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) = sqrt(R^2 - u^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 V_{n}[R] for n-dimensional spheres,
we can determine the formula V_{n+1}[R] for (n+1)-spheres as follows
R
/
V_{n+1}[R] = | V_n[r(u)] du
/
u=-R
Also, it's clear that the volume of an n-sphere must be proportional
to R^n, so for every n there is a constant k_n such that the volume
of an n-sphere of radius R is given by V_n[R] = k_n R^n. Thus we
have
R
/
V_{n+1}[R] = k_n | r(u)^n du
/
u=-R
Inserting the expression for r(u) = sqrt(R^2 - u^2) and factoring out
a constant R from the integrand gives
R
/ / 2 \ n/2
V_{n+1}[R] = k_n R^n | ( 1 - (u/R) ) du
/ \ /
u=-R
The outside factor on the integrand is simply V_n[R], and if we define
the variable s = u/R we have du = R ds, so the above formula can be
written as
1
/ / 2 \ n/2
V_{n+1}[R] = R V_n[R] | ( 1 - s ) ds (1)
/ \ /
s=-1
Using this formula, beginning with the (-1)-dimensional case
V_{-1}[R] = 1/(pi R), we can recursively compute the volumes of the
higher n-spheres
n V_n[R] k_n
----- ---------------- --------
-1 pi^-1 R^-1 0.3183
0 1 1.0000
1 2R 2.0000
2 pi R^2 3.1415
3 (4/3) pi R^3 4.1887
4 (1/2) pi^2 R^4 4.9348
5 (8/15) pi^2 R^5 5.2637
6 (1/6) pi^3 R^6 5.1677
7 (16/105) pi^3 R^7 4.7247
8 (1/24) pi^4 R^8 4.0587
9 (32/945) pi^4 R^9 3.2985
etc.
This shows that the integer dimension with maximum integer coefficient
is n = 5. Also, the power of pi 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
n
pi 2n
V_{2n}[R] = ----- R (2)
n!
2n+1
2 n! n 2n+1
V_{2n+1}[R] = -------- pi R (3)
(2n+1)!
By equation (1) we also have
1
/ / 2 \ n
V_{2n+1}[R] = R V_{2n}[R] | ( 1 - s ) ds
/ \ /
s=-1
Substituting from the previous expressions for V_{2n+1} and V_{2n},
cancelling terms, and solving for n! gives
_______________________
/ +1
/ (2n+1)! / n
n! = / --------- | (1-u^2) du (4)
/ 2^(2n+1) /
/ u=-1
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
__
(-1/2)! = /pi
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 (1/2)! = (1/2)sqrt(pi). 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
1/x -1/x^2 2/x^3 -6/x^4 24/x^5 -120/x^6 ...
In general, the magnitude of the nth derivative of 1/x is n!/x^(n-1).
Now, we can also express 1/x as the definite integral
oo
/ -xt 1
| e dt = ---
/ x
t=0
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
_ _
| v | v
d | / | dv du / Df
-- | | f(x,t) dt | = f(x,v) -- - f(x,u) -- + | -- dt
dx | / | dx dv / Dx
|_ u _| u
Since our limits of integration are constants (i.e., independent of
x), we can simply bring the partial differentiation inside the
integrand. Hence the nth derivative is
oo
/ n -xt n!
| t e dt = -------
/ x^(n-1)
t=0
Setting x = 1 gives our alternative definition of the factorial
function
oo
/ n -t
| t e dt = n!
/
t=0
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", whose argument is offset
by 1 from the factorial, i.e.,
gamma(x) = (x-1)!
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
u/2
pi u
V_{u}[R] = ------ R
(u/2)!
u
2 ((u-1)/2)! u/2 u
V_{u}[R] = -------------- pi R
1/2
(u)! pi
Either one of these formulas can be used to give V_{u}[R]. Equating
them gives the relation
1/2 u
u! pi = 2 ((u-1)/2)! (u/2)!
Writing u! = u(u-1)! and (u/2)! = (u/2)(u/2 - 1)!, this becomes
1/2 u-1
(u-1)! pi = 2 (u/2 - 1/2)! (u/2 - 1)!
Setting u-1 = 2v, this can also be written as
2v 1/2
2 (v)! (v - 1/2)! = pi (2v)! (5)
This is called the duplication formula. If we let I(x) denote the
definite integral
+1 1 x
/ x / w
I(x) = | (1 - u^2) du = | --------- dw
/ / sqrt(1-w)
u=-1 w=0
then equation (4) can be written as
2v+1 2
2 (v!) = (2v+1)! I(v) = (2v+1)(2v)! I(v) (6)
Solving for (2v)! and substituting into (5) gives
v! 1
---------- = -------- I(v)
(v + 1/2)! sqrt(pi)
Also, equation (6) can be re-arranged to the form
2v+1 2v
(2v)! 2 2
-------- = ----------- = ---- I(v - 1/2)
(v!)(v!) (2v+1) I(v) pi
which gives a formula for the "middle" binomial coefficients. We
also have the relation
pi
I(w) I(w + 1/2) = -----
w + 1
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