Question

does anyone know about the Riemann zeta function?

Answer 1

The Riemann zeta function or Euler–Riemann zeta function, ζ(s), is a function of a complex variable s that analytically continues the sum of the infinite series. https://en.wikipedia.org/wiki/Riemann_zeta_function

Answer 2

i have been researching multidimensional spheres and have found a 1/2 invariant and was wondering if there was a correlation between the RZF and multidimensional spheres. note also that these spheres are isomorphic to the general linear group of the dimensionality of the sphere in question

Answer 3

I can't answer your question, but out of interest, what is 1/2 invariant with respect to?

Answer 4

A unique multi-dimensional sphere through the zero point (the origin) can be defined by a set of linearly independent points. the number of points required is equal to the dimensionality of the sphere. These points will have a given tangent displacement from the sphere. the invariant is defined by the sum of the products of the components of the center of the sphere and any of the points divided by difference of the distance from the origin to the point and the tangent displacement of the point. Symbolically, \[\sum^{\delta}_{k=0} \frac{x_k(\omega\rho)_{k}}{c^2-d^2}=\frac{1}{2}\]
\[x_k:k^{th}\;component\;of\;the\;point\]
\[(\omega\rho)_k :k^{th}\;component\;of\;the\;center\]
\[c^2:the\;square\;of\;the\;distance\;from\;the\;origin\;to\;the\;point\]
\[d^2:the\;square\;of\;the\;tangent\;displacement\;of\;the\;point\;to\;the\;sphere\]
\[\delta:dimensionality\;of\;the\;sphere\]

Answer 5

Just to clarify, you're saying that the origin is one of the points on the sphere and that the origin is not the centre of the sphere? Also, what does "tangent displacement of a point to the sphere" mean?

Answer 6

yes the sphere must pass through the origin. there must be n additional, linearly independent points. through each point, there is a line tangent to the sphere. the tangent displacement is the distance from the chosen point to the intersection of the tangent through the point and the sphere

Answer 7

Why do you suspect there's some correlation between the RZF and these multidimensional spheres?

Answer 8

\[The\;RFZ \;is \;defined\;as
\\ \zeta(s) = \sum_{k=1}^\infty \frac{1}{k^s}\]
\[This\;can\;be\;rewritten\;as
\\ \zeta(s)=\lim_{\alpha\rightarrow 0} \sum_{k=1}^\infty\frac{1}{k^s+\alpha^s}\]
\[A\;multidimensional\;sphere\;is\;defined\;by\;the\;equation
\\ \sum^{\delta}_{k=1} (x_{kj} - x_{k0})^2 = R^2 + d^2_j\]
\[\delta: dimensionality\;of\;the\;surface\;of\;sphere
\\R^2:square\;of\;the\;radius
\\d_j^2:square\;of\;the\;tangent\;displacement
\\x_{kj}:k^{th}\;component\;of\;the\;j^{th}\;point
\\x_{k0}:k^{th}\;component\;of\;the\;center\]
\[Given\;x_{kj}\;and\;d_j^2,\;x_{k0}\;and\;R^2\;can\;be\;determined.
\\The\;points,\;x_{kj},\;form\;a\;matrix\;with\;indices\;k\;and\;j.
\\The\;displacements,\;d_j^2,\;form\;a\;vector;\;index\;j.\]
\[The\;center\;can\;be\;found\;by\;the\;following:
\\matrix:\Phi(x_{kj})=2x_{kj}
\\vector:\Psi(d_j^2)= \left[\sum^{\delta}_{k=1}x_{kj}^{2}\right]-d^2_j
\\center: \Omega_k(\Phi,\Psi)=(\Phi^{-1}\Psi)_k
\\radius\;squared:\rho^2=\sum^{\delta}_{k=1}\Omega_k^2\]
\[The\;center\;can\;be\;geometrically\;defined\;as\;the\;intersection\;of\;n\;hyperplanes.
\\The\;radius\;squared\;is\;the\;sum\;of\;the\;squares\;of\;the\;components\;of\;the\;center.\]
\[The\;following\;must\;be\;true\;for\;any\;\delta-sphere\;through\;the\;origin:
\\\sum^{\delta}_{k=1}\frac{x_{kj}\Omega_k(\Phi,\Psi)}{c_j^2-d_j^2}=\frac{1}{2}
\\If\;c_j^2=d_j^2,\;then\;the\;line\;joining\;x_{kj}\;and\;the\;origin\;is\;tangent\;to\;the\;sphere.\]

Answer 9

following this post..

Answer 10

\[The\;spheres\;that\;I\;introduced\;above\;have\;many\;important\;properties:
\\\
\\1.\;\;\;The\;center\;of\;an\;n\;dimensional\;sphere\;is\;isomorphic\;to\;the
\\\;\;\;\;\;\;intersection\;of\;a\;particular\;set\;of\;linear\;equations.
\\\
\\2.\;\;\;The\;set\;of\;points\;used\;to\;define\;the\;sphere\;form\;a\;unique
\\\;\;\;\;\;\;basis\;of\;the\;vector\;field\;in\;which\;the\;sphere\;is\;embedded.
\\\
\\3.\;\;\;The\;difference\;between\;the\;tangent\;distance\;of\;each\;point
\\\;\;\;\;\;\;to\;the\;sphere\;and\;the\;distance\;of\;each\;point\;from\;the\;origin
\\\;\;\;\;\;\;form\;a\;vector\;
\\\
\\4.\;\;\;The\;sphere\;passes\;through\;the\;origin
\\\
\\5.\;\;\;The\;product\;of\;any\;invertible\;matrix\;with\;a\;vector\;can\;be
\\\;\;\;\;\;\;represented\;by\;a\;sphere
\]

Answer 11

\[Another\;project\;that\;I\;have\;been\;working\;on\;involves\;prime\;state\;logic
\\and\;polynomials\;with\;coefficients\;in\;Z_p\;where\;p\;is\;prime.
\\\
\\The\;connection\;between\;this,\;the\;spheres,\;and\;the\;zeta\;function
\\lies\;in\;the\;conversion\;matrix\;from\;algebra\;(polynomials)
\\to\;logic\;(states)\;and\;its\;inverse.
\\\
\\\alpha_{kj}\:=\delta_{0j}\delta_{k0}+j^k(1-\delta_{0j})\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,(mod\;p)
\\\alpha_{kj}^{-1}=\delta_{0j}-\delta_{k0}\delta_{(p-1)j}-k^{-j}(1-\delta_{k0})\;\;(mod\;p)
\\where\;\delta_{kj}\;is\;the\;kronecker\;delta\;function
\\\
\\It\;is\;important\;to\;note\;that\;\alpha_{k1}=1\;for\;all\;k.
\\Thus,\;\lim_{p\rightarrow\infty}\left[\Omega_1(\Phi,\Psi)=\sum^{p}_{t=1}\frac{1}{t^s}\right]=\zeta(s)\;with\;appropriate\;\Phi\;and\;\Psi.
\\Furthermore,\;\Omega_k(\Phi,\Psi)=\sum^p_{t=1}\frac{k^t\;(mod\;p)}{t^s}
\\Also,\;recall\;that\;\Omega_k\;varies\;inversely\;with\;\Phi\;and\;directly\;with\;\Psi;
\\therefore,\alpha^{-1}_{kj}\;should\;be\;used\;to\;determine\;the\;points
\]

Answer 12

@ganeshie8 what is your opinion about above wrote please ?
ty. in advance

Answer 13

@adrimit Woah, that's awesome. Is there a paper/pdf I can read in my own time?

Answer 14

yes, I have a group on facebook that I post some of my work. it is invitation only, and I can give you access to it. you can find me by my email, adrian.mitrea@yahoo.com on facebook. send a friend request, and I will add you to the group. Thank you for your interest :P

Answer 15

I have quit facebook for a while. I'll save your email and then I'll be back on in about a month. Please don't think it's a random stranger when it happens :P

Answer 16

lol, or you could just email me