Wallis product

The Wallis product is the infinite product representation of π:

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

)

⋅

(

8

7

⋅

8

9

)

⋅

⋯

{\displaystyle {\begin{aligned}{\frac {\pi }{2}}&=\prod _{n=1}^{\infty }{\frac {4n^{2}}{4n^{2}-1}}=\prod _{n=1}^{\infty }\left({\frac {2n}{2n-1}}\cdot {\frac {2n}{2n+1}}\right)\\[6pt]&={\Big (}{\frac {2}{1}}\cdot {\frac {2}{3}}{\Big )}\cdot {\Big (}{\frac {4}{3}}\cdot {\frac {4}{5}}{\Big )}\cdot {\Big (}{\frac {6}{5}}\cdot {\frac {6}{7}}{\Big )}\cdot {\Big (}{\frac {8}{7}}\cdot {\frac {8}{9}}{\Big )}\cdot \;\cdots \\\end{aligned}}}

It was published in 1656 by John Wallis.

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

∫

0

π

sin

n

⁡

x

d

x

{\displaystyle \int _{0}^{\pi }\sin ^{n}x\,dx}

for even and odd values of

n

{\displaystyle n}

, and noting that for large

n

{\displaystyle n}

, increasing

n

{\displaystyle n}

by 1 results in a change that becomes ever smaller as

n

{\displaystyle n}

increases. Let

I

(

n

)

=

∫

0

π

sin

n

⁡

x

d

x

.

{\displaystyle I(n)=\int _{0}^{\pi }\sin ^{n}x\,dx.}

(This is 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

{\displaystyle {\begin{aligned}u&=\sin ^{n-1}x\\\Rightarrow du&=(n-1)\sin ^{n-2}x\cos x\,dx\\dv&=\sin x\,dx\\\Rightarrow v&=-\cos x\end{aligned}}}

⇒

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

{\displaystyle {\begin{aligned}\Rightarrow I(n)&=\int _{0}^{\pi }\sin ^{n}x\,dx\\[6pt]{}&=-\sin ^{n-1}x\cos x{\Biggl |}_{0}^{\pi }-\int _{0}^{\pi }(-\cos x)(n-1)\sin ^{n-2}x\cos x\,dx\\[6pt]{}&=0+(n-1)\int _{0}^{\pi }\cos ^{2}x\sin ^{n-2}x\,dx,\qquad n>1\\[6pt]{}&=(n-1)\int _{0}^{\pi }(1-\sin ^{2}x)\sin ^{n-2}x\,dx\\[6pt]{}&=(n-1)\int _{0}^{\pi }\sin ^{n-2}x\,dx-(n-1)\int _{0}^{\pi }\sin ^{n}x\,dx\\[6pt]{}&=(n-1)I(n-2)-(n-1)I(n)\\[6pt]{}&={\frac {n-1}{n}}I(n-2)\\[6pt]\Rightarrow {\frac {I(n)}{I(n-2)}}&={\frac {n-1}{n}}\\[6pt]\end{aligned}}}

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

I

(

2

n

)

=

2

n

−

1

2

n

I

(

2

n

−

2

)

{\displaystyle I(2n)={\frac {2n-1}{2n}}I(2n-2)}

I

(

2

n

+

1

)

=

2

n

2

n

+

1

I

(

2

n

−

1

)

{\displaystyle I(2n+1)={\frac {2n}{2n+1}}I(2n-1)}

We obtain values for

I

(

0

)

{\displaystyle I(0)}

and

I

(

1

)

{\displaystyle I(1)}

for later use.

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

{\displaystyle {\begin{aligned}I(0)&=\int _{0}^{\pi }dx=x{\Biggl |}_{0}^{\pi }=\pi \\[6pt]I(1)&=\int _{0}^{\pi }\sin x\,dx=-\cos x{\Biggl |}_{0}^{\pi }=(-\cos \pi )-(-\cos 0)=-(-1)-(-1)=2\\[6pt]\end{aligned}}}

Now, we calculate for even values

I

(

2

n

)

{\displaystyle I(2n)}

by repeatedly applying the recurrence relation result from the integration by parts. Eventually, we end get down to

I

(

0

)

{\displaystyle I(0)}

, which we have calculated.

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

)

{\displaystyle I(2n)=\int _{0}^{\pi }\sin ^{2n}x\,dx={\frac {2n-1}{2n}}I(2n-2)={\frac {2n-1}{2n}}\cdot {\frac {2n-3}{2n-2}}I(2n-4)}

=

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

{\displaystyle ={\frac {2n-1}{2n}}\cdot {\frac {2n-3}{2n-2}}\cdot {\frac {2n-5}{2n-4}}\cdot \cdots \cdot {\frac {5}{6}}\cdot {\frac {3}{4}}\cdot {\frac {1}{2}}I(0)=\pi \prod _{k=1}^{n}{\frac {2k-1}{2k}}}

Repeating the process for odd values

I

(

2

n

+

1

)

{\displaystyle I(2n+1)}

,

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

)

{\displaystyle I(2n+1)=\int _{0}^{\pi }\sin ^{2n+1}x\,dx={\frac {2n}{2n+1}}I(2n-1)={\frac {2n}{2n+1}}\cdot {\frac {2n-2}{2n-1}}I(2n-3)}

=

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

{\displaystyle ={\frac {2n}{2n+1}}\cdot {\frac {2n-2}{2n-1}}\cdot {\frac {2n-4}{2n-3}}\cdot \cdots \cdot {\frac {6}{7}}\cdot {\frac {4}{5}}\cdot {\frac {2}{3}}I(1)=2\prod _{k=1}^{n}{\frac {2k}{2k+1}}}

We make the following observation, based on the fact that

sin

⁡

x

≤

1

{\displaystyle \sin {x}\leq 1}

sin

2

n

+

1

⁡

x

≤

sin

2

n

⁡

x

≤

sin

2

n

−

1

⁡

x

,

0

≤

x

≤

π

{\displaystyle \sin ^{2n+1}x\leq \sin ^{2n}x\leq \sin ^{2n-1}x,0\leq x\leq \pi }

⇒

I

(

2

n

+

1

)

≤

I

(

2

n

)

≤

I

(

2

n

−

1

)

{\displaystyle \Rightarrow I(2n+1)\leq I(2n)\leq I(2n-1)}

Dividing by

I

(

2

n

+

1

)

{\displaystyle I(2n+1)}

:

⇒

1

≤

I

(

2

n

)

I

(

2

n

+

1

)

≤

I

(

2

n

−

1

)

I

(

2

n

+

1

)

=

2

n

+

1

2

n

{\displaystyle \Rightarrow 1\leq {\frac {I(2n)}{I(2n+1)}}\leq {\frac {I(2n-1)}{I(2n+1)}}={\frac {2n+1}{2n}}}

, where the equality comes from our recurrence relation.

By the squeeze theorem,

⇒

lim

n

→

∞

I

(

2

n

)

I

(

2

n

+

1

)

=

1

{\displaystyle \Rightarrow \lim _{n\rightarrow \infty }{\frac {I(2n)}{I(2n+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

{\displaystyle \lim _{n\rightarrow \infty }{\frac {I(2n)}{I(2n+1)}}={\frac {\pi }{2}}\lim _{n\rightarrow \infty }\prod _{k=1}^{n}\left({\frac {2k-1}{2k}}\cdot {\frac {2k+1}{2k}}\right)=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

⋅

⋯

{\displaystyle \Rightarrow {\frac {\pi }{2}}=\prod _{k=1}^{\infty }\left({\frac {2k}{2k-1}}\cdot {\frac {2k}{2k+1}}\right)={\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot {\frac {6}{5}}\cdot {\frac {6}{7}}\cdot \cdots }

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.