Published December 2000,April 2000,December 2011,February 2011.

The discussion about fractional calculus, which led to this series of articles, was started by a couple of sixth form students on the askNRICH webboard. Before reading this article you may like to read the first two articles Fractional Calculus I and Fractional Calculus II and see the conversation on the webboard.

Derivatives and integrals

Given a function $f(x)$ we can differentiate it once, twice, and so on in the usual way. We can also integrate the function once, twice, and so on to get $(If)(x), (I^2f)(x),\ldots$, and now we can even get $(I^af)(x)$ for any positive $a$.

We are used to the idea that differentiation 'reverses' the process of integration, and this is just the formula $$\frac{d}{dx}(If)(x) = \frac{d}{dx}\int_0^x f(t)\,dt = f(x).$$ Note however, that integrating from $0$ to $x$ (that is, calculating $(If)(x)$) does NOT reverse differentiation. For example, if $f(x)=e^x$ then $df/dx = f'(x) = f(x)$ and $$(If')(x) = \int_0^x e^t\,dt = e^x-1 \neq f(x).$$ In short, differentiating and then integrating the derivative from $0$ to $x$ does not (in general) return us to the same function ; this reversal only holds if we integrate first .

Derivatives of
integrals

Suppose that $p$ and $q$ are integers, and that $p> q$. If
we integrate a function $p$ times (from $0$ to $x$), and then
differentiate the resulting function $q$ times we obtain the same
result as integrating the function $p-q$ times; this is because
$$\frac{d}{dx}(I^pf)(x) = \frac{d}{dx}\Big( I(I^{p-1}f)\Big)(x)
=(I^{p-1}f)(x),$$

and repeating this process $q-1$ more times gives the result.
As $$I(I^af)(x) = (I^{1+a}f)(x)$$

for every positive $a$ (we commented on this at the end of the
last article), the same argument holds in general so that if $a>
0$, $k$ is a positive integer and $a> k$, then
$$\frac{d^k}{dx^k}(I^af)(x) = (I^{a-k}f)(x). \quad (3.1)$$

Fractional
derivatives

We have seen that (3.1) holds when $a> k$, but what happens
if $k> a$? Our intuition tells us that integrating $-2$ times
should be the same as differentiating twice so, based entirely on
our intuition, we shall now DEFINE the $a\!$-th derivative of $f$
to be $(I^{-a}f)(x)$, where this is given by the formula (3.1). To
be more explicit, given any positive number $a$, we choose any
integer $k$ such that $k> a$, and then define
$$\frac{d^a}{dx^a}f(x) = \frac{d^k}{dx^k}\big(I^{k-a}f\big)(x).$$
Let us consider an example.

Example 1 What is the
$1/2$-derivative of $x$? According to our definition (with $k=1$
and $a=1/2$) we have

\begin{eqnarray} \\
\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}}x &=& \frac{d}{dx}
\left( \frac{1}{\Gamma (\frac{1}{2})}\int_0^x (x-t)^{-\frac{1}{2}}
dt \right) \\ &=& \frac{1}{\sqrt{\pi}} \frac{d}{dx} \left(
\int_0^x u^{\frac{1}{2}}(x-u)du\right) & (u=x-t) \\ &=&
\frac{1}{\sqrt{\pi}}\frac{d}{dx}\left( x \int_0^x
u^{-\frac{1}{2}}du - \int_0^x u^{\frac{1}{2}} du\right) \\
&=& \frac{1}{\sqrt{\pi}} \left(\int_0^x u^{-\frac{1}{2}}du
+ x.x^{-\frac{1}{2}} - x^{\frac{1}{2}} \right) \\ &=&
\frac{2\sqrt{x}}{\sqrt{\pi}} \end{eqnarray}

This seems great : we now know how to differentiate functions
a fractional number of times. However, there are some problems you should be aware
of .

Example 2 What is the
$1/2$-derivative of $x^{-1/2}$? According to our definition (with
$k=1$ and $a=1/2$) we have

\begin{eqnarray} \\
\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}} &=& \frac{d}{dx}
\left( \frac{1}{\Gamma (\frac{1}{2})} \int_0^x (x-t)^{-\frac{1}{2}}
t^{-\frac{1}{2}} dt \right) \\ &=& \frac{1}{\sqrt{\pi}}
\frac{d}{dx} \left( \int_0^1 (1-u)^{-\frac{1}{2}}u^{-\frac{1}{2}}
du \right) & (t=xu) \\ &=& 0 \end{eqnarray}

because the integral here does not depend on $x$. It is clear
that if $g(x)=0$ for all $x$, then any integral of $g$ is zero,
hence so is any derivative of $g$. It follows from this that if
$f(x)=x^{-1/2}$, then
$$\frac{d^{1/2}}{dx^{1/2}}\left(\frac{d^{1/2}}{dx^{1/2}}\right)
f(x) = \frac{d^{1/2}}{dx^{1/2}}0 = 0 \neq \frac{d}{dx}f(x).$$

Thus it is NOT always true that
$$\frac{d^{a}}{dx^{a}}\left(\frac{d^{b}}{dx^{b}}\right) =
\frac{d^{a+b}}{dx^{a+b}}.$$

Example 3 Suppose that
$f(x)=1$ for every $x$. What is the $1/2$-derivative of $f(x)$?
You should be prepared for a
surprise here . Using the same argument as above, we see
that

\begin{eqnarray}
\frac{d}{dx}(I^{\frac{1}{2}}f)(x) &=& \frac{d}{dx} \left(
\frac{1}{\Gamma (\frac{1}{2})} \int_0^x (x-t)^{-\frac{1}{2}}.1 dt
\right) \\ &=& \frac{1}{\sqrt{\pi}}\frac{d}{dx} \left(
\int_0^x (x-t)^{-\frac{1}{2}} dt \right) \\ &=&
\frac{1}{\sqrt{\pi}}\frac{d}{dx} \left( \left[
-2(x-t)^{\frac{1}{2}}\right]^x_0 \right) \\ &=&
\frac{1}{\sqrt{\pi}}\frac{d}{dx} (2\sqrt{x}) \\ &=&
\frac{1}{\sqrt{\pi}\sqrt{x}} \end{eqnarray}

We have reached the rather surprising result that the
$1/2$-derivative of the constant function $f(x)=1$ is NOT zero. But
is this really surprising? We might expect that if we start with
$f(x) = x^r$ and take the $p$-th derivative of this then we obtain
a constant times $x^{r-p}$. In fact, this is all that has happened
here because we started with $x^0$ and arrived at $cx^{-1/2}$; it
just happens that in this case $c\neq 0$. It is clear
that we should now try and really understand why the derivative
of a constant function is zero. What once seemed obvious now seems
a problem to be overcome, and this is a common experience in higher
mathematics! One way to see this is as follows. We know that
$$\frac{d}{dx} x^k=kx^{k-1} = \frac{k!}{(k-1)!}x^{k-1}
=\frac{\Gamma (k+1)}{\Gamma(k)}x^{k-1}.$$

If we now put $k=0$ we get an answer $0$ because the only
sensible definition of $\Gamma(0)$ is $\infty$, and $1/\infty$
should be $0$. All this can be justified, but not here.

More generally, working in the way we have indicated above, we
get $$\frac{d^a}{dx^a}\big(x^k\big) =
\frac{\Gamma(k+1)}{\Gamma(k+1-a)} x^{k-a},$$ and if $a=k+1$ we get
the answer $0$. In short, for any $k$,
$$\frac{d^{k+1}}{dx^{k+1}}x^k = 0.$$

Difference
quotients

Naturally, we want to try to think of the first derivative of
$f$ as $$\lim_{h\to 0}\ \frac{f(x+h)-f(x)}{h}$$ and try to
generalise this too. This can be done, and we end with a brief
description of this process.

First we introduce an operator $E^t$ on functions by saying
that this takes $f(x)$ to $f(x+t)$. The operator $E^0$ has no
effect on $f(x)$ and we prefer to write this as ${\bf I}$ (the
identity operator). We can now see that as $h\to0$, $$\frac{({\bf
I}-E^{-h})f(x)}{h}=\frac{{\bf
I}(f(x))-E^{-h}(f(x))}{h}=\frac{f(x)-f(x-h)}{h} \to
\frac{d}{dx}f(x).$$

Similarly, one can show that if $n$ is a positive integer
then, as $h\to0$, $$\frac{({\bf I}-E^{-h})^nf(x)}{h^n}\to
\frac{d^n}{dx^n}f(x),$$

where $({\bf I} - E^{-h})^n$ is obtained by using the Binomial
Theorem and noting that $(E^{-h})^m=E^{-mh}$. For example,
$$\frac{f(x) - 2f(x-h) + f(x-2h)}{h^2}\to f''(x) \quad
(3.2)$$

as $h\to0$. You can easily check this by expressing $f$ as a
Taylor series $$f(x+t) = \sum_{k=0}^\infty
\frac{f^{(k)}(x)}{k!}t^k$$

and substituting this in the expression in the left hand side
of (3.2).

Finally, one can prove that if $a> 0$ then, as $h\to 0$,
$$\frac{({\bf I}-E^{-h})^af(x)}{h^a}\to
\frac{d^a}{dx^a}f(x).$$

What exactly does this mean? Recalling that for $|x|< 1$ we
have the general Binomial Theorem $$(1+x)^a =
\sum_{k=0}^\infty{a\choose k}x^k,$$

we now take $$({\bf I}-E^{-h})^af(x)= \sum_{k=0}^\infty
{a\choose k}(-1)^kE^{-kh} f(x). $$

Final
comments

Perhaps the most important observation we can make now is that
the familiar topics of factorials, binomial coefficients, and the
Binomial Theorem, which are usually regarded as discrete
mathematics, all generalise (through the Gamma function) to
continuous situations.

Where do we go next? The Gamma function itself can be extended
to be defined on the whole complex plane (taking the value $\infty$
at $0,-1,-2,\ldots $), so eventually, but not here, one can even
define $z!$ for complex numbers $z$.