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Does the Mellin transform have a half-iterate ?
After all these half-iterates of functions , integrals and derivatives, I wonder ( again ) about half-iterates of integral transforms.
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tommy1729
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Well I posted a reply to this but it deleted it :/
I have looked at this for so long Tommy. It's very closely related to tetration. I'll tell you how I can do it for some functions.
\( M(f) = \int_0^\infty f(x)x^{s-1}\,dx \)
define
\( \vartheta(w) = \sum_{n=0}^\infty M^n(f)(s) \frac{w^n}{n!} \)
Then if
\( \phi(z) = [\frac{d^z}{dw^z} \vartheta(w)]_{w=0} \)
then
\( \phi(z) = M^z (f)(s) \)
However we have to show lots of conditions on convergence and what not.
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I notice that Im not so comfortable with many iteration of the Mellin transform.
Convergeance issues seem to pop up out of nowhere.
For instance if we start with f(x) = exp(x) or f(x) = exp(-x) we get in trouble before we reach the 3rd iteration of the mellin transform.
And if we pick an elementary function between exp(x) and exp(-x), I seem to have trouble finding a closed form for every n th mellin transform.
Maybe I need to consider using hypergeometrics ?
What are standard tables of n th mellin transforms (for s,x > 0) , if that has even been made yet ?
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tommy1729
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Also of intrest to me are integral transforms that are cyclic :
Like for almost all f,s and some transform M :
M^[3](f)(s) = f(s) but M^[1](f)(s) =/= f(s) and M^[2](f)(s) =/= f(s).
I assume this is consistant with Schwartz kernel theorem.
As Kernel I assume abelian functions could be used. And maybe some others too ?
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tommy1729
Posts: 1,924
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Ok this has given me an idea for tetration ...
The James-tommy method is in progress
regards
tommy1729