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You may recall that one of the first things we learn in calculus is simple derivatives, $$\frac{d}{dx} (x^2) = 2x.$$

We quickly learn a pattern forms. The general method for differentiating a monomial is $$\frac{d}{dx} (x^k) = kx^{k-1}.$$

Of course, if we differentiate once, we can do it again. These second derivatives are written $\frac{d^2}{dx^2}$ and using our monomial we get $$\frac{d^2}{dx^2} (x^k) = k(k-1)x^{k-2}.$$

We continue this process and find we can write the $a^{\mathrm th}$ derivative easily

$$\frac{d^a}{dx^a} (x^k) = k(k-1)(k-2)...(k-a+1)x^{k-a} = \frac{k!}{(k-a)!}x^{k-a}$$

This formula works wonderfully when $a \in \mathbb N$, that is, when $a$ is a positive integer. The natural question is then: "What if we have a non-integer $a$?". At first glance it seems hopeless, the factorial is only defined for integer arguments. Luckily, there exists a natural extension of the factorial in the form of the Gamma function,

$$k! = \Gamma(k+1).$$

We can use this relation directly as a drop in replacement in our differentiation formula,

$$\frac{d^a}{dx^a} (x^k) = \frac{\Gamma(k+1)}{\Gamma(k-a+1)}x^{k-a},$$

and magically the formula supports non-integer $a$ (note, however, that this formula is only well behaved for $k>0$. The gamma function can act weird with negative arguments). Of course one might wonder does this formula provide a meaningful result?

## The Half Derivative

One property that would be nice for our fractional derivative to have would be
$$\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}}\left [\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}} f(x) \right] = \frac{d}{dx} f(x),$$

forming an analogue to similar expressions; e.g. $x^{\frac{1}{2}}x^{\frac{1}{2}} = x.$ It turns out our fractional derivatives do indeed work in this way!

Let's look at an example when k=1 and $f(x) = x$. The first derivative is $$\frac{d}{dx} (x) = 1.$$

We can find the half derivative using our formula above, $$\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}} (x) = \frac{\Gamma(2)}{\Gamma\left (\frac{3}{2} \right )}x^{\frac{1}{2}} = \frac{2}{\sqrt{\pi}}x^{\frac{1}{2}}.$$

Repeating the process we find $$\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}} \left [\frac{2}{\sqrt{\pi}}x^{\frac{1}{2}}\right ] = \frac{2}{\sqrt{\pi}}\dfrac{\Gamma(\frac{1}{2}+1)}{\Gamma(\frac{1}{2}-\frac{1}{2}+1)}x^{\frac{1}{2}-\frac{1}{2}}=\frac{2}{\sqrt{\pi}}\dfrac{\Gamma(\frac{3}{2})}{\Gamma(1)}x^{0}=\dfrac{2 \sqrt{\pi}x^0}{2 \sqrt{\pi}0!}=1,$$

as we required!

Now let us finally look at what a half derivative looks like:

See how the half derivative creates a line sort of half way between no derivative (or the $0$th derivative) and the first derivative. Incredibly, as shown before, if we once again half derivate the strange curved line, we magically end up back at the straight line $x=1$.

Note that this is just a single example of a half derivative, one where $a=\frac{1}{2}$. It's only half of the story...

## Interpolating through Derivatives

We can use any positive $a$ in the fractional derivative formula, it need not be $a<1$. A fun exercise one can do is plot the fractional derivative while slowly increasing $a$. As $a$ passes through the integers we see the usual derivatives, and what happens inbetween is very interesting.

As $a$ increases, the line smoothly transforms from derivative to derivative , almost as if the line is being interpolated via some trick. A continuous semigroup is formed, with parameter $a$, where the original derivatives are recovered when $a$ is a positive integer.

## Fractional Integration

Recall our fractional derivative formula for monomials,

$$\frac{d^a}{dx^a} (x^k) = \frac{\Gamma(k+1)}{\Gamma(k-a+1)}x^{k-a}.$$

So far we have only considered values of $a \geq 0$, but if we use $a<0$ we recover integration! Almost to be expected when you consider that the process of integrating the monomial is simply the reverse of the derivative.

The effect is demonstrated here by oscillating around $x^2$, integrating and differentiating in a fractional way. Once again the process is smooth.

## Coming Soon

Hopefully I've shown that fractional calculus can have an intersting effect on simple functions! In a later blog post: Calculus with more complicated functions, and letting $a\in\mathbb C$.

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