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 operatorwhich sends functions from themselves to their continuum sum.
What's truly even more amazing is that I have found an inverse expression forThis 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