13
$\begingroup$

If I want to take the derivative of $ax^n$, I will get $anx^{n-1}$. If I were to take the derivative again, I get $an(n-1)x^{n-2}$.
We can generalize this for integer k easily to get the kth derivative $a\frac{n!}{(n-k)!} x ^{n-k}$. But what about for a more general k?

Does this have some name? Has it been widely studied? If so, can you show how to generalize this formula for kth derivative of $ax^n$, and explain how it works? If not, is there a good reason it is impossible?

  • 1
    Nobody has posted this yet, so here: check out [this post](http://www.johndcook.com/blog/2009/03/13/fractional-derivatives/) on [John D. Cook](http://math.stackexchange.com/users/247/john-d-cook)'s blog.2010-09-14

2 Answers 2

12

To expand on Jonas's comment: Yes, it makes sense. For the case of the power function, one can consider

$\frac{\Gamma(n+1)}{\Gamma(n-\alpha+1)}x^{n-\alpha}$

as the $\alpha$-th derivative of the power function $x^n$, where $\Gamma(z)$ is the gamma function, the generalization of the factorial to the complex plane.

In general, one has a number of definitions for so-called "fractional derivatives", or, as Spanier and Oldham prefer to call it, the "differintegral". Negative values of $\alpha$ in expressions like the one given above correspond to integration, positive values correspond to differentiation, and in general $\alpha$ can be complex.

There's a lot of things to look at (Caputo derivatives, Riemann-Liouville integrals, Grunwald-Lednikov series), and I suggest you look at the book I linked to first, and then search around the web. Have fun!

  • 0
    @Rajesh: Right; in particular if $\alpha$ isn't$a$nonnegative integer, a full determination of the differintegral requires the specification of a "lower limit" $a$, so the full notation ought to be something like ${}_a D_x^{(\alpha)} f(x)$...2011-08-21
1

Here's a blog post that motivates a definition of fractional derivatives in terms of Fourier transforms.