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

n=1(2n2n12n2n+1)=2123434565678789=π2

Derivation

Wallis did not derive this infinite product as it is done in calculus books today, by comparing 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. Since infinitesimal calculus as we know it 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.

Proof using Euler's infinite product for the sine function[1]

sinxx=n=1(1x2n2π2)

Let x = Template:Frac:

2π=n=1(114n2)π2=n=1(4n24n21)=n=1(2n2n12n2n+1)=212343456567

Proof using integration[2]

Let:

I(n)=0πsinnxdx

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

u=sinn1xdu=(n1)sinn2xcosxdxdv=sinxdxv=cosx
I(n)=0πsinnxdx=0πudv=uv|x=0x=π0πvdu=sinn1xcosx|x=0x=π0πcosx(n1)sinn2xcosxdx=0(n1)0πcos2xsinn2xdx,n>1=(n1)0π(1sin2x)sinn2xdx=(n1)0πsinn2xdx(n1)0πsinnxdx=(n1)I(n2)(n1)I(n)=n1nI(n2)I(n)I(n2)=n1nI(2n1)I(2n+1)=2n+12n

This result will be used below:

I(0)=0πdx=x|0π=πI(1)=0πsinxdx=cosx|0π=(cosπ)(cos0)=(1)(1)=2I(2n)=0πsin2nxdx=2n12nI(2n2)=2n12n2n32n2I(2n4)

Repeating the process,

=2n12n2n32n22n52n4563412I(0)=πk=1n2k12k
I(2n+1)=0πsin2n+1xdx=2n2n+1I(2n1)=2n2n+12n22n1I(2n3)

Repeating the process,

=2n2n+12n22n12n42n3674523I(1)=2k=1n2k2k+1
sin2n+1xsin2nxsin2n1x,0xπ
I(2n+1)I(2n)I(2n1)
1I(2n)I(2n+1)I(2n1)I(2n+1)=2n+12n, from above results.

By the squeeze theorem,

limnI(2n)I(2n+1)=1
limnI(2n)I(2n+1)=π2limnk=1n(2k12k2k+12k)=1
π2=k=1(2k2k12k2k+1)=212343456567

Relation to Stirling's approximation

Stirling's approximation for n! asserts that

n!=2πn(ne)n[1+O(1n)]

as n → ∞. Consider now the finite approximations to the Wallis product, obtained by taking the first k terms in the product:

pk=n=1k2n2n12n2n+1

pk can be written as

pk=12k+1n=1k(2n)4[(2n)(2n1)]2=12k+124k(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 Template:Frac as k → ∞.

ζ'(0)[1]

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

ζ(s)=n=11ns,(s)>1η(s)=(121s)ζ(s)=n=1(1)n1ns,(s)>0

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

η(s)=12+12n=1(1)n1[1ns1(n+1)s],(s)>1η(s)=(121s)ζ(s)+21s(ln2)ζ(s)=12n=1(1)n1[lnnnsln(n+1)(n+1)s],(s)>1
η(0)=ζ(0)ln2=12n=1(1)n1[lnnln(n+1)]=12n=1(1)n1lnnn+1=12(ln12ln23+ln34ln45+ln56)=12(ln21+ln23+ln43+ln45+ln65+)=12ln(21234345)=12lnπ2ζ(0)=12ln(2π)

See also

External links

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