a probabilistic approach to vieta’s formula by: mark osegard

A Probabilistic Approach toVieta’s Formula

by: Mark Osegard

Points of Discussion

I. Vieta’s FormulaII. Rademacher FunctionsIII. Vieta’s Formula in terms of

Rademacher FunctionsIV. A Classical Analysis Proof

V. Random Variables & Statistical Independence

VI. Probabilistic ProofVII. 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

n

kkn

n xxx

xxxx

xxx

xxx

1

3

2

2cos

2sin2 sin

2

cos4

cos8

cos8

sin2

2cos

4cos

4sin2

2cos

2sin2 sin

The ideas all begin withsome simple trigonometry.Using the double angleformula.

xx

x

xx

x

xfor

nn

n

nn

n

n

n

n

2sin2lim

2sin2lim

1

2

2sin

lim1

,0 From calculus we find.

Vieta’s Formula

Combining the equations derived

on the past two slides we see:

xxn

n

n

2sin2lim

n

kkn

n xxx

1 2cos

2sin2 sin &

1 2cos

sin

kk

x

x

x

Vieta’s Formula

This trigonometric identity that leads to vieta’s formula

Setting we see an interestingcase, which yields this formula.

2 x

112

cos2

nn

,2

222

2

22

2

2

Vieta’s Formula

Using the half-angle formula for cosine repeatedly

Binary Expansion

Taking another approachis usually helpful, andthis case is no different.

1or 0either is ε where

,2

)(ε

2

)(ε t

221

tt

We know that every numberbetween can be represented in the form:

10 t

Rademacher Functions

Using the same ideas we find it more convenient touse the function defined below, which uses the interval [0,1]. of instead ]1,1[

,3, 2, 1,k

)(ε21)(

ttr kk

Rademacher Function

Rademacher Functions

These are the dyadic subintervals of the Rademacher Functions.

2

1

0

1

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:

1 2

)(2t-1

kk

k tr

,2

)(ε

2

)(εt 21

tt

as

Notice

Now we shall see that the term

can be written :

x

xdt

tix sine

)21(1

0

x

xsin

The Proofx

xdte

tix sin)21(1

0

?

du edu edte ixuixutix

1

1

1

1

)21(1

02/1)2/1(

dudtdtdutuN 2/1 , 2 ,21 :ote

We start with, substitute ‘u’, change parameter

)sin()cos( xuixueixu Recall Euler’s Formula,

Replacing in the equation we get.ixue

du xuixu

du eixu

)]sin()cos([2/1

2/1

1

1

1

1

The Proof x

xdte

tix sin)21(1

0

?

Splitting up the integral,

duxuixu )sin(2/du )cos(2/11

1

1

1

Evaluating the integral,

1

1

)sin(2/1

u

ux

xu

)]sin()[sin(2/1 xxx

The Proof x

xdte

tix sin)21(1

0

?

Recall the equation from the previous slide,

)]sin()[sin(2/1 xxx

dtx

x tix )21(1

0e

sin

QED

The Proof x

xdte

tix sin)21(1

0

pull out -1

x

xsin

Also Notice

written in terms of Rademacher Functions :

Now we shall see that can be k

x

2cos

kkk x

dttr

ix2

cos 2

)(exp

1

0

The Proof kkk xtr

ix2

cosdt 2

)(exp

1

0

?

dttr

ixk

k 2

)(exp

1

0

dt dt odd

2)1(

2m

2

even

2)1(

2m

2

m

m ix

m

m ixk

k

kk

k

k

ee

Notice that we can break up the integral

into this sum of integrals over dyadic subintervals

Note: (2-k) represents the length of a subinterval

The Proof kkk xtr

ix2

cosdt 2

)(exp

1

0

?

There is a total of 2k subintervals. We define the end

points of the EVEN subintervals to be represented by

, also we define the end points of the ODD

subintervals to be represented by . (as seen below)

kkj

jjI

2

12,

2

22

jI

jJ

kkj

jjJ

2

2,

2

12

The Proof kkk xtr

ix2

cosdt 2

)(exp

1

0

?

The Rademacher functions or are defined asfollows:

jk

jk

Jtr

Itr

on 1- )(

on 1 )(

)(trk

If we union these subintervals we see:

1,0

jj IJ

The Proof kkk xtr

ix2

cosdt 2

)(exp

1

0

?

Using and we get:jI

jJ

1-k1-k 2

1

22

1

2 dt dtj J

ix

j I

ix

j

k

j

k

ee

Inserting the end points,

1-k1-k 2

1

2

2

2

122

2

1

2

12

2

222 dt dt

j

j

jix

j

j

jix k

k

kk

k

k

ee

The Proof kkk xtr

ix2

cosdt 2

)(exp

1

0

?

1-k1-k 2

1

2

2

2

122

2

1

2

12

2

222 dt 1 dt 1

j

j

jix

j

j

jix k

k

kk

k

k

ee

We then pull out the power of e:

Evaluating this we find it equals

1-k1-k 2

1

22

1

2

2

1

2

1

j

ixk

j

ixk

kk

ee

The Proof kkk xtr

ix2

cosdt 2

)(exp

1

0

?

1-k

1-k

2

1

2

1

22

2cos2

2

1

2

1

jkk

j

ixixk

x

eekk

Simplifying:

ii ee cos2where:

We can now pull out the

k

x

2cos2

The Proof kkk xtr

ix2

cosdt 2

)(exp

1

0

k

jkk

xx

2cos1

2cos

2

11-k2

11

This gives the desired solution:

kkk xtr

ix2

cosdt 2

)(exp

1

0

QED

Interesting Development

Recall the formula,

1 2

cossin

kk

x

x

x

which may now be seen as:

dt 2

)(expe

sin 1

01

)21(1

0

kk

ktix tr

ixdtx

x

1k

1

01

dt 2

)(exp

2cos

kk

kk

trix

x

Proved by def

Trig proof by prev obs.

Astonishing Discovery

dt 2

)(exp

2cos

1

011

kk

k

kk

trix

xSince we can now say:

Clearly, we see that an integral of products is a product of integrals!

1

1

0

1

01

dt 2

)(expdt

2

)(exp

kk

k

kk

k trix

trix

Vieta’s formula has this interesting expression interms of Rademacher Functions

Analyzing the following function may show us.

n

kkk trc

1

)(

We realize this is a step function over the intervals:

12,0,1,s ,2

1,

2

n

nn

ss

Can we prove Vieta’s formula using Rademacher’sFunctions as a starting point?

Classical Analysis Proof

Since every sequence of +1’s and –1’s corresponds

to one and only one interval

nn

ss

2

1,

2By breaking the integral as before into dyadic subintervalsWe can say that:

n

kn

n

kkk cidttrci

1

1

01

exp2

1 )(exp

sequences possible allover


n

kk

n

k

icic

n

kn

cee

ci

kk

11

1

cos2

exp2

1 Now,

So,

n

kk

n

kkk cdttrci

1

1

01

cos )(exp

n

k

tric kke1

1

0

)(


Note: This proof is similar to the last

Classical Analysis ProofUsing the derived equation from the previous page and

setting we get this familiar equation:kk

xc

2

dt 2

)(exp

2cos

1

011

n

kk

kn

kk

trix

x

When we take the limit,

ttrn

kk

k

n21

2

)(lim

1

So now we have proven that uniformly on the interval (0,1), we have,

11

1

01

)21(1

0

2cos

2coslim

dt 2

)(explime

sin

kk

n

kkn

n

kk

k

n

tix

xx

trixdt

x

x



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?

Consider the set of t’s:r1(t)=+1, r2(t)=-1, r3(t)=-1

) (2

1

2

1

2

1timesn

n


The set of t’s (possibly excluding the end

points) is the interval and the

length of the interval is clearly , and

8

4,

8

3

8

1

2

1

2

1

2

1

8

1


This observation can be written: 1)(1)(1)(

1)( ,1)( ,1)(

321

321

trtrtr

trtrtr

Note: is the length of the set inside the bracesThis can be generalized to this form:

nn

nn

trtrtr

trtr

)()()(

)( , ,)(

2211

11

Independence &Rademacher Functions

Rademacher Functions are independent Bernoulli random variables (-1 or 1).

represents the length of a dyadic

subinterval.


You may see this as only a more complicated way of writing

) (2

1

2

1

2

1

2

1timesn

n

but this means much more. It shows a deep property of rk(t) and subsequentlybinary digits. This now leads us into the Probabilistic Proof.

Probabilistic Proof

We start with the equation:

x

xsin )(exp

1

01

dttrci

n

kk

Trying to prove that

dtedttrcin

k

tricn

kkkk

1

1

0

)(1

01

)(exp

Splitting the integral over the dyadic subintervals corresponding to we find:

nn

n

kk trtrtrcin

)()()(exp 11,, 11

n

Probabilistic Proof

n

kk

nic tre

n

kk

1,, 1

)(1

The inner sum evaluates to:

Probabilistic Proof

kk

nic tre

n

kk

)(,, 11

Remove the second product

n

kkk

ic

k

kk tre1

)(

Interchange the product and sum:

dttrcidten

kk

n

k

tric kk )(exp1

011

1

0

)(

Replacing the sum with the integralwe find the desired outcome:

Probabilistic Proof

The vieta identity follows when you take x/2k

Conclusion

So we have seen three ways to proveVieta’s formula and it’s connection tostatistical 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

Dr. Deckelman

“Statistical Independence in Probability, Analysis and Number Theory” by Mark Kac, published by the MAA Carus Mongraph Series, 1959.

a probabilistic approach to vieta’s formula by: mark osegard

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