Yesterday I started to write a paper about the reformulation of the Riemann Hypothesis.

My idea was to map the function such that all of the trivial zeros are outside of the unit disk, and the non-trivial zeros are on the circle. Iff RH is true, then the radius of convergence (the distance to the closest singularity from the origin of the taylor series) of the taylor series representing reciprocal of the function is $1$.

After some manipulations, I have got 2 conjectures: https://mathoverflow.net/questions/212289/riemann-hypothesis-reformulation-lim-n-to-infty-sum-k-lnka-kn-over-n-s. (Topic deleted from MO.)

I would like to know if they really imply RH, or I went wrong somewhere.

EDIT: I post the reformulation here:

$\zeta(s)$ has its non-trivial zeros on the line $Re(s)=0.5$. It means, that the the taylor series of $$Z(s)={1\over\zeta\left(\frac{1}{2}+\frac{1+s}{1-s}\right)}$$ have its radius of convergence of $1$. (I mapped the right half plane to the unit disc, so trivial zeros are outside of the disk.)

Its derivates given by Cauchy's integral formula, and taking the right contour $C$ such that $C(t)=f^{-1}(t-(a-1/2)i)$ with $f(z)=i(z+1)/(z-1)$ the map from $\mathbb{D}\to\mathbb{\overline{H}}$ becames $${Z}^{(n)}(0)=\frac{n!}{2\pi i}\int_{-\infty}^{\infty}{1\over \zeta(a+it)}C_n(t)\;dt$$ with $C_n(t)=C'(t)/C(t)^{n+1}$. WLG letting $1.5>a>1$, and using the dirichlet series for the reciprocal of the zeta function, I got \begin{align*}{Z}^{(n)}(0)&=\frac{n!}{2\pi i}\int_{-\infty}^{\infty}\left[\sum_{k=1}^\infty \frac{\mu(k)}{k^{a+it}}\right]C_n(t)\;dt\\&=\sum_{k=1}^\infty \frac{\mu(k)}{k^a}\int_{-\infty}^{\infty}\frac{n!}{2\pi i}\frac{C_n(t)}{k^{it}}\;dt\\&=\sum_{k=1}^\infty \frac{\mu(k)}{k^a}\int_{-\infty}^{\infty}\frac{n!}{2\pi i}g_k(C(t))C_n(t)\;dt\\&=\sum_{k=1}^\infty \frac{\mu(k)}{k^a}g^{(n)}_{k}(0).\end{align*}


In the last 2 steps, I changed the contour integral to the derivates of a function series, noticing that $g\circ C(t)=1/(k^{it})$. The only singularity of $g$ is at $1$, but $g$ is bounded inside the contour, so I tought the integral and the derivates are the same.

For later to have the limits defined, define the function $d\colon\mathbb{N}\mapsto \mathbb{N}$ such that $d(n)$ gives the $n$th square-free integer. $$Z^{(n)}(0)=\sum_{k=1}^{\infty}\frac{\mu(d(k))}{d(k)^2}g^{(n)}_{d(k)}(0)$$

1. conjecture: Using the ratio test led me to my first question here, such that given a series $$A(n)=\sum_{k=1}^{\infty}a_k(n),$$ $$a_k(n)=\frac{\mu(d(k))}{n!d(k)^2}g^{(n)}_{d(k)}$$ with $|a_k(n)/a_k(n+1)|\to 1$ (Taylor series of $g$ about the origin have its radius of conv. 1) as $n\to \infty$, it is true that $$\lim_{n\to \infty}\left|\frac{A(n)}{A(n+1)}\right|=1.$$ I suppose it is true for some series satisfying certain conditions, but I cannot prove it.

2. conjecture: Using the recurrence relation of the coefficients (due to WolframAlpha): $na_k(n)+(n+2)a_k(n+2)-2(n+1-\ln(k))a_k(n+1)=0$, $$\lim_{n\to \infty}{\sum_k\ln(k)a_k(n)\over\sum_kna_k(n)}=0$$ would imply RH.

Does proving the 2 conjectures above prove the Riemann hypothesis?

I think it would also prove GRH for Dirichlet L-function with a little change, and with a good choice of $a$ to ensure convergence.

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    $\begingroup$ Why does this have a downvote? $\endgroup$ – BCLC Jul 25 '15 at 14:12
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    $\begingroup$ I have read somewhere that "RH is resisting all proofs". It seems true to me, as questions about them are instantly downvoted instead of helping eachother out. Thats collaboration. $\endgroup$ – David Szalai Jul 25 '15 at 14:14
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    $\begingroup$ That would mean it is closed $\endgroup$ – ClassicStyle Jul 25 '15 at 14:39
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    $\begingroup$ There looks to be a basic flaw in the argument here: "If the Riemann Hypothesis is true, then blah". You're investigating blah and hoping to draw conclusions about the Riemann Hypothesis - that's doing it backwards. $\endgroup$ – Milo Brandt Jul 25 '15 at 16:16
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    $\begingroup$ Please stop edit your post! If you want that your post be "on top", write clear and interesting post. Continuous edition is really not a good way. $\endgroup$ – Michael Galuza Jul 25 '15 at 16:23

As far as RH is concerned, your "method" does not bring anything new to the table. It is just RH folklore in disguise: assuming your calculation is sound (sorry, I did not bother to check b/c the contour description is a mess), it is 'equivalent' to investigating the analytic radii of $1/\zeta(s)$ with origins along the abscissa $s=1+\varepsilon+it,t\in\mathbb{R}$, for some (and any) small $\varepsilon$. Then RH is equivalent to all those radii being $\geq 1/2+\varepsilon$. Notice that the Dirichlet series of $1/\zeta(s)$ is absolute convergent on that abscissa. It is straightforward to see that these radii being $\geq 1/2+\varepsilon$ is equivalent to the Merten's function $M(x) := \sum_{n\leq x}\mu(n)$ being $O(x^{1/2+\varepsilon})$, which is nothing new really since the latter is already known to be equivalent to RH by elementary facts about Dirichlet series.

In other words, your "method" is just an overkill reformulation of the equivalence between RH and $M(x)=O(x^{1/2+\varepsilon})$. Overkill because there is a priori no reason to pull back the reciprocal zeta function to the unit disc, nor there is a reason for the whole contour integration/infinite sum yoga since we already know what the derivatives of the Dirichlet series of $1/\zeta(s)$ look like - they are twisted by $(-\ln{k})^n$.

Regarding your 1st conjecture, I believe it has been addressed in another thread of yours.

Regarding your 2nd conjecture, the answer is no, it does not imply RH in any way. Here is why. Notice that the recursion involves only the variable $n$ and not $k$ (the latter is used by $\mu(k)$) and that your coefficients $a_k(n)$ depend on the function $\mu(-)$ linearly. In other words, the recursion formula bears no information regarding the Möbious function and thus no relevance for the behavior of $1/\zeta(s)$ (you can just factor out the Möbius term b/c of linearity of the recursion relation).

  • $\begingroup$ I understand what you say about the mertens function. However I dont see how my radius should be at least 1. I mapped the line $Re(s)=0.5$ to the unit circle, the half plane $Re(s)>0.5$ to the unit disk. Is RH not equvivalent that the rad. of conv. $r=1$? If it would be smaller, there would be zeros off the circle, and because of the symmetry of the zeros, one of them would be inside the disk, so $r<1$ means RH is not true. Since there is at least one zero on the circle, $r>1$ is impossible. On Z(s), all singularities should have the form $\exp(it)$ for real $t$. $\endgroup$ – David Szalai Jul 25 '15 at 18:57
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    $\begingroup$ Thats not true. Look at the Cayley transform. What you not consider in your contradiction, is that I mapped singularities due to nt-zeros to the unit circle. Plot $Z$ and you will see. The closest singularity to the origin have its absolute value $1$. $1/\zeta(0.5+(1+s)/(1−s))$ Use wolfram alpha to see this. $\endgroup$ – David Szalai Jul 25 '15 at 19:21
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    $\begingroup$ In a general setting, your method starts with the inverse functions of L-candidates for RH. These L-funct. have Dirichlet series in some suitable half-plane of abs. conv. The Dirichlet inverses necessarily involve the Möbius function. If a general form of your 1st conjecture is true in a way that it goes around the Möbius function, then it forgets about inverses of Dirichlet series and is too general ("RH for things that are not even L-functions?!"). On the other hand, not all Dirichlet series have RH. => The appropriate conjecture should involve $\mu$, but without being too specific about it. $\endgroup$ – M.G. Jul 25 '15 at 19:53
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    $\begingroup$ That's the thing. You can do same integral yoga with other functions that are known to have no RH, and you'd still arrive at the general stuff with the $A$-s and $a$-s (even incl. some $\mu$-s like above). $\endgroup$ – M.G. Jul 25 '15 at 20:00
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    $\begingroup$ Dirichlet series with proper half-planes of absolute convergence is still way too general. Only those that come from the automorphic world are of interest, and that is somehow much more subtle. So, if some general conjecture of the form like yours do in fact hold, it is going to be much more intricate since it will have to filter out false positives while not being too restrictive (b/c it is general) ;) $\endgroup$ – M.G. Jul 25 '15 at 20:09

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