Eta function of period four
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From: Raymond Manzoni on 27 Feb 2010 19:49 Akira Bergman a �crit : > 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
From: Akira Bergman on 27 Feb 2010 20:58 On Feb 28, 11:49 am, Raymond Manzoni <raym...(a)free.fr> wrote: > Akira Bergman a crit : > > > > > > > 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 Thanks for the help. So it seems there may be analytic expressions for all these functions. As the period increases, it may be possible to use the previous results like a fractal. A picture of these would be nice to look at, to see what kind of transformations the zeros go through, and what kind of new zeros are introduced.
From: Akira Bergman on 27 Feb 2010 23:40 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. At this level the eight cycles should also be done; 1, (1+i)/s2, i, (i-1)/s2, -1, (-1-i)/s2, -i, (-i+1)/s2 It is a mediation sequence, like the Stern-Brocot tree with the mediation rule changing at every step. The deeper we get into it, the longer wavelengths we can cover, thus exhausting the entire bandwidth of the Zeta function. This may imply that the zeros of the Riemann's Zeta function are in fact the smallest set of resonances in the primes. At every wavelength there may be a whole new set of resonances requiring a whole new function. An infinite fractal of infinite analytical expressions. This could explain why it has been so difficult to prove Riemann's Hypothesis. It was not the whole picture.
From: Gerry Myerson on 28 Feb 2010 17:13 In article <53784e8b-1801-42e7-bcb5-29403d798f72(a)o16g2000prh.googlegroups.com>, Akira Bergman <akirabergman(a)gmail.com> wrote: > I just calculated that > > pi*sum(1/2^n; n=2 to infinity) = o.5*pi Which is to say, you just calculated that (1/4) + (1/8) + (1/16) + ... = 1/2. I'm sure the Riemann Hypothesis is just minutes away. -- Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email)
From: Akira Bergman on 28 Feb 2010 17:36
On Mar 1, 9:13 am, Gerry Myerson <ge...(a)maths.mq.edi.ai.i2u4email> wrote: > In article > <53784e8b-1801-42e7-bcb5-29403d798...(a)o16g2000prh.googlegroups.com>, > Akira Bergman <akiraberg...(a)gmail.com> wrote: > > > I just calculated that > > > pi*sum(1/2^n; n=2 to infinity) = o.5*pi > > Which is to say, > you just calculated that (1/4) + (1/8) + (1/16) + ... = 1/2. > > I'm sure the Riemann Hypothesis > is just minutes away. > > -- > Gerry Myerson (ge...(a)maths.mq.edi.ai) (i -> u for email) Ok got your point. |