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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 dierence 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.

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