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Powerful way to perform continuum sum
#3
Actually I wasn't thinking about Ramanujan. I was thinking about Euler and the integral representation for the Gamma function. I wanted to exploit integration by parts as beautifully as he did.

Why yes the difference operator is how I found the inverse, but this gives us an integral transform:



Is an expression for one of the inverses. We also have:



Each inverse operator works on different classes of functions. I'm having a little trouble finding the exact restrictions on the functions we can use.

Being more general, we can change the Riemann-liouville differintegral to work on different functions by changing the limits of integration in the integral expression. We can solve the following continuum sum using different limits:







Therefore:






I'm wondering how to apply this to tetration or hyper operators. This performs a fair amount of mathematical work and solves a nice iteration problem--maybe its related to hyper operators *fingers crossed*
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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 JmsNxn - 08/11/2013, 04:05 PM

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