Probabilistic number theory : A report on work done. What is the probability that a randomly chosen integer has no square factors?

Transcription

1 Pobabilistic numbe theoy : A eot on wo done What is the obability that a andomly chosen intege has no squae factos? We can constuct an initial fomula to give us this value as follows: If a numbe is to have no squae factos, then it must not be a multile of the squae of any ime. P(intege is multile of a squae ime, ) = (if integes ae to be chosen andomly). Theefoe, P(numbe is not a multile of a squae ime, ) = - In alying this to all integes, we must find the obability, P, that a numbe is not a multile of o o o o n. Fo eamle: The obability that a numbe is not divisible by 4, 9, o 5 (ie. the squaes of the fist thee imes), is c - 4 mc - 9 mc - 5 m = 6 5 A simle numeical calculation eveals that the oduct of the fist tems is.64 This aoach taes account of numbes which have the squae of a comosite numbe, as a facto. This comosite will, in accodance with the fundamental theoem of aithmetic (see age ), be evaluable as the oduct of imes, and thus such a numbe would be included in this method. Clealy, % P = - e o Fo all imes, fom to Each tem in this oduct can be evaluated as a geometic seies, summed to infinity. Page of 7

2 Deivation of the fomula fo the summation of an infinite seies Conside the evaluation of the oduct: Q =( - )( ) =( ) - ( ) Fom visualizing how these bacets will eand, it is clea that: Q= - 5 Now in fact: ( - )( n (n +) )= - Dividing by ( - ): n = - n + - So suose it was elaced by whee: - <<. Then, in the limit, As n + ", and so: Theefoe, as n" - n + - " - = = - when - << Conside the fomula fo the sum to infinity of the infinite seies: - = Now let = (clealy -<<) Substituting this into the fomula gives: e o = = - So it seems the oiginal comonent of the oduct P aeas as the denominato of the RHS. Theefoe: P = % = - ime % ime = e o Page of 7

3 Conside the eansion of P as oducts: c m: c m: c m: c m: Clealy the comutation of this oduct is going to involve the summing of many oducts, each of which will contain a tem fom each of the seies baceted above. One such oduct will fo eamle be: and in geneal each denominato will be the oduct of squaes of imes. h 6 : :5 : 7 8 _Conside the Fundamental Theoem of Aithmetic, which states that evey intege n can be eessed as the oduct of imes, in a unique way, such that: e n = e e e... And theefoe: n e = e e e... as in the denominatos. Due to the fact that each eesentation of n is unique, this means that each unique denominato can be eessed as n. Theefoe: P = n = n Clealy this seies is not geometic, as the atio between the tems is not constant. Theefoe anothe method is necessay to evaluate the sum of the eciocals of the squaes. The seach fo this sum is not new. Since the time of Eule, a numbe of methods have been suggested, one of which is given hee. It tuns out, ehas athe suisingly, that the sum: 6 n n = Page of 7

4 The Evaluation of: = n = n Giesy, D. Mathematics Magazine 97 This sum is easily calculable to a finite degee of accuacy though basic aithmetic, but fo centuies mathematicians have been finding methods of evaluating the eact sum. Hee is one such method. Tigonomety, as indicated by the aeaance of π, is a majo comonent. Conside the following tigonometic sum A: Using the identity: sin Acos B = sin(a + B) + sin(a - B) Combined with the fact that : The sum A can be e-witten as: Clealy this sum will telescoe to: Theefoe: sin + sin cos + sin cos + sin cos sin cos n sin(- )=- sin sin + (sin - sin ) + (sin 5 - sin ) + (sin 7 - sin 5 ) (sin( n + )) - (sin( n - )) sin(n + ) sin + sin cos + sin cos + sin cos sin cos n = sin(n + ) Dividing both sides by sin Gives: Conside the esult of multilying both sides by and integating with esect to : Evaluating the LHS: + cos + cos + cos cos n = sin(n + ) sin # d + # cos d + # cosd + # cosd # cos nd = # sin(n + ) sin # d = 4 Page 4 of 7

5 Each of the net integals can be evaluated though integation by ats: Theefoe, the sum of the integals on the evious age will not change fo even values of n. Hence we may assume n is odd, say n = t + Maing this substitution gives: - ( sin(t + ) (t + ) ) = # d sin As t" the LHS is equal to 4 minus twice the sum to infinity of the eciocals of the odd squaes. The RHS will tend to zeo, given the following theoem of Riemann. Riemann's Lemma Fo any function, f, which is continuous, and diffeentiable on (, π), the integals: # f()sin t d and # f()cos t d both go to as t" The integal, sin(t + ) # d may be witten as: sin # sin(t + ) d sin Page 5 of 7

6 Since is continuous and diffeentiable on (, π), this can be tansfomed into the sum of sin two integals lie those in Riemann's lemma by alying the addition fomula fo sine to sin(t + ) Theefoe: 4 - ( )= O: 8 = The 'missing tems', i.e. the sum of the eciocals of the even owes ae a multile of the entie sum of the eciocals of squaes: = 4 ( ) Also: = 8 Theefoe, adding the two equations above we have: = ( ) O in sigma notation: n n = n= n That is: 4 n 8 n = Theefoe: = n 6 n = Page 6 of 7

7 Theefoe: P = 6 = The obability that an intege has no squae factos Taing ie to 4 sf, this gives the value of P as.677 not fa fom the.64 we obtained fom the ealie numeical simulation. Conclusion This was by fa the most hefty iece of mathematics I had eve attemted Woing though this oblem involved a mitue of sills. My own mathematics maniulation layed an imotant at in the fist section of my wo. The whole concet of obabilistic numbe theoy was new to me, and the guidance of my mento was imotant in homing my effots, and heling my out when I was stuc. The summation of / n involved a change in what my wo involved. I began to ead the mathematics that was sent to me ove , and wo though it myself, tying to undestand it. The dialogue with my mento focussed on unavelling the oblems in what I had ead, as new concets wee elained to me. Oveall, wo on this oject led me to discove and eloe an aea of mathematics that was entiely new to me, and intoduced me to a ange of new ideas. I would be hay to discuss anything witten hee. Do me a line at: Joenoss@bitishlibay.net Page 7 of 7