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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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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 $\eta(s)$. –  Peter Humphries May 11 '11 at 9:03
    
Many thanks Peter! I'll try that this coming weekend. –  Luca May 11 '11 at 9:21
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2 Answers 2

up vote 1 down vote accepted

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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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. –  Luca Jun 3 '11 at 16:58
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Hi,
following Peter's advice I have got an answer, albeit raising more questions.
I have tried the case $ x \ = \ y \ = \ 2N $, and then used the equalities
$ \sum_{n=1}^{2N}\frac{(-1)^{n-1}}{n^s} \ = \ \sum_{n=1}^{2N}\frac{1}{n^s} \ - \ 2^{1-s} \sum_{n=1}^{N}\frac{1}{n^s} $
and
$ \sum_{n=1}^{2N}\frac{(-1)^{n-1}}{n^{1-s}} \ = \ \sum_{n=1}^{2N}\frac{1}{n^{1-s}} \ - \ 2^s \sum_{n=1}^{N}\frac{1}{n^{1-s}} $
to finally obtain
$ \eta(s) = \sum_{1}^{2N}\frac{(-1)^{n-1}}{n^s} \ - 2^{1-s} \chi(s) \ \sum_{1}^{2N}\frac{(-1)^{n-1}}{n^{1-s}} \ - 2^{1-s} \sum_{N+1}^{2N}\frac{1}{n^s} \ + $
$ + \ \chi(s) \ ( \ \sum_{1}^{2N}\frac{1}{n^{1-s}} \ - \ 2 \sum_{1}^{N}\frac{1}{n^{1-s}} \ ) \ + \ O((2N)^{-\sigma}+ \ |t|^{\frac{1}{2}-\sigma}(2N)^{\sigma - 1}) $
differing in a fundamental way from the Riemann-Siegel approximate functional equation: take s inside the critical strip, then for any given t the partial sums of the eta function do converge (and with hindsight, I am now realising that this is also the reason why my question was not so interesting after all ...), while the partial sums of the zeta function do not.
Incidentally, because the first term on the right, for N->infinity, converges to the eta function itself, and the Big-O term vanishes, I am led to conclude that:
$ - 2^{1-s} \chi(s) \ \sum_{1}^{2N}\frac{(-1)^{n-1}}{n^{1-s}} \ - 2^{1-s} \sum_{N+1}^{2N}\frac{1}{n^s} \ + \ \chi(s) \ ( \ \sum_{1}^{2N}\frac{1}{n^{1-s}} \ - \ 2 \sum_{1}^{N}\frac{1}{n^{1-s}} \ ) $
must also vanish, but as big-O of what ?

the related question being:
inside the critical strip, for any given t and for N->infinity, does the answer to my question incidentally prove that

$ 2^{s} \chi(s) \ ( \ \frac{1}{2} \sum_{1}^{2N}\frac{1}{n^{1-s}} \ - \ \sum_{1}^{N}\frac{1}{n^{1-s}} \ ) \ - \ \frac{1}{\chi(s)} \sum_{N+1}^{2N}\frac{1}{n^s} \ \sim \ \eta(1-s) \; \; \; \; ? $
or, in a perhaps more elegant form
$ \eta(1-s) \ (1-\chi(s)) \ = \ \chi(s) \ (2^{s-1}-1) \sum_{1}^{2N}\frac{1}{n^{1-s}} \ - \ \frac{1}{\chi(s)} \sum_{N+1}^{2N}\frac{1}{n^s} \ + \ \ O( ... of what ? ...)$

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Can you edit your TeX to break the lines up? –  Stopple May 12 '11 at 22:55
    
OK, I guess that the troublesome reading was with this one: $ \eta(s) = \sum_{1}^{2N}\frac{(-1)^{n-1}}{n^s} \ - 2^{1-s} \chi(s) \ \sum_{1}^{2N}\frac{(-1)^{n-1}}{n^{1-s}} \ - 2^{1-s} \sum_{N+1}^{2N}\frac{1}{n^s} \ + $ $ + \ \chi(s) \ ( \ \sum_{1}^{2N}\frac{1}{n^{1-s}} \ - \ 2 \sum_{1}^{N}\frac{1}{n^{1-s}} \ ) \ + \ O((2N)^{-\sigma}+ \ |t|^{\frac{1}{2}-\sigma}(2N)^{\sigma - 1}) $ –  Luca May 13 '11 at 8:02
    
uupps ... that is even worse. In fact the trouble is that I cannot get the preview working at all, so first verify with TeXnicCenter, and then I paste here. –  Luca May 13 '11 at 8:05
    
finally fixed ! –  Luca May 13 '11 at 8:22
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