A definition of the fractional derivative I was investigating the idea of fractional derivatives and devised the following definition. WHich definition is it equivalent to and can I have a reference for it?
$$\frac{d^n}{dx^n}f(x) = \lim_{h \to 0} \frac{\sum_{i = 0}^\infty (-1)^i\binom{n}{i} f(x - ih)}{h^n} $$
 A: Your version seems to coincide with the Grünwald–Letnikov derivative
PS As L Spice points out in the comment below, the main definition in the Wikipedia article that I linked is different. That article contains several other versions, the one coinciding with the one in question is the very last one in that article, it is called direct Grünwald–Letnikov derivative there.
A: 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.
A: I believe my way corresponds to the Grunwald-Letnikov definition. Here is a nice article explaining the motivation behind fractional derivatives: http://www3.nd.edu/~msen/Teaching/UnderRes/FracCalc.pdf
