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301:. The version I have removed was a later attempt by Carifio24 to simplify the expression. However, he must have simplified incorrectly because the two equations give different results. At any rate, I don't think either of them are correct, because eta for a positive integer should always yield a positive value. If anyone knows the correct formula, please add it in!
422:) 00:53, 2 April 2011 (UTC). Say we use half-lines to group zeros, assuming RH. The eta function then has one infinity of zeros on the negative real axis, one infinity on the critical half-line above the real axis, a third on the critical half-line below the real axis, a fourth on the half-line {s | Re(s)=1, Im(s): -->
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Could we just enclose the proof of the relationship between the
Riemann zeta function and the Dirichlet eta function? It's a very simple proof, one that wouldn't take up much space. Furthermore, when the eta function is defined in the Dirichlet series expansion, couldn't the first couple terms of
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I don't think this is a good phrase, and if there is no objection I can rewrite the paragraph and rework sentences like "Thus the eta function has five countable infinities of zeros, while the zeta function has only three", which don't really make much sense. One problem is that if RH is true, the
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The proof above could go in a section labelled "Relation with the zeta function", and I will do that after finding a reference for it (Euler, Titchmarsh ?). But the emphasis should remain on the eta function itself since this page is about the
Dirichlet eta function, not the Riemann zeta function,
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1, eta(s)= 1 - 2^(-s) + 3^(-s) - 4^(-s) + ... = 1 + 2^(-s) + 3^(-s) + 4^(-s) + ... - 2( 2^(-s) + 4^(-s) + ...) = zeta(s) - 2^1*2^(-s)(1 + 2^(-s) + ... ) = ( 1 - 2^(1-s) ) zeta(s). This is an equation that holds true for alls complex values of s (even s=1) by analytic continuation. Note that the
394:) 23:40, 15 August 2010 (UTC) Found a reference in Henrici Applied Complex Analysis book, p. 295-6, which has a very good section on Dirichlet series. Henrici writes zeta(s) = odd(s) + even(s), eta(s) = odd(s) - even(s) and then deduces the relation, pointing out the case when 2^(1-s)=1.
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factor (1 - 2^(1-s)) is zero for s=1 and also for an infinity of points on the line Re(s)=1, s= 1 + 2 i Pi k/log(2), where the equation zeta(s) = eta(s) /(1-2^(1-s)) breaks down formally (a problem often unoticed, unfortunately).
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But to a mathematician, the word "analytic" *always* means finite. If a function has only poles at its singularities it would be called "meromorphic" (and can be viewed as a holomorphic function to the
Riemann sphere
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1. Then the
Dirichlet series for the zeta(s) function converges absolutely by the integral test, and we can reorder the terms at will without changing the sum. The same goes for the eta function, but only for Re(s):
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That is an interesting and possibly useful concept to associate to an analytic function f: the smallest number of closed half-lines in the complex plane such that all zeroes of f lie in their union.
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It helps no one to include an illustration whose colors' meaning is described only in terms of some feature of some commercial product most readers surely have never heard of.
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However, in the equation η must be zero at all the points , where the denominator is zero, if the
Riemann zeta function is analytic and finite there.
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The
Dirichlet eta function can still be used to define zeta(s) at those values where s = 1 + 2Kπi, K ∊ ℤ - {0}, just by taking the limit of the ratio
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Okay, I think I've found the correct form for even positive integers. I've added it into the article, post here if there are any problems with it.
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1) The eta function has zeros on Re(s)=1, which do not show up very clearly on the color plot (which looks like a plot of zeta(s) instead).
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my guess would be that n is an arbitrary number, and the algorithm is improved in the limit as n -: -->
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Yeah, that's the proof I'm talking about. Do we want to put it into the article, or create a link?
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on
Knowledge. If you would like to participate, please visit the project page, where you can join
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2) The eta function cannot be used to define the zeta function at s=1+2*k*Pi*i, for k integer.
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If you agree with this, please edit the page, and if possible provide a clearer plot.
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0}, and a fifth on {s | Re(s)=1, Im(s)<0}. Thanks for clearing out the ambiguity.
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The factor 1-2^(1-s) is zero for s=1 AND also for s=1+2*k*Pi*i, k integer. So ...
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have zeroes where the zeta runction doesn't, because 1 - 2 = 0 when s = s
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would it be possible to explain, what the "n" means in the
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the series be given? It makes it much easier to get a feel of the series.
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are not on the line Re(s) = 1/2, but instead on the line Re(s) = 1.
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originally posted a different version of the equation
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http://fr.wikipedia.org/Fonction_z%C3%AAta_de_Riemann
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I think there is a mistake in this image. Since the
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