Is it possible to "calculate" / simplify this expression? If it is, how can it be done?
$$ \frac{d^\beta}{d\alpha^\beta}\frac{d^\alpha}{dx^\alpha}\sin(x) $$
for
$ \alpha,\beta\in\mathbb{R}_{\ge0} $
I think it is equal to $\left(\frac{\pi}{2}\right)^\beta \sin\left(x+\frac{\pi}{2}\left(\alpha+\beta\right)\right)$, but i am absolutely not sure if this is true.
Thank you,
Best regards
Kevin
-edit-
Here is the desription how i got my result (which maybe helps to verify it):
https://en.wikipedia.org/wiki/Differintegral#A_selection_of_basic_formul.C3.A6 $$\Rightarrow \frac{d^\alpha}{dx^\alpha}\sin(x)=\sin(x+\alpha\frac{\pi}{2})$$
Because $\frac{d^n}{dx^n}f(ax)=a^nf^{\left(n\right)}(ax), n\in \mathbb{N}$ i think $\frac{d^\alpha}{dx^\alpha}f(ax)=a^\alpha f^{\left(\alpha\right)}(ax), \alpha\in\mathbb{R}_{\ge 0}$ is also true. Unfortunately i'm not sure if this is true.
Because $\frac{d^n}{dx^n}f(x+a)=f^{\left(n\right)}(x+a), n\in\mathbb{N}$ is true, i also hope that $\frac{d^\alpha}{dx^\alpha}f(x+a)=f^{\left(\alpha\right)}(x+a), \alpha\in\mathbb{R}_{\ge 0}$ is true.
With these new rules the expression can be calculated: $$ \frac{d^\beta}{d\alpha^\beta}\frac{d^\alpha}{dx^\alpha}\sin(x) $$ $$ =\frac{d^\beta}{d\alpha^\beta}\sin(x+\alpha\frac{\pi}{2}) $$ $$ =\frac{d^\beta}{d\alpha^\beta}\sin(\frac{\pi}{2}\left(\alpha+\frac{2}{\pi}x\right)) $$ This is just a shifted $\sin(\frac{\pi}{2}\gamma)$. $$ \gamma=\alpha+\frac{2}{\pi}x $$ $$ \Rightarrow\frac{d^\beta}{d\alpha^\beta}\sin(\frac{\pi}{2}\gamma) $$ $$ =\left(\frac{\pi}{2}\right)^{\beta}\sin^{\left(\beta\right)}(\frac{\pi}{2}\gamma) $$ $$ =\left(\frac{\pi}{2}\right)^{\beta}\sin(\frac{\pi}{2}\gamma+\frac{\pi}{2}\beta) $$ $$ =\left(\frac{\pi}{2}\right)^{\beta}\sin(\frac{\pi}{2}\left(\alpha+\frac{2}{\pi}x\right)+\frac{\pi}{2}\beta) $$ $$ =\left(\frac{\pi}{2}\right)^\beta \sin\left(x+\frac{\pi}{2}\left(\alpha+\beta\right)\right) $$
For shorthand, $\frac{d^n}{dx^n}=D^n_x$
Consider $D^n_xe^{cx}=c^ne^{cx}$. Seems true through induction.
Now, have $c=i$, and we can use Euler's formula.
$$e^{ix}=\cos(x)+i\sin(x)$$
$$D^n_xe^{ix}=i^ne^{ix}=e^{i(x+\frac{n\pi}{2})}=\cos(x+\frac{n\pi}2)+i\sin(x+\frac{n\pi}2)$$
Of course, you could have some doubt in this, since this not directly imply the fractional derivative of the trig functions, but we can work our way there.
More directly, consider $\sin(x)=\frac{e^{ix}-e^{ix}}{2i}$ and differentiate like so
$$D^n_x\sin(x)=D^n_x\frac{e^{ix}-e^{-ix}}{2i}=\frac{i^ne^{ix}-(-i)^ne^{-ix}}{2i}$$
$$=\frac{e^{i(x+\frac{n\pi}2)}-e^{-i(x+\frac{n\pi}2)}}{2i}$$
$$=\sin(x+\frac{n\pi}2)$$
I would consider that the easiest way to showing it, and then the rest you've done right.