(05/06/2014, 12:11 AM)JmsNxn Wrote: If I understand your question correctly the answer is yes to both questions. I'll write it out. Using the operators from my paper we can completely represent the iterated difference:
\( F(z) =[\frac{d^{-z}}{dx^{-z}}e^x \beta(-x)]_{x=0} \)
then:
\( \bigtriangledown^s F(z) = [\frac{d^{-z}}{dx^{-z}}e^x \frac{d^s}{d(-x)^s}\beta(-x)]_{x=0} \)
I'm working on writing this all up. So far all I have is a bunch of notes and papers compiled together unorganized.
And the fractional derivatives can be written using the formula for the Weyl differintegral, thus an integral transform. So how do you get \( \beta(x) \) from \( F(z) \)?
(05/06/2014, 12:11 AM)JmsNxn Wrote: Now for the first question, to work on these periodic functions define:
\( [\frac{d^z}{dw^z} f(w)]_{w=0} = \frac{1}{\G(-z)} (\sum_{n=0} f^{(n)}(0) \frac{(-1)^n}n!(n-z)} + \int_1^\infty f(-x)x^{-z-1}\,dx) \)
And then we can generate the differintegral using taylor series. now if \( p(w) = \sum_{n=1} a_n e^{inw} \)
then
\( \frac{d^{-z}}{dw^{-z}} p(w) = \frac{i^{-z}}{\G(z)} \int_0^\infty p(w + ix)x^{z-1}\,dx \)
But I was wondering if it was possible to work in the other direction, starting with the definition for periodic functions and then expanding it to the integral-transform definition, and so if something similar could be done for the forward difference operator.