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 Powerful way to perform continuum sum JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 08/10/2013, 09:06 PM (This post was last modified: 08/11/2013, 04:30 PM by JmsNxn.) Hello everyone. I've recently come across a way to perform continuum sums and I was wondering if anyone has any suggestions on how to formalize this or if they have any nice comments about how it works. We start by defining an operator: $\mathcal{J} f(s) = \frac{1}{\Gamma(s)}\int_0^\infty f(-X)X^{s-1}dX$ Thanks to Riemann and Liouville, if $\mathcal{J} f$ and $\mathcal{J} \frac{df}{ds}$ converge then: $\mathcal{J} \frac{df}{ds} = (\mathcal{J} f)(s-1)$ And, neatly $(\mathcal{J}\frac{df}{ds})(1) = f(0)$ therefore, by induction: $(\mathcal{J} f)(N) = \frac{d^{-N}}{ds^{-N}}f(s)|_{s=0}$ It is noted that this operator can be inverted using Mellin inversion and Taylor series for certain analytic functions, and for some continuum Taylor transformation. So it makes sense to talk about $\mathcal{J}^{-1}$ However, we must use Riemann-Liouville differintegration. Define: $\frac{d^{-s}f(t)}{dt^{-s}} = \frac{1}{\Gamma(s)}\int_{-\infty}^{t}f(u)(t-u)^{s-1}du$ And we find when $t=0$ the operator reduces to $\mathcal{J}$ We then define our continuum sum operator which works beautifully: $\mathcal{Z}f(s) = \int_0^{\infty} e^{-t}\frac{d^{-s}f(t)}{dt^{-s}}dt$ $\mathcal{Z}f(s) = [-e^{-t}\frac{d^{-s}}{dt^{-s}}f(t)]_{t=0}^{\infty} + \int_0^{\infty} e^{-t}\frac{d^{1-s}f(t)}{dt^{1-s}}dt$ Performing integration by parts, which is easy, and using the fact that $\frac{d}{dt}\frac{d^{-s}f(t)}{dt^{-s}} = \frac{d^{1-s}f(t)}{dt^{1-s}}$ We get: $\mathcal{Z}f(s) = \mathcal{J}f(s) + (\mathcal{Z}f)(s-1)$ And if we take $f = \mathcal{J}^{-1}g$ and say: $\phi(s) = \mathcal{Z}\mathcal{J}^{-1} g(s)$ $\phi(s) = g(s) + \phi(s-1)$ Which is the glory of the continuum sum! The real problem now is finding what kinds of functions g(s) does this work on $\mathcal{J}^{-1}g(s) = \int_{\sigma + -i\infty}^{\sigma + i\infty}\Gamma(t)g(t) e^{- \pi i t}s^{-t} dt$ with $\sigma$ chosen appropriately for g(s). We also have: $\mathcal{J}^{-1}g(s) = \sum_{n=0}^{\infty} g(-n)\frac{s^n}{n!}$ for some functions, this is remembering because $g(-n) = \frac{d^n}{dt^n}f(t)|_{t=0}$ for some functions. For example we can find a continuum sum, $\lambda \in \mathbb{R}$ $\mathcal{J}e^{\lambda s} = \lambda^{-s}$ therefore for $|\lambda| < 1$ $\Re \lambda < 1$: $\phi(s) = \int_0^{\infty} e^{-t} \frac{d^{-s}e^{\lambda t}}{dt^{-s}}dt$ $\phi(s) = \lambda^{-s} \int_0^{\infty}e^{(\lambda-1)t}dt = \frac{\lambda^{-s}}{1-\lambda} = \sum_{n=0}^{\infty} \lambda^{n-s}$ which satisfies the continuum sum rule. We can do the same trick for $e^{-s^2}$ we know that $\mathcal{J} e^{-s^2} = \frac{ \Gamma(s/2)}{\Gamma(s)}$ Therefore the continuum sum, or function that satisfies: $\phi(s) = \frac{\Gamma(s/2)}{\Gamma(s)} + \phi(s-1)$ Is: $\phi(s) = \frac{1}{\Gamma(s)}\int_0^\infty e^{-t} \int_{-\infty}^t e^{-u^2}(t-u)^{s-1}dudt$ I thought I would just write out some examples that I have worked out. $\phi(s) = \sin(\frac{\pi s}{2}) + \phi(s-1)$ $\phi(s) = \int_{0}^{\infty} e^{-t}\sin(t - \frac{\pi s}{2})dt$ Same for cosine. $\psi(s) = \int_{0}^{\infty} e^{-t}\cos(t - \frac{\pi s}{2})dt$ And miraculously: $\frac{d\phi}{ds} = -\frac{\pi}{2}\psi(s)$ Whats truly amazing is that this is a linear operator. Therefore we have an operator $\mathcal{S}f(s) = \mathcal{Z}\mathcal{J}^{-1} f(s)$ which sends functions from themselves to their continuum sum. What's truly even more amazing is that I have found an inverse expression for $\mathcal{Z}$ This was an extra trick. « 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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