1
$\begingroup$

In this question, I went quite a bit over the top, so I now tried to rephrase it in a much simpler way.

Take $a \in \mathbb{R}, s \in \mathbb{C}$ and:

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

with $\xi_{c}(z) = \frac{\sinh(\pi z)}{z}$.

And then also:

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

with $\xi_{r}(z) = \frac{1}{z\, \Gamma(z \, i)}$.

Questions:

For the infinite product of the conjugated zeros $C(s,a)$, the closed form has only two factors, however the closed form for the product of the reflexive zeros $R(s,a)$ does require four factors.

1) Is there a way to also reduce the closed form for $R(s,a)$ to two $\xi$-factors by altering $\xi_r(z)$?

Additional clarification:

I do believe that it is impossible to reduce these four factors:

$$\displaystyle \frac{\xi_{r}(0 -a+s)}{\xi_{r}(0-a)}\frac{\xi_{r}(1 -a-s)}{\xi_{r}(1-a)}$$

into a division of only two factors.

The most logical extension of $\xi_{r}(z)$ would be to assume: $\xi_{r}(z) = \frac{1}{z\, \Gamma(z \, i)} \, \, \frac{1}{(1-z)\, \Gamma((1-z) \, i)}$

and then to reduce the division to: $\displaystyle \frac{\xi_{r}(0-a+s)}{\xi_{r}(0-a)}$.

However this only correctly induces the original four factors for the nominators only and it gives the wrong results for the two denominators. Any tweak I tried to make to $\xi_r(z)$ fails on this apparently fundamental conflict, hence my conjecture that it is impossible. Keen to learn if this could be proven and/or which branch of Mathematics deals with this type of problems.

2) If not (per my conjecture), is this four factor closed form with $\xi_r(z)$ then dependent on the choice of $n$ or does it simply originate from the reflexive nature of the zeros in the product? To illustrate what I am after with an example; the relations $C(s,a) = C(2a-s,a)$ and $R(s,a) = R(1-s,a)$ are fully independent of the choice of $n$ and the product being finite or infinite, and just stem from the conjugated/reflexive relation between the zeros.

Thanks.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.