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Powerful way to perform continuum sum
#5
(08/11/2013, 07:31 PM)tommy1729 Wrote: This is probably one of your best posts imho.
Although I havent checked all. It seems a bit weird , your integrals seem to lack a variable by being very selfreferential. This might not be a problem but it might be restrictive ... such as L-series or exponential sums ... maybe it can be solved by adding variables.
Thanks for the comment. I feel like this is finally a pay off after a good year of work in fractional calculus. It definitely feels like the first nice result. I've certainly come a long way since I first joined this forum. Nice 3 years of toning my mathematical muscles ^_^

It is quite restricted but if you fiddle with the lower limit you can mix up the types of functions that appear. Any fractional integral works so long as it satisfies: and works better if it has an inverse over s at any t.

Quote:A problem might be that the n-th derivative is not just the formula of the nth integral with a minus sign added. I assume you knew that already from reading/lectures/exercises about fractional calculus.

Exactly! I've found that it works for exponentials and the closure of exponential functions, so double exponentials. We can take any exponential and infinite combinations of them.

The types of functions this works very beautifully on are functions I call distributed. where f is distributed iff



These functions have lots of beautiful properties, insofar as their derivatives and integrals are very easy to calculate. And if is distributed then so is and and .

I have been working on conditions for a function to be distributed and I've found a few. is distributed (note the minus sign)



We get that functions are distributed if

This is because I have shown that if converges it equals f.

I'm a little stuck on my proof, all I have to do is show that this operator (continuum taylor transformation) is continuous and the theorem is perfect. Once I have this I'll work on trying to prove the continuity of the continuum sum transformation and other little nitpickys.


Also, if a function is distributed, by the fourier inversion theorem, we get:



Which I'm sure you can see how I got. This is my own representation for the iterated derivative. It has the property of being non convergent for integration values and convergent for derivative values. The exact opposite of the mellin transform expression.
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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, 08:18 PM

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