
    – Marc Palm
    Commented Apr 6, 2012 at 15:23
  • $\begingroup$ @Agno there are known zeros in (0,1/2) for higher derivatives, check mathoverflow.net/questions/90577/… $\endgroup$
    – joro
    Commented Apr 7, 2012 at 4:20
  • $\begingroup$ @Agno regarding derivatives of gamma. This may be a counter example for k=1: 2.4822983600814302743+0.90095059275474156519i. Finding exact roots is not easy because the derivatives are very small as Im increases, so I suspect numerical instability. The given counterexample might be correct, check it. $\endgroup$
    – joro
    Commented Apr 9, 2012 at 5:42
  • $\begingroup$ @Agno, might have found counterexample to Gamma(s)-Gamma(1-s), checkhttp://mathoverflow.net/questions/89324/are-all-zeros-of-gammas-pm-gamma1-s-on-a-line-with-real-part-frac12/93548#93548 check it. $\endgroup$
    – joro
    Commented Apr 9, 2012 at 7:38

2 Answers 2


Agno, I suspect your zeros finding algorithms are suboptimal. Attached sage program found 39 complex counterexamples for derivatives <= 5 and 57 purely real zeros, probably there are infinitely many purely real zeros of $(\zeta^{(k)}(s)+\zeta^{(k)}(1-s))(\zeta^{(k)}(s)-\zeta^{(k)}(1-s))$.

import mpmath

def agno1():
    numerically seraching for zeros of (zeta^(k)(s))^2 - (zeta^(k)(1-s))^2
    pre=20 #precision
    global DE

    def L(x):
        global DE #derivative
        return mpmath.zeta(x,derivative=DE)**2-mpmath.zeta(1-x,derivative=DE)**2

    for D in xrange(1,6): #derivative
      for k in xrange(1,30): #imaginary range
        for r0 in xrange(-5,3): #real range
            r=mpmath.findroot(L,[a],solver="muller") #may fail
            print 'r=',r,'found=',len(f),'f=',f
            #v complex zeros
            if abs(a-1/2)>0.0001 and abs(r.imag)>0.0001:
                print 'found'
                if not s in cac:  f += [(DE,r)]
        except KeyboardInterrupt:   return f    
    return f

Here are the first few zeros of $(\zeta^{(k)}(s)+\zeta^{(k)}(1-s))(\zeta^{(k)}(s)-\zeta^{(k)}(1-s))$ found.

1, -4.3598720412304466086 + 1.3472660066799204586i
1, -4.3598720412304466086 - 1.3472660066799204586i
1, -1.4790601896163449093 + 2.4524390104493105696i
1, 2.4790601896163449093 + 2.4524390104493105696i
1, 5.3598720412304466086 + 1.3472660066799204586i
1, 5.3598720412304466086 - 1.3472660066799204586i
2, -5.238008341582134426 - 0.23390576482129954322i
2, -2.9216469510099648289 + 1.8759500821314771318i
2, -1.0479308378014667797 + 4.34696069590639551i
2, 6.238008341582134426 - 0.23390576482129954322i
2, 6.238008341582134426 + 0.23390576482129954322i
2, 2.0479308378014667797 + 4.34696069590639551i
2, 3.9216469510099648289 + 1.8759500821314771318i
3, -4.0672366129294800445 + 1.0559044658884738519i
3, -2.8061657314035651174 + 2.9523424448287208926i
3, -1.3543734560710258045 + 3.3044686695414223579i
3, 2.3543734560710258045 + 3.3044686695414223579i

Added later Since you appear interested in the critical strip, in the critical strip for k=8 and k=11 $(\zeta^{(k)}(s)+\zeta^{(k)}(1-s))(\zeta^{(k)}(s)-\zeta^{(k)}(1-s))$ has the quadruple of zeros $\rho,1-\rho,\overline{\rho},\overline{1-\rho}$ for $\rho_8=0.762670158543295459229480665406 + 5.79824402154402061398733349266i$ and $\rho_{11}=0.90531956105932089396089746035 + 6.28835450211871487823193399274i$.

  • $\begingroup$ Joro, thanks for your work and the program. I am indeed mostly interested in $k = 1,2,3$ and the critical strip, since finding a zero left of the critical line would falsify RH. You have found clear counterexamples for $k=8$ and $k=11$ (I tested up to k=7...), so the claim does not seem valid for all $k$. However, if it can be proven that all zeros for $k=1,2$ or $3$ are on the critical line and $\zeta^{(k)}(s) \ne \zeta^{(k)}(1-s)$ when either of them is zero, RH is true. $\endgroup$
    – Agno
    Commented Apr 7, 2012 at 9:05
  • $\begingroup$ This article arxiv.org/abs/1002.0362 claims Yildirim recently proved that both $\zeta^{(2)}(s)$ and $\zeta^{(3)}(s)$ have exactly one pair of non-trivial zeros with $\sigma < 0$. $\zeta^{(2)}(s)$ has a zero at approximately $-0.35508433021 \pm 3.590839324398i$. Saw that your program already found that (in the other thread). I feel encouraged that there is no $'(1-s)'$ equivalent for this unique pair and I am now looking at explicit formula for the higher order derivatives (e.g. from Spira and Apostol), that all seem to be dependent on sums of lower order derivatives. $\endgroup$
    – Agno
    Commented Apr 7, 2012 at 9:19
  • $\begingroup$ @Agno for k=1,2,3 certainly not all zeros are on the critical line. Check the red zeros and there a lot more real zeros. Probably you need to be more precise. $\endgroup$
    – joro
    Commented Apr 7, 2012 at 9:35
  • $\begingroup$ @Joro. Apologies for slow response. We are fully aligned on the fact that not ALL zeros of $\zeta^{(k)}(s) \pm \zeta^{(k)}(1-s)$ are on the critical line. There are of course many real and complex zeros, however my claim is that for k=1,2,3 (maybe up to k=7), all zeros in the critical strip do reside on the critical line. You have not yet published a counter example (other than for k=8 and k=11) of this conjecture, right? $\endgroup$
    – Agno
    Commented Apr 7, 2012 at 21:34
  • $\begingroup$ @Agno don't have counterexample to your latest edit and comment, but this is different from the original conjecture as far as I can tell. $\endgroup$
    – joro
    Commented Apr 8, 2012 at 6:53

Just to share what I have found about the second part of my question.

From Tom Apostol's paper I derived the following formula for $\zeta^{(1)}(1-s)$:

$$A(s)= \Gamma \left( s \right) \cos \left(\frac12\pi s \right) 2 \left( 2\pi \right) ^{-s}$$

$$\zeta^{(1)}(1-s) = A(s)\left( \zeta \left( s \right) \left( \ln \left( 2\pi \right) +\frac12\pi \tan \left(\frac12\pi s \right) -\Psi \left( s \right)\right) -\zeta^{(1)}(s) \right)$$

It is easy to isolate $\zeta(s)$ and to obtain a closed form related to its derivatives:

$$\zeta(s) = \frac{\frac {\zeta^{(1)}(1-s)}{A(s)} +\zeta^{(1)}(s)} {\ln \left( 2\pi \right) +\frac12\pi \tan \left(\frac12\pi s \right) -\Psi(s)}$$

When $s=\rho$, and knowing that $\zeta(\rho) = \zeta(1-\rho)$ then from: $$\zeta(\rho)=\frac{\zeta^{(1)}(1-\rho)}{A(\rho)} +\zeta^{(1)}(\rho)=0$$

it follows that when $\zeta^{(1)}(1-\rho)=\zeta^{(1)}(\rho)=0$, then $\zeta(\rho) = \zeta(1-\rho) =0$. This obviously doesn't answer my second question, but maybe does provide a small clue.

Since it is also true that $\zeta(s) A(s) = \zeta(1-s)$, the following equation can be derived:

$$\frac {\zeta^{(1)}(1-\rho)}{A(\rho)} +\zeta^{(1)}(\rho)=\zeta^{(1)}(1-\rho)+A(\rho)\zeta^{(1)}(\rho)$$

that can be simplified into:

$$\frac{\zeta^{(1)}(1-\rho)}{\zeta^{(1)}(\rho)} + A(\rho)=0$$

This equation correctly reproduces all the known $\rho$, but also adds a single new (actually quite natural) one at $\rho'=\frac12 \pm 6.2898359888369027796...$. This is a value I did encounter before see other question and I believe it is the point where:

$$\displaystyle \lim_{\sigma \to \frac12} |\dfrac{\zeta(\sigma+ti)}{\zeta(1-(\sigma+ti))}|=1$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.