Eta function of period four
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From: Akira Bergman on 22 Feb 2010 19:50 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?
From: Gerry Myerson on 22 Feb 2010 20:36 In article <64a49230-abaf-47e0-9af4-741207830707(a)y7g2000prc.googlegroups.com>, Akira Bergman <akirabergman(a)gmail.com> wrote: > 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? Looks something like Dirichlet L-functions, q.v. -- Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email)
From: Akira Bergman on 22 Feb 2010 21:47 On Feb 23, 12:36 pm, Gerry Myerson <ge...(a)maths.mq.edi.ai.i2u4email> wrote: > In article > <64a49230-abaf-47e0-9af4-741207830...(a)y7g2000prc.googlegroups.com>, > Akira Bergman <akiraberg...(a)gmail.com> wrote: > > > > > > > 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? > > Looks something like Dirichlet L-functions, q.v. > > -- > Gerry Myerson (ge...(a)maths.mq.edi.ai) (i -> u for email) Thanks for the reply. Yes, my list look like the Dirichlet Character but not included in it, according to Wikipedia; http://en.wikipedia.org/wiki/Dirichlet_character which lists them explicitly until mod-10. Mod-5 kind of includes two of mine, but mine is mod-4.
From: Akira Bergman on 25 Feb 2010 19:22 On Feb 23, 11:50 am, Akira Bergman <akiraberg...(a)gmail.com> wrote: > 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? and then we can define four more by rotating the unit circle pi/4 degrees, and doing the same cycles on that one; Etaf5 = (1+i)/s2 + ((i-1)/s2)/2^z - ((i+1)/s2)/3^z - ((i-1)/s2)/4^z + ... Etaf6 = (1+i)/s2 - ((i-1)/s2)/2^z - ((i-1)/s2)/3^z + ((i-1)/s2)/4^z + ... Etaf7 = (1+i)/s2 + ((i-1)/s2)/2^z - ((i+1)/s2)/3^z - ((i+1)/s2)/4^z + ... Etaf8 = (1+i)/s2 - ((i-1)/s2)/2^z + ((i-1)/s2)/3^z - ((i+1)/s2)/4^z + ... where s2 = sqrt(2). And then rotate all eight of them pi/8 degrees and get a set of 16 of them. So on until the full circle at infinity; pi/4 + pi/8 + pi/16 + ... pi*sum(1/2^n; n=2 to infinity) If the starting points also change then we get an enormous set of points on the unit complex circle. If we unleash this monster gradually starting with the first four onto the Zeta function, who knows what kind of resonances in the primes would be found? We may find that the complex unit circle is isomorphic to the primes, and that they have is infinite complexity. Therefore they would be isomorphic to the H-fractal, or the entire binary tree. This may imply that the Riemann's Hypothesis is true, since all branches would be independent and the amplitude and phase of the zeros would be independent. That no matter how you divide this monster, you get totally independent trees. It is fully sick; self similar yet different.
From: Akira Bergman on 25 Feb 2010 20:27
On Feb 26, 11:22 am, Akira Bergman <akiraberg...(a)gmail.com> wrote: > On Feb 23, 11:50 am, Akira Bergman <akiraberg...(a)gmail.com> wrote: > > > > > > > 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? > > and then we can define four more by rotating the unit circle pi/4 > degrees, and doing the same cycles on that one; > > Etaf5 = (1+i)/s2 + ((i-1)/s2)/2^z - ((i+1)/s2)/3^z - ((i-1)/s2)/4^z > + ... > Etaf6 = (1+i)/s2 - ((i-1)/s2)/2^z - ((i-1)/s2)/3^z + ((i-1)/s2)/4^z > + ... > Etaf7 = (1+i)/s2 + ((i-1)/s2)/2^z - ((i+1)/s2)/3^z - ((i+1)/s2)/4^z > + ... > Etaf8 = (1+i)/s2 - ((i-1)/s2)/2^z + ((i-1)/s2)/3^z - ((i+1)/s2)/4^z > + ... > > where s2 = sqrt(2). > > And then rotate all eight of them pi/8 degrees and get a set of 16 of > them. So on until the full circle at infinity; > > pi/4 + pi/8 + pi/16 + ... pi*sum(1/2^n; n=2 to infinity) > > If the starting points also change then we get an enormous set of > points on the unit complex circle. If we unleash this monster > gradually starting with the first four onto the Zeta function, who > knows what kind of resonances in the primes would be found? > > We may find that the complex unit circle is isomorphic to the primes, > and that they have is infinite complexity. Therefore they would be > isomorphic to the H-fractal, or the entire binary tree. > > This may imply that the Riemann's Hypothesis is true, since all > branches would be independent and the amplitude and phase of the zeros > would be independent. That no matter how you divide this monster, you > get totally independent trees. > > It is fully sick; self similar yet different. I just calculated that pi*sum(1/2^n; n=2 to infinity) = o.5*pi |