What does a “half derivative” mean?

I was looking at fractional calculus on Wikipedia, specifically this section and came across the half derivative of the function $$y=x$$ which is $$\frac{d^{1/2}y}{dx^{1/2}}=\frac{2\sqrt{x}}{\sqrt{\pi}}$$ . The derivative tells the slope at any point on the curve, but what does the "half derivative" mean - it's obviously not $$\frac{1}{2}$$ the derivative of $$y=x$$ which would be just $$\frac{1}{2}$$.

I do not have a very deep understanding of calculus - I have just taken Calc 1 & 2 out of a 4 series, but anything helps!

Also, I have checked similar questions, but they did not seem to answer my question that I have bolded.

The half derivative itself doesn't have much physical interpretation (though I believe there is a field called fractional quantum mechanics which may use it)

So why does it exist if its not any real physical thing.

I will explain.

Lets consider the idea of counting children at a school. We use whole numbers (positive integers) to count children. The statements are 5,6...201992 children each are meaningful in the sense that they exist mathematically AND have a physical interpretation.

But the set of Numbers isn't just whole numbers. It includes numbers like $1/2$ and $2^{1/2}$. So we could try to ask well what's half a child or square root of 2 children?

These are meaningless questions in the sense that you can't have half a child (contrary to popular belief disassembly and reassembly of children is not an easy or practical thing to do). Irrational quantities are even harder to produce. Put simply, they just DO NOT appear in that context.

So why am I telling you this? Here's why, lets ask the question not what 1/2 means in therms of children but how it came along. It came along because we wanted to generalize the set of numbers to include stuff in between the integers. It came along for applications besides counting children and is in fact most specifically an "accidental-byproduct" of the existence of division.

So what's a fractional derivative? We can easily answer the question that the nth derivative is the "rate of change of the rate of change ... (Repeat n times) of the rate of change of the function". This like children is a discrete structure. Only whole numbers (and if you include integrals then negative integers as well (like a backwards derivative)) work.

The fractional derivative is a consequence of the question "what is the function whom I apply twice to get a first derivative". Rather than "what is the rate of... Rate of change of the function"

So in short. It's an interesting question where we extend our level of control and understanding of calculus but it shares little similarity with the more physical forms that calculus originally had.

• I see someone down voted this ( or two people actually since it's -2) I would be more than willing to answer any questions if you'd like to bring them up – frogeyedpeas Jun 16 '14 at 15:40
• Not all appreciate what one does in life, and this is just another instance; forget the downvotes. Well written answer, brings up the concept quite nicely; + 1. – Kugelblitz Mar 15 '15 at 3:35
• I didn't downvote, but I didn't find this answer very enlightening. – Mehrdad Aug 31 '15 at 4:00
• Don't take offense to this, but I'm downvoting because I don't believe this answer is very informative and has received more upvotes than its quality deserves. – Simply Beautiful Art Jul 15 '17 at 14:10
• what a great explanation. if only my understanding of math was 1/10000000th of his. – Chase CB Oct 25 '17 at 22:04

Short answer: The half-derivative $H$ is some sort of operator (it isn't uniquely defined by this property) such that $H(Hf) = f'$.

Long answer: We can think of the derivative as a linear operator $D:X \to X$, where $X$ is some convenient (say, smooth) space of functions. The $n$th order derivative is then, by definition, the $n$-fold composition $D^n = D\circ \cdots \circ D:X \to X$. Clearly $D^n D^m = D^{n+m}$. Here we've restricted the index $n$ to an integer, but what if we allowed it to be a real number? That is, we want a family of operators $D_t$, with $t\geq 0$ real, such that

• $D_t$ behaves nicely with respect to $t$;
• $D_1$ is just the ordinary derivative $D$;
• $D_t D_s = D_{t + s}$.

(I'm not going to make the first point more precise here, but we ideally want something analogous to continuity or smoothness in $t$. I haven't defined what exactly the space $X$ is or what its geometry looks like, so I'm going to evade the point for now.) Thus, for example, we get an operator $D_{1/2}$ with $D_{1/2} D_{1/2} f = D_1 f = f'$ for suitable $f$.

Where do we get such an operator $D_t$? One place to start is Cauchy's integral formula: \begin{align*} f^{(-n)}(x) = \frac{1}{(n-1)!}\int_0^x (x - \xi)^{n-1} f(\xi)\, d\xi, \end{align*} where $f^{(-n)}(x)$ denotes the antiderivative of $f$, all normalized to have $f^{(-n)}(0)$ for $n > 0$. The factorial above is only defined for positive integers $n$, but we can use the relation $\Gamma(n) = (n - 1)!$ to define something similar for arbitrary $t\geq 0$: \begin{align*} I_t f(x) = \frac{1}{\Gamma(t)}\int_0^x (x - \xi)^{t-1} f(\xi)\, d\xi, \end{align*} Clearly $I_t f(x)$ is just $f^{(-t)}$ if $t$ is a positive integer, and we can show with a bit of work that $I_t(I_s f) = I_{t+s} f$.

Now, that's for an antiderivative. In order to get to the derivative $D_t$ for non-integer $t$, we can use the definition above to get rid of the fractional part. Since $D(If) = f$, we can define \begin{align*} D_t f= \frac{d^n}{dx^n} \left(I_\tau f\right) \end{align*} for $t = n - \tau$ with $n$ an integer and $\tau\in [0, 1)$. This is not the only possible construction of a fractional derivative $D_t$, though.

• Note here that we specifically have $f^{(-1)}(x)=\int_0^xf(t)~\mathrm dt$ (note the lower bound is $0$) so technically, its not simply the antiderivative (which would be a class of functions differing by a constant) but a definite integral. – Simply Beautiful Art Jul 15 '17 at 14:12