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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?

$$\frac{d^n}{dx^n}f(x) = \lim_{h \to 0} \frac{\sum_{i = 0}^\infty (-1)^i\binom{n}{i} f(x - ih)}{h^n} $$

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    $\begingroup$ I think my way is the Grünwald–Letnikov derivative. The motivation for it was exactly the same as that given in the wikipedia article. $\endgroup$ – Halbort Oct 14 '15 at 19:05
  • $\begingroup$ The function $\binom{x}{k}$ is the polynomial of 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
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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.

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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.

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    $\begingroup$ That definition has $f(x + (n - i)h)$ where the proposed definition has $f(x - i h)$; and this may be significant rather than a notational quirk, as @LiviuNicolaescu argues above that this definition does not give a fractional derivative for $x \mapsto x$. $\endgroup$ – LSpice Oct 14 '15 at 18:41
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    $\begingroup$ @LSpice There are several versions of the Grünwald-Letnikov derivative around, and unless I confuse something, the one from OP definitely coincides with one of them, although frankly speaking it is not easy for me to identify the particular name. The reason for all these different versions (the way I understand it) seems to be exactly that each of them fails to generalize some particular feature of the "ordinary" derivative $\endgroup$ – მამუკა ჯიბლაძე Oct 14 '15 at 19:18
  • $\begingroup$ How can you compute the GL derivative of a polynomial? $\endgroup$ – Halbort Oct 14 '15 at 19:40
  • $\begingroup$ Where does this derivative fail $\endgroup$ – Halbort Oct 14 '15 at 19:54
  • $\begingroup$ @Halbort Sorry I don't really know. Some time ago I tried to use fractional derivatives for one of my questions here (see a comment to that question by rlo) but I could not find out which one of the many would be useful and then abandoned it. $\endgroup$ – მამუკა ჯიბლაძე Oct 14 '15 at 22:28
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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

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