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 Powerful way to perform continuum sum JmsNxn Long Time Fellow Posts: 339 Threads: 73 Joined: Dec 2010 08/12/2013, 07:17 PM (This post was last modified: 08/12/2013, 07:54 PM by JmsNxn.) 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. « Next Oldest | Next Newest »

 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 tommy1729 - 08/11/2013, 02:00 AM RE: Powerful way to perform continuum sum - by JmsNxn - 08/11/2013, 04:05 PM RE: Powerful way to perform continuum sum - by tommy1729 - 08/11/2013, 07:31 PM RE: Powerful way to perform continuum sum - by JmsNxn - 08/11/2013, 08:18 PM RE: Powerful way to perform continuum sum - by tommy1729 - 08/11/2013, 10:29 PM RE: Powerful way to perform continuum sum - by tommy1729 - 08/11/2013, 11:05 PM RE: Powerful way to perform continuum sum - by JmsNxn - 08/12/2013, 07:17 PM

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