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Let $Y_{n, m} (t) = \left\{ \begin{array}{ll} t & m = 0\\ t + \cos (\pi n) \tanh (Z (Y_{n, m - 1} (t))) & m \geqslant 1 \end{array} \right.$

where $\begin{array}{ll} Z (t) & = e^{i \vartheta (t)} \zeta \left( \frac{1}{2} + i t \right) \end{array}$ is the Hardy Z function and $\vartheta (t) = - \frac{i}{2} \left( \ln \Gamma \left( \frac{1}{4} + \frac{i t}{2} \right) - \ln \Gamma \left( \frac{1}{4} - \frac{i t}{2} \right) \right) - \frac{\ln (\pi) t}{2} \label{vartheta}$ is the Riemann-Siegel function and $\zeta$ is the Riemann zeta function.

The function $Y_{n, m} (t)$ has fixed-points at the zeros $Z(t)=0$ which is easy to prove.

Can it be shown that when $n$ is an odd number, $Y_n (t)$ has attractive fixed-points at the odd-numbered roots $y_{2 k - 1} \forall 2 k - 1 \geqslant n$ and repulsive fixed-points at the even-numbered roots $y_{2 k} \forall 2 k \geqslant n$ ?

conversely

Can it be shown that when $n$ is an even number, $Y_n (t)$ has attractive fixed-points at the even-numbered roots $y_{2 k} \forall 2 k \geqslant n$ and repulsive fixed-points at the odd-numbered roots $y_{2 k - 1} \forall 2 k - 1 \geqslant n$ ?

The problem is equivalent to showing that the signs of $Z'$ alternates at the zeros of $Z$

Let $T_f^{\pm} (t)$ be the tanh fixed-point functional of $f (t)$ defined by $T^{_{} \pm}_f (t) = t \pm \tanh (f (t))$

Theorem:

$T^{_{} \pm}_f (t)$ has indifferent fixed-points at multiple-roots of $f (t)$

Proof.

If $f (t)$ has a multiple root at $t = \alpha$ it corresponds to an indifferent fixed-point since the first-derivative $\dot{f} (t)$ vanishes at multiple roots and $f (\alpha) = 0$ here therefore $\cosh (f (t)) = \cosh (0) = 1$ which means $\lambda_{T_f^{_{} \pm}} (t) = \pm \frac{\cosh (f (t))^2 + \dot{f} (t)}{\cosh (f (t))^2} = \pm \frac{1 + 0}{1} = \pm 1$ so $\left| \lambda_{T_f^{_{} \pm}} (t) \right| = 1$.

It needs to be determined if the subtracted roots are weakly atttracting or weakly repelling.

crow
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  • I don't think $\zeta(s)$ has anything special here. Can you replace it by any analytic real function ? – reuns Feb 26 '17 at 03:42
  • I dont think so, the function needs to have the property described here . https://en.wikipedia.org/wiki/Steffensen's_method

    "It is only for the purpose of finding h for this auxiliary point that the value of the function f must be an adequate correction to get closer to its own solution, and for that reason fulfill the requirement that − 1 < f′( x ) < 0 . For all other parts of the calculation, Steffensen's method only requires the function f f to be continuous and to actually have a nearby solution." Z has this property, so one can avoid Newtons method and problem when Z' is 0

     – crow Feb 26 '17 at 03:47
  • The analytic functions have isolated zeros, and since $\zeta(s)$ doesn't have anything special here, can you replace it by a (much) simpler example ? – reuns Feb 26 '17 at 03:49
  • is an isolated root necessarily a simple root? It seems thats not the case otherwise someone would have already proved the roots of zeta are simple but its still a conjecture.

    I dont know if another function besides zeta will work because the Riemann-Siegel function is specially paired with it.. the function would have to grow at a rate proportional in the same way for it to make sense.

    I was actually trying to find examples of analytic functions which have sequences of zeros in a line, but I couldn't think of any off the top of my head besides trivial functions like sin or cos

     – crow Feb 26 '17 at 03:53
  • generally this tanh method does pair well with Newtons method though, I described it here, http://math.stackexchange.com/questions/2142892/does-this-newton-like-iterative-root-finding-method-based-on-the-hyperbolic-tang – crow Feb 26 '17 at 03:54
  • $a$ is an isolated zero of $f$ means that $f(x) \ne 0$ on $0 < |x-a| < \epsilon$. You have an example here of a non-identically zero smooth function vanishing on $[1,\infty)$ – reuns Feb 26 '17 at 03:56
  • Bessel functions of the 1st or 2nd kind look like good candidates – crow Feb 26 '17 at 04:11
  • I see what you are saying. It seems like it would work generally for any analytic function. However, I'm specifically interested in the case of zeta, precisely because its so hard to prove for that case. A slight modification of this method, I've used to calculate the zeta zeros up to n=779640 so far. Its not enough to know that the zeros can be switched between attracting and repelling.. proving that they alternate is more important, because that would prove that the roots of Z and Z' interlace and that would prove the RH – crow Feb 26 '17 at 04:23
  • No isolated zeros don't have to be simple zeros. $z^3$ has and isolated zero of order $3$ at $z=0$. Take so much simpler examples before playing with $\zeta(s),Z(t)$ which is very... complicated. – reuns Feb 26 '17 at 06:16
  • thanks for your comments. ill try that – crow Feb 26 '17 at 16:02
  • @user1952009 This method will not converge to multiple-zeros if multiplicity greater than 1 implies the vanishing of the derivative at that point. Its true for polynomials, is it true for all differentiable functions Let $T_f^{\pm} (t)$ be the tanh fixed-point functional of $f (t)$ defined by $T^{_{} \pm}_f (t) = t \pm \tanh (f (t))$ – crow Mar 04 '17 at 04:45
  • Theorem:

    $T^{_{} \pm}_f (t)$ has indifferent fixed-points at multiple-roots of $f (t)$

     – crow Mar 04 '17 at 04:46
  • Tf(t) also has fixed-points at the points where $Re(f(t))=0$ and $Im(f(t)) mod \Pi=0$ – crow Mar 04 '17 at 07:10

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