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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?

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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 $0< \Re s <1/2$. –  plusepsilon.de Apr 6 '12 at 15:13
    
Jus commenting, since GH's answer has used that $\zeta$ does not vanish off the critcal line in the critical region, whereas $\zeta'$ does... –  plusepsilon.de Apr 6 '12 at 15:23
    
@Agno there are known zeros in (0,1/2) for higher derivatives, check mathoverflow.net/questions/90577/… –  joro Apr 7 '12 at 4:20
    
@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. –  joro Apr 9 '12 at 5:42
    
@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. –  joro Apr 9 '12 at 7:38
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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
    mpmath.mp.pretty=True
    mpmath.mp.dps=pre
    global DE
    DE=0

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

    cac={}
    f=[]
    for D in xrange(1,6): #derivative
      DE=D
      for k in xrange(1,30): #imaginary range
        for r0 in xrange(-5,3): #real range
        a=r0+I*k
        try:
            r=mpmath.findroot(L,[a],solver="muller") #may fail
            a=r.real
            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'
                s=str(r)
                if not s in cac:  f += [(DE,r)]
                cac[s]=1
        except KeyboardInterrupt:   return f    
        except: 
            pass
    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 $\rho_{11}=0.90531956105932089396089746035 + 6.28835450211871487823193399274i$.

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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. –  Agno Apr 7 '12 at 9:05
    
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. –  Agno Apr 7 '12 at 9:19
    
@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. –  joro Apr 7 '12 at 9:35
    
@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? –  Agno Apr 7 '12 at 21:34
    
@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. –  joro Apr 8 '12 at 6:53
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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:

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

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