VIETÈ’S PRODUCT PROVED IN THE FINEST
ANCIENT STYLE
ÓSCAR CIAURRI, EMILIO FERNÁNDEZ, RODOLFO LARREA,
AND LUZ RONCAL
Vietè’s product is the identity
q
p
√
√
√ p
2
+
2+ 2
2 2+ 2
2
··· = .
2
2
2
π
As might be expected, the first known proof of this identity was by
Francois Vietè in 1593. The usual proof, in elementary calculus, uses
trigonometric identities and a limit argument. In fact, if we let
n radicals
zr
q
2+
}|
{
p
√
2 + 2 + ··· 2
cn ≡
then from the identity cos
,
2
x
2
=
q
1+cos x
,
2
cn = cos
it follows that
π
2n+1
.
Using the formula for the sine of the double-angle, it can be checked
that
N
Y
sin π2
cn = N
.
2 sin 2Nπ+1
n=1
Finally
q
p
√
√
√ p
N
Y
2 2+ 2 2+ 2+ 2
· · · = lim
cn
N →∞
2
2
2
n=1
sin π2
2
= ,
π
N
N →∞ 2 sin N +1
π
2
= lim
THIS PAPER HAS BEEN PUBLISHED IN: College Math. J. 43 (2012),
291–296.
Research of the first and third authors supported by grant MTM2009-12740C03-03 of the DGI.
1
2
Ó. CIAURRI, E. FERNÁNDEZ, R. LARREA, AND L. RONCAL
where in the last step we have used that sin x → x, as x → 0. The argument is not very deep, but does require a good knowledge of trigonometry and evaluation of a non-elementary limit.
This paper aims to evaluate Vietè’s product using arguments as elementary as possible. Our proof is essentially of visual nature. We use
only Pythagoras’ theorem to construct c1 , similarity of triangles, and
exhaustion. All these were well known to the ancient Greek mathematicians; hence we think of this as a proof in the finest ancient style.
The original proof by Vietè (see [1]) also used exhaustion. He applied
trigonometric identities to express the products pN in terms of the area
of a 2N -gon inscribed in a circle of radius one. Our approach is similar,
but we consider perimeters of polygons instead of areas.
Two identities
Consider a regular n-gon and a regular 2n-gon circumscribed about
it. In Figure 1 we show a sector of amplitude 2π/n of the bigger
polygon.
C
Β
Α
D
Α
Β
O
Α
A
Figure 1.
B
VIETÈ’S PRODUCT PROVED IN THE FINEST ANCIENT STYLE
3
Following the notation in the Figure 1, we see that triangles 4ABC
and 4OCD are similar. Then
AC
OD
=
.
CB
OC
From this, denoting by An = OA and Ln = CD the apothem and the
side of the n-gon, respectively, and by An+1 = OD and Ln+1 = CB the
same elements of the 2n-gon, we conclude that
2Ln+1 An+1
.
OC
(From Figure 1m, we could observe that Ln = 2OC sin 2α, Ln+1 =
2OC sin α and An+1 = OC cos α. In this way, (1) can be read as
sin 2α = 2 sin α cos α; i.e., the double-angle formula for the sine.)
Now, we entend segment OA to obtain a point C 0 such that OC 0 =
OC, as in Figure 2. The triangles 4OCD and 4CC 0 A are now similar,
(1)
Ln =
C
Β
Β
C'
D
Α
Α
Α
O
A
Figure 2.
and so
AC 0
OD
=
.
AC
DC
Then, using the notation in terms of the apothem and the side of the
polygons, and AC 0 = AO + OC 0 = AO + OC = An + OC, we get
An + OC
An+1
=
.
Ln /2
Ln+1 /2
4
Ó. CIAURRI, E. FERNÁNDEZ, R. LARREA, AND L. RONCAL
So, from the identity (1), we obtain
(2)
2
A2n+1
= An + OC.
OC
(If we didn’t want to avoid trigonometry, we might observe that
An = OC cos 2α and An+1 = OC cos α. From these relations and (2)
we obtain the double-angle relation cos 2α = 2 cos2 α − 1.)
The proof
2
1
c1 =
2
2
1
Figure 3.
For n ≥ 1, let Pn be the polygon of 2n+1 sides inscribed in a circumference of radius one. From Figure 3 and Pythagoras’ theorem, we see
that the apothem of P1 is exactly c1 . Each element of the sequence cn
can be written in terms of the previous one as
2c2n+1 = cn + 1.
Then, from (2), we check that each cn is the apothem of the polygon
Pn . Moreover, with (1) and taking `n as the length of the side of Pn ,
we have
`n = 2`n+1 cn+1 ,
and this fact implies that
Perimeter(Pn ) = cn+1 Perimeter(Pn+1 ).
VIETÈ’S PRODUCT PROVED IN THE FINEST ANCIENT STYLE
5
Then
Perimeter(Pn )
Perimeter(Pn−1 )
=
= ···
cn+1
cn+1 cn
Perimeter(P1 )
4
=
=
,
cn+1 cn · · · c3 c2
cn+1 cn · · · c3 c2 c1
√
because Perimeter(P1 ) = 4 2 = c41 .
As n → ∞ the polygons Pn cover completely the circumference of
radius one and
Perimeter(Pn+1 ) =
Perimeter(Pn ) → Length of the circumference = 2π.
So
q
p
√
√ p
√
2
+
2+ 2
2 2+ 2
· · · = lim c1 c2 · · · cn
n→∞
2
2
2
4
2
= lim
= .
n→∞ Perimeter(Pn )
π
In the end, the beginning
2c4
2c3
2
2c2
2c1
2
Figure 4.
6
Ó. CIAURRI, E. FERNÁNDEZ, R. LARREA, AND L. RONCAL
Our proof of Vietè’s product began with a proposal in the classic “A course of Pure Mathematics” [2, p.20]. There p
G. H. Hardy
√
√
asks
for
geometrical
constructions
of
the
numbers
2,
2
+
2 and
q
p
√
2 + 2 + 2. These real numbers correspond to 2c1 , 2c2 and 2c3 ,
respectively. After obtaining the geometrical constructions requested
by Hardy, we found an elementary extension to the sequence 2cn . Each
number of this sequence of numbers is the apothem of the 2n+1 -gon inscribed in a circumference of radius two and they can be constructed
as we show in Figure 4. Moreover, it is clear that the sequence 2cn
tends to the radius of the circumference; i.e.,
lim 2cn = 2,
n→∞
which is the necessary condition to get the convergence of Vietè’s product. And so this turned out to be the starting point of our proof...
References
[1] E. Maor, The Pythagorian theorem, Princeton University Press, Princeton,
2007.
[2] G. H. Hardy, A course of Pure Mathematics, ninth edition, Cambridge University Press, Cambridge, 1946.
CIME and Departamento de Matemáticas y Computación, Universidad de La Rioja, Edificio J. L. Vives, Calle Luis de Ulloa s/n, 26004
Logroño, Spain
E-mail address: [email protected]
I. E. S. Práxedes Mateo Sagasta, Glorieta del Dr. Zubı́a s/n, 26003,
Logroño, Spain
E-mail address: [email protected]
I. E. S. Hermanos D’Elhuyar, Calle Albia de Castro 9, 26003 Logroño,
Spain
E-mail address: [email protected]
Departamento de Matemáticas y Computación, Universidad de La
Rioja, Edificio J. L. Vives, Calle Luis de Ulloa s/n, 26004 Logroño,
Spain
E-mail address: [email protected]