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 Powerful way to perform continuum sum tommy1729 Ultimate Fellow Posts: 1,380 Threads: 338 Joined: Feb 2009 08/11/2013, 02:00 AM (08/10/2013, 09:06 PM)JmsNxn Wrote: Whats truly amazing is that this is a linear operator. Therefore we have an operator $\mathcal{S}f(s) = \mathcal{Z}\mathcal{J}^{-1} f(s)$ which sends functions from themselves to their continuum sum. What's truly even more amazing is that I have found an inverse expression for $\mathcal{Z}$ This was an extra trick. Since S = Z J^-1, the difference operator D satisfies D = J Z^-1. Is this your inverse expression for Z or do you have an integral transform for Z^-1 ? I assume you were influenced by Ramanujan ('s master theorem). Maybe this too : http://en.wikipedia.org/wiki/Mertens_function You seem to like integrals Regards tommy1729 « Next Oldest | Next Newest »

 Messages In This Thread Powerful way to perform continuum sum - by JmsNxn - 08/10/2013, 09:06 PM RE: Powerful way to perform continuum sum - by tommy1729 - 08/11/2013, 02:00 AM RE: Powerful way to perform continuum sum - by JmsNxn - 08/11/2013, 04:05 PM RE: Powerful way to perform continuum sum - by tommy1729 - 08/11/2013, 07:31 PM RE: Powerful way to perform continuum sum - by JmsNxn - 08/11/2013, 08:18 PM RE: Powerful way to perform continuum sum - by tommy1729 - 08/11/2013, 10:29 PM RE: Powerful way to perform continuum sum - by tommy1729 - 08/11/2013, 11:05 PM RE: Powerful way to perform continuum sum - by JmsNxn - 08/12/2013, 07:17 PM

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