Volume of nSpheres 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 1sphere is simply 2R. A sphere of radius R in two dimensions can be regarded as consisting of a continuous sequence of onedimensional spheres at a perpendicular distance u from some fixed 1sphere, and such that the radius of the 1sphere at a distance u is r(u) = (R^{2}  u^{2})^{1/2}, where u varies uniformly from R to +R. In general, an nsphere consists of a sequence of (n1)spheres of radius r(u) as u varies uniformly from R to +R. 

Therefore, given the volume formula V_{n}[R] for ndimensional spheres, we can determine the formula V_{n+1}[R] for (n+1)spheres as follows 

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Also, it's clear that the volume of an nsphere must be proportional to R^{n}, so for every n there is a constant k_{n} such that the volume of an nsphere of radius R is given by V_{n}[R] = k_{n} R^{n}. Thus we have 

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

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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 

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Using this formula, beginning with the (1)dimensional case V_{1}[R] = 1/(pR), we can recursively compute the volumes of the higher nspheres 

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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 

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By equation (1) we also have 

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Substituting from the previous expressions for V_{2n+1} and V_{2n}, canceling terms, and solving for n! gives 

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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 

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

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Furthermore, given all the factorials of the whole and half integers, equation (4) enables us to compute the factorials of all the quarterintegers, because if we set n to a quarterinteger 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 8thinteger, 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 

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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 

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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 

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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 

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

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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 ndimensional 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., 

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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(x1)! 

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 

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Either one of these formulas can be used to give V_{u}[R]. Equating them gives the relation 

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Writing u! = u(u1)! and (u/2)! = (u/2)(u/2  1)!, this becomes 

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Setting u1 = 2v, this can also be written as 

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

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then equation (4) can be written as 

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

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Also, equation (6) can be rearranged to give 

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

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