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How could a half derivative be computed? Not that I have found a use for what I wish I could call partial derivatives, but they are still interesting. Taking the nth derivative of the $m^{th}$ derivative of $f(x)$ is the $(n+m)^{th}$ derivative of $f(x)$. My real question is how to compute the nth derivative where $n$ is any real number, but I’ll settle for just half derivatives for now. How do you find the half derivative of $x^p$?

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  • $\begingroup$ See:en.wikipedia.org/wiki/… $\endgroup$ – Mariusz Iwaniuk Jul 17 '19 at 17:38
  • $\begingroup$ Well, what do you even mean by a half derivative? There are some extensions of the notion of derivative to extend them to rationals, but they all have at least some of the following pitfalls: 1. They only work on certain types of differentiable functions (i.e. polynomials). 2. They aren't particularly useful. 3. They don't translate well at all to multi-variate calculus. $\endgroup$ – Don Thousand Jul 17 '19 at 17:38
  • $\begingroup$ Some videos: youtube.com/results?search_query=fractional+derivatives $\endgroup$ – md2perpe Jul 18 '19 at 13:52
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A natural requirement for a half-derivative operator $H : \mathcal{F} \to \mathcal{F}$ on some suitable space $\mathcal F$ of functions on some interval in $\Bbb R$ is that $$H^2 = D ,$$ where $D$ denotes the usual derivativce.

Here's a conceptual approach to constructing such an operator: Formally writing $H = D^{1 / 2}$ and demanding that the Laplace transform identity $\mathcal L \{D^k f\}(s) = s^k \mathcal L \{f\}(s)$ for integer $k$ hold also for noninteger values imposes that $$\mathcal L \{Hf\}(s) = s^{1 / 2} \mathcal L \{f\}(s) ,$$ and so we can set---again for the functions to which we can apply $\mathcal L$--- $$(Hf)(t) := \mathcal L^{-1} \{s^{1 / 2} \mathcal L\{f\}\}(t) .$$ Then using that $\mathcal L$ maps convolutions to products lets us rewrite this formula, at least formally, as $$(Hf)(t) := \frac{1}{\sqrt{\pi}} \frac{d}{dx} \int_0^x \frac{f(t) \,dt}{\sqrt{x - t}} ,$$ which recovers a special case of a formula due to Cauchy. (The coefficient is a consequence of the fact that $\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}$, where $\Gamma$ is the Gamma function.) The map $f \mapsto Hf$ is evidently linear (over $\Bbb R$).

Example For a power function $p(x) = x^k$ evaluating the latter formula gives $$(H p)(x) = \frac{\Gamma(k + 1)}{\Gamma(k + \frac{1}{2})} x^{k - \frac{1}{2}}.$$ Computing then gives $$(H^2 p)(x) = (H(Hp))(x) = \frac{\Gamma(k + 1)}{\Gamma\left(k + \frac{1}{2}\right)} \cdot \frac{\Gamma\left(k + \frac{1}{2}\right)}{\Gamma(k)} x^{((k - \frac{1}{2}) - \frac{1}{2})} = k x^{k - 1} ,$$ so $H^2 p = D p$ as expected. Linearity then implies that $H^2 q = D q$ for all polynomials $q$.

It's evident how to extend this at least some formally to fractions (or even real numbers) other than $\frac{1}{2}$, but formulas must be interpreted with some care must be taken for "powers" outside $(0, 1)$. For negative "powers" one recovers fractional integral operations, which in some ways are better behaved than fractional derivatives.

The formal Laplace transform approach is not nearly the only possible one, nor is it (I think) the most general. See the Wikipedia article on Fractional calculus for much more.

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