Volume of n-Spheres and the Gamma Function

 

A "sphere" of radius R in n dimensions is defined as the locus of points with a distance less than R from a given point. This implies that a sphere in n = 1 dimension is just a line segment of length 2R, so the volume (or "content") of a 1-sphere is simply 2R. A sphere of radius R in two dimensions can be regarded as consisting of a continuous sequence of one-dimensional spheres at a perpendicular distance u from some fixed 1-sphere, and such that the radius of the 1-sphere at a distance u is r(u) = (R2 - u2)1/2, where u varies uniformly from -R to +R. In general, an n-sphere consists of a sequence of (n-1)-spheres of radius r(u) as u varies uniformly from -R to +R.

 

Therefore, given the volume formula Vn[R] for n-dimensional spheres, we can determine the formula Vn+1[R] for (n+1)-spheres as follows

 

 

Also, it's clear that the volume of an n-sphere must be proportional to Rn, so for every n there is a constant kn such that the volume of an n-sphere of radius R is given by Vn[R] = kn Rn.  Thus we have

 

 

Inserting the expression for r(u) and factoring out a constant R from the integrand gives

 

 

The outside factor on the integrand is simply Vn[R], and if we define the variable s = u/R we have du = R ds, so the above formula can be written as

 

 

Using this formula, beginning with the (-1)-dimensional case V-1[R] = 1/(pR), we can recursively compute the volumes of the higher n-spheres

 

 

This shows that the integer dimension with maximum integer coefficient is n = 5. Also, the power of p increases by 1 at each even dimension, and the rational coefficients of the formulas for the sequence of even dimensions (1, 1/2, 1/6, 1/24,...) change by a factor of 1, 1/2, 1/3, 1/4,...etc., whereas the coefficients for the sequence of odd dimensions (2, 4/3, 8/15, 16/105,...) change by factors of 2/3, 2/5, 2/7, 2/9,...etc. Hence we have

 

 

By equation (1) we also have

 

   

Substituting from the previous expressions for V2n+1 and V2n, canceling terms, and solving for n! gives

 

 

Notice that if we set n = -1/2 we can evaluate the right hand side, because the factorial in the numerator has the argument 2(-1/2) + 1 = 0. So, evaluating the integral gives the result

 

 

This fact, together with the general relation (x+1)! = (x+1)(x!), enables us to easily compute the factorial of any positive half-integer. For example, we have

 

 

Furthermore, given all the factorials of the whole and half integers, equation (4) enables us to compute the factorials of all the quarter-integers, because if we set n to a quarter-integer the argument of the factorial on the right hand side will be a half or whole integer. Likewise it follows that we can compute the factorial of any positive 8th-integer, i.e., any number of the form k/8 for positive integer k. From these we can compute the factorials of all numbers of the form k/16, and so on. Thus we have the factorials of all numbers whose binary expansions are finite, and from these we can define the factorial of any positive real number x as the limit of the factorials of finite binary numbers as they approach arbitrarily close to x.

 

Another approach to defining a continuous function whose value equals the factorial of n for integer arguments n is to consider the sequence of derivatives of 1/x, which gives

 

 

In general, the magnitude of the nth derivative of 1/x is n!/xn-1. Now, we can also express 1/x as the definite integral

 

 

Therefore, the successive derivatives of the left hand side equal the derivatives of the right hand side. Recall Leibniz's rule for the derivative of an integral

 

 

Since the limits of integration u and v are constants (i.e., independent of x), the first two terms on the right side are zero, so this shows that we can simply bring the partial differentiation inside the integrand. Hence the nth derivative is

 

 

Setting x = 1 gives an alternative definition of the factorial function

 

 

It's easy to show that this definition satisfies all the same recursive relations as our previous definition (inferred from the formulas for the volumes of n-dimensional spheres), and it has the same "boundary values", so the definitions are equivalent. Notice that both of these definitions are applicable to arbitrary values of n, not just integers. It's customary to express this factorial function in terms of the "gamma function", denoted by the Greek symbol G, whose argument is offset by 1 from the factorial, i.e.,

 

 

and this function is tabulated in many reference books. Of course, given the values of x! over any unit interval, we can easily compute all other values by means of the relation x! = x(x-1)!

 

Returning to equations (2) and (3) for the volumes of spheres of even and odd dimensions, we can assert now that those two formulas are equivalent, i.e., if we set 2n in equation (2) and 2n+1 in equation (3) to any positive real value u, we have

 

 

Either one of these formulas can be used to give Vu[R]. Equating them gives the relation

 

 

Writing u! = u(u-1)! and (u/2)! = (u/2)(u/2 - 1)!, this becomes

 

 

Setting u-1 = 2v, this can also be written as

 

 

This is called the duplication formula. If we let I(x) denote the definite integral

 

 

then equation (4) can be written as

 

 

Solving for (2v)! and substituting into (5) gives

 

 

Also, equation (6) can be re-arranged to give

 

 

which, for integer values of v, furnishes a formula for the “middle” binomial coefficients.

Lastly, we have the relation

 

 

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