Fractional Calculus III

Age 16 to 18
Article by Alan Beardon

Published 2000 Revised 2008

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 $z$.