0
$\begingroup$

This question is related to my question here, I have used The Riemann-Siegle theta function particulary its taylor series arround $t=0$ to check wether Riemann zeta function has an explicite inversion formula near $s=0$ may it exist as a formal power series using lagrange inversion theorem since it is converges for $|t|< \frac12 $ then :

What is the compositional inverse of Riemann zeta function near $s=0$ if it exists?

Note: The motivation of this question is to compute $\zeta^{-1}(\dfrac{-1}{2})$

$\endgroup$
0

1 Answer 1

2
$\begingroup$

Let $U$ be the image by $\zeta$ of $|s| < 1/2$ and for $z \in U$ $$F(z) = \frac{1}{2i\pi} \int_{|s| = 1/2} \frac{\zeta'(s) s}{\zeta(s)-z}ds$$

Then $F$ is analytic on $U$ and $F(\zeta(s)) = s$.

Proof : for $z\in U, \exists |a| < 1/2, z = \zeta(a)$ then for $|s| \le 1/2$, $s \mapsto \frac{\zeta'(s) s}{z-\zeta(s)}$ has a unique simple pole at $s=a$ of residue $a$ so that $F(z) = Res(\frac{\zeta'(s) s}{\zeta(s)-z},a)= a$.

Since $\zeta(0) = -1/2$ then $F(-1/2) = 0$.

When extending this local branch $F$ of $\zeta^{-1}$ by analytic continuation it will have a branch point at each $z= \zeta(a), \zeta'(a)=0$.

$\zeta$ isn't injective and for any $a\ne 0$ there are infinitely many $s$ such that $\zeta(s)=a$, infinitely many of them are on the strip $\Re(s) \in (\sigma-\epsilon,\sigma+\epsilon)$ for any $\sigma > 1$ such that $1/\zeta(\sigma) < |a| < \zeta(\sigma)$.

$\endgroup$
1

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .