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:

\( \mathcal{Z}^{-1} f(s) = \int_{\sigma - i\infty}^{\sigma + i\infty }e^{-\pi i t}\Gamma(t) (f(t) - f(t-1))s^{-t}dt \)

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

\( \mathcal{Z}^{-1} f(s) = \sum_{n=0}^{\infty} (f(-n) - f(-n-1))\frac{s^n}{n!} = \int_{-\infty}^{\infty} (f(-y) -f(-y-1))\frac{s^y}{y!}dy \)

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:

\( \frac{d^{-s}}{dt^{-s}_0} f(t) = \frac{1}{(s-1)!}\int_0^t f(u)(t-u)^{s-1}du \)

\( Rf(s) = \frac{d^{-s}}{dt^{-s}_0}f(t) |_{t=1} \)

\( R t^n = \frac{n!}{(s+n)!} \)

Therefore:

\( \phi(s) = \int_0^\infty e^{-t} \frac{d^{-s}}{dt^{-s}_0}(t+1)^ndt \)

\( \phi(s) = \frac{n!}{(s+n)!} + \phi(s-1) \)


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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