So, I'm very new to Fractional Calculus, and I don't know too much about series, but I'm very interested in Fractional Calculus, so I wanted to know if you could find the solution to this problem.
Using the general formula for derivatives:$$\frac{d^n}{dx^n}f(x)=\lim_{\epsilon \to 0} \frac{\Gamma(n+1)}{\epsilon^n}\sum_{j=0}^{\lfloor\frac{x}{\epsilon}\rfloor}\frac{f(x-j\epsilon)(-1)^j}{j!\Gamma(n+1-j)}$$ Where $\Gamma(x)$ is the gamma function. The upper limit to the summation (because it's a little hard to see) is $\lfloor{\frac{x}{\epsilon}\rfloor}$, which is the floor function. I want to find:$$\frac{d^{\frac{1}{3}}}{dx^{\frac{1}{3}}}(x)$$
The place where I get stuck is interpreting the upper limit of the summation. Am I supposed to treat it as an infinite series since it has a variable?
The place I got this formula is: Link w/ Formula
Generally, one extends the formula
$$\frac{d^n}{dx^n} x^m = m (m-1) \cdots (m-n+1) x^{m-n} = \frac{m!}{(m-n)!}x^{m-n}$$
to
$$\frac{d^n}{dx^n} x^m = \frac{\Gamma(m+1)}{\Gamma(m-n+1)} x^{m-n}$$
Thus,
$$\frac{d^{1/3}}{dx^{1/3}} x= \frac{\Gamma(2)}{\Gamma(5/3)} x^{2/3} = \frac{1}{\Gamma(5/3)} x^{2/3}$$