Volume of the $n$-ball

The following is a simple and beautiful proof, shown to me to my great delight while I was in high school, that the ${n}$ -ball of radius ${r}$ has ${n}$ -volume

$\displaystyle \frac{\pi^{n/2} r^n}{\Gamma(\frac{n}{2}+1)}$

Although I expect most readers will know it, I believe that everybody should see it. I don’t know the history of it, and would be interested in learning.

Suppose that the ${n}$ -ball ${B^n}$ of radius ${1}$ has ${n}$ -volume ${V_n(1)}$ . Then, by considering linear transformations, the ${n}$ -ball ${rB^n}$ of radius ${r}$ has ${n}$ -volume ${V_n(r) = V_n(1) r^n}$ . Moreover, differentiating this with respect to ${r}$ should produce the surface ${(n-1)}$ -volume of the ${(n-1)}$ -sphere ${S^{n-1} = \partial B^n}$ : thus we expect ${\text{vol}^{(n-1)}(S^{n-1}) = V_n(1) n r^{n-1}}$ .

Now consider the integral

$\displaystyle I = \int_{\mathbf{R}^n} e^{-\pi|x|^2}\,dx.$

We will compute ${I}$ in two different ways. On the one hand,

$\displaystyle I = \int_{-\infty}^{+\infty}\cdots\int_{-\infty}^{+\infty} e^{-\pi x_1^2 -\cdots - \pi x_n^2}\,dx_1\cdots dx_n = \left(\int_{-\infty}^{+\infty} e^{-\pi x^2}\,dx\right)^n = 1.$

On the other hand, the form of ${I}$ suggests introducing a radial coordinate ${r=|x|}$ . Computing this way,

$\displaystyle I = \int_0^\infty e^{-\pi r^2} (V_n(1) n r^{n-1})\,dr = \frac{n V_n(1)}{\pi^{n/2}} \int_0^\infty e^{-t^2} t^{n-1}\, dt\qquad\quad$

$\displaystyle \qquad\quad= \frac{n V_n(1)}{2 \pi^{n/2}} \int_0^\infty e^{-s} s^{n/2-1}\,ds = \frac{n V_n(1)}{2 \pi^{n/2}} \Gamma(n/2) = \frac{V_n(1)\Gamma(n/2+1)}{\pi^{n/2}}.$

Rationals are repeating p-adics

I’ve been amusing myself with ${p}$ -adic arithmetic lately: I’ve never really got to know it until now.

We are all familiar with the fact that every ${q\in\mathbf{Q}}$ gets represented in ${\mathbf{R}}$ as a repeating decimal, or whatever your favourite base is. (Here I count terminating decimals as decimals which eventually repeat ${000...}$ .) The converse is of course true as well: repeating decimals are rationals.

Is this true in ${\mathbf{Q}_p}$ as well? That is, are the rationals precisely those ${p}$ -adics which are eventually repeating (in the other direction, of course)? One direction, that repeating ${p}$ -adics are rational, is pretty obvious: if ${a\in p^{-t}\mathbf{Z}}$ and ${b\in\mathbf{Z}}$ then

$\displaystyle a + b p^r + b p^{r+s} + b p^{r+2s} + \cdots = a + b \frac{p^r}{1 - p^s}$

is rational. What about the converse?

The converse seems trickier. How again did we do it in ${\mathbf{R}}$ ? I don’t even remember: it’s one of those things that we know so fundamentally (until very recently, ${\mathbf{Q}}$ was almost defined in my brain as the reals which eventually repeat) that we forget how to prove it.

Who cares how to prove it? It is true, and it says, in base ${p}$ , that if ${q\in\mathbf{Q}}$ then there exists ${a\in p^{-t}\mathbf{Z}}$ and ${b\in\mathbf{Z}}$ such that

$\displaystyle q = a + b p^r + b p^{r-s} + b p^{r-2s} + \cdots = a + b \frac{p^r}{1 - p^{-s}}.$

But hey, this implies that

$\displaystyle q = a - b \frac{p^{r+s}}{1 - p^s},$

which we already know is a repeating ${p}$ -adic.