For $\alpha\in (0,1)$ the derivative of order $\alpha$ of $f(x)$ is defined to be (see Section I.5.5 in the first volume of the book *Generalized Functions* by Gelfand and Shilov)

$$\frac{d^\alpha}{dx^\alpha} f(x):=\frac{1}{\Gamma(1-\alpha)} \int_0^xf'(\xi)(x-\xi)^{-\alpha} d\xi. $$

One can define derivatives of arbitrary orders, but their definition is a bit more involved. Suffices to say that for $\alpha \in(0,1)$ and $n$ a positive integer one has

$$\frac{d^{n\alpha}}{d x^{n\alpha}}= \underbrace{\frac{d^\alpha}{dx^\alpha}\circ \cdots \circ \frac{d^\alpha}{dx^\alpha}}_n. $$

For $\alpha=\frac{1}{2}$ and $f(x)=x$ the above definition yields

$$ \frac{d^\alpha}{dx^\alpha} f(x)=\frac{1}{\Gamma(1/2)} \int_0^x(x-\xi)^{-1/2} d\xi $$

($\xi=tx$)

$$= \frac{x^{1/2}}{\Gamma(1/2)}\int_0^1 (1-t)^{-1/2} dt =\sqrt{\frac{t}{\pi}} \int_0^1 t^{1-1}(1-t)^{1/2-1} dt $$

$$= \sqrt{\frac{x}{\pi}}\cdot \frac{\Gamma(1)\Gamma(1/2)}{\Gamma(3/2)}=\frac{1}{2} \sqrt{\frac{x}{\pi}}. $$
In this case, using the formula you suggested where $n=1/2$, we have

$$\sum_{i=0}^\infty (-1)^i\binom{1/2}{i}(x-ih)= x\sum_{i=0}^\infty (-1)^i\binom{1/2}{i}-h\sum_{i=0}^\infty (-1)^ii\binom{1/2}{i}.$$

We have a Taylor expansion

$$\sqrt{1-t}=\sum_{i=0}^\infty(-1)^i\binom{1/2}{i} t^i,\;\;|t|<1. $$

Observe that

$$ (-1)^i\binom{1/2}{i} <0, \;\;\forall i>0. $$

One can prove that the above series converges for $t=1$ (this is tricky and relies on the rarely used Gauss' criterion of convergence) and

$$ \sum_{i=0}^\infty(-1)^i\binom{1/2}{i} =0. $$

Thus the $1/2$-th derivative of $f(x)$, according to your definition does not depend on $x$.

$$\sum_{i=0}^\infty(-1)^ii\binom{1/2}{i} t^i=t\frac{d}{dt}\sqrt{1-t}=-\frac{t}{2\sqrt{1-t}}, \; t\in (0,1).$$

If in the above equality we formally set $t=1$ to deduce see that the series $\sum_{i=0}^\infty (-1)^ii\binom{1/2}{i}$ sums-up to $-\infty$. One can verify directly using Gauss' criterion of convergence that the series is indeed divergent. To conclude, I think that there is a problem with the definition you suggested.

polynomialof degree $k$, $x(x-1)\cdots (x-k+1)/k!$ whereas the definition of Grunwald derivative uses the so called Euler $B$-function. These two functions agree for $x$ positive integer. That is why I have detected a problem with your definition. $\endgroup$ – Liviu Nicolaescu Oct 15 '15 at 10:31