# Half of a derivative?

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 $$x^p$$?

• – Mariusz Iwaniuk Jul 17 '19 at 17:38
• 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. – Don Thousand Jul 17 '19 at 17:38
• – md2perpe Jul 18 '19 at 13:52

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.