I'm a little confused about what you're doing but I understand your arguments about sexp as a continuum product.
We have A beautiful result I would like to show:
We start with the two identities that are already proven by others working on fractional calculus.
\( \mathcal{J} f = \frac{d^{-s}f(t)}{dt^{-s}} |_{t=0} \)
\( \mathcal{J}( f \cdot g )= \mathcal{J}f * \mathcal{J}g= \sum_{n=0}^{\infty} \frac{\Gamma(1-s)}{\Gamma(n + s+1) n!}(\mathcal{J} f)(n) (\mathcal{J} g)(-s-n) \)
And even more generally:
\( \frac{d^{-s}f(t)g(t)}{dt^{-s}} = \frac{d^{-s}f(t)}{dt^{-s}} * \frac{d^{-s}g(t)}{dt^{-s}} \)
whree the convolution is done over s, and the values at t are the same for both f and g.
Therefore: if \( \mathcal{Z} f = \phi \) and \( \mathcal{Z} g = \psi \)
\( \mathcal{Z} f \cdot g = \int_0^\infty e^{-t} \frac{d^{-s}}{dt^{-s}} f(t) g(t) dt= \phi * \psi \)
That means even more remarkably
\( \mathcal{Z} \mathcal{J}^{-1} (f * g) = \mathcal{Z} ((\mathcal{J}^{-1} f) \cdot (\mathcal{J}^{-1} g)) = (\mathcal{Z} \mathcal{J}^{-1} f) * (\mathcal{Z}\mathcal{J^{-1}}g) \)
That means \( S( f * g) = (Sf) * (Sg) \)
This has so much value for continuum sums. This is remarkable!
I have to properly justify this using the continuity of these operators over some hilbert space. That's the only way I can think of.
I would also like to standardize a notation that is very intuitive. If we take the continuum sum over the interval [a,b] we say:
\( \sum_a^b f(y) \, \sigma y = S f(b) - Sf (a) \)
This has all the linearity rules of the integral, and some own unique rules of its own.
We have A beautiful result I would like to show:
We start with the two identities that are already proven by others working on fractional calculus.
\( \mathcal{J} f = \frac{d^{-s}f(t)}{dt^{-s}} |_{t=0} \)
\( \mathcal{J}( f \cdot g )= \mathcal{J}f * \mathcal{J}g= \sum_{n=0}^{\infty} \frac{\Gamma(1-s)}{\Gamma(n + s+1) n!}(\mathcal{J} f)(n) (\mathcal{J} g)(-s-n) \)
And even more generally:
\( \frac{d^{-s}f(t)g(t)}{dt^{-s}} = \frac{d^{-s}f(t)}{dt^{-s}} * \frac{d^{-s}g(t)}{dt^{-s}} \)
whree the convolution is done over s, and the values at t are the same for both f and g.
Therefore: if \( \mathcal{Z} f = \phi \) and \( \mathcal{Z} g = \psi \)
\( \mathcal{Z} f \cdot g = \int_0^\infty e^{-t} \frac{d^{-s}}{dt^{-s}} f(t) g(t) dt= \phi * \psi \)
That means even more remarkably
\( \mathcal{Z} \mathcal{J}^{-1} (f * g) = \mathcal{Z} ((\mathcal{J}^{-1} f) \cdot (\mathcal{J}^{-1} g)) = (\mathcal{Z} \mathcal{J}^{-1} f) * (\mathcal{Z}\mathcal{J^{-1}}g) \)
That means \( S( f * g) = (Sf) * (Sg) \)
This has so much value for continuum sums. This is remarkable!
I have to properly justify this using the continuity of these operators over some hilbert space. That's the only way I can think of.
I would also like to standardize a notation that is very intuitive. If we take the continuum sum over the interval [a,b] we say:
\( \sum_a^b f(y) \, \sigma y = S f(b) - Sf (a) \)
This has all the linearity rules of the integral, and some own unique rules of its own.