Arithmetic 1: Prime numbers and factorization (with Solutions)

Transcription

1 Bachelor of Ecole Polytechique Computatioal Mathematics, year 2, semester 1 Arithmetic 1: Prime umbers ad factorizatio (with Solutios) P. Tchebychev (Russia, ). Amog may thigs, he was the rst mathematicia to obtai the asymptotics for the umber of prime umbers less tha (see the rst Exercise below). Table of cotets Prime umbers ad divisibility Factorizatio The Euclid algorithm # execute this part to modify the css style from IPytho.core.display import HTML def css_stylig(): styles = ope("./style/custom2.css").read() retur HTML(styles) css_stylig()

2 ## loadig pytho libraries # ecessary to display plots ilie: %matplotlib ilie # load the libraries import matplotlib.pyplot as plt # 2D plottig library import umpy as p # package for scietific computig from math import * # package for mathematics (pi, arcta, sqrt, Prime umbers ad divisibility We aim to ivestigate the distributio of primes amog itegers. Namely, how may prime umbers are there (approximately) betwee ad? 1 Do it yourself. Write a boolea fuctio IsPrime() which returs True if ad oly is prime. a (mod p) (Recall that is obtaied with a%p.) def IsPrime(): # iput: iteger # output: True or False depedig o whether is prime or ot if ==1: retur False if ==2: retur True elif %2==0: retur False factor=3 while factor**2 < +1: if %factor == 0: retur False factor=factor+2 retur True prit(isprime(2)) True 2 P() P(11) = 5 2, 3, 5, 7, 11 Now we are ready for experimet. For, let deote the umber of primes less tha. For example, sice are prime.

3 Do it yourself. Write a script which takes as iput [P(2), P(3),, P()]. ad returs the list Plot the fuctio (try ). P() = 100, 1000, CoutigPrimes=[0] Primes=[] T=1000 for i rage(2,t): if IsPrime(): CoutigPrimes.apped(CoutigPrimes[-1]+1) Primes.apped() else: CoutigPrimes.apped(CoutigPrimes[-1]) X=rage(1,T) plt.plot(rage(1,t),coutigprimes,label='pi') #plt.plot(rage(1,t),x/p.log(x+1)) plt.xlabel('number $$'),plt.ylabel('primes less tha $$') plt.title('coutig Primes') plt show() Do it yourself. Modify your previous plot to guess (by trials ad errors) what is the asymptotic behaviour of F() whe + : d a sequece a such that. P() a I order to improve your guess you ca plot i some iterval (istead (0, T) of ). P a (T/2, T)

4 # By trial ad errors we fially guess that pi() is close to /log() Guess=[] for i i rage(1,t-1): Guess.apped(CoutigPrimes[i]/(i/(0.01+p.log(i)))) R=9*T/10 # We plot the R last steps #plt.plot(rage(t-r,t-1),guess[t-r-1:t-2]) plt.plot(rage(1,t),coutigprimes*p.log(x)/x) plt show() Factorizatio Do it yourself. Write a fuctio Factorize() which returs the factorizatio of ito primes. For example your fuctio should retur: Factorize(12) [2,2,3] Hit: Thik recursive! def Factorize(): # iput: iteger # output: list of factors of for factor i rage(2,): # Tests divisio by 2,3,..., if %factor == 0: retur [factor]+factorize(//factor) retur [] prit(factorize( )) IsPrime(659) [2, 2, 3, 3, 7, 13, 659] True

5 For 2 we itroduce F() = Number of prime factors of, couted with multiplicity. F(12) = 3 For example,. Do it yourself. Plot the fuctio F() (try = 100, 1000, 5000). =100 F=[le(Factorize(k)) for k i rage(2,)] X=rage(2,) plt.plot(x,f,'o') plt.plot(x,p.log(x)/p.log(2)) plt show() Do it yourself. (Theory) What ca you say about the asymptotic behaviour of F() (whe + )? I particular: 1. What is? 2. Fid a simple sequece a such that F() lim sup = 1. + a (You ca check your claim with the previous script.) By de itio lim if F() lim ifu = lim if ( ), lim supu = lim sup u ( k ). k u k k

6 Aswers. 1. If is prime the. As there are i itely may primes, F() = 1 lim iff() = lim if F(k) = lim 1 = 1. + k 2. Heuristic: We expect F()/ to be large if is very friable, i.e. if it has may small factors. The worst case seems to be whe is a power of two. Proof of lower boud: Whe Thus, for i itely may itegers (the powers of two),. Proof of lower boud: O the other had, we always have that F() log 2 (): if Factorize() = [ a 1, a 2, a F() ] the ecessarily i.e.. As a coclusio, + NB: O the previous plot we have also show with at powers of two. = 2 k is a power of two, F( 2 k ) = k. F() log 2 () = a 1 a 2 a F() 2 F(), F() log 2 () F() F() lim sup = 1. log 2 () log 2 (), which coicides The Euclid algorithm We recall that Euclid's algorithm (which computes the gcd of two o-egative itegers) relies o the fact that for every we have a%b a, b gcd(a, b) = gcd(b, a%b), a/b where is the remaider of the euclidea divisio. Do it yourself. Write a fuctio GreatestCommoDivisor(a,b) which returs gcd(a, b) usig the Euclid algorithm.

7 def GreatestCommoDivisor(a,b): # iput: a,b: o-egative itegers # output: returs the gcd of a ad b if b==0: retur a else: retur GreatestCommoDivisor(b,a%b) GreatestCommoDivisor(67 938) 67 Do it yourself. Write a fuctio GreatestCommoDivisor_3(a,b,c) which returs the gcd of three umbers. def GreatestCommoDivisor_3(a,b,c): # iput: a,b,c: o-egative itegers # output: returs the gcd of a,b,c retur GreatestCommoDivisor(b,GreatestCommoDivisor(b,c)) GreatestCommoDivisor_3(11*4*10*7 4*10*3 5*4*10*17) 40 m, gcd(m, ) = 1 14, 9 Itegers are said to be coprime if. For example, are coprime. I may refereces (see e.g. /probability-that-two-radom-umbers-are-coprime-is-frac6-pi2it ( it is said that 6 "The probability that two umbers radomly chose are coprime is." Yet there is o obvious way to rigorously de e what are "two umbers radomly chose". π 2 Do it yourself. 1. Try to give a rigorous formulatio to the above assertio. 2. Use your fuctio GreatestCommoDivisor to check the claim. 6 ( is approximately.) π %

8 Aswers. The above assertio ca be iterpreted as follows: if we pick uiformly (i, j) [1, ] 2 gcd(i, j) = 1 6/π 2 at radom a pair of itegers i (for large ) the the probability that goes to. Let I, J be two idepedet radom variables i {1, 2,, } we thus have to estimate: card {(i, j) [1, such that gcd(i, j) = 1} P(gcd( I, J ) = 1) =. card {(i, j) [1, ] 2 } ] 2 N=400 # Fial value for the test PairsOfCoprime=[0] for i rage(2,n): NewCoprimeWith_=0 for k i rage(1,): if GreatestCommoDivisor(k,)==1: NewCoprimeWith_=NewCoprimeWith_+1 CoprimeWith_=PairsOfCoprime[-1]+2*NewCoprimeWith_ PairsOfCoprime.apped(CoprimeWith_) X=p.arage(1,N) prit(pairsofcoprime[-1]/(n**2+0.0)) Claim=6/(p.pi**2) plt.plot(x,pairsofcoprime/(x**2+0.0)) plt.plot([0,],[claim,claim],label='6/pi^2') plt.xlabel('number $$'),plt.ylabel('probability') plt.title('probability of beig coprime') plt.leged() plt show()

9 Do it yourself. 1. Modify your previous script to estimate the probability that three itegers are pairwise coprime. Warig: "pairwise coprime" is ot the same as "coprime": (2, 4, 5) are ot pairwise coprime but (2, 4, 5) is a coprime triple: gcd(2, 4, 5) = You ca compare your umerical estimate with this lik ( ow.et/questios/119416/probability-of-allcombiatios-of-k-umbers-amog--beig-coprime). N=100 TriplesOfCoprime=[0] for i rage(2, N): CoprimeTriples_lead = 0 for k i rage(1, ): for m i rage(1, k): if ((GreatestCommoDivisor(,k)==1) ad (GreatestCommoDivisor(, (GreatestCommoDivisor(k,m)==1)): CoprimeTriples_lead = CoprimeTriples_lead + 1 CoprimeTriples_ = TriplesOfCoprime[-1] + 6*CoprimeTriples_lead TriplesOfCoprime.apped(CoprimeTriples_) X=p.arage(1,N) prit(triplesofcoprime[-1]/((n)**3+0.0)) #prit(triplesofcoprime) # prit the array of umber of triples where pairwise elemest are coprime # it is importat that 'pairwise coprime' is ot the same as 'coprime' ((2 4 # however (2 4 5) are coprime sice g.commo divisor is 1) plt.plot(x,triplesofcoprime/(x**3+0.0)) plt.xlabel('number $$'),plt.ylabel('probability') plt.title('probability of a triple beig pairwise coprime') plt.show()

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