# Fractional Calculus III

##### Age 16 to 18

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

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 $z$.