Question

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?

Answer 1

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 $\rho_{11}=0.90531956105932089396089746035 + 6.28835450211871487823193399274i$.

Answer 2

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:

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