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I have a question regarding the definition of a fractional derivative. I've searched, but I can't find a definition of fractional derivatives that explain the concept in terms of an operator on some function. I guess what I'm trying to ask is the following:

Let f(z) be an analytic function. Let $_0D_z^v(f(z))$ denote the $v^{th}$ fractional derivative of $f$ in the Riemann-Liouville sense. Let $D_z^v(f(z))$ denote the $v^{th}$ fractional derivative of $f$ in the Weyl sense. Assume $f(z)$ can be written as $f(z)=g(z)+h(z)$, where $g$ and $h$ are analytic (NOTE: one of $g$ or $h$ could be the zero function). Let $D^v$ be the operator defined by $D^v(f(z))= _0D_z^v(g(z)) + D_z^v(h(z))$, whenever this makes sense i.e. when each part is defined.

Could one say that $D^v(f(z))$ here is a fractional derivative in the sense that the operator satisfies:

  1. This operator is linear
  2. The operator preserves analytic properties, i.e. the result of applying the operator is again analytic.
  3. When $v$ is a positive integer, the operator corresponds to the $n^{th}$ derivative of $f$ with respect to $z$. Furthermore, when $n$ is a negative integer the operator corresponds to the $n$ fold integral of $f$.
  4. When $v=0$ applying $D^v$ leaves $f(z)$ unchanged.
  5. $D^{v}D^r(f(z))= D^{v+r}(f(z))$.

I believe at first glance (unless I'm making a mistake) that each property described above is satisfied by virtue that the Riemann-Liouville and Weyl fractional derivatives satisfy these properties as well.

I guess the above definition could give two different values (by switching the order of $g$ and $h$) but there are multiple ways of defining a fractional derivative as well. I don't think this is necessarily a problem. Could be wrong though.

I hope this question makes sense. Let me know if I need to be clearer about something.

Thank You for Your Help,

Rick

P.S. I'm not sure that "operator" is the correct wording, but hopefully you understand what I mean by the above definition anyway.


Edit:

$D^v$ is not well-defined. Since I'm having an issue with a specific function, let $g(z)=1$ and $h(z) = f(z)-1$.

To be more specific, I'm trying to take the fractional derivative of:

$f(x) = 1 + n^{-x}$ where $n$ is an integer and $n\geq 2$.

I need the fractional derivative of $f(x)$ defined above when $0<v<1$.

In a perfect world, I'd really like for the derivative of $1$ to SLOWLY go to $0$ as $v$ gets closer to 1 and the derivative of the second piece to go to $\log(n)n^{-x}$ as $v$ gets close to 1. Just to suit my specific needs. But, I don't know if I can.

I really like the Riemann-Liouville fractional derivative of the constant term and the Weyl fractional derivative of the second piece, but it seems like I can't take the Weyl fractional derivative of $1$ and the Riemann-Liouville fractional derivative of $n^{-x}$ is "nasty" for my specific goal/desire.

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    $\begingroup$ It looks like $D^v$ is not well-defined: given $f$, how do you determing $g$ and $h$? $\endgroup$
    – Ben Barber
    Feb 25 '14 at 10:10
  • $\begingroup$ Ben, thanks for the response. See the edited question. $\endgroup$
    – Rick Farr
    Feb 26 '14 at 16:33

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