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Fractional Calculus II

Here explore some ideas of how the definitions and methods of calculus change if you integrate or differentiate n times when n is not a whole number.

Fractional Calculus I

Stage: 5
Article by Alan Beardon

Introduction

There has been a lot of correspondence recently on the Ask Nrich web-board about fractional derivatives. We know how to differentiate a function once, twice and so on, but can we differentiate the function 3/2 times? Similarly, we know how to integrate a function once, twice, and so on, but can we integrate it 1/2 times? This is the first of three articles which will introduce you to the main ideas in this new and rather strange world of fractional integrals and derivatives. Questions about the existence of such things were asked not long after calculus was created; for example, in 1695 Leibnitz wrote "Thus it follows that $d^{1/2}x$ will be equal to $\ldots$ from which one day useful consequences will be drawn ." Also, Euler (1738) wrote "When $n$ is an integer, the ratio $d^np$, $p$ a function of $x$, to $dx^n$ can always be expressed algebraically. Now it is asked: what kind of ratio can be made if $n$ is a fraction?"

If we differentiate $x^n$ $n$ times, where $n$ is a positive integer, we get $n!$ ; thus $${d^n\over dx^n}x^n = n!\ .$$ Later, we are going to see that this is true for all positive numbers $a$; that is, for $a> 0$, $${d^a\over dx^a}x^a = a!\ .$$ In order to understand this, we need to understand what $a!$ means when $a$ is not an integer. In this article we shall discuss the factorial function $a!$ for positive numbers $a$, even when $a$ is not an integer. In the second article we shall discuss fractional integrals; these are easier to define than fractional derivatives, and their definition requires the use of the factorial function. In the third article we shall discuss fractional derivatives.

The Factorial function

Let $F(n)$ be the factorial function; then for every positive integer $n$ we have $F(n) = 1.2.3\cdots n$. Of course, we usually write $n!$ instead of $F(n)$, and we can define $F(n)$ (and therefore $n!$) by the conditions $$F(1)=1, \quad F(n) = nF(n-1), \quad n=2,3,\ldots. \quad (1.1) $$

In order to define $y!$ for every positive $y$ we need to discuss an extremely important function in mathematics known as the Gamma function $\Gamma(x)$. This function is defined for every positive $x$, and it satisfies the intruiging formulae $$\Gamma(1)=1, \quad \Gamma(x+1) = x\Gamma(x), \quad x> 0. \quad (1.2)$$ Let us begin by exploring some of the consequences of (1.2) without (at first) worrying about how the Gamma function is defined. First, (1.2) implies that $\Gamma(2)=\Gamma(1)=1$. It also implies that $\Gamma(x)$ is not defined when $x=0$ for otherwise we would have $0 = 0\Gamma(0)=\Gamma(1)=1$ which is false.

Next, if we write $G(x) = \Gamma(x+1)$ we obtain $$G(1)=1, \quad G(x) = xG(x-1),$$ and a comparison with (1.1) shows that $G(n)=F(n) = n!$ for every positive integer $n$; thus for all positive integers $n$, $$n! = \Gamma (n+1).$$ Assuming that we have already defined the Gamma function, it is now quite natural to define $$x! = \Gamma (x+1) \quad (1.3)$$ whenever $x> -1$ (because $\Gamma(t)$ is only defined for $t> 0$). The Gamma function plays a crucial role in many parts of mathematics; particularly in probability theory, applied mathematics and complex analysis.

Binomial coefficients

For the moment, we continue with our discussion of the consequences of (1.2) and (1.3) even though we have still not defined the Gamma function. First we note that (1.3) gives the curious formula $$0! = \Gamma(1)=1.$$ This formula is used a lot in working with Binomial coefficients and this is the justification for it.

Speaking of binomial coeffients, we recall that if $k$ and $n$ are positive integers with $n> k$, then the corresponding binomial coefficient is defined to be $${n\choose k} = {n!\over k!\,(n-k)!}.$$ Of course we can now write this as $${n\choose k} = {\Gamma(n+1)\over \Gamma(k+1)\Gamma(n-k+1)},$$ which can now be generalised to give $${a\choose b} = {\Gamma(a+1)\over \Gamma(b+1)\Gamma(a-b+1)},$$ whenever $a> b> 0$. Let us do just one calculation here. We have $${3\,^1\!/_2\choose 2} = {\Gamma (9/2)\over \Gamma(3)\Gamma(5/2)} = {(7/2)(5/2)\Gamma(5/2)\over 2!\,\Gamma(5/2)} = {35\over 8}.$$

Binomial coefficients with non-integral entries are used in the Binomial expansion with non-integral powers. For example, we have the Binomial Theorem $$(1+x)^n = \sum_{k=0}^n {n\choose k}x^k$$ where $n$ is a positive integer, and also $$(1+x)^a = \sum_{k=0}^\infty {a\choose k}x^k \quad (1.4)$$ whenever |x|\leq1 (for convergence) and $a> 0$.

Quite generally, if $n$ is a positive integer and 0\leq\theta \leq 1 , from (1.2) and (1.3) we have $$(n+\theta)! = \Gamma(n+\theta +1) =(n+\theta)(n-1+\theta)\cdots (1+\theta)\theta\Gamma(\theta).$$ Notice that if $n+\theta$ is an integer, then $\theta =1$ and the formula ends with $\Gamma(1)$ which is 1; thus the Gamma function no longer appears explicitly in this formula. It is for this reason that the Gamma function does not appear in the binomial coefficients with integer entries.

The Gamma function

All of this discussion depends on having the Gamma function available so how, then, do we define the Gamma function? The definition is this : $$\Gamma (x) = \int_0^\infty t^{x-1}e^{-t}\,dt.$$ Notice that $x$ appears as a parameter in the integral, and that the integration is with respect to $t$ not $x$. This looks complicated but in fact, it is easy to verify (1.2). First, $$\Gamma(1) = \int_0^\infty e^{-t}\, dt = 1.$$ Next, we note that $${d\over dt}\Big(t^xe^{-t}\Big) = xt^{x-1}e^{-t} - t^xe^{-t},$$ and if we integrate both sides of this equation from $t=0$ to $t= \infty$ we obtain $$\big[t^xe^{-t}\big]_0^\infty = x\Gamma(x)-\Gamma(x+1)$$ (this is just integration by parts). As the left hand side is zero (because $x> 0$), this completes the proof of (1.2).

The formula $\Gamma(x+1) =x\Gamma(x)$ enables us to calculate $\Gamma(x)$ in terms of the value of $\Gamma$ at the fractional part of $x$. The following illustrative example will show what we mean here : $$(7/2)! = \Gamma (9/2) = {7\over 2}\Gamma(7/2) = {7\over 2}{5\over 2}\Gamma(5/2) = \cdots = {7\over 2}{5\over 2}{3\over 2}{1\over 2}\Gamma (1/2) ={105\over 16}\Gamma(1/2),$$ and in this example we cannot make further progress unless we know what $\Gamma(1/2)$ is. There is a beautiful, but deep, formula that relates the Gamma function to the trigonometric function $\sin x$, and this formula enables us to calculate $\Gamma(1/2)$. The formula is, for $0 \leq x \leq 1$, $$\Gamma(x)\Gamma(1-x) = {\pi \over \sin \pi x}.$$ We cannot prove this here, but note that when $x=1/2$ we get $$\Gamma(1/2)^2 = {\pi\over \sin \pi /2} = \pi,$$ so that $$\Gamma (1/2) = \sqrt{\pi}.$$ This means, for example, that $$(5/2)! = {5\over 2}{3\over 2}{1\over 2}\Gamma (1/2) ={15\sqrt{\pi}\over 8} = 3.3234\cdots. $$

Evaluating the Gamma function

The values of the Gamma function are given in tables (just as the values of $\sin x$ are), and using these tables we can calculate, for example, $(9/4)!$: its value is $\Gamma(13/4) = 2.5493\cdots$.

The following table of values of $\Gamma(0.1),\ldots ,\Gamma(0.9)$ will enable you to find some values of $y!$

$k$ $\Gamma(k/10)$
$1$ 9.5135
$2$ 4.5908
$3$ 2.9916
$4$ 2.2182
$5$ 1.7725
$6$ 1.4892
$7$ 1.2981
$8$ 1.1642
$9$ 1.0686

Using this you should be able to see, for example, that $$ (3/2)! = 1.3293, \quad (3.7)! = 15.431,\quad (4.2)! = 32.578, \quad (5.2)! = 169.41, \quad (6.7)! = 2769.8.$$ You know that $3! = 6$, and you can show (using the table above) that $(2.9)!$ is a little less than 6, and that $(3.1)!$ is a little greater than 6. Finally, two more curiosities are $$\pi! = \Gamma(1+\pi) = 7.1881\cdots , \quad e! = \Gamma(1+e) = 4.2608\cdots .$$

The next article in the series .