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Another question! - JmsNxn - 08/22/2013
Let's take some function and some fractional differentiation method such that Now create the function: Integrate by parts, and for we get the spectacular identity that: What is going on here? RE: Another question! - tommy1729 - 08/22/2013
(08/22/2013, 05:54 PM)JmsNxn Wrote: I do not even need to use integrate by parts to see a problem. Its funny you say 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 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 " if you take the derivative times. This is similar to . Hence by the very definition of the gamma function you also get 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 RE: Another question! - JmsNxn - 08/25/2013
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: RE: Another question! - mike3 - 08/25/2013
(08/22/2013, 05:54 PM)JmsNxn Wrote: Let's take some function and some fractional differentiation method such that I'm not sure why this is necessarily bizarre. The functional equation you mention has infinitely many solutions. In general, is a solution of for any 1-cyclic function . If you take and use the Riemann-Liouville with lower bound , then and you recover the gamma function. I bet if you use another , you'll just get for some 1-cyclic function which is not just equal to 1. RE: Another question! - JmsNxn - 08/27/2013
OH! That makes a lot of sense. That's very interesting. |