Eta function of period four
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From: Akira Bergman on 1 Mar 2010 20:59 On Feb 28, 11:49 am, Raymond Manzoni <raym...(a)free.fr> wrote: > Akira Bergman a écrit : > > > > > > > Dirichlet Eta function has period two, as it separates odds and evens > > of the Zeta function by assigning them different signs; > > > Eta(z) = Sum[((-1)^(n-1)) / n^z, n=1 to infinity] > > > It plays a direct role in extending the Zeta function to Re(z) < 1 and > > exposing the non-trivial zeros. > > > What would happen if new Eta functions of period four are defined like > > this?; > > > Etaf1 = 1 + i/2^z - 1/3^z - i/4^z + ... > > Etaf2 = 1 - i/2^z - 1/3^z + i/4^z + ... > > Etaf3 = 1 + i/2^z - i/3^z - 1/4^z + ... > > Etaf4 = 1 - i/2^z + i/3^z - 1/4^z + ... > > > Where i^2 = -1, and the summations are taken on four different cycles > > on the complex plane, starting from +1. One of them is the 4/4 rhythm > > sign conductors use, and another is the popular cross sign. The other > > two are opposite loops around the plane. > > > Has this been done before? > > For Etaf1 and Etaf2 you may use directly the polylogarithm : > Li_z(x) = sum_{k=1}^oo x^k/k^z > <http://en.wikipedia.org/wiki/Polylogarithm> > > Etaf1(z)= Li_z( i)/i = (2^(-z) Eta(z) + i Beta(z))/i > Etaf2(z)= Li_z(-i)/i = (2^(-z) Eta(z) - i Beta(z))/i > > with Beta(z) the Dirichlet beta function : > Beta(z)= sum_{n=0}^oo (-1)^n/(2n+1)^z > <http://en.wikipedia.org/wiki/Dirichlet_beta_function> > > The Clausen function could be of interest too : > <http://en.wikipedia.org/wiki/Clausen_function> > > For Etaf3 and Etaf4 you may rewrite them with a little more work > using for example (I think you may write them too as sum of > polylogarithms...) > sum_{k=0}^oo 1/(4*k+1)^z = zeta(z, 1/4)/4^z > sum_{k=0}^oo 1/(4*k+2)^z = (2^z-1) zeta(z)/4^z > sum_{k=0}^oo 1/(4*k+3)^z = zeta(z, 3/4)/4^z > sum_{k=0}^oo 1/(4*k+4)^z = zeta(z)/4^z > (with zeta(z, q) the Hurwitz zeta function > <http://en.wikipedia.org/wiki/Hurwitz_zeta_function>) > > Hoping it helped, > Raymond Polylogarithm all the way. They add up to a form similar to the other definition of Li; sum_{n=1}^oo Li_z(e^(i*2*pi/2^n)) / e^(i*2*pi/2^n) On Wikipedia the similar expression is; http://en.wikipedia.org/wiki/Polylogarithm Li_z+1(s) = integral[Li_z(t) dt/t; t=0 to s] |