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Powerful way to perform continuum sum

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Powerful way to perform continuum sum
tommy1729 Offline
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#2
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 Smile

Regards

tommy1729
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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

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