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 Another question! JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 08/22/2013, 05:54 PM Let's take some function $f(t)$ and some fractional differentiation method $\frac{d^s}{dt^s}$ such that $\frac{d^s f}{dt^s}(-t) < e^{-t}$ Now create the function: $\phi(s) = \int_0^\infty t^{s-1} \frac{d^s f}{dt^s}(-t) dt$ Integrate by parts, and for $\Re(s) > 0$ we get the spectacular identity that: $s \phi(s) = \phi(s+1)$ What is going on here? tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 08/22/2013, 10:46 PM (08/22/2013, 05:54 PM)JmsNxn Wrote: $\frac{d^s f}{dt^s}(-t) < e^{-t}$ $\phi(s) = \int_0^\infty t^{s-1} \frac{d^s f}{dt^s}(-t) dt$ Integrate by parts, and for $\Re(s) > 0$ we get the spectacular identity that: $s \phi(s) = \phi(s+1)$ I do not even need to use integrate by parts to see a problem. Its funny you say $\frac{d^s f}{dt^s}(-t) < e^{-t}$ because its more like an equality when we differentiate a given amount of times with respect to t. You see : s is considered a constant with respect to t since s is not a function OF t NOR f. There is big difference between a function , an operator , a variable and a constant. ALthough that may sound belittling or trivial , your example shows this is an important concept !! If you consider $\frac{d^s f}{dt^s}(-t)$ as a function F(s,f) then it is no surprise that taking the derivative with respect to f leaves s unchanged. By the chain rule you then get the " wrong " / " correct " $\frac{d^s f}{dt^s}(-t) (-1)^{-M}$ if you take the derivative $M$ times. This is similar to $D^m [ s e^{-t}] dt$. Hence by the very definition of the gamma function you also get $s \phi(s) = \phi(s+1)$ here which you already showed yourself with the - overkill - method integrate by parts. This might not answer all your questions yet but I assume it helps. It not completely formal either sorry. It might affect your other posts about integral representations for fractional calculus , tetration and continuum sum. Im still optimistic though and hope I did not discourage you to much. regards tommy1729 JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 08/25/2013, 03:00 PM I just read that over today and it makes a lot more sense a second time through. I've been finding a lot of interesting paradoxes with fractional calculus and it must be my lack of rigor. This one and the function which if converges is its own derivative: i.e: $\phi(s) = \int_{-\infty}^{\infty} cos(2\pi t) \frac{d^t}{ds^t}f(s) \, dt$ mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 08/25/2013, 10:21 PM (This post was last modified: 08/25/2013, 10:35 PM by mike3.) (08/22/2013, 05:54 PM)JmsNxn Wrote: Let's take some function $f(t)$ and some fractional differentiation method $\frac{d^s}{dt^s}$ such that $\frac{d^s f}{dt^s}(-t) < e^{-t}$ Now create the function: $\phi(s) = \int_0^\infty t^{s-1} \frac{d^s f}{dt^s}(-t) dt$ Integrate by parts, and for $\Re(s) > 0$ we get the spectacular identity that: $s \phi(s) = \phi(s+1)$ What is going on here? I'm not sure why this is necessarily bizarre. The functional equation you mention has infinitely many solutions. In general, $\phi(x) = \Gamma(x) \theta(x)$ is a solution of $x \phi(x) = \phi(x+1)$ for any 1-cyclic function $\theta(x)$. If you take $f(t) = e^t$ and use the Riemann-Liouville with lower bound $-\infty$, then $\frac{d^s f}{dt^s} = e^t$ and you recover the gamma function. I bet if you use another $f$, you'll just get $\Gamma(x) \theta(x)$ for some 1-cyclic function $\theta(x)$ which is not just equal to 1. JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 08/27/2013, 06:57 PM OH! That makes a lot of sense. That's very interesting. « Next Oldest | Next Newest »

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