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 which sends functions from themselves to their continuum sum.

What's truly even more amazing is that I have found an inverse expression for 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