User agno - MathOverflow

What happens when infinite values of $\zeta_{H}(s,z)$ approach $\zeta(s)$ ?

2012-05-27T19:16:51Z

Take the following Hurwitz zeta:

$$\zeta_{H}(s,z)$$

with $s=\sigma \pm ti$ and $\displaystyle z=1 \pm \frac{i}{a}$ and $t,a \in \mathbb{R}$.

In the critical strip $0 \lt \sigma \lt 1$, this Hurwitz zeta appears to have zeros $\rho_H$ for combinations of $(s,a)$ for all values of $\sigma$, with the exception of $\sigma = \frac12$. However, contrary to $\zeta(s)$, these zeros do not seem to obey the rule that when $s=\rho_H$ then the function must vanish also at $1-\rho_H$ .

Example:

$$\zeta_{H}(0.8-24.96910...i, 1+ \frac{i}{44.82238...} = 0 \ne \zeta_{H}(0.2+24.96910...i, 1+\frac{i}{44.82238...})$$

$$\zeta_{H}(0.2+25.05217...i, 1+ \frac{i}{44.84243...} = 0 \ne \zeta_{H}(0.8-25.05217...i, 1+\frac{i}{44.84243...})$$

Would a proof for all $\rho_H$ that:

$$\zeta_{H}(\rho_H,z) \ne \zeta_{H}(1- \rho_H,z)$$

for all $\sigma \ne \frac12$ in the critical strip, imply the RH when $\displaystyle \lim_{a \to +\infty}$?

My logic towards a 'yes' to this question, would be that the contradiction between 'all zeros off the critical line cannot be symmetrical for values of $a < \infty$' and 'all non trivial zeros of $\zeta(s)$ must be symmetrical', can only be solved when 'there cannot be zeros of $\zeta(s)$ lying off the critical line'. This would also imply that both functions quite naturally complement each other (i.e. all zeros lie off the critical line versus all zeros are on it). Not sure though about what happens to the properties of both functions at the tilting point when $\displaystyle \lim_{a \to +\infty} \zeta_{H}(s, 1 \pm \frac{i}{a})$ 'morphs' into $\zeta(s)$.

Is there information about the $\rho$'s hidden in the zeros of $\Re(\chi(s))$ ?

2012-01-22T19:41:09Z

Take the symmetrical form of the completed Zeta-function:

$\displaystyle \chi(s) \zeta(s) = \chi(1-s) \zeta(1-s)$

with

$\chi(s)=\pi^{-(\frac{s}{2})} \Gamma(\frac{s}{2})$.

For $s=\sigma + ti$, I conjecture that only for $\sigma=\frac12$:

$\Re(\chi(s)) = \Re(\zeta(s)) =0$ is always true, except when $s=\rho_n$ (assuming RH).

If $\sigma=\frac12$ and $t$ is real then $\Re(\chi(s))=0$ can be rewritten as:

$A(t) = \cot \left( \frac12 t\ln \left( \pi \right) \right) \dfrac{\Re \left( \Gamma \left(1/4+\frac{t i}{2} \right) \right)}{\Im \left(\Gamma \left( 1/4+\frac{t i}{2} \right) \right)} +1 = 0$

Similar to $s=\rho$ for non trivial zeros, let's call $s=\alpha$ when $A(t)=0$.

There seems to be a strong connection between the $\Re(\zeta(\alpha_m))$ and $\Re(\zeta(\rho_n))$. They exclusively come in an adjacent pair $(\alpha_m,\rho_n)$ or $(\rho_n,\alpha_m)$ that is connected via a 'sharp trough' of $\Re(\zeta(s))$ through the x-axis. The distance between the paired values appears to become smaller when $t$ grows.

If this paired pattern is true, then it would imply that when $\Re(\zeta(s))$ dives into a 'trough' through the x-axis and we find that $s =\alpha_m$, then one could predict with certainty that $s=\rho$ for the next time $\Re(\zeta(s)) = 0$ and that it must be located before $s=\alpha_{m+1}$. In that sense there would be information hidden in $\chi(s)$ that constrains the location of the $\rho$'s.

A bit complicated question, I know, but hope the picture below illustrates the thought.

http://imageshack.us/photo/my-images/855/riemannzero.jpg/

Questions:

Has it been proven that $\Re(\chi(s)) = \Re(\zeta(s)) =0$ (and/or $\Im(\chi(s)) = \Im(\zeta(s)) =0$) only when $\sigma=\frac12$ and $s \ne \rho$ ?
Is there anything known about $\alpha_m$ and $\rho_n$ always coming in connected pairs $(\alpha_m,\rho_n)$ or $(\rho_n,\alpha_m$) and/or that they converge when $t \rightarrow \infty$?

Thanks.

One additional afterthought:

Assuming RH and the adjacent paired values of $Re(\zeta(s))=0$ always having the shape:

$(\alpha,\rho)$ or $(\rho,\alpha)$,

then it would only require counting $\alpha$'s (from $t>1$) to establish the exact number of $\rho$'s $\pm1$ below a certain number $N$ (i.e. without even looking at $\zeta(s)$).

Values where infinite products of primes and composites are equal

2011-01-25T19:22:02Z

Highly grateful for your help/steers on the following question (at the end):

Take the infinite product:

$$\displaystyle T(s) = \prod _{n=2}^{\infty } \left( \dfrac{{n}^{s}} {{n}^{s}-1}\right)$$

for $\Re(s) > 1$ it is equal to:

$$\displaystyle \prod _{primes}^{\infty } \left( \dfrac{{p}^{s}} {{p}^{s}-1}\right) * \prod _{composites}^{\infty } \left( \dfrac{{c}^{s}} {{c}^{s}-1}\right)$$

I.e. the Euler-product (equal to $\zeta(s)$) multiplied by its composite "equivalent" ( excluding 1 since that is a bit of a strange composite anyway).

Why my interest? I wanted to learn more about the composite infinite product (and see if it had a 'zeta' like version). Soon became clear to me that the only way to learn more about this product, is to concentrate on $T(s)$ and then divide it by $\zeta(s)$.

I searched the web but there is hardly anything known about $T(s)$. F.i. Wolfram math only shows (formula 20) two different solutions (note: both need to be raised to $^{-1}$ to get $T(s)$ !) for odd and even integers and by reading through some arxiv math pre-prints the best I could find was a single, but still integer only formula that is:

$$\prod _{k=1}^{s-1}\Gamma \left( 2- {{\rm e}} ^{{\frac {2 i \pi k}{s}}} \right), ( \Re(s) > 1, s \in \mathbb{N})$$

I then decided to explore ways to extend the domain for $s$ and derived the following formula:

$$\displaystyle \ln \left( T\left( s \right) \right) = \ln \prod_{n=2}^{\infty } \left( \left( -1+{n}^{-s} \right) ^{-1} \right) = \sum_{n=2}^{\infty } \ln \left( \left( 1-{n}^{-s} \right) ^{-1} \right)$$

$$\displaystyle = \sum_{m=1}^\infty \sum_{n=2}^{\infty } \frac{1}{mn^{ms}} = \sum_{n=1}^\infty \frac{\zeta(n s)-1}{n}$$

And this brings us to:

$$ T(s)={\rm e}^{\left( \displaystyle \sum_{n=1}^\infty \frac{\zeta(n s)-1}{n} \right)}$$

Yep, there's always a $\zeta(s)$ hiding around the corner somewhere...

So, let's see what the plot looks like for $s>0$ ($T(s)$ diverges for $s<0$).

$T(s)=\displaystyle {\rm e}^{\left( \displaystyle \sum_{n=1}^\infty \frac{\zeta(n s)-1}{n} \right)} \text{ blue}$

$\displaystyle \zeta(s) = \prod _{primes}^{\infty } \left( \dfrac{{p}^{s}} {{p}^{s}-1}\right) \text{purple}$

$\displaystyle \frac{T(s)}{\zeta(s)} = \prod _{composites}^{\infty } \left( \dfrac{{c}^{s}} {{c}^{s}-1} \right) \text{ brownish}$

graph

For $s>1$ I could numerically solve the following equation:

$$\displaystyle \prod _{primes}^{\infty } \left( \dfrac{{p}^{s}} {{p}^{s}-1}\right) = \prod _{composites}^{\infty } \left( \dfrac{{c}^{s}} {{c}^{s}-1} \right)$$

giving this interesting number $s = 1.397737620...$ (there is only one for $\Re(s) > 1$ )

I obviously took a deep dive with this number on Google and Plouffe's inverter, but have not found anything 'beautiful' or related to other constants as yet.

Then the domain $0 < s < 1$. It is easy to see in the graph that $T(s)$, and therefore also $\dfrac{T(s)}{\zeta(s)}$, have 'trivial' poles for $s= \dfrac{1}{k}, k \in \mathbb{N}$ that are induced by the fact that for each $s= \dfrac{1}{k}$ there always is a $n s = 1$ that makes at least one term in the infinite sum equal to the pole $\zeta(1)$ (hence the whole sum turns into a pole).

But I'm actually mostly intrigued by what happens under the x-axis and especially where:

$$\zeta(s) = \dfrac{T(s)}{\zeta(s)}$$ or

$$|\zeta(s)| = {\rm e}^{\displaystyle \left(\frac12 \sum_{n=1}^\infty \frac{\zeta(n s)-1}{n} \right)}$$.

If I have done my analysis correctly, this result would imply that there are an infinite number of values for $0 < s < 1$, where the (analytically continued) infinite products of primes and composites are equal (since $\zeta(s)$ remains negative between $0 < s < 1$ and there are an infinite number of poles separating the intersection points). And that would imply/reveal an infinite amount of tiny bits of information about how the primes 'grow like weed between the composites'.

Of course I checked $T(s)$ also for $s \in \mathbb{C}$, however, any graph I've produced sofar for $s=a+bi$ of $T(s)={\rm e}^{\left(\displaystyle \sum_{n=1}^\infty \frac{\zeta(n s)-1}{n} \right)}$ did not reveal any non-trivial zeroes (nope, not even at $a=\frac12$...), although the curves do seem to be trending towards a large number of very chaotically distributed zeroes when $a \rightarrow 0$.

So, apologies for the relatively long intro to my question:

Since $\zeta(s)$ has been analytically continued throughout the entire complex domain, is it allowed to also analytically continue the division of $\dfrac{T(s)}{\zeta(s)}$ into the domain $s<1$? Or do the nominator and denominator each require an individual continuation and does the concept of division get 'lost in continuation'?

Does there exist a closed form for the factors of this infinite product ?

2013-03-28T21:34:02Z

Assume $s,a \in \mathbb{C}, a \pm in \ne 0$.

The following infinite product nicely converges and can be expressed in a closed form:

$$\displaystyle \prod_{n=1}^\infty \left(1- \frac{s}{a+i n} \right) \left(1- \frac{s}{{a- i n}} \right) = {\frac {a\sinh \left( \pi \left( a-s \right) \right) }{ \left( a-s \right) \sinh \left( \pi a \right) }}$$

The individual factors however diverge, so I tried to exchange the sub factors for each $n$ and found that:

$$\displaystyle \prod_{n=1}^\infty \left(1- \frac{s}{a+ (-1)^n i n} \right)$$

and

$$\displaystyle \prod_{n=1}^\infty \left(1- \frac{s}{a+ (-1)^{n+1} i n} \right)$$

actually do (slowly but surely) converge and when multiplied together yield the closed form above.

Is there a closed form for these two individual factors?

EDIT (and follow up question):

Many thanks to Carlo for answering the question so quickly.

A (maybe too) provocative follow up question deals with a similar product that also has a closed form (when assuming RH is true):

$$\prod_{n=1}^\infty \left(1- \frac{s}{a + i \Im(\rho_n)} \right) \left(1- \frac{s}{{a - i \Im(\rho_n)}} \right) = \frac{\xi_{rie}(\frac12 - a+s)}{\xi_{rie}(\frac12 - a)}$$

so runs through the non-trivial zeros $\rho_n$ with:$\xi_{rie}(s)= \frac12 s(s-1) \pi^{-\frac{s}{2}} \Gamma(\frac{s}{2}) \zeta(s)$

Again, the individual factors diverge, but after exchanging the sub factors for each $n$, I again found that:

$$\displaystyle \prod_{n=1}^\infty \left(1- \frac{s}{a+ (-1)^n i \Im(\rho_n)} \right)$$

and

$$\displaystyle \prod_{n=1}^\infty \left(1- \frac{s}{a+ (-1)^{n+1} i \Im(\rho_n)} \right)$$

actually do converge.

A closed form for these would obviously be quite spectacular... Could it exist?

P.S.:

In another context, I asked about the similarity between these products here: closed forms infinite products

Answer by Agno for Does there exist a closed form for the factors of this infinite product ?

2013-04-05T22:45:38Z

I did explore the second question a bit further and now have a follow up question.

Assuming the RH, the following product:

$$\displaystyle \displaystyle \prod_{n=1}^\infty \left(1- \frac{s}{a+ (-1)^n i \Im(\rho_n)} \right) \left(1- \frac{s}{a+ (-1)^{n+1} i \Im(\rho_n)} \right) = \frac{\xi_{rie}(\frac12 - a+s)}{\xi_{rie}(\frac12 - a)}$$

runs through the alternating non-trivial zeros $\rho_n$, with $\xi_{rie}(s)= \frac12 s(s-1) \pi^{-\frac{s}{2}} \Gamma(\frac{s}{2}) \zeta(s)$.

Based on $\sinh(s)$ in the original question being factored into closed forms containing $\Gamma(s)$ and $\Gamma(1-s)$, I more and more believe that similar closed form factors could exist for $\frac{\xi_{rie}(\frac12 - a+s)}{\xi_{rie}(\frac12 - a)}$.

Assume $a=\frac12$. This makes the function to be split into two factors:

$$s(s-1) \pi^{-\frac{s}{2}} \Gamma(\frac{s}{2}) \zeta(s)$$

For $s$ and $s-1$ factoring seems trivial, and also for $\Gamma(\frac{s}{2})$ and $\pi^{-\frac{s}{2}}$, further factoring might be done by for instance using the function: $\Gamma(\frac{s}{2}) = \Gamma(\frac{s}{4})\Gamma(\frac{s}{4}+\frac12)\pi^{-\frac12}2^{\frac{s}{2}-1}$.

The tricky part lies in how to factor $\zeta(s)$ into two complementary functions that each generate alternating non-trivial zeros; i.e. for the first function: $\frac12+14.13...i$, $\frac12-21.02...i$, $\frac12+25.01...i$, etc.

Such a function would still require $\zeta(s)$ itself, since it is the only known function that induces $\rho_n$'s, but it now also requires additional information on each subsequent zero about whether it should be generated or be 'suppressed'. Such information does not seem to exist in the real or the imaginary parts of $\zeta(s)$ nor does it in its derivatives. However I did find that when defining $Ξ(t) = \xi_{rie}(s)$ when $s=\frac12 + ti$:

$\displaystyle \frac{|\Xi'(t)|}{\Xi'(t)}$ alternates between $-1$ and $1$ at each subsequent $\rho_n$.

And this brought me to the following factors for $\zeta(s)$, that do what they should do, but I do find very ugly:

$$\displaystyle \zeta\left(1 \pm \frac{|\Xi'(t)|}{2\Xi'(t)}+it \right)$$

There must be a smarter way to produce the alternating $\rho_n$'s and I'd appreciate any help/steers you have.

Thanks!

Alternating sums of the non-trivial zeros of $\zeta(s)$.

2013-04-01T13:41:22Z

It is known that the infinite sum of the non-trivial zeros $\rho_n =\beta + \gamma_ni$ of $\zeta(z)$, when taken in pairs that are either conjugated or reflexive (they give the same outcome), converges:

$$Z_1(s):=\displaystyle \sum_{n=1}^\infty \frac{1}{\rho_n^s} + \sum_{n=1}^\infty \frac{1}{\overline{\rho_n^s}}$$

For $s \ge 1, s \in \mathbb{R}$ the sum can be described analytically for integer values e.g. $Z(1)= 1 + \frac{\gamma}{2}-\frac{\ln(4\pi)}{2}$. For $s<1$ the sum diverges.

However the individual sums $\dfrac{1}{\rho_n^s}$ and $\dfrac{1}{\overline{\rho_n^s}}$ do not converge. Even though the real parts do, their imaginary parts diverge and are annihilated by adding both sums together.

I believe there exists an alternative 'alternating' way to split up the total sum into subtotals that each do converge:

Take $\mu_n =\beta + (-1)^n \gamma_ni$ and: $$Z_2(s):= \sum_{n=1}^\infty \frac{1}{\mu_n^s}+\sum_{n=1}^\infty \frac{1}{\overline{\mu_n^s}}$$

and the individual sums seem to nicely converge for all $\beta \in \mathbb{R}$ and $s \ge 1$. Obviously $Z_1(s)=Z_2(s)$, however I found that contrary to $Z_1(s)$ the domain $Z_2(s)$ can be expanded towards $0 \lt s \lt 1$ by taking:

$$Z_{2-}(s):= \sum_{n=1}^\infty \frac{1}{\mu_n^s}-\sum_{n=1}^\infty \frac{1}{\overline{\mu_n^s}}$$

The difference converges towards an imaginary value and the diverging real parts are annihilated.

Question:

1) Can it be proven that $\displaystyle \sum_{n=1}^\infty \frac{1}{\mu_n^s}$ converges for all $s\ge1$ ?

2) Is there anything known about analytic (closed form) values of $\displaystyle \sum_{n=1}^\infty \frac{1}{\mu_n^s}$ ?

Is there a connection between the closed forms of these two infinite products?

2013-02-21T23:35:02Z

Take the following two infinite products that have closed forms.

Assume: $\gamma_n > 0 \in \mathbb{R};s,a,x \in \mathbb{C}; x \ne 0,a \pm ix\gamma_n \ne 0$

The first product:

$$\displaystyle H_{int}(s,a,x) := \prod_{n=1}^\infty \left(1- \frac{s}{a + i x \gamma_n} \right) \left(1- \frac{s}{{a - i x \gamma_n}} \right) = \frac{\xi_{int}(0 -\frac{a}{x}+\frac{s}{x})}{\xi_{int}(0-\frac{a}{x})}$$

has $\gamma_n = n$, so runs through the integers with: $\xi_{int}(s) = \frac{\sinh(\pi s)}{s}$.

The second product, for which the closed form can be derived assuming RH is true,

$$\displaystyle H_{rie}(s,a,x) := \prod_{n=1}^\infty \left(1- \frac{s}{a + i x \gamma_n} \right) \left(1- \frac{s}{{a - i x \gamma_n}} \right) = \frac{\xi_{rie}(\frac12 - \frac{a}{x}+\frac{s}{x})}{\xi_{rie}(\frac12 - \frac{a}{x})}$$

has $\gamma_n = \Im(\rho_n)$, so runs through the non-trivial zeros $\rho_n$ with: $\xi_{rie}(s)= \frac12 s(s-1) \pi^{-\frac{s}{2}} \Gamma(\frac{s}{2}) \zeta(s)$.

The products clearly have a comparable structure and share the following characteristics:

1) They are both reflexive: $H(s,a,x) = H(2a-s,a,x)$ (since this is fully independent of $\gamma_n$).

2) Their $\xi(s)$-functions are reflexive as well:

$$\xi_{int}(s)=\xi_{int}(2(0)-s)$$

$$\xi_{rie}(s)=\xi_{rie}(2(\frac12)-s)$$

3) They both have a similar 'base' Hadamard product (for $a=0$ and $a=\frac12$ respectively):

$$\displaystyle \xi_{int}(s) = \xi_{int}(0) \prod_{n=1}^\infty \left(1- \frac{s}{0+i n} \right) \left(1- \frac{s}{{0- i n}} \right) \rightarrow \xi_{int}(0)=\pi$$

$$\displaystyle \xi_{rie}(s) = \xi_{rie}(0) \prod_{n=1}^\infty \left(1- \frac{s}{\frac12 + i \Im(\rho_n)} \right) \left(1- \frac{s}{{\frac12 - i \Im(\rho_n)}} \right) \rightarrow \xi_{rie}(0)=\frac12$$

4) They are both entire functions and the "undesired" zeros/poles from their meromorphic components $\sinh(s)$ and $\zeta(s)$ are annihilated via $s$ and $\frac12 s(s-1), \Gamma(\frac{s}{2})$ respectively. Both meromorphic elements can be expressed as an infinite product as well as an infinite sum (over a certain domain):

$$\sinh(s) = s \prod_{k=1}^\infty \left(1+ \frac{s^2}{k^2\pi^2)} \right)=\sum_{n=1}^\infty \frac{s^{2n-1}}{\Gamma(2n)}$$

$$\zeta(s) = \prod_{p} \left(\frac{1}{1-p^{-s}} \right)=\sum_{n=1}^\infty \frac{1}{n^s}$$

5) And both of the above can be analytically continued throughout the entire complex plane via:

$$\sinh(0-s) = -\sinh(s)$$

$$\zeta(1-s) = \chi(s)\zeta(s)$$

My questions:

1) Is there any other possible (or known) set of values for $\gamma_n$ that could yield yet another closed form? Or is this all there is, i.e. are the ("via parameter $x$ linearly scalable") integers $\gamma_n=n$ at $a=0$ and $\gamma_n =\Im(\rho_n)$ at $a=\frac12$, the only possible choices?

2) Since the $\rho_n$'s contain information about the (distribution of) primes and the primes in turn can induce the integers via unique multiplication, could there be a connection made between (the closed forms of) these two products?

Thanks.

Answer by Agno for Is there a connection between the closed forms of these two infinite products?

2013-03-12T15:41:59Z

In my quest for similar closed forms as the above, I did find:

$$\displaystyle H_{int^2}(s,a,x) := \prod_{n=1}^\infty \left(1- \frac{s}{a + i x \gamma_n} \right) \left(1- \frac{s}{{a - i x \gamma_n}} \right) = \frac{\xi_{int^2}(0 -\frac{a}{x}+\frac{s}{x})}{\xi_{int^2}(0-\frac{a}{x})}$$

with $\gamma_n = n^2$, running through the squared integers and $\xi_{int^2} = {\frac {\sin \left( \left(\frac12-\frac12i \right) \sqrt{2s} \ \pi \right) \sin \left( \left(\frac12+\frac12i \right) \sqrt{2s} \ \pi \right)}{s}}$

I do believe more closed forms exist, however only for $\gamma_n = n^{2k}$ with ($k= 1,2,3...)$ and would like to conjecture that for the 'base' Hadamard product it is true that:

$$\displaystyle \xi_{int^{2k}}(s) = \xi_{int^{2k}}(0) \prod_{n=1}^\infty \left(1- \frac{s}{0+i n^{2k}} \right) \left(1- \frac{s}{{0- i n^{2k}}} \right) \rightarrow \xi_{int^{2k}}(0)=\pi^{2k}$$

However, my (maybe too big) dream was to find a closed form for $\gamma_n = p_n$ with $p_n$ being the n-th prime. Via brute force calculations, I did find a few (all explainable) 'leads' for:

$$\displaystyle H_{prime}(s,a,x) := \prod_{n=1}^\infty \left(1- \frac{s}{a + i x p_n} \right) \left(1- \frac{s}{{a - i x p_n}} \right)$$

$$H_{prime}(1,1,1) = \frac{\pi^2}{15} = \frac{\zeta(4)}{\zeta(2)}$$ $$H_{prime}(1,\frac12,1) = 1$$ $$H_{prime}(i,i,1) = \frac{\pi^2}{6} = \zeta(2)$$

and the general rules, that also apply to the above: $$H_{prime}(s,a,x) = H_{prime}(2a-s,a,x)$$ $$H_{prime}(s,a,x) = \frac{1}{H_{prime}(s,1-a,x)}$$

I can also prove that a closed (entire) form must have the shape: $\dfrac{\xi_{prime}(\beta -\frac{a}{x}+\frac{s}{x})}{\xi_{prime}(\beta-\frac{a}{x})}$, however the key snag is that the meromorphic component of $\xi_{prime}(s)$ that generates the zeros (similar to $\zeta(s)$), must do so only at the primes, whilst all extra "undesired and maybe more trivial" zeros need to annihilated by other components of the entire function $\xi_{prime}(s)$.

This brought me immediately to Wilson's formula, where I used the $\Gamma$-function instead of the factorial and replaced the $mod$ by the $cos$ to produce a function that generates zeros only at integers that are prime:

$$\cos\left(\dfrac{\pi}{2}\dfrac{\Gamma(s)+1}{s}\right)$$

however it unfortunately also generates a tremendous amount of 'undesired, but hopefully more trivial' (non-integer) zeros that I need to annihilate. Is there any known function (other than using the floor function) that could accomplish this? Any other ideas\steers?

Thanks.

The distribution of balls in a Bean Machine that omits all the "prime pegs"?

2013-02-14T22:10:03Z

The Bean Machine, also known as a quincunx or the Galton box, is a well known triangular board that contains several rows ($n$) of staggered, but equally spaced pegs. Balls are dropped from the top one by one to avoid interference. They bounce off the pegs and stack up at the bottom of the triangle in bins. The resulting stacks of balls approach the characteristic Bell curve shape for large $n$.

Consider the following altered Bean Machine that now has all the 'prime numbered pegs' (counting top to bottom, left to right) removed:

Based on what is known about the primes, is there anything that could be predicted about the resulting distribution of balls for large $n$? Is it expected to be skewed to the left or the right? Or does it nicely balance out like the normal distribution?

Thanks.

A closed form of infinite products of complex zeros involving $\Im(\rho_n)$. Does a proof of this closed form imply RH?

2013-01-02T15:32:55Z

Building on this question scaling the imaginary part of $\rho$s in infinite products, I like to conjecture that:

$$\displaystyle \prod_{n=1}^\infty \left(1- \frac{s}{\mu_n} \right) \left(1- \frac{s}{1-\mu_n} \right)\left(1- \frac{s}{\overline{\mu_n}} \right) \left(1- \frac{s}{\overline{1-\mu_n}} \right)$$

with $\mu_n = a + \Im(\rho_n)x i$ and $a,x \in \mathbb{R},x \ne 0, s \in \mathbb{C}$ and $\rho_n$ the n-th non-trivial zero of $\zeta(z)$,

has the following closed form:

$$\displaystyle H(s,a,x):= \frac{\xi(\frac12 - \frac{a}{x} + \frac{s}{x})}{\xi(\frac12 - \frac{a}{x})} \frac{\xi(\frac12 - \frac{a}{x} + \frac{1}{x} - \frac{s}{x})}{\xi(\frac12 - \frac{a}{x}+ \frac{1}{x})}$$

where $\xi(z) = \frac12 z(z-1) \pi^{-\frac{z}{2}} \Gamma(\frac{z}{2}) \zeta(z)$ is the Riemann xi-function.

If this formula is correct, the 'constructed' zeros $\mu_n$ can be stretched/condensed vertically via $x$ on the imaginary axis and shifted left/right on the real line via $a$. In all cases they would yield an entire function expressed by this closed form (think of it as a reversed application of the Weierstrass factorization theorem, i.e. starting with products of 'constructed' zeros).

Further factorization also seems possible with:

$$\frac{\xi(\frac12 - \frac{a}{x} + \frac{s}{x})}{\xi(\frac12 - \frac{a}{x})} = \prod_{n=1}^\infty \left(1- \frac{s}{\mu_n} \right) \left(1-\frac{s}{\overline{\mu_n}} \right)$$

and

$$\frac{\xi(\frac12 - \frac{a}{x} + \frac{1}{x} - \frac{s}{x})}{\xi(\frac12 - \frac{a}{x}+ \frac{1}{x})} = \prod_{n=1}^\infty \left(1- \frac{s}{1-\mu_n} \right) \left(1- \frac{s}{\overline{1-\mu_n}} \right)$$

When $a=\frac12$ and $x=1$, the formula correctly reduces to:

$$\frac{\xi(s)}{\xi(0)} = \prod_{\rho} \left(1- \frac{s}{\rho} \right) \left(1- \frac{s}{1-\rho} \right)$$

from which the known Hadamard product for $\zeta(s)$ can been derived.

Unfortunately I do not have a proof for this formula, however rigorously checked it against many 'brute force' calculations using the first 2mln $\rho$s (all correct results, but accurate up to 5 decimals max). I manufactured the formula by replicating the symmetry of the closed form for $\mu_n = a + n x i$ (i.e. running through the integers rather than $\Im(\rho_n)$, see the linked question). Since until today, all non-trivial zeros appear to be lying on the critical line, I have used $\frac12$ as the "source" for all zeros for different $a$'s i.e.: $\frac12 - \frac{a}{x} + \frac{s}{x}$ just inserts $\frac12$ when $\Re(s)=a$. I guess I have thereby implicitly assumed the RH in constructing the formula.

My questions:

Is this a known closed form?
Does a proof of this closed form imply the RH, i.e. does it "force" the Hadamard product into a "straight jacket" that only allows it to be valid when all $a=\Re(\rho_n)=\frac12$ ?

UPDATE:

Assuming RH is true, I believe that I have found a nice proof for the equation in the OP. Since "the comments section is too small to contain it", I decided to put it as an answer to my own question to round it up.

Answer by Agno for A closed form of infinite products of complex zeros involving $\Im(\rho_n)$. Does a proof of this closed form imply RH?

2013-01-10T06:47:14Z

Below is a proposed proof, that assuming the RH, the following equation is true:

$$\displaystyle \frac{\xi(\frac12 - \frac{a}{x} + \frac{s}{x})}{\xi(\frac12 - \frac{a}{x})} = \prod_{n=1}^\infty \left(1- \frac{s}{\mu_n} \right) \left(1- \frac{s}{\overline{\mu_n}} \right)$$

where $\mu_n = a + i x \gamma_n$ and $\gamma_n = \Im(\rho_n)$, with $\rho_n$ the n-th non-trivial zero of $\zeta(s)$.

Take $t, \gamma_n \in \mathbb{R},a,x,s \in \mathbb{C}$ with $x \ne 0$ and $\gamma_n > 0$.

Starting from Hadamard's proof that:

$$\xi(s) = \xi(0) \prod_{n=1}^\infty \left(1- \frac{s}{\rho_n} \right) \left(1- \frac{s}{1-\rho_n} \right) \qquad (1)$$

with $\xi(s) = \frac12 s(s-1) \pi^{-\frac{s}{2}} \Gamma(\frac{s}{2}) \zeta(s)$ being the Riemann xi-function.

Now we assume the RH and all $\Re(\rho_n)=\frac12$. Take $s=\dfrac{a+ i t}{x}$.

This gives:

$$\xi\left(\dfrac{a+ i t}{x}\right) = \xi(0) \prod_{n=1}^\infty \left(1- \frac{\frac{a}{x} +\frac{i t}{x}}{\frac12 + i \gamma_n} \right) \left(1- \frac{\frac{a}{x} +\frac{i t}{x}}{\frac12 - i \gamma_n} \right)$$

and can be expanded into:

$$\xi\left(\dfrac{a+ i t}{x}\right) = \xi(0) \prod_{n=1}^\infty \left(\frac{\frac12 - \frac{a}{x} + i \gamma_n -\frac{i t}{x}}{\frac12 + i \gamma_n} \right) \left(\frac{\frac12 - \frac{a}{x} - i \gamma_n - \frac{i t}{x}}{\frac12 - i \gamma_n} \right)$$

By multiplying each factor with $\dfrac{(\frac{a}{x} + i \gamma_n)}{(\frac{a}{x} + i \gamma_n)}$ and $\dfrac{(\frac{a}{x} - i \gamma_n)}{(\frac{a}{x} - i \gamma_n)}$ respectively, we get:

$$\displaystyle \xi\left(\dfrac{a+ i t}{x}\right) = \xi(0) \prod_{n=1}^\infty \left(\frac{\frac{a}{x} + i \gamma_n}{\frac12 + i \gamma_n} \right) \left(\frac{\frac{a}{x} - i \gamma_n}{\frac12 - i \gamma_n} \right) \prod_{n=1}^\infty \left(\frac{\frac12 - \frac{a}{x} + i \gamma_n -\frac{i t}{x}}{\frac{a}{x} + i \gamma_n} \right) \left(\frac{\frac12 - \frac{a}{x} - i \gamma_n - \frac{i t}{x}}{\frac{a}{x} - i \gamma_n} \right) $$

$$\displaystyle = \xi(0) \prod_{n=1}^\infty \left(1- \frac{\frac12 -\frac{a}{x}}{\frac12 + i \gamma_n)} \right) \left(1- \frac{\frac12-\frac{a}{x}}{\frac12 - i \gamma_n} \right) \prod_{n=1}^\infty \left(1- \frac{\frac{2a}{x}-\frac12+\frac{i t}{x}}{\frac{a}{x} + i \gamma_n)} \right) \left(1- \frac{\frac{2a}{x}-\frac12+\frac{i t}{x}}{\frac{a}{x}- i \gamma_n} \right)$$

By now injecting $\dfrac12 - \dfrac{a}{x}$ into equation (1), this can be simplified as:

$$\xi\left(\dfrac{a+ i t}{x}\right) = \xi\left(\frac12 - \frac{a}{x}\right) \prod_{n=1}^\infty \left(1- \frac{\frac{2a}{x}-\frac12+\frac{i t}{x}}{\frac{a}{x} + i \gamma_n}\right) \left(1- \frac{\frac{2a}{x}-\frac12+\frac{i t}{x}}{\frac{a}{x}- i \gamma_n} \right)$$

and since $\dfrac{2a}{x} + \dfrac{i t}{x} = s + \dfrac{a}{x}$ this can be rewritten into:

$$\xi\left(\dfrac{a+ i t}{x}\right) = \xi\left(\frac12 - \frac{a}{x}\right) = \prod_{n=1}^\infty \left(1- \frac{x(s+\frac{a}{x}-\frac12)}{a+ i x \gamma_n)}\right) \left(1- \frac{x (s+\frac{a}{x}-\frac12)}{a- i x \gamma_n} \right)$$

so that we can now say that:

$$\xi\left(\frac12 - \frac{a}{x} + \frac{s}{x}\right) = \xi\left(\frac12 - \frac{a}{x}\right) = \prod_{n=1}^\infty \left(1- \frac{s}{a+ i x \gamma_n)} \right) \left(1- \frac{s}{a- i x \gamma_n} \right)$$

and the desired result is obtained:

$$\displaystyle \frac{\xi(\frac12 - \frac{a}{x} + \frac{s}{x})}{\xi(\frac12 - \frac{a}{x})} = \prod_{n=1}^\infty \left(1- \frac{s}{\mu_n} \right) \left(1- \frac{s}{\overline{\mu_n}} \right)$$

When starting from $s = \dfrac{1-(a+i t)}{x}$ the equivalent result is $\displaystyle \prod_{n=1}^\infty \left(1- \frac{s}{1- \mu_n} \right) \left(1- \frac{s}{\overline{1-\mu_n}} \right)$

Answer by Agno for A closed form of infinite products of complex zeros involving $\Im(\rho_n)$. Does a proof of this closed form imply RH?

2013-01-04T15:49:30Z

The conjecture above:

$H(s,a,x) =\prod_{n=1}^\infty \left(1- \frac{s}{\mu_n} \right) \left(1- \frac{s}{1-\mu_n} \right)\left(1- \frac{s}{\overline{\mu_n}} \right) \left(1- \frac{s}{\overline{1-\mu_n}} \right)= \frac{\xi(\frac12 - \frac{a}{x} + \frac{s}{x})}{\xi(\frac12 - \frac{a}{x})} \frac{\xi(\frac12 - \frac{a}{x} + \frac{1}{x} - \frac{s}{x})}{\xi(\frac12 - \frac{a}{x}+ \frac{1}{x})}$

with $\mu_n = a + \Im(\rho_n)x i$ might be expanded even further.

When the domains for $a$ and $x$ are extended to $\mathbb{C}_{/(x=0)}$, the formula for $H(s,a,x)$ still appears to work (tested at 5 decimals accuracy) and for instance induces the following result (assume $\gamma_n = \Im(\rho_n)$):

$H(s,0,i) =\prod_{n=1}^\infty \left(1 + \frac{s}{\gamma_n} \right) \left(1- \frac{s}{\gamma_n} \right)\left(1- \frac{s}{{1+\gamma_n}} \right) \left(1- \frac{s}{{1-\gamma_n}} \right)= \frac{\xi(\frac12 + \frac{s}{i})}{\xi(\frac12)} \frac{\xi(\frac12 + \frac{1}{i} - \frac{s}{i})}{\xi(\frac12 + \frac{1}{i})}$

i.e. all imaginary parts of $\rho_n$ become real and their infinite product still has a closed form that, as expected, induces a zero at e.g. $H(\gamma_n,0,i)$.

Or, when we take one of the two sub-factors from the OP, this gives:

$$\displaystyle \xi\left(\frac12+\frac{s}{i}\right) = \xi\left(\frac12 \right)\prod_{n=1}^\infty \left(1 - \frac{s^2}{\gamma_n^2} \right)$$

and for $s=i$, this result emerges:

$$\displaystyle \xi\left(\frac32\right) = \xi\left(\frac12 \right)\prod_{n=1}^\infty \left(1 + \frac{1}{\gamma_n^2} \right)$$

and $s=\frac{-i}{2}$ or $s=\frac{i}{2}$ yields this quite interesting outcome:

$$\displaystyle \xi\left(\frac12 \right) = \frac12 \prod_{n=1}^\infty \left(\frac{1}{1 + \frac{1}{(2 \gamma_n)^2} }\right) = \Xi(0)$$

So that:

$$\Xi\left(\frac12\right) = \frac12 \prod_{n=1}^\infty \left(\frac{(2 \gamma_n)^2 -1} {(2 \gamma_n)^2 +1}\right)$$

Here's my 'Overflow' follow-up question: I have tested the conjecture in Maple up to $\gamma_{2.000.000}$, but would really appreciate it when somebody could replicate and confirm the outcomes (I do not possess Mathematica, but do know the $\rho$s are already coming with the package so not much 'programming' should be required).

Many thanks!

Answer by Agno for What happens to $\zeta(s)$ when all its $\Im(\rho_n)$ are "scaled" linearly?

2012-12-16T13:11:33Z

Have not given up yet on whether or not there exists a closed form for:

$$Had(s, \sigma, x):=\displaystyle \prod_\rho \left(1- \frac{s}{\sigma + xti} \right) \left(1- \frac{s}{1-(\sigma + xti)} \right)$$

that, as Juan proved above, reduces to (assuming RH):

$$Had(s,\frac12,x):=\frac{x^2-4(s-\frac12)^2}{x^2-1}\pi^{-s/2x} \frac{\Gamma(\frac{1}{4}-\frac{1}{4x}+\frac{s}{2x})} {\Gamma(\frac14-\frac{1}{4x})}\frac{\zeta(\frac12+\frac{s}{x}-\frac{1}{2x})} {\zeta(\frac12-\frac{1}{2x})}$$

and for $x=1$, further reduces to the Hadamard product:

$$Had(s, \frac12, 1):=\dfrac{2(s-1)\Gamma(1+\frac{s}{2}){\zeta(s)}}{ \pi^{\frac{s}{2}}}$$

Assuming RH, a closed form for $Had(s, \sigma, x)$ requires:

$Had(0, \sigma, x)=1$ and $Had(1, \sigma, x)=1$.
$Had(s, \sigma, x)= Had(s, 1-\sigma, x)$
$Had(\frac12, \sigma, x)$ is the function's minimum.
$Had(s, \sigma, x)$ to reduce to the closed forms for $Had(s,\frac12,x)$ and $Had(s, \frac12, 1)$
the Zeta function's non-trivial zeros to be the 'source' for all (horizontally shifted) zeros.
the function to be entire (all poles annihilated by zeros).

The following function does meet all the criteria, except for the second:

$$\displaystyle {\frac {{x}^{2}-4 \left( \sigma-s \right) ^{2}}{{x}^{2}-4 \left( 2s\sigma- s-\sigma \right) ^{2}}}{\pi }^{{\frac {s \left( \sigma-1 \right) }{x}}} \dfrac{\Gamma \left( \dfrac{\frac12-{\frac {\sigma}{x}}+{\frac {s}{x}}}{2}\right)}{\Gamma \left( \dfrac{\frac12-{\frac {\sigma}{x}}+{\frac {s(2\sigma-1)}{x}}}{2}\right)} \dfrac{\zeta \left( \frac12-{\frac {\sigma}{x}}+{\frac {s}{x}} \right)}{\zeta \left( \frac12-{\frac {\sigma}{x}}+{\frac {s(2\sigma-1)}{x}} \right)}$$

My 'brute force' infinite product calculations (based on the first 2 mln zeros) show that the shapes of the curves are close (but not equal), however the results for $Had(s, \sigma, x)$ and $Had(s, 1-\sigma, x)$ differ (slightly, yet consistently) from each other.

Could there be any way to improve this?

What happens to $\zeta(s)$ when all its $\Im(\rho_n)$ are "scaled" linearly?

2012-12-02T00:32:03Z

I found that the following infinite product with $\mu = a +n b i$ and a,b real, $s \in \mathbb{C}$:

$$\displaystyle \prod_{n=1}^\infty \left(1- \frac{s}{\mu} \right) \left(1- \frac{s}{1-\mu} \right)$$

can be expressed in a closed form (with poles at $a,b = 0$ or $a=1$ and when $a=s$):

$$\dfrac{\left( {a}^{2}-a \right)} {\left( {a} ^{2}-a+s-{s}^{2} \right)} \dfrac{\Gamma \left( {\frac {-ia}{b}} \right) \Gamma \left( {\frac {-i \left( a-1 \right) }{b}} \right)}{\Gamma \left( {\frac {-i \left( a-s \right) }{b}} \right) \Gamma \left( {\frac {-i \left( a+s-1 \right) }{b}} \right)}$$

When $a=\frac12$ this could be further reduced to (poles at $s=\frac12$ and $b=0$):

$$ \dfrac{1}{(2s-1)} \dfrac{\sinh \left( {\frac { \left( 2s-1 \right) \pi }{2b}} \right)} { \sinh \left({\frac {\pi }{2b}} \right)}$$

Encouraged by this result, my wish was to use it to find new hints about the Hadamard product:

$$\displaystyle \prod_\rho \left(1- \frac{s}{\rho} \right) \left(1- \frac{s}{1-\rho} \right) = \dfrac{2(s-1)\Gamma(1+\frac{s}{2}){\zeta(s)}}{ \pi^{\frac{s}{2}}}$$

but it is quite obviously an impossible task to transform the linear element $nb$ into to very random imaginary parts of the $\rho$s. However, it still triggered a follow up question:

With $\rho = \sigma + ti$ and $t,x$ real, the following product:

$$Had(s,x):=\displaystyle \prod_\rho \left(1- \frac{s}{\sigma + xti} \right) \left(1- \frac{s}{1-(\sigma + xti)} \right)$$

allows for "scaling" of the imaginary parts of the $\rho$s.

Since $Had(s,1)$ has a closed form and is entire, does this imply that the (linearly) scaled $\Im(\rho_n)$ must also induce entire functions and have closed forms (possibly related to $\zeta(s)$ and assuming RH is true)?

Edit: Extra question:

To take it a step further: similar for the infinite products with $n$ above, could the known closed form:

$$\dfrac{2(s-1)\Gamma(1+\frac{s}{2}){\zeta(s)}}{ \pi^{\frac{s}{2}}}$$

just be the 'reduced' version for $\sigma=\frac12$ and be extended with $\sigma$ and $x$ to express $Had(s,x,\sigma)$?

P.S.:

I wrote a program to calculate $Had(s,x)$ by using the first 2 mln $\rho$s from Andrew Odlyzko's table, however when calibrating the results with the known $Had(2,1) =\dfrac{\pi}{3}$, I found that the accuracy is limited to 5 decimals max. (i.e. too few to link it to known constants). Are there any larger $\rho$-files available on the web?

Question about the function $\zeta(s) \pm \dfrac{1}{\zeta(1-s)}$

2012-11-27T22:55:16Z

It is easy to see that the function:

$$\zeta(s) \pm \dfrac{1}{\zeta(1-s)}$$

has a pole at each non-trivial zero $s=\rho_n$.

However, after some experiments with this function, I would like to conjecture that:

$$|\zeta(s) - \dfrac{1}{\zeta(1-s)}|$$

only has zeros in the critical strip on the line $\Re(s)=\frac12$, but also that:

$$|\zeta(s) + \dfrac{1}{\zeta(1-s)}|-2$$

always has at least a zero in the critical strip for each $\Re(s) \ne \frac12$ (i.e. all zeros lie off the critical line).

Since both conjectures complement each other, I guess they must be connected in some way.

Using the reflection formula with $\chi(s)=2^s \pi^{s-1} \sin(\frac{\pi s}{2}) \phantom. \Gamma(1-s)$, I rewrote the functions as:

$$\left| \dfrac{\zeta(1-s)^2 \chi(s) - 1}{\zeta(1-s)\chi(s)} \right| =0 \text { and } \left| \dfrac{\zeta(1-s)^2 \chi(s) + 1}{\zeta(1-s)\chi(s)}\right| = 2$$

however this didn't help much solving e.g. the first conjecture that implies $\zeta(1-s)^2 \chi(s) = 1$ only when $\Re(s)=\frac12$ in the critical strip.

Grateful for any steers/hints on how I could best approach this problem.

Thanks!

$\zeta(2k+1)$ expressed in a product of two infinite products of non-trivial zeros.

2012-11-21T00:12:36Z

Take the Hadamard product for $\zeta(s)$:

$$\displaystyle \zeta(s) = \pi^{\frac{s}{2}} \dfrac{\prod_\rho \left(1- \frac{s}{\rho} \right) \left(1- \frac{s}{1-\rho} \right)}{2(s-1)\Gamma(1+\frac{s}{2})}$$

and reshuffle it into:

$$\displaystyle \dfrac{\prod_\rho \left(1- \frac{s}{\rho} \right) \left(1- \frac{s}{1-\rho} \right)} {\zeta(s)} = \dfrac{2(s-1)\Gamma(1+\frac{s}{2})}{ \pi^{\frac{s}{2}}}$$

After experimenting with some values for $s$, I found that f.i. $\zeta(3)$ can be simply written as:

$$\zeta(3) = \prod_\rho \left(1- \frac{2}{\rho} \right) \left(1- \frac{2}{1-\rho} \right)\left(1- \frac{3}{\rho} \right) \left(1- \frac{3}{1-\rho} \right)$$

and believe this can be generalized into ($k= 1,2,3...$):

$$\displaystyle \zeta(2k+1) = a[2k+1] \prod_\rho \left(1- \frac{2k}{\rho} \right) \left(1- \frac{2k}{1-\rho} \right)\left(1- \frac{2k+1}{\rho} \right) \left(1- \frac{2k+1}{1-\rho} \right)$$

with $a[3]=1, a[5]=\frac12, a[7]=\frac15, a[9]=\frac{5}{84}, \dots$

1) Is this a known result? Could the $a[2k+1]$ sequence be derived from the Bernoulli numbers?

2) Could it be proven that when solving $\rho$ for one factor of the infinite product, i.e. only equate the 4 subfactors to $\zeta(2k+1)$, that the complex roots always must have $\Re(s)=\frac12$ ?

The influence of $\chi(s)$ on complex zeros of $\frac{\zeta(\overline{s})}{\zeta(1-s)} \pm \frac{\zeta(s)}{\zeta(1-\overline{s})}$

2012-11-01T19:42:40Z

I was exploring the formula:

$$g(s)_{\pm} := \displaystyle \frac{\zeta(\overline{s})}{\zeta(1-s)} \pm \frac{\zeta(s)}{\zeta(1-\overline{s})}$$

and found that for all $\Re(s) \ne \frac12$:

$|g(s)_{+}|$ has pairs of complex zeros that always encapsulate a $\rho$ between them.

$|g(s)_{-}|$ has complex zeros near the known $\rho$'s (call them $\mu$'s).

Similar to the $\rho$'s (if they would exist for $\Re(s) \ne \frac12)$, all $\mu$-roots are equal for ${\mu,1-\mu,\overline{\mu},1-\overline{\mu}}$.

The $\mu$'s obviously vanish without a trace when $\Re(s)=\frac12$, however the $\mu$-zeros for $\displaystyle \lim_{\Re(s) \to \frac12}$ clearly and smoothly converge towards the known $\rho$'s.

The $\mu$'s can also be computed by putting $|g(s)_{+}| - 2|\chi(s)|=0$ with $\chi(s)=2^s \pi^{s-1} \sin(\frac{\pi s}{2}) \phantom. \Gamma(1-s)$.

It is conjectured (or maybe already proven?), that $\chi(s)$ does not contain any information about the $\rho$'s, however the formula above suggests that for all $\Re(s) \ne \frac12$, there actually is information in $\chi(s)$ about the $\mu$'s. $\chi(s)$ apparently plays a complementary role for $\Re(s)=\frac12$ and $\Re(s) \ne \frac12$.

I realise this is a very broad question, but could this complement in any way help explain that the only way to make $\chi(s)$ fully independent from the $\rho$'s, is when $\Re(s)=\frac12$?

Does there exist a Weierstrass/Hadamard factorization for $\chi(s)-1$ ?

2012-09-01T18:36:54Z

Would like to build once more on this question.

Take $s=\sigma + ti, s \in \mathbb{C}, 0<\Re(\sigma)<1$.

Let's assume it is proven that:

$$\zeta(1-s) - \zeta(s)$$

has all its zeros on the critical line when:

$$\chi(s)=2^s \pi^{s-1} \sin(\frac{\pi s}{2}) \phantom. \Gamma(1-s) = \pm 1$$

The other zeros are the $\rho$'s and they simply emerge when $0 - 0 = 0$, and obviously no further information can be derived about the validity of the RH. Searching for ways around this (unsuccesfully), I stumbled on another question.

From:

$$\zeta_H(s,a) = \zeta_H(s,a+1) + a^{-s}$$

it follows that:

$$\zeta(1-s) - \zeta(s) = \zeta_H(1-s,2) - \zeta_H(s,2)$$

And the latter term can be factored into:

$$ \left( \sqrt{(\zeta_H(1-s,2)} - \sqrt {\zeta_H(s,2)} \right) \left( \sqrt{(\zeta_H(1-s,2)} + \sqrt {\zeta_H(s,2)} \right)$$

A plot of both factors for $\sigma = \frac12$ revealed:

that all $\rho$ are produced as zeros of the second factor only.
the first factor shows discontinuities at the $\rho$'s, but also for a few additional values.
the discontinuities vanish when taking the absolute value (this only works when $\sigma = \frac12$).
this absolute function has zeros, but all $\rho$'s will now be residing on the line $y=2$.

Assume: $f(t) = |\left( \sqrt{(\zeta_H(\frac12-t i,2)} - \sqrt {\zeta_H(\frac12+t i,2)} \right)|$.

Note that $f(t)$ shares all its zeros with $|\chi(\frac12+t i)-1|$, but not the other way around. There appear to be 'missing zeros' in $f(t)$ that emerge at random spots. Given the similarities and overlaps between the two plots, I wondered if $f(t)$ might be a 'distorted' version of $|\chi(\frac12+t i)-1|$. A distortion potentially caused by randomly missing zeros in a "Hadamard"-type infinite product of the zeros of $|\chi(\frac12+t i)-1|$?

Question:

The function $\chi(s)-1$ is a meromorphic function, could the Weierstrass/Hadamard factorisation theorem be used to express $\chi(s)-1$ as an infinite product of its zeros?

Deriving the Riemann non-trivial zeros from $\zeta_{H}(s,a) + \zeta_{H}(s,1-a)$

2012-06-05T18:55:47Z

The Hurwitz zeta function:

$$\zeta_{H}(s,a)$$

reduces to $\zeta(s)$ when $a=1$ and to $(2^s-1)\zeta(s)$ when $a=\frac12$.

However, I stumbled upon a peculiar third connection:

$$\zeta_{H}(s,a) + \zeta_{H}(s,1-a)$$

that seems to exactly produce the non-trivial zeros of $\zeta(s)$,

when $a=\frac12$ (obviously), but also (and apparently only) when $a=\frac13, \frac14$ or $\frac16.$

Why does it only work for these values? Is there any reference to this in the literature?

Thanks.

Answer by Agno for Deriving the Riemann non-trivial zeros from $\zeta_{H}(s,a) + \zeta_{H}(s,1-a)$

2012-06-05T22:23:02Z

Found the answer, the question can be closed. answer

It boils down to:

$$\zeta_{H}(s,a) + \zeta_{H}(s,1-a) = \frac{4}{(2\pi)^{1-s}}\Gamma(1-s)C(1-s,a)$$

$$C(s,a)=\sum_{n=1}^\infty \frac{\cos(2n\pi a)}{n^s}$$

And $C(s,a)$ reducing to:

$a=\frac12 \rightarrow$ $(2^{1-s}-1)\zeta(s)$

$a=\frac13 \rightarrow$$\dfrac12(3^{1-s}-1)\zeta(s)$

$a=\frac14 \rightarrow$$2^{-s}(2^{1-s}-1)\zeta(s)$

$a=\frac16 \rightarrow$$\dfrac12(1-2^{1-s})(1-3^{1-s})\zeta(s)$

hence the non-trivial zeros.

Are all zeros of $\Gamma(s) \pm \Gamma(1-s)$ on a line with real part = $\frac12$ ?

2012-02-23T19:46:34Z

The function $\Gamma(s)$ does not have zeros, but $\Gamma(s)\pm \Gamma(1-s)$ does.

Ignoring the real solutions for now and assuming $s \in \mathbb{C}$ then:

$\Gamma(s)-\Gamma(1-s)$ yields zeros at:

$\frac12 \pm 2.70269111740240387016556585336 i$ $\frac12 \pm 5.05334476784919736779735104686 i$ $\frac12 \pm 6.82188969510663531320292827393 i$ $\frac12 \pm 8.37303293891455628139008877004 i$ $\frac12 \pm 9.79770751746885191388078483695 i$ $\frac12 \pm 11.1361746342106720656243966380 i$ $\frac12 \pm 12.4106273718343980402685363665 i$

$\dots$

and

$\Gamma(s)+\Gamma(1-s)$ gives zeros at:

$\frac12 \pm 4.01094805906156869492043027819 i$ $\frac12 \pm 5.97476992595365858561703252235 i$ $\frac12 \pm 7.61704024553573658642606787126 i$ $\frac12 \pm 9.09805003388841581320246381948 i$ $\frac12 \pm 10.4760650707765536619292369200 i$ $\frac12 \pm 11.7804020877663106830617193188 i$ $\frac12 \pm 13.0283749883477570386353012761 i$

$\dots$

By multiplication, both functions can be combined into: $\Gamma(s)^2 - \Gamma(1-s)^2$

After playing with the domain of $s$ and inspecting the associated 3D output charts, I now dare to conjecture that all 'complex' zeros of this function must have a real part of $\frac12$.

Has this been proven? If not, appreciate any thoughts on possible approaches.

Thanks!

Zeros of the function $\zeta(s) \pm \zeta(\overline s)$

2012-04-01T22:01:58Z

Building on this question: Zeros of $\zeta(s) \pm \zeta(1-s)$, I experimented further with:

$$\zeta(s) \pm \zeta(\overline s)$$

Assuming $s=\sigma + ti$, I observed that this function also has many "semi-trivial" as well as "non-trivial" zeros for each $\sigma$. Furthermore these "non-trivial" zeros all seem to reside very close to the Riemann non trivial zeros at $\sigma=\frac12$.

However, what I found curious is that only when $\frac12 < \sigma < 1$ the function:

$$\zeta(s) + \zeta(\overline s)$$

suddenly loses all of its "non-trivial" zeros (i.e. the ones near the Riemann zeros), whilst still retaining all of its "semi-trivial" zeros (they disappear when $\sigma >1$). Is there a logical explanation or even proof for this?

P.S.:

In an attempt to find out more, I used the alternating zeta-function $\eta(s)$ and rewrote it as:

$$\eta(s) - \eta(\overline s) =\displaystyle 2i \sum _{n=1}^{\infty } \frac{e^{\pi i n} \sin(t \ln(n))}{n^\sigma}$$

and

$$\eta(s) + \eta(\overline s) =\displaystyle 2i \sum _{n=1}^{\infty } \frac{e^{\pi i n} \cos(t \ln(n))}{n^\sigma}$$

These functions look very symmetrical, but it seems that the denominator $n^\sigma$ drives the infinite alternating sum of the cosines to always positive near the non-trivial zeros whilst keeping the semi-trivial ones intact, when $\sigma > \frac12$. Could these functions be rewritten as an integral?

Definite integral of $\zeta(s)$ over the critical strip

2012-04-14T21:55:06Z

Take the following definite integral:

$$f(s):=\int_s^{1-s} \zeta(x) \mathrm{d} x$$

with $s \in \mathbb{C}$, $s=\sigma \pm ti$, $0<\sigma<1$ and $t,\sigma \in \mathbb{R}$.

The graph of $|f(s)|$ shows a monotonically increasing function for $\sigma=\frac12$ (as expected, it 'plateaus' exactly at the $\rho$s) and an apparently strictly increasing function when $\sigma\ne\frac12$.

There is however a small 'dip' in the area $1 < t < 3$, that unexpectedly induces a zero at $\frac12 \pm 2.50056818181399528638615277529..i$. For $\sigma\ne\frac12$ there are no zeros.

Is there anything known about this zero? Could it be proven that it only exists for $\sigma=\frac12$?

Thanks!

Are all zeros of ζ^{k}(s)±ζ^{k}(1−s) on the critical line (k=k-th derivative)?

2012-04-06T15:01:44Z

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 by Agno for Are all zeros of ζ^{k}(s)±ζ^{k}(1−s) on the critical line (k=k-th derivative)?

2012-04-09T18:17:32Z

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:

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

Are the 'semi' trivial zeros of $\zeta(s) \pm \zeta(1-s)$ all on the critical line?

2012-02-25T21:08:55Z

The proof that $\Gamma(z)\pm \Gamma(1-z)$ only has zeros for $z \in \mathbb{R}$ or $z= \frac12 +i \mathbb{R}$ has been given here:

http://mathoverflow.net/questions/89324/are-all-zeros-of-gammas-pm-gamma1-s-on-a-line-with-real-part-frac12

An obvious follow up question is whether $\zeta(s) \pm \zeta(1-s)$ also has zeros (other than its non-trivial ones that would induce 0+0 or 0-0).

This is indeed the case and $\zeta(s)^2 - \zeta(1-s)^2$ has the following zeros:

$\frac12 \pm 0.819545329 i$

$\frac12 \pm 3.436218226 i$

$\frac12 \pm 9.666908056 i$

$\frac12 \pm 14.13472514 i$ (the first non trivial)

$\frac12 \pm 14.51791963 i$

$\frac12 \pm 17.84559954 i$

$\dots$

These 'semi' trivial zeros appear to all lie on the critical line. I wonder if anything is known or proven about their location (I guess not, since a proof that they must have real part of $\frac12$ would automatically imply RH, right?).

EDIT: Two counterexamples found by Joro in the answers below. Both have real parts outside the critical strip, so I would like to rephrase my question as:

Are the 'semi' trivial zeros that are located within the critical strip all on the critical line?

Answer by Agno for Are the 'semi' trivial zeros of $\zeta(s) \pm \zeta(1-s)$ all on the critical line?

2012-02-28T22:53:57Z

We now have strong results from GH for $\Gamma(s) \pm \Gamma(1-s)$ and $\zeta(s) \pm \zeta(1-s)$ (when both are nonzero), as well as a more generalized claim outlined in this article http://arxiv.org/abs/0712.1266, that zeros for $f(s) = h(s) - h(2a-s)$, with $h$ being a meromorphic function satisfying appropriate growth conditions, are all inclined to lie on a critical line $a$, with only a finite number of exceptions.

However, despite the encouraging outcomes that all zeros are on the critical line and only a finite number of exceptions residing outside the critical strip, there doesn't seem to be any way to extrapolate these results to the non trivial zeros ($\rho$). No new constraints seem to be imposed on their location and they really seem to originate "from deep down the unaccessible cave" of $\zeta(s)$ itself. They therefore could still reside anywhere on the critical strip.

Just to share some 'futile attempts' I made to make a link. Firstly I had hoped that somehow the results would force the following outcome for all $\Re s$ in the strip:

$\lim_{s\to\rho}\left|\frac{\zeta(s)}{\zeta(1-s)}\right|=1$

and thereby seriously limit the possible values $\Re(\rho)$ could assume. But they don't.

In an attempt to find another relation between $\zeta(s)2$ and $\zeta(1-s)$, I converted both of them to the alternating $\eta(s)$ function and then paired up the individual terms for each $n$ (this is allowed since $\eta(s)$ is valid for $\Re s>0$). This gives:

$\displaystyle \sum _{n=1}^{\infty } \left({\frac { \left( -1 \right) ^{n-1}}{(1-{2}^{1-s}) {n}^{s}}} \pm{\frac { \left( -1 \right) ^{n-1}}{ (1-{2}^{s}) {n}^{1-s}}}\right)$

For each individual $n$, this yields a wave with a fixed frequency and amplitude, that only has zeros when $\Re s=1/2$. These waves nicely sum up to a curve that produces all the 'non & semi' trivial zeros from the OP. However, the $\rho$s obviously do arise from summing up the individual terms as well and I could not find a meaningful way to smartly swap out left and right terms, so that maybe new information about the non trivial zeros would be revealed.

As a last attempt, I also experimented with the PrimeZeta function $P(s)$.

For $P(s)-P(1-s)$ I found:

$\displaystyle \sum _{k=1}^{\infty } \frac{\mu \left( k \right)}{k} \ln \left( {\frac {\zeta \left( ks \right) }{\zeta \left( k \left( 1-s \right) \right) }} \right)$

and for $P(s)+P(1-s)$:

$\displaystyle \sum _{k=1}^{\infty } \frac{\mu \left( k \right)}{k} \ln \left( {\zeta \left( ks \right) \zeta \left( k \left( 1-s \right) \right) } \right)$

For both functions, all zeros appear to lie on the critical line $\Re s=\frac12$, but for the first function the non-trivial zeros have now turned into poles. Interesting to note that the Wolfram site on the PrimeZeta states: "According to Fröberg (1968), very little is known about the roots", so maybe there is something new here to proof for $P(s) \pm P(1-s)$... :-)

Possible locations for non trivial zeroes lying off the critical line

2012-01-10T18:39:25Z

It has been proven that:

1) if $s$ is a non trivial zero $\rho$ of $\zeta(s)$ then so is $1−s$.

2) $\zeta(s) = 2^s \pi^{s-1} \sin(\frac{\pi s}{2}) \Gamma(1-s) \zeta(1-s)$

3) $ 0 < \Re(\rho) <1$

From this it follows that when $s \to \rho$:

$\displaystyle \lim_{s \to \rho} |\dfrac{\zeta(s)}{\zeta(1-s)}| = |2^s \pi^{s-1} \sin(\frac{\pi s}{2}) \Gamma(1-s)|=1$

It is easy to see that the outcome will be $1$ for all $y$ in $s=\frac12 + y i$.

But if a $\rho$ would lie off this critical line, it also must reside in 'spots' where $\displaystyle \lim_{s \to \rho} |\dfrac{\zeta(s)}{\zeta(1-s)}|=1$.

On which points off the critical line could this occur? I found a surprisingly small domain (no proof).

The blue line shows the only values where:

$\displaystyle |2^s \pi^{s-1} \sin(\frac{\pi s}{2}) \Gamma(1-s)|=1$, $s=x + y i$, $ 0 \le x \le 1$.

Note that $y \to 2\pi$ for both $x=0$ and $x=1$. The $y$ rises only a little in the middle.

This doesn't say anything about whether or not off-line $\rho$'s are actually hiding on this curve. There still is an infinite number to check. However, I wondered if anything more is known about this curve?

[IMG]http://img822.imageshack.us/img822/3065/riemanntest.jpg[/IMG]

Please click here for the picture

Viète's generalized infinite product yielding other converging values?

2011-12-03T15:11:43Z

I took Viète's infinite product for $\frac{2}{\pi}$:

$\displaystyle \dfrac{2}{\pi} = \dfrac{\sqrt2}{2} . \dfrac{\sqrt{2+\sqrt2}}{2} . \dfrac{\sqrt{2+\sqrt{2+ \sqrt2}}}{2} \dots$

and made it generic:

$\displaystyle v = |\dfrac{\sqrt{z}}{z} . \dfrac{\sqrt{z+\sqrt{z}}}{z} . \dfrac{\sqrt{z+\sqrt{z+ \sqrt{z}}}}{z} . \dots |$

Checked whether other converging values for $v$ exist and found three more (by trial & error):

$z=-1$ yielding $1.29425...$

$z=i-1$ yielding $0.92741...$

$\displaystyle z=i+ \frac13 (1+ \sqrt[3]{(28-3 \sqrt{87})}+\sqrt[3]{(28+3 \sqrt{87})})$ yielding $0.64801...$

Tested the necessary (but not sufficient) convergence of the 'final' factor to 1 by using:

$\displaystyle a = \sqrt{z+\sqrt{z+ \sqrt{z} \dots}}$

then solving $a$ via:

$a = \sqrt{z + a}$

and indeed $\displaystyle |\frac{a}{z}|=1$ for all these three values.

How could this be proven? Are there more converging values $v$?

Are these values for $v$ connected in any way e.g. on a circle in the complex plane or to other mathematical constants (did check Plouffe's, Wolfram Math and Sloane's but without any result)?

Thanks!

Answer by Agno for Non trivial zeros of the Zeta function

2011-08-09T20:20:45Z

My last thought on this one (I promise).

To be more precise, I indexed the non-trivial zeros in the formula in the opening post but now changed the $\rho$ in the second product into $1-\rho$ (that is in line with Riemann's observation that when a $\rho$ is a non-trivial zero, also $1-\rho$ must be one):

$\displaystyle \prod_{\rho_1}^{\rho_\infty} |\left(1- \dfrac{s}{\rho_n} \right)|=\prod_{\rho_1}^{\rho_\infty} | \left(1- \dfrac{1-s}{1-\rho_n} \right)|$

This means that each term in the product with the same $\rho_n$ can be equated as follows:

$|\left(\dfrac{{\rho_n} - s}{\rho_n} \right)| = |\left(\dfrac{s-\rho_n}{1-\rho_n} \right)|$

EDIT: This step is cleary not allowed and implicitly already assumes $\Re(\rho_n) =\frac12$. The terms in both products can be different. Dividing out all terms with equal $\rho_n$ is allowed (making the infinite product $1$) as I did in my previous post, but this doesn't yield any additional info about the individual $\rho_n$. So, back to the drawing board.

This equation is valid for all $s$ when $\Re(\rho_n) =\frac12$, but it is also valid for all $s=\rho_n$. And that could be for any complex number $\rho_n$ in the critical strip. However, to better see what happens when $s$ approaches $\rho_n$, the following equation gives an indeterminate form of type $0/0$:

$|\left(\dfrac{{\rho_n} - s}{s-\rho_n} \right)| = |\left(\dfrac{\rho_n}{1-\rho_n} \right)|$ or $|\left(\dfrac{{s-\rho_n}}{\rho_n -s} \right)|= |\left(\dfrac{1-\rho_n}{\rho_n} \right)|$

but by applying L'Hôpital's rule we find,

$\displaystyle \lim_{s \to \rho_n} |\left(\dfrac{{\rho_n} - s}{s-\rho_n} \right)| =1$

which implies that when $s$ approaches $\rho_n$ infinitely close, the following equation must be true:

$|\left(\dfrac{\rho_n}{1-\rho_n} \right)| = |\left(\dfrac{1-\rho_n}{\rho_n} \right)| =1$

And this equation only has solutions when $\Re(\rho_n) =\frac12$. It also implies (if the logic is correct) that each term in the following infinite products:

$\displaystyle \prod_{\rho_n} |\left(\frac{1- \rho_n}{\rho_n} \right)| = 1$

$\displaystyle \prod_{\rho_n} |\left(\frac{\rho_n}{1-\rho_n} \right)| = 1$

is equal to $1$ and therefore all $\rho$'s contribute independently from each other to the overall product (and are therefore simple?).

Comment by Agno

2013-06-06T21:31:37Z

Thanks Joro. This is a question I had almost forgotten about (or I did suppress it since it received two down votes...). You have convincingly shown that my conjecture is false, however I might restrict the claim to the critical strip only i.e.: $\Re(\chi(s)) = \Re(\zeta(s)) =0$ can only happen in the critical strip for $\sigma=\frac12$ and $s\ne\rho$ (the crossings in your graph between $0$ and $1$ appear to support this conjecture).

Comment by Agno

2013-05-19T15:07:06Z

@unFortunately. Went through all my old notes, pdf's and links, but have so far not managed to retrieve my source. I do recall that the formula was only mentioned as a premise (in an on-line book or arxiv). It was used for a different topic and I am certain there was no proof of this formula provided. In those days I googled on "infinite product", but also recall I searched on specific values of this particular product at the integers > 1. Will keep digging a bit further and post it when I find more.

Comment by Agno

2013-04-02T15:21:15Z

@Joro, I don't believe these sums are the same. The secondzeta-function in mpmath sums $\displaystyle \sum_{n=1}^\infty \frac{1}{\gamma_n^s}$ and can indeed be analytically continued (via four components). However, $Z_1(s)$ sums over all (paired) $\rho_n$'s. The secondzeta-function for instance has a pole at $s=1$ whereas $Z_1(1)$ doesn't.

Comment by Agno

2013-04-01T16:31:19Z

Hi Joro. Complex $s$ do not seem to converge in my Maple program. What does work though is to make $\beta \in \mathbb{C}$ as long as $\beta \pm \gamma_n \ne 0$ since that would induce a pole.

Comment by Agno

2013-03-28T23:04:51Z

Brilliant. Many thanks, Carlo! I have followed up with a (very) provocative second question, that might not have an answer like this, but maybe it does...

Comment by Agno

2013-03-26T10:09:22Z

@joro. Interesting thought for which I don't have an immediate answer and need to explore further. P.S. In the mean time I have found the closed form (and a proof assuming RH) for $Had(s,\sigma,x)$ and posted it here: mathoverflow.net/questions/117874/… with an additional observation here: mathoverflow.net/questions/122582/…

Comment by Agno

2013-03-26T09:57:21Z

In the mean time I have found the closed form for the above and posted it here: mathoverflow.net/questions/117874/… with an additional observation here: mathoverflow.net/questions/122582/…

Comment by Agno

2013-02-16T11:10:21Z

@Aaron. Many thanks, this is really great. Nice approach with the Pascal Triangle. Is there anything to say about the ratio between the total number of permanently empty bins and the total number of bins $r+1$ (as function of $-y$). Will that ratio converge?

Comment by Agno

2013-01-14T06:21:10Z

Have now added the proof that when assuming RH, the formula can be logically derived.

Comment by Agno

2013-01-14T06:16:53Z

To be complete I should have added the constraint: $a \pm ix\gamma_n \ne 0$

Comment by Agno

2013-01-03T16:56:52Z

@Juan. Many thanks! The outcome is also in line with my expectations :-) Now that there is evidence that "The proposed formula is not true if RH is not true", a logical follow up question would be: Can the formula actually be derived assuming RH (i.e. similar to what you did prove for $x$ in the previous question)? (\sloppy math mode on) Have again tested the formula a lot today against the brute force calculations with 2mln $\rho$s and I believe it just works too well to not be correct.(\sloppy math mode off).

Comment by Agno

2013-01-03T16:15:40Z

@Juan. I believe each $H(s,a,x)$ generates a unique zero when $(s-a)=\frac12+i\gamma$ via the $\zeta$s in either $\xi(\frac12 - \frac{a}{x} + \frac{s}{x})$ or $\xi(\frac12 - \frac{a}{x} + \frac{1}{x} - \frac{s}{x})$. This also implies that when $H(s,a,x)=0$, then also $H(s,1-a,x)=H(1-s,a,x)=H(1-s,1-a,x)=0$. The values correspond to $s=\mu, s=\overline{1-\mu},1-s=1-\mu, 1-s=\overline{\mu}$ respectively. Also when $a=\frac12$ there will be 4 zeros associated with it, of which two pairs are equal.

Comment by Agno

2012-12-02T14:33:19Z

Juan, you're right. I was way too quick. This is not the correct formula $Had(s,x,\sigma)$.

Comment by Agno

2012-12-02T12:31:30Z

@Juan. It is indeed easy to expand and: $$Had(s,x, \sigma)=\frac{x^2-4(s-\sigma)^2}{x^2-1}\pi^{-s/2x} \frac{\Gamma(\sigma^2-\frac{1}{4x}+\frac{s}{2x})} {\Gamma(\sigma^2-\frac{1}{4x})}\frac{\zeta(\sigma+\frac{s}{x}-\frac{1}{2x})} {\zeta(\sigma-\frac{1}{2x})}$$ For $x \rightarrow 1$ it reduces to the known form for $\sigma=\frac12$. What does this mean?

Comment by Agno

2012-12-02T11:45:56Z

Many thank Juan. Checked your formula against the numbers I have computed with 'brute force' and it all looks correct! The structure of your closed form also looks very similar to the ones for the infinite products with $n$ and therefore might be easily expandable to $Had(s,x, \sigma)$. Could the RH just be that $Had(s,1, \frac12) = \dfrac{2(s-1)\Gamma(1+\frac{s}{2}){\zeta(s)}}{ \pi^{\frac{s}{2}}}$ (i.e. reduced form of a more general product formula)?