Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

(I asked this in MSE before but there was only a general reference which did not help for my specific question)

I think I understood the concept of fractional derivatives applied to monomials to consider some general cases and also applied to exponential terms as they occur in the formal expression of the Riemann zeta as Dirichlet series and their integer or fractionally indexed derivatives.

For integer indexed derivatives of the zeta at zero I can numerically evaluate the alternating zeta (Dirichlet's eta) instead using some common version of divergent summation, say Euler-summation ($\mathfrak E$) and then formally refer to $$ {d^m \over dx^m } \eta (0) =\eta^{(m)}(0) \underset{\mathfrak E}= \sum _{k=0}^\infty (-1)^k (-\log(1+k))^m$$
and then express the m'th derivatives of the zeta at zero recursively using a binomial formula (and writing $ \zeta_m =\zeta^{(m)}(0)$ and $ \eta_m =\eta^{(m)}(0) $ and $u=\log(1/2)$ for convenience) $$ \zeta_m = - \left(\eta_m + 2 \cdot \sum_{k=0}^{m-1} \left[ \binom{m}{k}u^{m-k} \zeta_k\right]\right) $$

However, I cannot generalize this to the according fractional derivatives, in this question the half-derivative, because the finite binomial sum would become an infinite series, likely also divergent and where I also would not know, in which direction I should re-interpret the binomial sum which is symmetric in its indexes for the case of integer exponents.

For the half-derivative of the $\eta$ I get by the Euler-summation and using the setting $\sqrt{-\log(1+k)} = i\sqrt{\log(1+k)} $ the approximation $$\eta_{1/2} \sim -0.347006596200 i $$

Q1: How could I express the half-derivative $\zeta_{0.5} = \zeta^{(0.5)}(0) $ formally
Q2: and what is a meaningful value?

share|improve this question
add comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.