$\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}}$, what does it mean?

Question

I would like to know what these fractional exponents means in a derivative $\frac{d}{dx}$ operator. Like, I've seen $\frac{d^2}{dx^2}$ but I don't know what $\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}}$ means.

Answer 1

$\frac{d^\frac{1}{2}}{dx^\frac{1}{2}}$ is the operator such that $\frac{d^\frac{1}{2}}{dx^\frac{1}{2}}$$\frac{d^\frac{1}{2}}{dx^\frac{1}{2}}$=$\frac{d}{dx}$, ie. $\frac{d^\frac{1}{2}}{dx^\frac{1}{2}}$$\frac{d^\frac{1}{2}}{dx^\frac{1}{2}}f$=$\frac{d}{dx}f$ for differentiable functions $f$, much like how $\frac{d^2}{dx^2}=\frac{d}{dx}\frac{d}{dx}$. It is basically the square root of $\frac{d}{dx}$. For example, $\frac{d^\frac{1}{2}}{dx^\frac{1}{2}}e^{ax}=\sqrt{a}e^{ax}$, since then $\frac{d^\frac{1}{2}}{dx^\frac{1}{2}}\sqrt{a}e^{ax}=ae^{ax}=\frac{d}{dx}e^{ax}$.

Answer 2

What you have is what is usually referred to as a semiderivative. I would say that this notation is a bit incomplete, since one also has to account for a lower limit, in the case of differintegrals whose orders are not nonnegative integers.

In particular, letting ${}_a D_x^{1/2}f(x)$ be the semiderivative, one possible definition is the Riemann-Liouville form,

$${}_a D_x^{1/2}f(x)=\frac{f(a)}{\sqrt{\pi(x-a)}}+\frac1{\sqrt\pi}\int_a^x \frac{f^\prime(t)}{\sqrt{x-t}}\mathrm dt$$

The semiderivative operator can be thought of this way: this is the operator that, when applied to a function twice, gives the ordinary (first order) derivative.

If you want to learn more about this, I would suggest picking up Oldham/Spanier's The Fractional Calculus.