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A Generalized Wallis Formula

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DOI: 10.4236/am.2018.93015    278 Downloads   579 Views  
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ABSTRACT

This article generalizes the famous Wallis’s formula for k ≥ 0 , to an integral over the unit sphere Sn-1. An application to the integral of polynomials over Sn-1 is discussed.

One of Wallis formulas is

0 sin 2 k θ d θ = 0 cos 2 k θ d θ = ( 2 k ) ! 2 π 2 2 k ( k ! ) 2

for k 0 . This formula can be proved by various methods [1] [2] [3] [4] including a repeated application of a reduction formula such as

0 sin k θ d θ = k 1 k 0 sin k 2 θ d θ . Note that s i n θ and c o s θ are coordinates

of a point on the unit sphere in R2. Since the above formula involves an integration over the unit circle in R2, its extension to higher dimensions is of interest.

For each x = ( x 1 , x 2 , , x n ) R n , let | x | = ( x i 2 ) 1 / 2 be its Euclidean norm. We call α = ( α 1 , α 2 , , α n ) , where α i 0 are non-negative integers, a multi-index, and | α | = | α i | its degree. Set α ! = α 1 ! α 2 ! α n ! and x α = x 1 α 1 x 2 α 2 x n α n . Let S n 1 = { ξ R n : | ξ | = 1 } be the unit sphere in Rn and d σ be its surface measure. Let B r ( a ) = { x R n : | x a | r } stand for the ball of radius r centered at a. The gamma function is defined as Γ ( s ) = 0 e t t s 1 d t , for s > 0 . The generalized Wallis’s formula is a special case of the following theorem.

Theorem 1 (i) S n 1 ξ α d σ = 0 , if any α i is odd. In particular, the integral equals zero if | α | is odd.

(ii) S n 1 ξ 2 α d σ = ( 2 α ) ! 2 π n / 2 2 2 | α | α ! Γ ( n / 2 + | α | ) , | α | 0 .

Setting α i = k and α j = 0 for j i in the theorem, the generalized Wallis’s formula follows

S n 1 ξ i 2 k d σ = ( 2 k ) ! 2 π n / 2 2 2 k k ! Γ ( n / 2 + k ) , k 0.

Note that for | α | = 0 , (ii) is equivalent to the well-known formula

ω n 1 = 2 π n / 2 Γ ( n / 2 ) (1)

where ω n 1 is the surface area of the unit sphere in Rn. Theorem 1 is interesting in its own right and has further applications. For example, for a polynomial

p ( x ) = | α | m b α x α of degree m, one may express B r ( 0 ) p ( x ) d x as a simple

polynomial of degree n + m in r. In the following we use polar coordinates x = ρ ξ , ρ = | x | , ξ S n 1 .

B r ( 0 ) p ( x ) d x = | α | m b α B r ( 0 ) x α d x = | α | m b α 0 r ρ | α | + n 1 d ρ S n 1 ξ α d σ = 2 | α | m b 2 α r 2 | α | + n 2 | α | + n S n 1 ξ 2 α d σ = | α | [ m / 2 ] b 2 α d α 2 | α | + n r 2 k + n = k = 0 [ m / 2 ] ( | α | = k b 2 α d α 2 k + n ) r 2 k + n = k = 0 [ m / 2 ] c k r 2 k + n .

Here d α = S n 1 ξ 2 α d σ as given by (ii), and [.] is the bracket function.

Proof of Theorem 1. (i) The proof is by induction on | α | .

If | α | = 1 then ξ = ξ i for some i. Therefore, S n 1 ξ α d σ = S n 1 ξ i d σ = 0 by the symmetry of the sphere.

Assume now the assertion is true for | α | m for some m 1 . Let | α | = m + 1 and assume, without loss of generality, that α 1 is odd. Applying the divergence theorem results in

S n 1 ξ α d σ = S n 1 ξ 1 ( ξ 1 α 1 1 ξ 2 α 2 ξ n α n ) d σ = B 1 ( 0 ) x 1 ( x 1 α 1 1 x 2 α 2 x n α n ) d x . (2)

If α 1 = 1 , the last integral in (2) is zero. Otherwise, a conversion to polar coordinates in (2), yields,

S n 1 ξ α d σ = ( α 1 1 ) B 1 ( 0 ) x 1 α 1 2 x 2 α 2 x n α n d x = ( α 1 1 ) 0 1 ρ m + n 2 d ρ S n 1 ξ β d σ = α 1 1 m + n 1 S n 1 ξ β d σ ,

where β = ( α 1 2 , α 2 , , α n ) . The last integral is now zero, by the induction hypothesis.

ii) The proof is by induction on | α | .

For | α | = 0 , we must establish (1). Let e n = R n e π | x | 2 d x . Writing e n as a product of integrals and using polar coordinates in R2 followed by a change of variables, one obtains

e n = i = 1 n R e π x i 2 d x i = ( e 1 ) n = ( e 2 ) n / 2 = ( 0 d θ 0 r e π r 2 d r ) n / 2 = ( 0 e u d u ) n / 2 = 1.

We used a change of variable u = π r 2 in the previous integral. Converting to polar coordinates for Rn results in

1 = e n = R n e π | x | 2 d x = ω n 1 0 r n 1 e π r 2 d r = ω n 1 2 π n / 2 0 u n / 2 1 e u d u = Γ ( n / 2 ) 2 π n / 2 ω n 1 .

Identity (1) follows immediately from the last equation.

Now suppose the claim is true for | α | = m . Let | α | = m + 1 . We may assume, without loss of generality, that α 1 1 . Applying the divergence theorem followed by a conversion to polar coordinates leads to

S n 1 ξ 2 α d σ = S n 1 ξ 1 ( ξ 1 2 α 1 1 ξ 2 2 α 2 ξ n 2 α n ) d σ = B 1 ( 0 ) x 1 ( x 1 2 α 1 1 x 2 2 α 2 x n 2 α n ) d x = ( 2 α 1 1 ) B 1 ( 0 ) x 1 2 α 1 2 x 2 2 α 2 x n 2 α n d x = 2 α 1 1 n + 2 m S n 1 ξ 2 β d σ ,

where β = ( α 1 1 , α 2 , , α n ) . Since | β | = m , and using the fact that Γ ( s + 1 ) = s Γ ( s ) along with the induction hypothesis, we get

S n 1 ξ 2 α d σ = 2 α 1 1 n + 2 m . ( 2 β ) ! 2 π n / 2 2 2 | β | β ! Γ ( n / 2 + m ) = ( 2 α ) ! 2 π n / 2 2 2 | α | α ! Γ ( n / 2 + | α | ) .

Conflicts of Interest

The authors declare no conflicts of interest.

Cite this paper

Namazi, J. (2018) A Generalized Wallis Formula. Applied Mathematics, 9, 207-209. doi: 10.4236/am.2018.93015.

References

[1] Muller, C. (1966) Spherical Harmonics. Springer-Verlag Berlin Heidelberg.
https://doi.org/10.1007/BFb0094775
[2] Macrobert, T.M. (1948) Spherical Harmonics: An Elementary Treatise on Harmonic Functions with Applications. Dover Publications, New York.
[3] Stein, E. and Weiss, G. (1971) Introduction to Fourier Analysis on Euclidean Spaces. Princeton University Press, Princeton.
[4] Jeffrey, H. and Jeffrey, B. (1999) Methods of Mathematical Physics. 3rd Edition, Cambridge University Press, Cambridge.

  
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