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Another question!
#1
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?
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#2
(08/22/2013, 05:54 PM)JmsNxn Wrote:



Integrate by parts, and for we get the spectacular identity that:


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
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#3
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:


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#4
(08/22/2013, 05:54 PM)JmsNxn Wrote: 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?

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.
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#5
OH! That makes a lot of sense. That's very interesting.
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