Generalization of Vieta’s Formula
AlJalila Al-Abri
Department of Mathematics and Statistics
Sultan Qaboos University
P.O.Box 36, Al-Khoud 123 Muscat
asalabri@hotmail.com
1
Introduction
In 1579 Francois Vieta (1540-1603) derived the following formula
q
p
p
p p
p
2
+
2+ 2
2
2
2+ 2
=
2
2
2
(1)
to approximate the value of . He obtained this formula, which subsequently
became known as Vieta’s formula using a geometric approach [3]. As it is often
the case, things can be achieved in many di¤erent ways in mathematics. Vieta’s
formula can be derived from the identity
x
sin x Y
=
cos k
x
2
k=1
1
(2)
by setting x = 2 , and identity (2) can be obtained by using the elementary
trigonometric identity
sin 2x = 2 sin x cos x
and elementary calculus.
In his beautiful monograph [1] Mark Kac (1914 - 1984) began with a proof of
Vieta’s formula using the Rademacher functions and their independence property,
and in the …rst chapter, a generalization of Vieta’s formula was suggested as an
exercise. In this paper we provide a proof following Kac’s idea of using the independence property of the Rademacher functions. To the best of our knowledge,
this generalization has only been achieved by Kent E. Morrison using the Fourier
transform and delta distributions (see [2]).
In the following section we introduce some facts about the expansion of numbers in the closed interval [0; 1] to any base g 2: In section 3 we brie‡y present
the Rademacher functions and their independence property. In section 4 we give
the generalization and its proof using the Rademacher functions and their independence property. In section 5 we use the general formulas to approximate the
value of and give a comparison of di¤erent formulas:
1
2
Preliminaries
Real numbers in the interval [0; 1] can be represented using the binary expansion.
In fact, each real number 0 t 1 can be written as
"1 "2 "3
t=
+ 2+ 3+
2
2
2
where each value of "i is either 0 or 1. For example
3
1
0
0
1
1
0
0
1
1
= + 2+ 3+ 4+ 5+ 6+ 7+ 8+ 9+ :
5
2 2
2
2
2
2
2
2
2
To ensure uniqueness we also impose a terminating expansion in which all the
digits from a certain point on are all equal to 0. For example, we write
3
1
1
0
0
0
= + 2+ 3+ 4+ 5+
4
2 2
2
2
2
instead of
3
1
0
1
1
1
= + 2+ 3+ 4+ 5+
4
2 2
2
2
2
The digits "i depend on t, so they are functions of t. Hence we write
"1 (t) "2 (t) "3 (t)
+ 2 + 3 +
2
2
2
This idea can of course be generalized to any integer base g
we can write
! 1 (t) ! 2 (t) ! 3 (t)
+
+
+
t=
g
g2
g3
where t 2 [0; 1] and ! k 2 f0; 1; ; g 1g.
t=
3
2, so in general
(3)
Rademacher functions
In this section we introduce the functions rk (t) which are de…ned in terms of the
binary digits "k (t) as follows
rk (t) = 1
2"k (t)
k = 1; 2; 3;
.
(4)
These functions were …rst introduced by Hans Rademacher (1892 - 1969) as a
system of orthogonal functions. He introduced them in a paper which was published in 1922 (more information can be found in [4] ), and these functions are
now known as the Rademacher functions. For illustrative purposes, the graphs
of the …rst four functions r1 (t) ; r2 (t) ; r3 (t) ; r4 (t) are given below.
r1 (t)
r2 (t)
2
r3 (t)
r4 (t)
Notice from the graphs above that rn (t), takes the values +1 or 1 alternatively
over the intervals ( 2sn ; s+1
); s = 0;
; 2n 1; and so the length of each small
2n
interval is 21n : Also we deduce that the function rn (t) has 2n intervals, half of
them have the value +1 and the other half have the value 1.
The Rademacher functions satisfy the independence property
fr1 (t) =
1 ; r2
(t) =
2;
ng
; rn (t) =
=
n
Y
frk (t) =
k=1
kg
(5)
where stands for the measure (length) of the set de…ned inside the braces and
n
; n is a sequence of +1 and 1: Thus the product in (5) is 12 .
1;
3.1
Generalization of the Rademacher functions
As we can have an expansion to any base for the numbers in the interval [0; 1],
we can also de…ne the Rademacher functions as
rk (t) = 1
2! k (t)
g 1
(6)
k = 1; 2; 3;
where ! k 2 f0; 1; ; g 1g and g is a positive integer greater than 1:
Moreover, the functions de…ned above can also be shown to be independent.
Therefore, we have
fr1 (t) =
where
1;
;
1 ; r2 (t) =
n
2 f1; gg
2;
3 g 5
;
;
1 g 1
ng =
; rn (t) =
; g11; g
even number and 1 ; ; n 2 f1; gg
when g is an odd number.
1
;
1
3 g 5
;
;
1 g 1
3
;
n
Y
k=1
frk (t) =
(g 5)
;
g 1
; g 2 1 ; 0; g
(g 3)
;
g 1
2
;
1
;
kg =
1
g
n
;
1g when g is an
(g 5)
;
g 1
(g 3)
;
g 1
1g
4
The general formulas
The following theorem generalizes formula (2).
Theorem 4.1 Let g be a positive integer 2: If g is even, then
0 g
1
1
2
Y
X
2@
sin x
(2m 1)x A
=
cos
:
x
g m=1
gk
k=1
(7)
If g is odd, then
0 0g 1
1
1
2
X
sin x
2mx A
@2 @
cos
=
+ 1A .
k
x
g
g
m=1
k=1
1
Y
1
(8)
Proof. Equality (8) can be proved by establishing the next …ve identities:
2t X rk (t)
=
1
gk
k=1
Z 1
Z 1
sin x
(g 1)ix( 1g 2t
)
1 dt =
e
eix(1 2t) dt =
x
0
0
0 g 1
1
Z 1
2
X
1
2mx
rk (t)
exp (g 1)ix k dt = @2
cos
+ 1A
k
g
g
g
0
m=1
"
#
Z 1
Z 1
n
X
1 2t
rk (t)
e(g 1)ix( g 1 ) dt =
exp (g 1)ix lim
dt
n!1
gk
0
0
k=1
1
1
g
Z
1
"
exp (g
0
1)ix lim
n!1
n
X
rk (t)
gk
k=1
#
dt =
=
lim
n!1
lim
n!1
Z
n
1Y
0 k=1
n Z 1
Y
k=1
0
(9)
(10)
(11)
(12)
exp (g
1)ix
rk (t)
dt
gk
exp (g
1)ix
rk (t)
dt:
gk
First, equation (9) can be proved as follows. From (3) and (6) we have
t =
=
1
X
! k (t)
k=1
1
X
gk
(g
1)(1 rk (t))
.
2:g k
k=1
Thus,
2t
g
1
X
1
=
1 k=1 g k
4
1
X
rk (t)
k=1
gk
.
(13)
But
1
X
1
gk
1
.
g 1
is a geometric series and converges to
Hence
k=1
2t
g
1
=
1
X
rk (t)
1
g
1
gk
k=1
which proves equation (9). Then equation (10) can be obtained by a simple
integration and the identity
eix
e
2i
ix
= sin x:
Similarly, equation (11) can be derived by a simple integration and noticing that
the function rk (t) is integrable over the intervals ( gsk ; s+1
); s = 0;
; gk 1
gk
5)
3)
and has the values 1; gg 13 ; gg 51 ; ; g 2 1 ; 0; g 21 ; ; (g
; (g
; 1 alternatively
g 1
g 1
k
over these intervals: Therefore we have g possible values and the set of points t
satisfying each value has a total length of g1 . Notice
Z
1
rk (t)
1)ix k dt = g k
g
exp (g
0
1
Z
1
gk 1
exp (g
0
1)ix
rk (t)
dt:
gk
and
Z
1
gk 1
rk (t)
dt
gk
0
Z 1k
Z gk
g
g
(g 1)
(g 1)
exp ix
dt
=
exp ix
dt
+
k
g
1
g
gk
0
gk
Z 2k
Z g k1
g
g
(g 3)
(g 3)
+
exp ix
dt +
exp ix
dt
k
g 2
1
gk
g
k
k
g
g
Z 3k
Z g k2
g
g
(g 5)
(g 5)
+
exp ix
dt +
exp ix
dt
k
g 3
2
g
gk
k
k
exp (g
1)ix
g
g
+
+
Z
+
Z
g 1
2
gk
g 3
2
gk
g+1
2
gk
g 1
2
gk
2(g 1)
exp ix
dt +
gk
exp ix
0(g 1)
dt:
gk
5
Z
g+3
2
gk
g+1
2
gk
exp ix
2(g 1)
dt
gk
Therefore
Z
1
rk (t)
exp (g 1)ix k dt
g
0
!#
"
ix(g 1)
ix(g 1)
ix(g 3)
ix(g 3)
gk
gk
gk
gk
1
e
+
e
+
e
+
e
+
= gk 1 k
ix(g 5)
ix(g 5)
2ix
2ix
g
e gk + e gk +
+ e gk + e gk + 1
0 g 1
1
2
1@ X
2mx
cos
+ 1A :
=
2
k
g
g
m=1
Equation (12) follows from a trivial substitution in equation (9), and …nally,
equation (13) is the where we use the independence property of the Rademacher
functions.
Since the exponential function is continuous, we have
"
#
"
#
Z 1
Z 1
n
n
X
X
rk (t)
rk (t)
exp (g 1)ix lim
dt =
lim exp (g 1)ix
dt:
k
k
n!1
n!1
g
g
0
0
k=1
k=1
"
#
n
X r (t)
k
Also since exp (g 1)ix
is a sequence of integrable functions on [0; 1]
gk
k=1
and converges uniformly on [0; 1] to the function exp [ix(1 2t)], we can interchange the limit with the integral. Therefore we have
"
#
"
#
Z 1
Z 1
n
n
X
X
rk (t)
rk (t)
lim exp (g 1)ix
dt = lim
exp (g 1)ix
dt:
k
k
n!1 0
g
g
0 n!1
k=1
k=1
Z 1Y
n
rk (t)
= lim
exp (g 1)ix k dt:
n!1 0
g
k=1
Now we show that
"
Z 1
lim
exp (g
n!1
0
1)ix
n
X
rk (t)
k=1
gk
#
dt = lim
n!1
n Z
Y
k=1
0
1
exp (g
1)ix
rk (t)
dt
gk
by evaluating the integral over all possible values of the function rk (t) as we did
to prove equation (11) and using the independence property of the Rademacher
functions. Taking 1 ;
; n to be the possible values of the function rk (t) we
have
"
#
Z 1
n
X
rk (t)
exp (g 1)ix
dt
gk
0
k=1
"
#
n
X
1 X
rk (t)
= n
exp ix
(g 1) k
g
g
k=1
"
#
n
X
X
k
=
exp ix
(g 1) k fr1 (t) = 1 ; r2 (t) = 2 ; ; rn (t) = n g
g
;
;
k=1
n
1
" n
# n
X
X
Y
k
=
exp i
(g 1) k
frk (t) = k g :
g
; ;
k=1
k=1
1
n
6
Simple calculations give
X
1;
=
=
exp ix
n
n
X Y
1;
=
;
"
;
n
k
1) k
g
(g
k=1
exp ix(g
1)
k=1
n X
Y
k=1 k
n Z 1
Y
k=1
n
X
exp ix(g
exp (g
1)
1)ix
0
k
gk
#
n
Y
k=1
k
frk (t) =
frk (t) =
gk
frk (t) =
kg
kg
kg
rk (t)
dt:
gk
Now using equation (11) we obtain formula (8). The formula for the even case can
be proved using very similar steps as in the odd case. The only di¤erence is that
in step (3) we need to consider the values of the function rk (t) in f1; gg 13 ; gg 51 ;
; g11; g
1
;
1
;
(g 5)
;
g 1
(g 3)
;
g 1
1g:
If we take the value of g from 2 to 10 alternatively in formula (7) and formula
(8) we get
1
Y
sin x
x
=
cos k
x
2
k=1
=
=
=
=
=
=
=
=
1
Y
1
k=1
1
Y
3
2 cos
2x
+1
3k
1
x
3x
cos k + cos k
2
4
4
k=1
1
Y
1
k=1
1
Y
5
2 cos
2x
4x
+
cos
+1
5k
5k
1
x
3x
5x
cos k + cos k + cos k
3
6
6
6
k=1
1
Y
1
k=1
1
Y
7
2 cos
2x
4x
6x
+ cos k + cos k + 1
k
7
7
7
1
x
3x
5x
7x
cos k + cos k + cos k + cos k
4
8
8
8
8
k=1
1
Y
1
k=1
1
Y
9
2 cos
2x
4x
6x
8x
+ cos k + cos k + cos k + 1
k
9
9
9
9
1
x
3x
5x
7x
9x
cos k + cos k + cos k + cos k + cos k
5
10
10
10
10
10
k=1
7
We notice that in general the formulas are in…nite products of sum of cosines.
The main di¤erence is that in the odd case we have the term "1" which appears
due to the value 0 of rk (t) :
In the next section we are going to use these formulas to approximate the
value of by setting x = 2 :
5
Approximating the value of
eral formulas
using the gen-
The following table is obtained using a computer program written in C++. The
program uses formulas (7) and (8) to get approximations of the value of up to
15 decimal places of di¤erent bases. In this table we report only the base value
from 2 to 13, but the program can give an approximation of the value of up
to 15 decimal places to any base g 2. The …rst row gives the base value to be
substituted in the formula. The second row gives the number of iterations needed
to compute up to 15 decimal places. The third row gives the total number of
cosines needed to compute up to 15 decimal places. The values of the third row
can be obtained by multiplying the number of cosines appearing in the formula
corresponding to the base value, by the number of iterations needed to compute
up to 15 decimal places. For example, when we use g = 4 in the formula (7),
the number of cosines are 2; and from the second row, the number of iterations
needed to compute up to 15 decimal places is 12: Therefore the number of
cosines needed to compute up to 15 decimal places are 2 12 = 24:
base value (g) 2 3 4 5 6 7 8 9 10 11 12 13
I(g)
24 15 12 11 10 9 8 8 8 7 7 7
C(g)
24 15 24 22 30 27 32 32 40 35 42 42
From the table we can observe the following:
1. As the base value g get larger, the number of iterations needed to compute
up to 15 decimal places decreases. This seems to be clear from the formulas, since as the base value gets larger the number of operations increases.
Therefore we will need fewer iterations to reach the approximated value of
up to 15 decimal places.
2. We see that among all base values appearing in the table, base 3 gives an
approximation to the value of up to 15 decimal places with least number
of cosines. So, we can say that the formula for base 3 is the best since it
gives the value of up to 15 decimal places with minimal cost in terms
of operations. Moreover, the odd base gives the approximation with fewer
number of cosines compared to the even base. In general, as we go higher
in the base (odd or even), the number of cosines increases.
8
Acknowledgement
Thanks to my advisor Dr. Mohamed B. Rhouma for his continuous encouragement and support while writing this paper. I also thank Dr. Stefan Veldsman
for his corrections and suggestions.
References
[1] M. Kac, Statistical Independence in Probability, Analysis and Number Theory,
Carus Mathematical Monograph, No 12, The Mathematical Association of
America, 1959.
[2] K. E. Morrison, Cosine Products, Fourier Transforms and Random Sums,
Amer. Math. Monthly, 102:716,1995.
[3] L. Berggren, J. Borwein and P. Borwein, Pi: A Source Book, 2nd ed, Springer,
2000.
[4] www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Rademacher.html.
9