Wallis product explained

In mathematics, the Wallis product for , published in 1656 by John Wallis,^[1] states that

\begin{align}	\pi
	2

	infty
\prod
	n=1

	4n²
	4n²-1

	infty
\prod		\left(
	n=1

	2n
	2n-1

⋅

	2n
	2n+1

\right)\\[6pt] &=(

	2
	1

⋅

	2
	3

) ⋅ (

	4
	3

⋅

	4
	5

) ⋅ (

	6
	5

⋅

	6
	7

) ⋅ (

	8
	7

⋅

	8
	9

) ⋅ … \\ \end{align}

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

	\pi
\int
	0

\sinⁿxdx

for even and odd values of

, and noting that for large

, increasing

by 1 results in a change that becomes ever smaller as

increases. Let^[2]

I(n)=

	\pi
\int
	0

\sinⁿxdx.

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

\begin{align} u&=\sin^n-1x\\ ⇒ du&=(n-1)\sin^n-2x\cosxdx\\ dv&=\sinxdx\\ ⇒ v&=-\cosx \end{align}

\begin{align} ⇒ I(n)&=

	\pi
\int
	0

\sinⁿxdx\\[6pt] {}&=-\sin^n-1x\cosx

	\pi
l\|
	0

	\pi
\int
	0

(-\cosx)(n-1)\sin^n-2x\cosxdx\\[6pt] {}&=0+(n-1)

	\pi
\int
	0

\cos^2x\sin^n-2xdx, n>1\\[6pt] {}&=(n-1)

	\pi
\int
	0

(1-\sin²x)\sin^n-2xdx\\[6pt] {}&=(n-1)

	\pi
\int
	0

\sin^n-2xdx-(n-1)

	\pi
\int
	0

\sinⁿxdx\\[6pt] {}&=(n-1)I(n-2)-(n-1)I(n)\\[6pt] {}&=

	n-1
	n

I(n-2)\\[6pt] ⇒

	I(n)
	I(n-2)

	n-1
	n

\\[6pt] \end{align}

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

I(2n)=

	2n-1
	2n

I(2n-2)

I(2n+1)=

	2n
	2n+1

I(2n-1)

We obtain values for

I(0)

and

I(1)

for later use.

\begin{align} I(0)&=

	\pi
\int
	0

dx=

	\pi
xl\|
	0

=\pi\\[6pt] I(1)&=

	\pi
\int
	0

\sinxdx=-\cosx

	\pi
l\|
	0

=(-\cos\pi)-(-\cos0)=-(-1)-(-1)=2\\[6pt] \end{align}

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.

	\pi
I(2n)=\int
	0

\sin²ⁿxdx=

	2n-1
	2n

I(2n-2)=

	2n-1
	2n

⋅

	2n-3
	2n-2

I(2n-4)

=	2n-1
	2n

⋅

	2n-3
	2n-2

⋅

	2n-5
	2n-4

⋅ … ⋅

	5
	6

⋅

	3
	4

⋅

	1
	2

I(0)=\pi

	n
\prod
	k=1

	2k-1
	2k

Repeating the process for odd values

I(2n+1)

	\pi
I(2n+1)=\int
	0

\sin²ⁿ⁺¹xdx=

	2n	I(2n-1)=
	2n+1

	2n
	2n+1

⋅

	2n-2
	2n-1

I(2n-3)

=	2n
	2n+1

⋅

	2n-2
	2n-1

⋅

	2n-4
	2n-3

⋅ … ⋅

	6
	7

⋅

	4
	5

⋅

	2
	3

I(1)=2

	n
\prod
	k=1

	2k
	2k+1

We make the following observation, based on the fact that

\sin{x}\leq1

\sin²ⁿ⁺¹x\le\sin²ⁿx\le\sin^2n-1x,0\lex\le\pi

⇒ I(2n+1)\leI(2n)\leI(2n-1)

Dividing by

I(2n+1)

⇒ 1\le

	I(2n)
	I(2n+1)

\le

	I(2n-1)	=
	I(2n+1)

	2n+1
	2n

, where the equality comes from our recurrence relation.

By the squeeze theorem,

⇒ \lim_{n → infty}

	I(2n)
	I(2n+1)

\lim_{n → infty}

	I(2n)	=
	I(2n+1)

	\pi
	2

\lim_{n → infty}

	n
\prod		\left(
	k=1

	2k-1
	2k

⋅

	2k+1
	2k

\right)=1

⇒

	\pi
	2

	infty
=\prod		\left(
	k=1

	2k
	2k-1

⋅

	2k	\right)=
	2k+1

	2
	1

⋅

	2
	3

⋅

	4
	3

⋅

	4
	5

⋅

	6
	5

⋅

	6
	7

⋅ …

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.

	\sinx
	x

	infty\left(1
\prod
	n=1

	x²
	n^2\pi²

\right)

Let

	\pi
	2

\begin{align} ⇒

	2
	\pi

	infty
\prod
	n=1

\left(1-

	1
	4n²

\right)\\[6pt] ⇒

	\pi
	2

	infty
\prod		\left(
	n=1

	4n²
	4n²-1

\right)\\[6pt] &=

	infty
\prod		\left(
	n=1

	2n	⋅
	2n-1

	2n
	2n+1

\right)=

	2
	1

⋅

	2
	3

⋅

	4
	3

⋅

	4
	5

⋅

	6
	5

⋅

	6
	7

… \end{align}

Relation to Stirling's approximation

Stirling's approximation for the factorial function

asserts that

n!=\sqrt{2\pin}{\left(

	n
	e

\right)}ⁿ\left[1+O\left(

	1
	n

\right)\right].

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

terms in the product

p_k=

	k
\prod
	n=1

	2n
	2n-1

	2n
	2n+1

where

p_k

can be written as

\begin{align} p_k&={1\over{2k+1}}

	k
\prod
	n=1

	(2n)⁴
	[(2n)(2n-1)]²

\\[6pt] &={1\over{2k+1}} ⋅ {{2^4k(k!)^4}\over{[(2k)!]^{2}}.
\end{align}}

Substituting Stirling's approximation in this expression (both for

and

(2k)!

) one can deduce (after a short calculation) that

p_k

converges to

	\pi
	2

k → infty

Derivative of the Riemann zeta function at zero

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

\begin{align} \zeta(s)&=

	infty
\sum
	n=1

	1
	n^s

,\Re(s)>1\\[6pt] η(s)&=(1-2^1-s)\zeta(s)\\[6pt] &=

	infty
\sum
	n=1

	(-1)^n-1
	n^s

,\Re(s)>0 \end{align}

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

\begin{align} η(s)&=

	1	+
	2

	1
	2

	infty
\sum
	n=1

(-1)^n-1\left[

	1	-
	n^s

	1
	(n+1)^s

\right],\Re(s)>-1\\[6pt] ⇒ η'(s)&=(1-2^1-s)\zeta'(s)+2^1-s(ln2)\zeta(s)\\[6pt] &=-

	1
	2

	infty
\sum
	n=1

(-1)^n-1\left[

	lnn	-
	n^s

	ln(n+1)
	(n+1)^s

\right],\Re(s)>-1 \end{align}

\begin{align} ⇒ η'(0)&=-\zeta'(0)-ln2=-

	1
	2

	infty
\sum
	n=1

(-1)^n-1\left[lnn-ln(n+1)\right]\\[6pt] &=-

	1
	2

	infty
\sum
	n=1

(-1)^n-1ln

	n
	n+1

\\[6pt] &=-

	1
	2

\left(ln

	1
	2

-ln

	2
	3

+ln

	3
	4

-ln

	4
	5

+ln

	5
	6

- … \right)\\[6pt] &=

	1
	2

\left(ln

	2
	1

+ln

	2
	3

+ln

	4
	3

+ln

	4
	5

+ln

	6
	5

+ … \right)\\[6pt] &=

	1	ln\left(
	2

	2	⋅
	1

	2	⋅
	3

	4	⋅
	3

	4
	5

⋅ … \right)=

	1	ln
	2

	\pi
	2

\\ ⇒ \zeta'(0)&=-

	1
	2

ln\left(2\pi\right) \end{align}

External links

- Web site: April 20, 2018 . Why does this product equal π/2? A new proof of the Wallis formula for π. . https://ghostarchive.org/varchive/youtube/20211212/8GPy_UMV-08. 2021-12-12 . live. YouTube. 3Blue1Brown.

Notes and References

Web site: Wallis Formula.
Web site: Integrating Powers and Product of Sines and Cosines: Challenging Problems.

Wallis product explained

Proof using integration

Proof using Euler's infinite product for the sine function

Relation to Stirling's approximation

Derivative of the Riemann zeta function at zero

See also

External links

Notes and References