Puneet Varma (Editor)

Wallis product

Updated on
Edit
Like
Comment
Share on FacebookTweet on TwitterShare on LinkedInShare on Reddit
Wallis product

In mathematics, Wallis' product for π, written down in 1655 by John Wallis, states that

Contents

∏ n = 1 ∞ ( 2 n 2 n − 1 ⋅ 2 n 2 n + 1 ) = 2 1 ⋅ 2 3 ⋅ 4 3 ⋅ 4 5 ⋅ 6 5 ⋅ 6 7 ⋅ 8 7 ⋅ 8 9 ⋯ = π 2

Derivation

Wallis derived this infinite product as it is done in calculus books today, by examining ∫ 0 π sin n ⁡ x d x for even and odd values of n, and noting that for large n, increasing n by 1 results in a change that becomes ever smaller as n increases. Since modern infinitesimal calculus did not yet exist then, and the mathematical analysis of the time was inadequate to discuss the convergence issues, this was a hard piece of research, and tentative as well.

Wallis' product is, in retrospect, an easy corollary of the later Euler formula for the sine function. In 2015 researchers C. R. Hagen and Tamar Friedmann, in a surprise discovery, found the same formula in quantum mechanical calculations of the energy levels of a hydrogen atom.

Proof using Euler's infinite product for the sine function

sin ⁡ x x = ∏ n = 1 ∞ ( 1 − x 2 n 2 π 2 )

Let x = π/2:

⇒ 2 π = ∏ n = 1 ∞ ( 1 − 1 4 n 2 ) ⇒ π 2 = ∏ n = 1 ∞ ( 4 n 2 4 n 2 − 1 ) = ∏ n = 1 ∞ ( 2 n 2 n − 1 ⋅ 2 n 2 n + 1 ) = 2 1 ⋅ 2 3 ⋅ 4 3 ⋅ 4 5 ⋅ 6 5 ⋅ 6 7 ⋯

Proof using integration

Let:

I ( n ) = ∫ 0 π sin n ⁡ x d x

(a form of Wallis' integrals). Integrate by parts:

u = sin n − 1 ⁡ x ⇒ d u = ( n − 1 ) sin n − 2 ⁡ x cos ⁡ x d x d v = sin ⁡ x d x ⇒ v = − cos ⁡ x ⇒ I ( n ) = ∫ 0 π sin n ⁡ x d x = − sin n − 1 ⁡ x cos ⁡ x | 0 π − ∫ 0 π ( − cos ⁡ x ) ( n − 1 ) sin n − 2 ⁡ x cos ⁡ x d x = 0 + ( n − 1 ) ∫ 0 π cos 2 ⁡ x sin n − 2 ⁡ x d x , n > 1 = ( n − 1 ) ∫ 0 π ( 1 − sin 2 ⁡ x ) sin n − 2 ⁡ x d x = ( n − 1 ) ∫ 0 π sin n − 2 ⁡ x d x − ( n − 1 ) ∫ 0 π sin n ⁡ x d x = ( n − 1 ) I ( n − 2 ) − ( n − 1 ) I ( n ) = n − 1 n I ( n − 2 ) ⇒ I ( n ) I ( n − 2 ) = n − 1 n ⇒ I ( 2 n − 1 ) I ( 2 n + 1 ) = 2 n + 1 2 n

This result will be used below:

I ( 0 ) = ∫ 0 π d x = x | 0 π = π I ( 1 ) = ∫ 0 π sin ⁡ x d x = − cos ⁡ x | 0 π = ( − cos ⁡ π ) − ( − cos ⁡ 0 ) = − ( − 1 ) − ( − 1 ) = 2 I ( 2 n ) = ∫ 0 π sin 2 n ⁡ x d x = 2 n − 1 2 n I ( 2 n − 2 ) = 2 n − 1 2 n ⋅ 2 n − 3 2 n − 2 I ( 2 n − 4 )

Repeating the process,

= 2 n − 1 2 n ⋅ 2 n − 3 2 n − 2 ⋅ 2 n − 5 2 n − 4 ⋅ ⋯ ⋅ 5 6 ⋅ 3 4 ⋅ 1 2 I ( 0 ) = π ∏ k = 1 n 2 k − 1 2 k I ( 2 n + 1 ) = ∫ 0 π sin 2 n + 1 ⁡ x d x = 2 n 2 n + 1 I ( 2 n − 1 ) = 2 n 2 n + 1 ⋅ 2 n − 2 2 n − 1 I ( 2 n − 3 )

Repeating the process,

= 2 n 2 n + 1 ⋅ 2 n − 2 2 n − 1 ⋅ 2 n − 4 2 n − 3 ⋅ ⋯ ⋅ 6 7 ⋅ 4 5 ⋅ 2 3 I ( 1 ) = 2 ∏ k = 1 n 2 k 2 k + 1 sin 2 n + 1 ⁡ x ≤ sin 2 n ⁡ x ≤ sin 2 n − 1 ⁡ x , 0 ≤ x ≤ π ⇒ I ( 2 n + 1 ) ≤ I ( 2 n ) ≤ I ( 2 n − 1 ) ⇒ 1 ≤ I ( 2 n ) I ( 2 n + 1 ) ≤ I ( 2 n − 1 ) I ( 2 n + 1 ) = 2 n + 1 2 n , from above results.

By the squeeze theorem,

⇒ lim n → ∞ I ( 2 n ) I ( 2 n + 1 ) = 1 lim n → ∞ I ( 2 n ) I ( 2 n + 1 ) = π 2 lim n → ∞ ∏ k = 1 n ( 2 k − 1 2 k ⋅ 2 k + 1 2 k ) = 1 ⇒ π 2 = ∏ k = 1 ∞ ( 2 k 2 k − 1 ⋅ 2 k 2 k + 1 ) = 2 1 ⋅ 2 3 ⋅ 4 3 ⋅ 4 5 ⋅ 6 5 ⋅ 6 7 ⋅ ⋯

Relation to Stirling's approximation

Stirling's approximation for n! asserts that

n ! = 2 π n ( n e ) n [ 1 + O ( 1 n ) ] .

Consider now the finite approximations to the Wallis product, obtained by taking the first k terms in the product:

p k = ∏ n = 1 k 2 n 2 n − 1 2 n 2 n + 1

pk can be written as

p k = 1 2 k + 1 ∏ n = 1 k ( 2 n ) 4 [ ( 2 n ) ( 2 n − 1 ) ] 2 = 1 2 k + 1 ⋅ 2 4 k ( k ! ) 4 [ ( 2 k ) ! ] 2

Substituting Stirling's approximation in this expression (both for k! and (2k)!) one can deduce (after a short calculation) that pk converges to  π⁄2 as k → ∞.

ζ'(0)

The Riemann zeta function and the Dirichlet eta function can be defined:

ζ ( s ) = ∑ n = 1 ∞ 1 n s , ℜ ( s ) > 1 η ( s ) = ( 1 − 2 1 − s ) ζ ( s ) = ∑ n = 1 ∞ ( − 1 ) n − 1 n s , ℜ ( s ) > 0

Applying an Euler transform to the latter series, the following is obtained:

η ( s ) = 1 2 + 1 2 ∑ n = 1 ∞ ( − 1 ) n − 1 [ 1 n s − 1 ( n + 1 ) s ] , ℜ ( s ) > − 1 ⇒ η ′ ( s ) = ( 1 − 2 1 − s ) ζ ′ ( s ) + 2 1 − s ( ln ⁡ 2 ) ζ ( s ) = − 1 2 ∑ n = 1 ∞ ( − 1 ) n − 1 [ ln ⁡ n n s − ln ⁡ ( n + 1 ) ( n + 1 ) s ] , ℜ ( s ) > − 1 ⇒ η ′ ( 0 ) = − ζ ′ ( 0 ) − ln ⁡ 2 = − 1 2 ∑ n = 1 ∞ ( − 1 ) n − 1 [ ln ⁡ n − ln ⁡ ( n + 1 ) ] = − 1 2 ∑ n = 1 ∞ ( − 1 ) n − 1 ln ⁡ n n + 1 = − 1 2 ( ln ⁡ 1 2 − ln ⁡ 2 3 + ln ⁡ 3 4 − ln ⁡ 4 5 + ln ⁡ 5 6 − ⋯ ) = 1 2 ( ln ⁡ 2 1 + ln ⁡ 2 3 + ln ⁡ 4 3 + ln ⁡ 4 5 + ln ⁡ 6 5 + ⋯ ) = 1 2 ln ⁡ ( 2 1 ⋅ 2 3 ⋅ 4 3 ⋅ 4 5 ⋅ ⋯ ) = 1 2 ln ⁡ π 2 ⇒ ζ ′ ( 0 ) = − 1 2 ln ⁡ ( 2 π )

References

Wallis product Wikipedia
(Text) CC BY-SA