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+--- Thread: Another question! (/showthread.php?tid=815)
Another question! - JmsNxn - 08/22/2013
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?
RE: Another question! - tommy1729 - 08/22/2013
(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
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:
\( \phi(s) = \int_{-\infty}^{\infty} cos(2\pi t) \frac{d^t}{ds^t}f(s) \, dt \)
RE: Another question! - mike3 - 08/25/2013
(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.
RE: Another question! - JmsNxn - 08/27/2013
OH! That makes a lot of sense. That's very interesting.