Double half-derivative of a constant Just as the title says, how does performing the half derivative of a constant twice work out? I haven't looked at the Riemann-Liouville method yet, but in terms of using the generalized power rule.
$D^\alpha (x^n)=\frac{\Gamma(n+1)}{\Gamma(n-\alpha+1)}x^{n-\alpha}$
and when applying it to 1 ($x^0$)
$D^\alpha (1)=\frac{1}{\sqrt{\pi x}}$
but applying that same power rule again gives
$D^\alpha (\frac{1}{\sqrt{\pi x}})=\left( \frac{\Gamma (\frac{1}{2})}{\Gamma (1)} \right) \frac{1}{\sqrt{\pi}} x^{-\frac{1}{2}-\frac{1}{2}}=\frac{1}{x}$
This could probably be a stupid error on my part, but I wanted to make sure. I'm thinking that maybe this is more of a limitation of how far this generalized derivative can differentiate. Again, I haven't looked at many other ways to differentiate (and the rule for $e^{cx}$ doesn't help much lol).
 A: That formula is a generalisation of the power rule, and not valid for $n<0$. Try the full integral definition for the RL derivative:
$$_a^{RL}D_t^n = \frac{d^p}{dt^p}\left(\frac{1}{\Gamma(m)}\int^t_a(t-x)^{m-1}f(x)dx\right)$$
Where $n$ is the fractional order of differentiation ($n\in\mathbb{R}_+$) and $f(x)$ is our integrand;
$p\in\mathbb{N}:={p=⌈n⌉}$,  $m\in\mathbb{R}:=m=1-n$ Function outputted in terms of t, $a$ and $t$ are our integration bounds.
You will find with these defined methods of finding fractional derivatives that differentiation order sums, and the "double half-derivative" will be zero for a constant.
The RL half derivative of a constant $c$ is therefore:
$$_0^{RL}D_t^{0.5}c = \frac{d}{dt}\left(\frac{1}{\Gamma(0.5)}\int^t_0(t-x)^{0.5-1}(c)dx\right)$$
$$= \frac{d}{dt}\left(\frac{c}{\sqrt\pi}\int^t_0(t-x)^{-0.5}dx\right)$$
$$= \frac{d}{dt}\left(\frac{c}{\sqrt\pi}\left[\frac{2x-2t}{\sqrt{t-x}}\right]^t_0\right)$$
$$= \frac{d}{dt}\left(\frac{2ct}{\sqrt{\pi{t}}}\right)=\frac{c}{\sqrt{\pi{t}}}$$
As you identified with the generalised power rule method, which worked as the differentiation order was 0. You can do this computation again on the answer, this time with the valid RL method:
$$_0^{RL}D_t^{0.5}\frac{c}{\sqrt{\pi{x}}} = \frac{d}{dt}\left(\frac{1}{\Gamma(0.5)}\int^t_0(t-x)^{0.5-1}(\frac{c}{\sqrt{\pi{t}}})dx\right)$$
$$=\frac{d}{dt}\left(\frac{c}{\sqrt{\pi}}\int^t_0\frac{1}{\sqrt{{\pi}xt-x^2{\pi}}}dx \right)$$
$$=\frac{d}{dt}\left(\frac{c}{\sqrt{\pi}}\left[\frac{2\left(\sqrt{tx(1-\frac{x}{t})}\right)arcsin\left(\frac{\sqrt{x}}{\sqrt{t}}\right)}{\sqrt{{\pi}tx-{\pi}x^2}}\right]^t_0 \right)$$
Which looks like a horrible equation, but all goes to zero when we input the limits due to the numerator.
Additionally, be careful with different definitions of the fractional differintegral, as many of them interact differently. For example, if you choose to look into the Caputo definition for the fractional differintegral, you will find any fractional derivative of a constant will be zero; with the Riemann-Liouville definition there will be a value.
