Volume of a n-dimensional Ball: Part 2

Volume of a n-dimensional Ball: Part 2

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1. Introduction

The first installment of this post investigated the question of computing the volume of a unit ball (and by extension any ball) of n dimensions. The main result we obtained was that the volume V_n can be computed recursively as V_n=a_nV_{n - 1}, where

    \begin{equation*} $a_n = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{n}(t) dt$ \end{equation*}

In this post, we will use this result to prove an explicit formula for V_n.

2. The Formula

A quick search online yields the following explicit formula for V_n:

    \begin{equation*} $V_n = \frac{\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2} + 1)}$ \end{equation*}

where the function \Gamma extends the factorial function to most of the real numbers. Note that \Gamma(x) actually equals (x - 1)!, so to keep things simple, we will replace the denominator \Gamma(\frac{n}{2} + 1) with (\frac{n}{2})!, and it will be understood that this factorial is extended to the real numbers in place of \Gamma. Note that it retains the important property that \frac{(n + 1)!}{n!} = n + 1.

3. More About an

Before we begin the proof, we will investigate the properties of a_n in greater detail. It seems quite difficult to obtain an explicit formula for a_n at first glance (if any of you know of a simple way to evaluate that integral, please let me know in the comments!), but we can attempt to apply some tools from calculus to simplify that integral. Specifically, because we can write \cos^n(t) as \cos(x) times a smaller term, we think of applying integration by parts (inverse of product rule). Let

  • u = \cos^{n - 1}(t)
  • v = \sin(t)
  • du = (n - 1)\cos^{n - 2}(t)*(-\sin(t)) dt (by the Chain Rule)
  • dv = \cos(t) dt

Then

    \begin{equation*} $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} u dv = [uv]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} - \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} v du \rightarrow$ \end{equation*}

    \begin{equation*} $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^n(t) dt = [\cos^{n - 1}(t)\sin(t) dt]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} - \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (n - 1)\cos^{n - 2}(t)(-\sin(t)) * \sin(t) dt$ \end{equation*}

A short computation shows that for n > 1, cos^{n - 1}(t) = 0 for t = -\frac{\pi}{2}, \frac{\pi}{2}. Thus, the first term on the right hand side equals 0. Furthermore, after cancelling the negative signs and factoring out n - 1 from the integral, we may collect the \sin^2(t) and replace it with 1 - \cos^2(t). Hence,

    \begin{equation*} $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^n(t) = (n - 1)\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{n - 2}(t)\left(1 - \cos^2(t)\right)dt =$ \end{equation*}

    \begin{equation*} $\left((n - 1)\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{n - 2}(t) dt\right) - \left((n - 1)\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^n(t) dt\right) \rightarrow$ \end{equation*}

    \begin{equation*} $a_n = (n - 1)a_{n - 2} - (n - 1)a_n \rightarrow a_n = \frac{n - 1}{n}a_{n - 2}$ \end{equation*}

Thus, we have obtained a recursive formula for the sequence a_n itself, in terms of a_{n - 2}.

4. The Proof of the Formula

Now, we have sufficient information to tackle the problem. We know that V_n = a_{n}V_{n - 1}; we can iterate this once more to get V_n = a_{n}(a_{n - 1}V_{n - 2}), and a third time, and so on. Thus, we have V_n = a_{n}a_{n - 1}a_{n - 2}\cdots a_{2}V_{1}. Note that if we were to extend the definition of a unit sphere as “the set of all points a distance of 1 unit away from the center” to a 1-dimensional space, we see that a 1-D sphere is simply a line segment of length 2 around its center. Thus, V_1 is the “volume” (length for 1-D) of a 1-D ball, so V_1 = 2. However, a quick computation shows that a_1 = 2 as well, so we may write

    \begin{equation*} $V_n = a_{n}a_{n - 1}a_{n - 2}\cdots a_{2}a_{1}$ \end{equation*}

We will prove that this is equivalent to the above known formula for V_n (from now on, assume V_n is our derived formula and F(n) = \frac{\pi^\frac{n}{2}}{{\left(\frac{n}{2}\right)!}} is the already-known formula. The following proof may be confusing, so here is an outline:

  • First, we will show that V and F have the same values at n = 1, 2.
  • Then, we will show that the ratios \frac{V_3}{V_1} = \frac{F(3)}{F(1)} and \frac{V_4}{V_2} = \frac{F(4)}{F(2)}.
  • Finally, we will prove that if the ratio \frac{V_n}{V_{n-2}} = \frac{F(n)}{F(n - 2)}, then that implies \frac{V_{n + 2}}{V_n} = \frac{F(n + 2)}{F(n)}.

Observe how these steps are sufficient to prove that V = F at all points. Essentially, we are proving that they start out the same way, and the scale factor between the first term and the term two steps ahead is the same in both sequences (and also for second and fourth terms), and finally that if this “two step ratio” is the same for both sequences at some term, then they are still the same after two more terms. The last two parts together prove that the “two step ratio” is always the same in both sequences, and that coupled with the first part shows that the terms themselves are always the same.

We will first show that V_n = F(n) for n = 1, 2. This can be done as a straightforward computation (note that a_1 = 2, a_2 = \frac{\pi}{2}, and \left(\frac{1}{2}\right)! = \frac{\sqrt{\pi}}{2}).

Now, we prove that \frac{V_3}{V_1} = \frac{F(3)}{F(1)} and \frac{V_4}{V_2} = \frac{F(4)}{F(2)}. Once more, these are simple computation problems and are left for the reader to verify.

Finally, we will prove that if \frac{V_n}{V_{n - 2}} = \frac{F(n)}{F(n - 2)}, then \frac{V_{n + 2}}{V_n} = \frac{F(n + 2)}{F(n)}. First of all, note that

    \begin{equation*} \frac{F(n)}{F(n - 2)} = \frac{\pi^{n/2}}{\left(\frac{n}{2}\right)!} * \frac{\left(\frac{n}{2}-1\right)!}{\pi^{\frac{n}{2} - 1}} = \frac{\pi}{\frac{n}{2}} = \frac{2\pi}{n} \end{equation*}

Also,

    \begin{equation*} \frac{V_n}{V_{n - 2}} = \frac{a_{n}a_{n - 1}\cdots a_{1}}{a_{n - 2}a_{n - 3}\cdots a_{1}} = a_{n}a_{n - 1} \end{equation*}

Now suppose that \frac{2\pi}{n} = a_{n}a_{n - 1}. We want to prove that \frac{2\pi}{n + 2} = a_{n + 2}a_{n + 1}. By our recursion for a_n, we have a_{n + 2} = \frac{n + 1}{n + 2}a_n and a_{n + 1} = \frac{n}{n + 1}a_{n - 1}. Plugging these in, we have

    \begin{equation*} a_{n + 2}a_{n + 1} = \left(\frac{n + 1}{n + 2}a_{n}\right)\left(\frac{n}{n + 1}a_{n - 1}\right) = \frac{n}{n + 2}a_{n}a_{n - 1} = \frac{n}{n + 2} * \frac{2\pi}{n} = \frac{2\pi}{n + 2} \end{equation*}

as claimed.

So we are finally done! We have fully proven the formula for the volume of a ball of any number of dimensions! Here are a few particular fruits of our labor:

  • V_4(r) = \frac{\pi^2}{2}r^4

  • V_5(r) = \frac{8\pi^2}{15}r^5

  • V_6(r) = \frac{\pi^3}{6}r^6

  • V_7(r) = \frac{16\pi^3}{105}r^7

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