# Fractional calculus: the calculus of witchcraft and wizardry

How can we differentiate a function 9¾ times? Differentiating a function is usually regarded as a discrete operation: we use the first derivative of a function to determine the slope of the line that is tangent to it, and we differentiate twice if we want to know the curvature. We can even differentiate a function negative times—ie integrate it—and thanks to that we measure the area under a curve. But why stop there? Is calculus limited to discrete operations, or is there a way to define the half derivative of a function? Is there even an interpretation or an application of the half derivative?

Fractional calculus is a concept as old as the traditional version of calculus, but if we have always thought about things using only whole numbers then suddenly using fractions might seem like taking the Hogwarts Express from King’s Cross station. However, fractional calculus opens up a whole new area of beautiful and magical maths.

How do we interpret the half derivative of a function? Since we are only halfway between the first derivative of a function and not differentiating it at all, then maybe the result should also be somewhere between the two? For example, if we have the function $\mathrm{e}^{cx}$, with $c>0$, then we can write any derivative of the function as
\begin{equation*}
\mathrm{D}^{n} \mathrm{e}^{cx} = c^{n} \mathrm{e}^{cx},
\end{equation*}
which works for $n=1, 2,\ldots$, but also works for integrating the function ($n = -1$) and doing nothing to it at all ($n=0$). As an aside, the symbol $\mathrm{D}^n$ might seem like a weird notation to represent the $n$-th derivative, or $\mathrm{D}^{-n}$ to represent the $n$-th integral of a function, but it’s just an easy way to represent derivatives and integrals at the same time. So if we have a real number $\nu$, why not express the fractional derivative of $\mathrm{e}^{\,cx}$ as
\begin{equation*}
\mathrm{D}^{\nu} \mathrm{e}^{cx} = c^{\nu} \mathrm{e}^{cx}?
\end{equation*}

With this new derivative we know that if $\nu$ is an integer then the fractional expression has the same result as the traditional version of the derivative or integral. That seems like an important thing, right? If we want to generalise something, then we cannot change what was already there.

If the fractional derivative is a linear operator (ie if $a$ is a constant then $\mathrm{D}^{1/2} a \hspace{1pt} f (x) = a\mathrm{D}^{1/2} f(x)$) , then we would also obtain that
\begin{equation*}
\mathrm{D}^{1/2} \left[\mathrm{D}^{1/2} \mathrm{e}^{cx} \right] = \mathrm{D}^{1/2} \left[c^{1/2} \mathrm{e}^{cx} \right] = c^{1/2} \mathrm{D}^{1/2} \left[ \mathrm{e}^{cx} \right] = c \mathrm{e}^{cx},
\end{equation*}
so half differentiating the half derivative gives us the same result as just applying the first derivative. In fact for this very first definition of a fractional derivative, we get that
\begin{equation*}
\mathrm{D}^{\nu} \left[\mathrm{D}^{\mu} \mathrm{e}^{cx} \right] = \mathrm{D}^{\nu + \mu} \mathrm{e}^{cx} = c^{\nu + \mu} \mathrm{e}^{cx}
\end{equation*}
for all real values of $\nu$ and $\mu$. Great: our fractional derivative has at least some properties that sound like necessary things. Differentiating a derivative or integrating an integral should just give us the expected derivative or appropriate integral.

This way of defining a fractional derivative for the exponential function is perhaps a good introductory example, but some important questions need to be asked. Firstly, is this the only way to define the half derivative for $\mathrm{e}^{cx}$ such that it has the above properties, or could we come up with a different definition? Secondly, what happens if $c<0$? For example, with $c = -1$, we would get that $\mathrm{D}^{1/2} \mathrm{e}^{-x} = \mathrm{i} \mathrm{e}^{-x}$, which is imaginary. So the fractional derivative of a real-valued function could be complex or imaginary? That sounds like dark arts to me. And finally, how does that $\mathrm{D}^{\nu} f(x)$ work if we are not talking about the exponential function, but if we have a polynomial or, even simpler, a constant function, like $f(x) = 4$?

If we start with $f(x)=4$ (a boring, horizontal line), we know that its first derivative is $\mathrm{D}^1 f(x)=0$, so should the half derivative be something like $\mathrm{D}^{1/2} f(x) = 2$? Then, if we half differentiate that expression again, we obtain a zero on the left-hand side (since $\mathrm{D}^{1} f(x)=0$) and the half derivative of a constant function (in this case $g(x)=2$), on the right-hand side. But this is certainly not right! We said that the half derivative of a constant function is half the value of that constant, but now we obtain that it is zero! There is nothing worse for a mathematician than a system that is not consistent.

The best way is to begin with a more formal definition. Perhaps after having to integrate a function thousands and thousands of times, Augustin-Louis Cauchy discovered in the 19th century a way in which he could write the repeated integral of a function in a very elegant way:
\begin{equation*}
\mathrm{D}^{-n}f(x)= \frac{1}{(n-1)!}\int^{x}_{a} (x-t)^{n-1} f(t) \, \mathrm{d}t.
\end{equation*}

Not only is this a beautiful and simple formula, it also gives us a way to write any iterated integration (although to actually solve it, we would usually need to do some not-so-beautiful integration by parts). So why don’t we just change that number $n$ to a fraction, like 9¾? Everything in that expression would work smoothly… except for that dodgy factorial! The value of $n$ factorial (written as $n!$, possibly the worst symbol ever used in maths since now we cannot express a number with surprise) is the product of the numbers from 1 through to $n$, ie $1 \times 2 \times \cdots \times n$. What, then, would the factorial of 9¾ be? Maybe close to 10! but not quite there yet?

Luckily for us, an expression for the factorial of a real number has intrigued mathematicians for centuries, and brilliant minds like Euler and Gauss, amongst others, have worked on this issue. They defined the gamma function, $\Gamma$, in such a way that it has the two properties we need: first, $\Gamma(n) = (n-1)!$, so we can use $\Gamma(n)$ instead of the factorial. Second—and even more importantly, given that we are dealing with fractions here—is that the function is well-defined and continuous for every positive real number, so we can now compute the factorial of 9¾, which is only 57% of the value of 10!. Now, we can write the repeated integral as
$\mathrm{D}^{-\nu}f(x)= \frac{1}{\Gamma(\nu)}\int^{x}_{a} (x-t)^{\nu-1} f(t) \, \mathrm{d}t,$
which gives the same result as before when $\nu$ is a positive integer and is well-defined when $\nu$ is not an integer. The integral above is known as the fractional integral of the function $f$. Awesome!

What happens if we take the derivative of the repeated integral? Easy! We get the fractional derivative, right? Not quite, since we have two options: we could either differentiate the original function first and then take the fractional integral, or we could fractionally integrate first and then take the derivative. Damn! Both definitions are equally valid and we mathematicians hate having two definitions for the same thing. But are they even the same thing? If we differentiate first and then take the fractional integral—known as the Caputo derivative—we don’t necessarily get the same result as if we fractionally integrate a function first and then take its derivative. The latter is called the Riemann–Liouville derivative or simply the fractional derivative since it is the one more frequently used.

As an example, let’s look at the 9¾ derivative of a polynomial, say $f(x) = x^{9}$. The 9¾ Caputo derivative is zero, since we first differentiate $x^{9}$ ten times, which is zero, and then integrate it; but the Riemann–Liouville derivative is
\begin{equation*}
\mathrm{D}^{9\text{¾}}f(x)= \frac{9!}{\Gamma(\frac{1}{4})} x^{-1/4},
\end{equation*}
which is clearly different to the Caputo derivative. Two things are to be noted here. The fractional part is only contained in the integral, so in order to obtain both of the 9¾ derivatives of a function we need to quarter integrate the tenth derivative or differentiate the ¼ integral ten times. Also, and very importantly in fractional calculus, the fractional integral depends on its integration limits (just as in the traditional version of calculus) but since the fractional derivative is defined in terms of the fractional integral, then the fractional derivatives also depend on the limits. The fractional derivatives of $\mathrm{D}^{\nu} x^{9}$, with $\nu$ between 0 and 10, and $x$ between 0 and 1.

There are many applications of fractional calculus in, for example, engineering and physics. Interestingly, most of the applications have emerged in the last twenty years or so, and it has allowed a different approach to topics such as viscoelastic damping, chaotic systems and even acoustic wave propagation in biological tissue.

Perhaps fractional calculus is a bit tricky to interpret, seeming at first to be a weird generalisation of calculus but for me, just thinking about the 9¾ derivative of a function was like discovering the entry into a whole new world between platforms 9 and 10. Certainly, there is some magic hidden behind fractional calculus! Rafael Prieto Curiel is doing a PhD in mathematics and crime. He is interested in mathematical modelling of any social issues, such as road accidents, migration, crime, fear and gossip.
@rafaelprietoc    rafaelprietocuriel.wordpress.com    + More articles by Rafael