John Wallis was an English mathematician who is given partial credi...

This was actually the way that Wallis himself derived the formula. ...

This is the proof present in most textbooks using integration and ...

A partial sum with an odd number of factors is for instance
$$
...

A partial sum with an even number of factors is for instance
$$
...

Note that $s_{n+1}=\frac{3}{2}\frac{5}{4}...\frac{2n+2-1}{2n+2-2}=s...

Note that
$$
\frac{2j+1}{2(j+1)}\frac{j+1}{i+j+1}+\frac{2i+1}{...

If we group all $R_{i,j}$ with the same $i+j$ we get the following ...

Note that
$$
\frac{\pi(n-1)}{2n}<W<\frac{\pi (n+1)}{2n}
$$
...

AN ELEMENTARY PROOF OF WALLIS’ PRODUCT FORMULA

FOR PI

JOHAN W

¨

ASTLUND

Abstract. We give an elementary proof of the Wallis product formula for pi.

The proof does not require any integr ation or trigon ome tric functio ns.

1. The Wallis product formula

In 1655, John Wallis wrote down the celebrated formula

(1)

2

1

·

2

3

·

4

3

·

4

5

···=

⇡

2

.

Most textbook proofs of (1) rely on evaluation of some deﬁnite integral like

Z

⇡/2

0

(sin x)

n

dx

by repeated partial integration. The topic is usually reserved for more advanced

calculus courses. The purpose of this note is to show that (1) can be derived

using only the mathematics taught in elementary school, that is, basic algebra, the

Pythagorean theorem, and the formula ⇡ · r

2

for the area of a circle of radius r.

Viggo Brun gives an account of Wallis’ method in [1] (in Norwegian). Yaglom

and Yaglom [2] give a beautiful proof of (1) which avoids integration but uses some

quite sophisticated trigonometric identities.

2. A number sequence

We denote the Wallis product by

(2) W =

2

1

·

2

3

·

4

3

·

4

5

··· .

The partial products involving an even number of factors form an increasing se-

quence, while those involving an odd number of factors form a decreasing sequence.

We let s

0

= 0, s

1

= 1, and in general,

s

n

=

3

2

·

5

4

···

2n 1

2n 2

.

The partial pro ducts of (2) with an odd number of factors can be written as

2n

s

2

n

=

2

2

· 4

2

···(2n)

1 · 3

2

···(2n 1)

2

>W,

while the partial pro ducts with an even number of factors are of the form

2n 1

s

2

n

=

2

2

· 4

2

···(2n 2)

2

1 · 3

2

···(2n 3)

2

· (2n 1)

<W.

It follows that

(3)

2n 1

W

<s

2

n

<

2n

W

.

Date: February 21, 2005.

1

2 JOHAN W

¨

ASTLUND

We denote the di↵erence s

n+1

s

n

by a

n

, and observe that

a

n

= s

n+1

s

n

= s

n

✓

2n +1

2n

1

◆

=

s

n

2n

=

1

2

·

3

4

···

2n 1

2n

.

We ﬁrst derive the identity

(4) a

i

a

j

=

j +1

i + j +1

a

i

a

j+1

+

i +1

i + j +1

a

i+1

a

j

.

Proof. After the substitutions

a

i+1

=

2i +1

2(i + 1)

a

i

and

a

j+1

=

2j +1

2(j + 1)

a

j

,

the right hand side of (4) becomes

a

i

a

j

✓

2j +1

2(j + 1)

·

j +1

i + j +1

+

2i +1

2(i + 1)

·

i +1

i + j +1

◆

= a

i

a

j

.

⇤

If we start from a

2

0

and repeatedly apply (4), we obtain the identities

(5) 1 = a

2

0

= a

0

a

1

+ a

1

a

0

= a

0

a

2

+ a

2

1

+ a

2

a

0

= ...

···= a

0

a

n

+ a

1

a

n1

+ ···+ a

n

a

0

.

Proof. By applying (4) to every term, the sum a

0

a

n1

+ ···+ a

n1

a

0

becomes

✓

a

0

a

n

+

1

n

a

1

a

n1

◆

+

✓

n 1

n

a

1

a

n1

+

2

n

a

2

a

n2

◆

+ ···+

✓

1

n

a

n1

a

1

+ a

n

a

0

◆

.

After collecting terms, this simpliﬁes to a

0

a

n

+ ···+ a

n

a

0

. ⇤

3. A geometric construction

We divide the positive quarter of the x-y-plane into rectangles by drawing the

straight lines x = s

n

and y = s

n

for all n. Let R

i,j

be the rectangle with lower

left corner (s

i

,s

j

) and upper right corner (s

i+1

,s

j+1

). The area of R

i,j

is a

i

a

j

.

Therefore the identity (5) states that the total area of the rectangles R

i,j

for which

i + j = n is 1. We let P

n

be the polygonal region consisting of all rectangles R

i,j

for which i + j<n. Hence the area of P

n

is n (see Figure 1).

The outer corners of P

n

are the points (s

i

,s

j

) for which i + j = n + 1. By the

Pythagorean theorem, the distance of such a point to the origin is

q

s

2

i

+ s

2

j

.

By (3), this is bounded from above by

r

2(i + j)

W

=

r

2(n + 1)

W

.

Similarly, the inner corners of P

n

are the points (s

i

+ s

j

) for which i + j = n. The

distance of such a point to the origin is bounded from below by

r

2(i + j 1)

W

=

r

2(n 1)

W

.

THE WALLIS PRODUCT FORMULA 3

0

1

3/2

15/8

35/16

.

.

.

01

3

2

15

8

35

16

···

R

0,0

R

0,1

R

1,0

R

1,1

R

2,0

R

0,2

R

3,0

R

2,1

R

1,2

R

0,3

Figure 1. The region P

4

of area 4.

Therefore P

n

contains a quarter circle of radius

p

2(n 1)/W , and is contained

in a quarter circle of radius

p

2(n + 1)/W . Since the area of a quarter circle of

radius r is equal to ⇡r

2

/4, we obtain the following bounds for the area of P

n

:

⇡(n 1)

2W

<n<

⇡(n + 1)

2W

.

Since this holds for every n, we c onclude that

W =

⇡

2

.

References

[1] Brun, Viggo., Wallis’s og Brounckers formler for ⇡ (in Norwegian), Norsk matematisk

tidskrift, 33 (1951) 73–81.

[2] Yaglom, A. M. and Yaglom, I. M., An elementary derivation of the formulas of Wallis,

Leibnitz and Euler for the number ⇡ (in Russi an), Uspechi matematiceskich nauk. (N.

S.) 8, no. 5 (57) (1953), 181–187.

Note that $s_{n+1}=\frac{3}{2}\frac{5}{4}...\frac{2n+2-1}{2n+2-2}=s_n\frac{2n+1}{2n}$
Note that
$$
\frac{\pi(n-1)}{2n}<W<\frac{\pi (n+1)}{2n}
$$
It is now clear that as $n \rightarrow \infty$, $W \rightarrow \frac{\pi}{2}$
It's worth taking a second look at the way $\pi$ appears in the proof. As we know $\pi$ is deeply interconnected with geometry and it's surprisingly refreshing how the author makes the bridge and by using the circle and its area brings $\pi$ into the game.
A partial sum with an even number of factors is for instance
$$
\frac{2}{1}\cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5}
$$
which has 4 factors. As we can see for even factors we always end up at the end with a term of the form
$$
\frac{2n-2}{2n-1}
$$
In the case above $n=3$, $\frac{2\cdot3-2}{2\cdot 3-1}=\frac{4}{5}$.
It's not difficult to see that for any even number of factors
$$
\frac{2^2\cdot4^2...(2n-2)}{1\cdot3^2...(2n-3)^2}\cdot \frac{2n-2}{2n-1}= \\
= \frac{2^2\cdot4^2...(2n-2)^2}{1\cdot3^2...(2n-3)^2}\cdot \frac{1}{2n-1} = \\
= \frac{2n-1}{s_n^2}
$$
John Wallis was an English mathematician who is given partial credit for the development of infinitesimal calculus. He is also credited with introducing the symbol $\infty$ for infinity. He similarly used $1/\infty$ for an infinitesimal.
![](https://upload.wikimedia.org/wikipedia/commons/8/89/John_Wallis_by_Sir_Godfrey_Kneller%2C_Bt.jpg)
Wallis made significant contributions to trigonometry, calculus, geometry, and the analysis of infinite series. In his Opera Mathematica I (1695) Wallis introduced the term "continued fraction".
This was actually the way that Wallis himself derived the formula. He compared $\int _{0}^{\pi }\sin ^{n}xdx$ for even and odd values of n, and noted 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.
Wallis' product is, in retrospect, an easy corollary of the later Euler formula for the sine function:
$$
\frac{\sin x}{x} = \prod_{n=1}^\infty\left(1 - \frac{x^2}{n^2\pi^2}\right)
$$
Let $x = \frac{π}{2}$
$$
\frac{2}{\pi} = \prod_{n=1}^{\infty} \left(1 - \frac{1}{4n^2}\right) \\
\frac{\pi}{2} = \prod_{n=1}^{\infty} \left(\frac{4n^2}{4n^2 - 1}\right) = \\
= \prod_{n=1}^{\infty} \left(\frac{2n}{2n-1}\cdot\frac{2n}{2n+1}\right) = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdots
$$
In 2015 researchers in a surprise discovery, found the same formula in [quantum mechanical calculations of the energy levels of a hydrogen atom](http://phys.org/news/2015-11-derivation-pi-links-quantum-physics.html).
This is the proof present in most textbooks using integration and recursion.
Consider $J_n=\int^{\pi/2}_0 \cos^n (x) dx$. Integrating by parts with $u=\cos^{n-1}(x)$ and $dv=\cos(x)$ we have
$$
\int_0^{\pi/2}\cos^n (x) dx = (n-1)\int_0^{\pi/2} \cos^{n-2} (x)\sin^2 (x) dx = \\
= (n-1)\int_0^{\pi/2} \cos^{n-2} (x)(1-\cos^2 (x)) dx= \\
= (n-1)\int_0^{\pi/2} \cos^{n-2} (x) dx - (n-1)\int_0^{\pi/2} \cos^{n} (x) dx
$$
Reorganizing the terms we have
$$
nJ_n=(n-1)J_{n-2}
$$
Now $J_1=1$ and $J_3=2/3$, $J_5=\frac{2\cdot4}{3\cdot5}$ which means that
$$
J_{2n+1}=\frac{2\cdot4...(2n-2)\cdot(2n)}{1\cdot 3 ... (2n-1)(2n+1)}
$$
At the same time $J_2=\frac{\pi}{2\cdot 2}$, $J_4=\frac{3 \cdot \pi}{2\cdot 4 \cdot 2}$ and $J_6=\frac{3\cdot 5 \cdot \pi}{2\cdot 4\cdot 6\cdot 2}$ which means that
$$
J_{2n}=\frac{3\cdot 5 ... (2n-3) \cdot (2n-1) \cdot \pi }{2 \cdot 4...(2n-2)\cdot (2n) \cdot 2}
$$
For $0\leq x \leq \frac{\pi}{2}$, $0\leq \cos (x) \leq 1$ so $\cos^{2n} (x) \geq \cos^{2n+1} (x)\geq \cos^{2n+2} (x)$
which implies that $J_{2n} \geq J_{2n+1}\geq J_{2n+2} $. Finally
$$
1 \geq \frac{J_{2n+1}}{J_{2n}} \geq \frac{J_{2n+2}}{J_{2n}} = \frac{2n+1}{2n+2}
$$
Note that
$$
\frac{2j+1}{2(j+1)}\frac{j+1}{i+j+1}+\frac{2i+1}{2(i+1)}\frac{i+1}{i+j+1}= \\
= \frac{2j+1}{2(j+1)}\frac{j+1}{i+j+1}\frac{i+1}{i+1}+\frac{2i+1}{2(i+1)}\frac{i+1}{i+j+1}\frac{j+1}{j+1}= \\ = \frac{2(j+1)(i+j+1)(i+1)}{2(j+1)(i+j+1)(i+1)}=1
$$
If we group all $R_{i,j}$ with the same $i+j$ we get the following "stripes"
![](http://i.imgur.com/Z3Y6kkC.png)
The area of each "stripe" is by identity (5) equal to 1
$$
R_{0,0}=1 \\
R_{1,0}+R_{0,1}=1 \\
R_{2,0}+R_{0,2}+R_{1,1}=1 \\
R_{3,0}+R_{0,3}+R_{2,1}+R_{1,2}=1 \\
$$
And so it is easy to conclude that the area of $P_n=n$
A partial sum with an odd number of factors is for instance
$$
\frac{2}{1}\cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5}
$$
which has 5 factors. As we can see for odd factors we always end up at the end with a term of the form
$$
\frac{2n}{2n-1}
$$
In the case above $n=3$, $\frac{2\cdot3}{2\cdot 3-1}=\frac{6}{5}$.
It's not difficult to see that for any odd number of factors
$$
\frac{2^2\cdot4^2...(2n-2)^2}{1\cdot3^2...(2n-1)}\cdot \frac{2n}{2n-1}= \\
= \frac{2^2\cdot4^2...(2n-2)^2}{1\cdot3^2...(2n-1)^2}\cdot 2n = \\
= \frac{2n}{s_n^2}
$$