Hi,
referring to the Riemann-Siegel approximate functional equation for Riemann's Zeta
$ \zeta(s) = \sum_{n\leq x}\frac{1}{n^s} \ + \ \chi(s) \ \sum_{n\leq y}\frac{1}{n^{1-s}} \ + \ O(x^{-\sigma}+ \ |t|^{\frac{1}{2}-\sigma}y^{\sigma - 1}) $
would anybody know of a similar result applying to the Dirichlet Eta function ?
$ \eta(s) = \sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s} = 1-\frac{1}{2^s}+\frac{1}{3^s}-\frac{1}{4^s}+-\ldots $
I am interested in expressing the Dirichlet Eta function in terms of its partial sums, as well as in terms of the partials sums of its critical line symmetrical "twin". So, I am looking for an expression of this kind
$ \eta(s) = !(s) \ \sum_{n\leq x}\frac{(-1)^{n-1}}{n^s} \ + \ ?(s) \ \sum_{n\leq y}\frac{(-1)^{n-1}}{n^{1-s}} \ + \ O( ........) $
where !(s) and ?(s) are functions yet unknown to me, and I am not even sure whether such an approximate functional equation might actually exist. I will greatly appreciate suggestions from anybody familiar with the subject. Many Thanks.
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2$\begingroup$ The usual relation between $\eta(s)$ and $\zeta(s)$ is the identity $\eta(s) = (1 - 2^{1-s}) \zeta(s)$. For partial sums, it is instead of the form $\sum_{n \leq x} \frac{(-1)^{n-1}}{n^s} = \sum_{n \leq x} \frac{1}{n^s} - 2^{1-s} \sum_{n \leq x/2} \frac{1}{n^s}$. Perhaps you might be able to use these two identities to convert the approximate functional equation for $\zeta(s)$ into one for $\eta(s)$. $\endgroup$– Peter HumphriesCommented May 11, 2011 at 9:03
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$\begingroup$ Many thanks Peter! I'll try that this coming weekend. $\endgroup$– LucaCommented May 11, 2011 at 9:21
1 Answer
First of all, if you want just a single sum up to T, then just like with the zeta function you have an approximation: $$\eta(s)\sim\sum_{n=1}^{T/2\pi}\frac{(-1)^{n-1}}{n^s}$$
It turns out additionally that the full approximate functional equation (aka Riemann-Siegel formula) holds for really anything where you have a functional equation. You can see, for example, "The approximate functional equation for a class of zeta-functions" by Chandrasekharan and Narasimhan.
In this case, you can write the functional equation as relating $1-2^{-s}+3^{-s}-\cdots$ and $0.5^{-s}-1.5^{-s}+2.5^{-s}-\cdots$ (this form does not have an annoying $1-2^s$ term in it) so this latter term is what you find in the final formula: $$\eta(s) \sim \sum_{n\le x}\frac{(-1)^{n-1}}{n^s} - \chi(s)\sum_{n\le y}\frac1{(n-1/2)^s}$$
Also, any of the methods for proving the formula for $\zeta$ carry over naturally to $\eta$ (sometimes more naturally), so let me know if you had a particular one in mind.
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$\begingroup$ Hi Ralph, thanks for the tips. In fact, I was rather looking for an expression in terms of partial sums of s and 1-s. In any case, the discussion on this topic has stimulated a perhaps a better focused question, in a way related to this one, and which I am going to post as a new question. $\endgroup$– LucaCommented Jun 3, 2011 at 16:58