A Probabilistic Approach to
Vieta’s Formula
by: Mark Osegard
Points of Discussion
I. Vieta’s Formula
II. Rademacher Functions
III. Vieta’s Formula in terms of
Rademacher Functions
IV. A Classical Analysis Proof
François Viéte
François Viéte was a Frenchman born in 1540. He
Commonly used the name Franciscus Vieta, the Latin
form of his given name. Until his death in 1603 he made
many significant contributions to the field of mathematics.
Some of the fields he was involved in were trigonometry,
algebra, arithmetic, and geometry.
Vieta’s Formula
x
x
sin x = 2 sin cos
2
2
x
x
2
= 2 sin cos cos
The ideas all begin with
4
4
some simple trigonometry.
x
x
3
Using the double angle
= 2 sin cos cos
8
8
formula.
x
n
sin x = 2 sin n
2
n
x
2
x
x
cos
4
2
x
cos k
∏
2
k =1
Vieta’s Formula
From calculus
≠
for
x
0
,
we find.
x
sin n
x
1
n
2
= lim 2 sin n
1 = lim
n→∞
x
x n→∞
2
n
2
x
n
lim 2 sin n = x
n→∞
2
Vieta’s Formula
Combining the equations derived
on the past two slides we see:
n
x
x
n
sin x = 2 sin n ∏ cos k
2 k =1
2
&
x
lim 2 sin n = x
n →∞
2
n
∞
x
sin x
= ∏ cos k
2
x
k =1
This trigonometric identity that leads to vieta’s formula
Vieta’s Formula
Setting x = π 2 we see an interesting
case, which yields this formula.
2
π
∞
= ∏ cos
n =1
π
2
n +1
Using the half-angle formula for cosine repeatedly
2
=
2
2+ 2
2
2+ 2+ 2
2
,
Binary Expansion
Taking another approach
is usually helpful, and
this case is no different.
We know that every number
between 0 ≤ t ≤ 1 can be
represented in the form:
(t )
2 (t )
+ 2 + ,
t=
2
2
where is either 0 or 1
1
Rademacher Functions
Using the same ideas we find it more convenient to
use the function defined below, which uses the
interval [1,−1] instead of [0,1].
rk (t ) = 1 − 2 k (t )
k = 1, 2, 3,
Rademacher Function
,
Rademacher Functions
These are the dyadic subintervals of
the Rademacher Functions.
1
0
1
2
Hans Rademacher
Hans Rademacher was born in 1892 near Hamburg,
Germany. He was most widely known for his work
in modular forms and analytic number theory. He
died in 1969 in Haverford, Pennsylvania.
Connecting
You may notice the relationship between the Rademacher
Functions and Binary Expansion. Subsequently, we
rewrite the equation:
(t )
2 (t )
t=
+
+
2
2
1
as
∞
rk (t )
1 - 2t =
k
k =1 2
,
Notice
sin x
Now we shall see that the term
x
can be written :
1
0
e
ix ( 1 − 2 t )
sin x
dt =
x
The Proof
We start with,
1
0
e
ix (1− 2 t )
dt =
1 ix (1− 2 t )
0
e
substitute ‘u’,
−1
1
? sin x
dt =
x
change parameter
1
e ( − 1 / 2 du ) = 1 / 2 e
ixu
ixu
−1
Note : u = 1 − 2t , du = −2 dt , dt = −1 / 2 du
du
1 ix (1− 2 t )
The Proof
0
e
? sin x
dt =
x
Recall Euler’s Formula,
eixu = cos( xu ) + i sin( xu )
Replacing eixu
1
1/ 2 e
−1
ixu
in the equation we get.
du
1
= 1 / 2 [ cos( xu ) + i sin( xu )] du
−1
The Proof
1 ix (1− 2 t )
0
e
? sin x
dt =
x
Splitting up the integral,
= 1/ 2
1
−1
cos( xu ) du + i / 2
Evaluating the integral,
sin( xu )
= 1/ 2
x
u =1
u = −1
= 1 / 2 x[sin( x ) − sin( − x )]
1
−1
sin( xu ) du
The Proof
1 ix (1− 2 t )
0
e
sin x
dt =
x
Recall the equation from the previous slide,
pull out -1
sin x
1 / 2 x[sin( x) − sin( − x)] =
x
1 ix (1− 2 t )
sin x
=
= e
dt
0
x
QED
Also Notice
x
Now we shall see that cos k
can be
2
written in terms of Rademacher Functions
rk (t )
x
exp ix k dt = cos k
0
2
2
1
:
rk (t )
x
?
exp ix k dt = cos k
0
2
2
1
The Proof
Notice that we can break up the integral
rk (t )
exp ix k dt
0
2
1
into this sum of integrals over dyadic subintervals
( m +1) 2 − k
m even
m2
−k
e
ix 2
k
( m +1) 2 − k
dt +
m odd
m2
−k
e
−ix 2 k
Note: (2-k) represents the length of a subinterval
dt
The Proof
rk (t )
x
?
exp ix k dt = cos k
0
2
2
1
There is a total of 2k subintervals. We define the end
points of the EVEN subintervals to be represented by
I j , also we define the end points of the ODD
subintervals to be represented by J j . (as seen below)
2 j − 2 2 j −1
, k
Ij =
k
2
2
2 j −1 2 j
Jj =
, k
k
2
2
The Proof
rk (t )
x
?
exp ix k dt = cos k
0
2
2
1
The Rademacher functions or rk (t )
follows:
are defined as
rk (t ) = 1 on I j
rk (t ) = - 1 on J j
If we union these subintervals we see:
Jj
I j = [0,1]
The Proof
Ij
Using
2 k-1
e
rk (t )
x
?
exp ix k dt = cos k
0
2
2
1
and
ix 2 k
dt +
j =1 I j
2 k -1
Jj
we get:
− ix 2 k
dt
2 k-1
2j
e
j =1 J j
Inserting the end points,
=
2 k-1
2 j −1
k
2
2 j −2
j =1
2k
e
ix 2 k
dt +
k
2
2 j −1
j =1 2 k
e
−ix 2 k
dt
rk (t )
x
?
exp ix k dt = cos k
0
2
2
1
The Proof
We then pull out the power of e:
=
2 k -1
j =1
e
ix 2
k
2 j −1
2k
2 j −2
1 dt +
2 k -1
e
−ix 2
j =1
2k
Evaluating this we find it equals
2 k -1
1 ix 2 k
=
e
+
k
j =1 2
2 k -1
1 −ix 2 k
e
k
j =1 2
k
2j
2k
2 j −1
2k
1 dt
The Proof
rk (t )
x
?
exp ix k dt = cos k
0
2
2
1
Simplifying:
1
= k
2
2 k -1
1
= k
2
2 k -1
j =1
(e
ix 2 k
+e
x
2 cos k
2
j =1
− ix 2 k
)
iα
− iα
2
cos
α
=
e
+
e
where:
rk (t )
x
exp ix k dt = cos k
0
2
2
1
The Proof
x
We can now pull out the 2 cos k
2
1
x
= k −1 cos k
2
2
2 k -1
x
1 = cos k
2
j =1
This gives the desired solution:
rk (t )
x
exp ix k dt = cos k
0
2
2
1
QED
Interesting Development
Recall the formula,
x
sin x ∞
= ∏ cos k
2
x
k =1
which may now be seen as:
sin x
=
x
1 ix (1− 2 t )
0
e
Proved
∞
rk (t )
dt
dt = exp ix
k
0
k =1 2
1
by def
∞
∞
1
rk (t )
x
= ∏ cos k = ∏ exp ix k dt
0
2
2
k =1
k =1
Trig proof
by prev obs.
Astonishing Discovery
Since we can now say:
∞
x
cos k =
∏
2
k =1
∞
rk (t )
dt
exp ix
k
0
k =1 2
1
Clearly, we see that an integral of products
is a product of integrals!
1 ∞
∞
1
rk (t )
rk (t )
exp ix k dt = ∏ exp ix k dt
∏
0
0
2
2
k =1
k =1
Vieta’s formula has this interesting expression in
terms of Rademacher Functions
Classical Analysis Proof
Can we prove Vieta’s formula using Rademacher’s
Functions as a starting point?
Analyzing the following function may show us.
n
k =1
ck rk (t )
We realize this is a step function over the intervals:
s s +1
, n ,
n
2 2
s = 0,1,
,2 − 1
n
Classical Analysis Proof
Since every sequence of +1’s and –1’s corresponds
to one and only one interval
s s +1
, n
n
2 2
By breaking the integral as before into dyadic subintervals
We can say that:
n
1
exp i ck rk (t ) dt = n
0
2
k =1
1
exp i
n
± ck
1
↑
over all possible sequences
Classical Analysis Proof
Now,
1
2n
exp i
=∏
k =1
1
0
± ck
1
n
So,
n
exp i
n
k =1
n
e ick + e −ick
= ∏ cos(ck )
2
k =1
n
ck rk (t ) dt = ∏ cos(ck )
Note: This proof is
similar to the last
k =1
n
1 ic r ( t )
k k
=∏ e
k =1
0
Classical Analysis Proof
Using the derived equation from the previous page and
x
setting ck = k
2
n
x
= ∏ cos k =
2
k =1
we get this familiar equation:
n
rk (t )
dt
exp ix
k
0
k =1 2
1
When we take the limit,
n
rk (t )
= 1 − 2t
lim
k
n →∞
k =1 2
Classical Analysis Proof
So now we have proven that uniformly
on the interval (0,1), we have,
sin x
=
x
1 ix (1− 2 t )
0
e
rk (t )
dt = lim exp ix
dt
k
n →∞ 0
k =1 2
1
∞
x
x
= lim ∏ cos k = ∏ cos k
n →∞
2
2
k =1
k =1
n
n
Classical Analysis Proof
The Classical Analysis Proof works, however,
the proof itself is not all that enlightening.
Sure it works, but why?
How are they Rademacher Functions involved?
We must now look at it in a different way, in
hopes of finding a better explanation.
Random Variables &
Statistical Independence
So what is it about the Rademacher
Functions or binary digits that makes
the previous proof work?
1
2
n
1
=
2
1
(n times )
2
Consider the set of t’s:
r1(t)=+1, r2(t)=-1, r3(t)=-1
The set of t’s (possibly excluding the end
points) is the interval
3 4
,
8 8
and the
1
length of the interval is clearly
, and
8
1 1 1 1
= ⋅ ⋅ ⋅
8 2 2 2
This observation can be written:
µ {r1 (t ) = +1, r2 (t ) = −1, r3 (t ) = −1}
= µ {r1 (t ) = +1}µ {r2 (t ) = −1}µ {r3 (t ) = −1}
Note: µ is the length of the set inside the braces
This can be generalized to this form:
µ {r1 (t ) = δ 1 , , rn (t ) = δ n }
= µ {r1 (t ) = δ 1}µ {r2 (t ) = δ 2 } µ {rn (t ) = δ n }
Independence &
Rademacher Functions
Rademacher Functions are
independent
Bernoulli random variables (-1 or 1).
µ
dyadic
represents the length of a
subinterval.
You may see this as only a more
complicated way of writing
1
2
n
1 1
= × ×
2 2
1
× (n times)
2
but this means much more. It shows a
deep property of rk(t) and subsequently
binary digits. This now leads us into
the Probabilistic Proof.
Probabilistic Proof
We start with the equation:
1
0
exp i
n
1
sin x
ck rk (t ) dt =
x
Trying to prove that
1
0
exp i
n
1
n
1 ic r ( t )
k k
ck rk (t ) dt = ∏ e
k =1
0
dt
Splitting the integral over the dyadic subintervals
corresponding to δ n we find:
exp i
δ 1 , ,δ n
n
ck rk (t ) µ {r1 (t ) = δ 1
rn (t ) = δ n }
1
The inner sum evaluates to:
n
=
δ 1 , ,δ n
n
∏ e ∏ µ {r (t ) = δ }
1
ick δ k
k
1
k
Remove the second product
n
=
δ 1 , ,δ n
∏e
ick δ k
µ {rk (t ) = δ k }
1
Interchange the product and sum:
n
=∏
k =1 δ k
e
ick δ k
µ {rk (t ) = δ k }
Replacing the sum with the integral
we find the desired outcome:
n
1 ic r ( t )
k k
=∏ e
k =1
0
dt =
1
0
exp i
n
ck rk (t ) dt
1
The vieta identity follows when you take x/2k
Conclusion
So we have seen three ways to prove
Vieta’s formula and it’s connection to
statistical independence.
Rademacher Functions are independent
random Bernoulli variables.
We have found that Vieta’s formula
does, in fact, depend upon random
variables and the statistical
independence of those variables.
References
“Statistical Independence in Probability, Analysis and
Number Theory” by Mark Kac, published by the MAA
Carus Mongraph Series, 1959.
Dr. Deckelman