1. A Probabilistic Approach to Vieta’s Formula by: Mark Osegard

2. 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

3. 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.

4. Vieta’s Formula The ideas all begin with some simple trigonometry. Using the double angle formula.

5. From calculus we find. Vieta’s Formula

6. Combining the equations derived on the past two slides we see: & Vieta’s Formula This trigonometric identity that leads to vieta’s formula

7. Setting we see an interesting case, which yields this formula. Vieta’s Formula Using the half-angle formula for cosine repeatedly

8. Binary Expansion Taking another approach is usually helpful, and this case is no different. We know that every number between can be represented in the form:

9. Rademacher Functions Using the same ideas we find it more convenient to use the function defined below, which uses the interval Rademacher Function

10. Rademacher Functions These are the dyadic subintervals of the Rademacher Functions. 0 1

11. 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.

12. Connecting You may notice the relationship between the Rademacher Functions and Binary Expansion. Subsequently, we rewrite the equation: as

13. Notice Now we shall see that the term can be written :

14. The Proof ? We start with, substitute ‘u’, change parameter

15. Recall Euler’s Formula, Replacing in the equation we get. The Proof ?

16. Splitting up the integral, Evaluating the integral, The Proof ?

17. Recall the equation from the previous slide, QED The Proof pull out -1

18. Also Notice written in terms of Rademacher Functions : Now we shall see that can be

19. The Proof ? 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

20. The Proof ? There is a total of 2 k 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)

21. The Proof ? The Rademacher functions or are defined as follows: If we union these subintervals we see:

22. The Proof ? Using and we get: Inserting the end points,

23. The Proof ? We then pull out the power of e: Evaluating this we find it equals

24. The Proof ? Simplifying: where:

25. We can now pull out the The Proof This gives the desired solution: QED

26. Interesting Development Recall the formula, which may now be seen as: Proved by def Trig proof by prev obs.

27. Astonishing Discovery Since we can now say: Clearly, we see that an integral of products is a product of integrals! Vieta’s formula has this interesting expression in terms of Rademacher Functions

28. Analyzing the following function may show us. We realize this is a step function over the intervals: Can we prove Vieta’s formula using Rademacher’s Functions as a starting point? Classical Analysis Proof

29. Since every sequence of +1’s and –1’s corresponds to one and only one interval By breaking the integral as before into dyadic subintervals We can say that: Classical Analysis Proof

30. Now, So, Classical Analysis Proof Note: This proof is similar to the last

31. Classical Analysis Proof Using the derived equation from the previous page and setting we get this familiar equation: When we take the limit,

32. So now we have proven that uniformly on the interval (0,1), we have, Classical Analysis Proof

33. 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.

34. 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: r 1 (t)=+1, r 2 (t)=-1, r 3 (t)=-1

35. Random Variables & Statistical Independence The set of t’s (possibly excluding the end points) is the interval and the length of the interval is clearly, and

36. Random Variables & Statistical Independence This observation can be written: Note: is the length of the set inside the braces This can be generalized to this form:

37. Independence & Rademacher Functions Rademacher Functions are independent Bernoulli random variables (-1 or 1). represents the length of a dyadic subinterval.

38. Random Variables & Statistical Independence You may see this as only a more complicated way of writing but this means much more. It shows a deep property of r k (t) and subsequently binary digits. This now leads us into the Probabilistic Proof.

39. Probabilistic Proof We start with the equation: Trying to prove that

40. Splitting the integral over the dyadic subintervals corresponding to we find: Probabilistic Proof The inner sum evaluates to:

41. Probabilistic Proof Remove the second product Interchange the product and sum:

42. Replacing the sum with the integral we find the desired outcome: Probabilistic Proof The vieta identity follows when you take x/2k

43. 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.

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