Wallis product

Short description: Infinite product for pi

Comparison of the convergence of the Wallis product (purple asterisks) and several historical infinite series for π. Sn is the approximation after taking n terms. Each subsequent subplot magnifies the shaded area horizontally by 10 times. (click for detail)

The Wallis product is the infinite product representation of π:

π2=∏n=1∞4n24n2−1=∏n=1∞(2n2n−1⋅2n2n+1)=(21⋅23)⋅(43⋅45)⋅(65⋅67)⋅(87⋅89)⋅⋯

It was published in 1656 by John Wallis.[1]

## Contents

* 1 Proof using integration
* 1.1 Proof using Laplace's method
* 2 Proof using Euler's infinite product for the sine function
* 3 Relation to Stirling's approximation
* 4 Derivative of the Riemann zeta function at zero
* 5 See also
* 6 Notes
* 7 External links

## Proof using integration

Wallis derived this infinite product using interpolation, though his method is not regarded as rigorous. A modern derivation can be found by examining ∫0πsinnxdx 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. Let[2]

I(n)=∫0πsinnxdx.

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

u=sinn−1x⇒du=(n−1)sinn−2xcos⁡xdxdv=sin⁡xdx⇒v=−cos⁡x

⇒I(n)=∫0πsinnxdx=−sinn−1xcos⁡x|0π−∫0π(−cos⁡x)(n−1)sinn−2xcos⁡xdx=0+(n−1)∫0πcos2xsinn−2xdx,n>1=(n−1)∫0π(1−sin2x)sinn−2xdx=(n−1)∫0πsinn−2xdx−(n−1)∫0πsinnxdx=(n−1)I(n−2)−(n−1)I(n)=n−1nI(n−2)⇒I(n)I(n−2)=n−1n

Now, we make two variable substitutions for convenience to obtain:

I(2n)=2n−12nI(2n−2)
I(2n+1)=2n2n+1I(2n−1)

We obtain values for I(0) and I(1) for later use.

I(0)=∫0πdx=x|0π=πI(1)=∫0πsin⁡xdx=−cos⁡x|0π=(−cos⁡π)−(−cos⁡0)=−(−1)−(−1)=2

Now, we calculate for even values I(2n) by repeatedly applying the recurrence relation result from the integration by parts. Eventually, we end get down to I(0), which we have calculated.

I(2n)=∫0πsin2nxdx=2n−12nI(2n−2)=2n−12n⋅2n−32n−2I(2n−4)

=2n−12n⋅2n−32n−2⋅2n−52n−4⋅⋯⋅56⋅34⋅12I(0)=π∏k=1n2k−12k

Repeating the process for odd values I(2n+1),

I(2n+1)=∫0πsin2n+1xdx=2n2n+1I(2n−1)=2n2n+1⋅2n−22n−1I(2n−3)

=2n2n+1⋅2n−22n−1⋅2n−42n−3⋅⋯⋅67⋅45⋅23I(1)=2∏k=1n2k2k+1

We make the following observation, based on the fact that sin⁡x≤1

sin2n+1x≤sin2nx≤sin2n−1x,0≤x≤π

⇒I(2n+1)≤I(2n)≤I(2n−1)

Dividing by I(2n+1):

⇒1≤I(2n)I(2n+1)≤I(2n−1)I(2n+1)=2n+12n, where the equality comes from our recurrence relation.

By the squeeze theorem,

⇒limn->∞I(2n)I(2n+1)=1

limn->∞I(2n)I(2n+1)=π2limn->∞∏k=1n(2k−12k⋅2k+12k)=1

⇒π2=∏k=1∞(2k2k−1⋅2k2k+1)=21⋅23⋅43⋅45⋅65⋅67⋅⋯

### Proof using Laplace's method

See the main page on Gaussian integral.

## Proof using Euler's infinite product for the sine function

While the proof above is typically featured in modern calculus textbooks, the Wallis product is, in retrospect, an easy corollary of the later Euler infinite product for the sine function.

sin⁡xx=∏n=1∞(1−x2n2π2)

Let x=π2:

⇒2π=∏n=1∞(1−14n2)⇒π2=∏n=1∞(4n24n2−1)=∏n=1∞(2n2n−1⋅2n2n+1)=21⋅23⋅43⋅45⋅65⋅67⋯ [1]

## Relation to Stirling's approximation

Stirling's approximation for the factorial function n! asserts that

n!=2πn(ne)n[1+O(1n)].

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

pk=∏n=1k2n2n−12n2n+1,

where pk can be written as

pk=12k+1∏n=1k(2n)4[(2n)(2n−1)]2=12k+1⋅24k(k!)4[(2k)!]2.

Substituting Stirling's approximation in this expression (both for k! and (2k)!) one can deduce (after a short calculation) that pk converges to π2 as k->∞.

## Derivative of the Riemann zeta function at zero

The Riemann zeta function and the Dirichlet eta function can be defined:[1]

ζ(s)=∑n=1∞1ns,ℜ(s)>1η(s)=(1−21−s)ζ(s)=∑n=1∞(−1)n−1ns,ℜ(s)>0

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

η(s)=12+12∑n=1∞(−1)n−1[1ns−1(n+1)s],ℜ(s)>−1⇒η′(s)=(1−21−s)ζ′(s)+21−s(ln⁡2)ζ(s)=−12∑n=1∞(−1)n−1[ln⁡nns−ln⁡(n+1)(n+1)s],ℜ(s)>−1

⇒η′(0)=−ζ′(0)−ln⁡2=−12∑n=1∞(−1)n−1[ln⁡n−ln⁡(n+1)]=−12∑n=1∞(−1)n−1ln⁡nn+1=−12(ln⁡12−ln⁡23+ln⁡34−ln⁡45+ln⁡56−⋯)=12(ln⁡21+ln⁡23+ln⁡43+ln⁡45+ln⁡65+⋯)=12ln⁡(21⋅23⋅43⋅45⋅⋯)=12ln⁡π2⇒ζ′(0)=−12ln⁡(2π)

## See also

* John Wallis, English mathematician who is given partial credit for the development of infinitesimal calculus and pi.
* Viète's formula, a different infinite product formula for π.
* Leibniz formula for π, an infinite sum that can be converted into an infinite Euler product for π.
* Wallis sieve
* The Pippenger product formula obtains e by taking roots of terms in the Wallis product.

## Notes

1. ↑ 1.0 1.1 1.2 "Wallis Formula". http://mathworld.wolfram.com/WallisFormula.html.
2. ↑ "Integrating Powers and Product of Sines and Cosines: Challenging Problems". http://www.sosmath.com/calculus/integration/powerproduct/problem/problem.html.

## External links

* Hazewinkel, Michiel, ed. (2001), "Wallis formula", Encyclopedia of Mathematics, Springer Science+Business Media B.V. / Kluwer Academic Publishers, ISBN 978-1-55608-010-4, https://www.encyclopediaofmath.org/index.php?title=p/w097040
* "Why does this product equal π/2? A new proof of the Wallis formula for π.". 3Blue1Brown. April 20, 2018. https://www.youtube.com/watch?v=8GPy_UMV-08.

0.00

(0 votes)

Original source: https://en.wikipedia.org/wiki/Wallis product. Read more

|

Retrieved from "https://handwiki.org/wiki/index.php?title=Wallis_product&oldid=3987170"