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
V. Random Variables &
Statistical Independence
VI. Probabilistic Proof
VII. Conclusion
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
x
2
 2 sin cos cos
The ideas all begin with
4
4
2
some simple trigonometry.
x
x
x
x
3
Using the double angle
 2 sin cos cos cos
8
8
4
2
formula.

x
n
sin x  2 sin n
2
n
x
cos k

2
k 1
Vieta’s Formula
From calculus
,
0

x
for
we find.
x
sin n
x
1
n
2
 lim 2 sin n
1  lim
n 
x
2
x n 
2n
x
n
lim 2 sin n  x
n 
2
Vieta’s Formula
Combining the equations derived
on the past two slides we see:
x
n
sin x  2 sin n
2
n
x
cos k

2
k 1
&
x
lim 2 sin n  x
n 
2
n


sin x
x
  cos k
x
2
k 1
This trigonometric identity that leads to vieta’s formula
Vieta’s Formula
Setting x   2 we see an interesting
case, which yields this formula.
2


  cos
n 1

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:
ε1 (t ) ε 2 (t )
t
 2  ,
2
2
where ε is either 0 or 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  2ε k (t )

Rademacher Function
k  1, 2, 3,,
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:
ε1 (t ) ε 2 (t )
t

 ,
2
2
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
ix (1 2 t )
e
sin x
dt 
x
The Proof
We start with,
1
e
0
ix (1 2 t )
1 ix (1 2 t )
e
0
substitute ‘u’,
1
? sin x
dt 
x
change parameter
1
dt   e (1 / 2 du )  1 / 2  e
1
ixu
ixu
1
Note : u  1  2t , du  2 dt , dt  1 / 2 du
du
1 ix (1 2 t )
e
The Proof
0
? sin x
dt 
x
Recall Euler’s Formula,
eixu  cos( xu)  i sin( xu)
Replacing e ixu
1
1 / 2 e
1
ixu
in the equation we get.
du
1
 1 / 2  [ cos( xu)  i sin( xu)] du
1
1 ix (1 2 t )
The Proof
e
0
? sin x
dt 
x
Splitting up the integral,
1
1
1
1
 1 / 2 cos( xu) du  i / 2 sin( xu) du
Evaluating the integral,
sin( xu)
 1/ 2
x
u 1
u  1
 1 / 2 x[sin( x)  sin(  x)]
1 ix (1 2 t )
The Proof
e
0
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
x
 rk (t ) 
0 exp  ix 2k  dt  cos 2k
1
:
x
 rk (t )  ?
0 exp  ix 2k  dt  cos 2k
1
The Proof
Notice that we can break up the integral
 rk (t ) 
0 exp  ix 2k  dt
1
into this sum of integrals over dyadic subintervals

m even
( m 1) 2  k
m2
k
e
ix 2
k
dt 

m odd
( m 1) 2 k
m2
k
e
ix 2 k
Note: (2-k) represents the length of a subinterval
dt
x
 rk (t )  ?
0 exp  ix 2k  dt  cos 2k
1
The Proof
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
 2 j  2 2 j  1
Ij   k , k 
2 
 2
j
. (as seen below)
 2 j 1 2 j 
Jj  k , k 
2 
 2
x
 rk (t )  ?
0 exp  ix 2k  dt  cos 2k
1
The Proof
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:

 

  J j    I j   0,1

 


 

x
 rk (t )  ?
0 exp  ix 2k  dt  cos 2k
1
The Proof
Ij
Using
2 k-1
 e
ix 2 k
j 1 I j
J
and
2 k -1
we get:
j
dt    e
ix 2 k
dt
j 1 J j
Inserting the end points,
2 k-1
2 j 1
k
k
ix
2
2
2 j 2
 
j 1
e
2k
2 k-1
2j
2k
2 j 1
dt   
j 1
2k
e
ix 2 k
dt
x
 rk (t )  ?
0 exp  ix 2k  dt  cos 2k
1
The Proof
We then pull out the power of e:
2 k-1
 e
j 1
ix 2
k
2 j 1
2k
2 j 2
2 k-1
2k
j 1

1 dt   e
ix 2
Evaluating this we find it equals
2 k -1
2 k -1
1 ix 2k
1 ix 2k
 k e
 k e
j 1 2
j 1 2
k
2j
2k
2 j 1

2k
1 dt
x
 rk (t )  ?
0 exp  ix 2k  dt  cos 2k
1
The Proof
Simplifying:
1
 k
2
 e
1
 k
2
 x
2 cos k 

2 
j 1
2 k -1
ix 2 k
e
 ix 2 k
j 1
2 k -1

i
 i
2
cos


e

e
where:
x
 rk (t ) 
0 exp  ix 2k  dt  cos 2k
1
The Proof
 x
We can now pull out the 2 cos k 
2 
1
 x
 x
 k 1 cos k 1  cos k 
2
 2  j 1
2 
2 k -1
This gives the desired solution:
x
 rk (t ) 
0 exp  ix 2k  dt  cos 2k
1
QED
Interesting Development
Recall the formula,

sin x
x
  cos k
x
2
k 1
which may now be seen as:
1 ix (1 2 t )
1
sin x
  rk (t ) 
 e
dt   exp  ix  k  dt
0
0
x
 k 1 2 
Proved
by def


1
x
 rk (t ) 
  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:
1
x
  rk (t ) 
cos k   exp  ix  k  dt

0
2
k 1
 k 1 2 

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
 c r (t )
k 1
k k
We realize this is a step function over the intervals:
 s s 1
 n , n ,
2 2 
s  0,1,  ,2 n  1
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:
1


ck rk (t ) dt  n
0 exp i
2
k 1

1
n


 exp  i1  ck 
n

over all possible sequences
Classical Analysis Proof
n


Now,
 exp  i1  ck 
n
 eick  e ick  n
   cosck 
  
2
k 1 
 k 1
1
2n
So,
n
 n

ck rk (t ) dt   cosck 
0 exp i
k 1
k 1

1
Note: This proof is
similar to the last
n
 e
k 1
1 ic r ( t )
k k
0
Classical Analysis Proof
Using the derived equation from the previous page and
x
setting ck  k
2
we get this familiar equation:
1
x
 n rk (t ) 
  cos k   exp  ix  k  dt
0
2
k 1
 k 1 2 
n
When we take the limit,
n
rk (t )
lim  k  1  2t
n 
k 1 2
Classical Analysis Proof
So now we have proven that uniformly
on the interval (0,1), we have,
1 ix (1 2 t )
1
sin x
 n rk (t ) 
 e
dt  lim  exp  ix  k  dt
0
n  0
x
 k 1 2 

x
x
 lim  cos k   cos k
n 
2
2
k 1
k 1
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?
n
1 1
1
    (n times)
2 2
 2
Consider the set of t’s:
r1(t)=+1, r2(t)=-1, r3(t)=-1
Random Variables &
Statistical Independence
The set of t’s (possibly excluding the end
3 4
points) is the interval  , 
8 8
and the
1
length of the interval is clearly 8 , and
1 1 1 1
   
8 2 2 2
Random Variables &
Statistical Independence
This observation can be written:
 r1 (t )  1, r2 (t )  1, r3 (t )  1
  r1 (t )  1 r2 (t )  1 r3 (t )  1
Note:  is the length of the set inside the braces
This can be generalized to this form:
 r1 (t )  1 , , rn (t )   n 
  r1 (t )  1 r2 (t )   2   rn (t )   n 
Independence &
Rademacher Functions
Rademacher Functions are independent
Bernoulli random variables (-1 or 1).

represents the length of a dyadic
subinterval.
Random Variables &
Statistical Independence
You may see this as only a more
complicated way of writing
n
1 1
1
1
       (n times)
2 2
2
 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:
sin x
 n

0 exp i1 ck rk (t ) dt  x
1
Trying to prove that
1


ick rk ( t )
e
dt
0 exp i1 ck rk (t ) dt  

0
k 1
1
n
n
Probabilistic Proof
Splitting the integral over the dyadic subintervals
corresponding to  n we find:


exp  i  ck rk (t )  r1 (t )   1  rn (t )   n 

 1 ,, n
 1

n
The inner sum evaluates to:

n
n
e   r (t )   




ick  k
k
1 ,, n
1
1
k
Probabilistic Proof
Remove the second product

n
e




1 ,, n
ick  k
 rk (t )   k 
1
Interchange the product and sum:
n
  e
k 1  k
ick  k
 rk (t )   k 
Probabilistic Proof
Replacing the sum with the integral
we find the desired outcome:
n
  e
k 1
1
0
ick rk ( t )


dt   exp i  ck rk (t ) dt
0
 1

1
n
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