The Prime Number Theorem

Transcription

1 The Prime Number Theorem

2 We study the distribution of primes via the function π(x) = the number of primes x

3 It s easier to draw this way: π(x) = the number of primes x

4 Clearly, π(x) < x (not every number is prime) and lim x π(x) = (there are infinitely many primes) How many primes are there among the first integers? By brute force, π( ) = Can we approximate π(x) using elementary functions? 4

5 Gauss s Prime Number Conjecture (792) π(x) x log(x) for large x. x π(x) x log(x) π(x) x log(x) π(x) x/ log(x) π(x)

6 In terms of limits, we may write Gauss s 792 conjecture as π(x) x/ log(x) lim x π(x) = 0 or lim x x/ log(x) π(x) =. Gauss s Conjecture of 849 or For large x, π(x) x 2 dt log t lim x x dt log t π(x) 2 =. Aside: Li(x) def = x µ dt log t. 6

7 ( x ) Exercise: lim x log(x) dt log t x 2 =. (Hint: l Hôpital) Corollary: lim x If either lim x ( ) x log(x) π(x) ( x ) log(x) π(x) or lim x x 2 dt log(t) π(x) = is equivalent to lim x exists, then x 2 dt log(t) π(x) =. 7

8 The Prime Number Theorem 40 lim x π(x) ( ) =. x log(x) The convergence is very slow: At x = 0 6, π(x)/(x/ log(x)) At x = 0 9, π(x)/(x/ log(x)) At x = 0 2, π(x)/(x/ log(x))

9 It s also messy. y = π(x) ( ) x log(x)

10 The first proofs of the Prime Number Theorem (896) use the Riemann Zeta Function ζ(s) = n= n s = s + 2 s + 3 s + From calculus, we know Vallée-Poussin ζ() = ζ(2) = ζ(4) = diverges π2 converges to 6 π4 converges to 90 Hadamard 0

11 The function ζ(s) is defined for 2 all s = x + iy in C, with a simple pole at s = What s the connection with primes? Riemann

12 Fix a number M. Let Let Λ M P M = {p, p 2,..., p r : p i M} = set of primes M. = { n Z + : prime factors of n are in P M }. Note that Λ M is infinite and Example: M = 8. Then P 8 = {2, 3, 5, 7} and Λ 8 = {, 2, 3, 4, 5, 6, 7, 8, 9, 0, 2, 4, 5, 6,...} = {2 α 3 α 2 5 α 3 7 α 4 : each α i is a non-negative integer} contains {, 2, 3,..., M}. 2

13 Recall that P 8 = {2, 3, 5, 7}. Consider the product ( ) ( ) 3 + ( ) ( ) Each term in the expansion has the form 2 α 3 α 2 5 α 3 7 α 4 where p P 8 ( α, α 2, α 3, α 4 are are non-negative integers These denominators are exactly the elements of Λ 8, and we get ( p + 0 p + ) p + = 2 n. n Λ 8 ) 3

14 Furthermore, by the geometric series theorem, we know that ( p + 0 p + ) ( ) r p + = = 2 p. p So we get n Λ 8 n = p P 8 r=0 ( p + 0 p + ) p + 2 = p P 8 p ( ) ( ) ( ) ( ) = =

15 More generally, we have Lemma: p P M p = n Λ M n Extension of Lemma: = n p s s all p n= = ζ(s) Now Λ M {, 2, 3,..., M}, so This gives us the n Λ M n > Corollary: M n= p P M p n. > M n= n 5

16 As M, p P M p all primes p and M n= n so that all primes p, and we get Corollary: all p ( ) p = 0 That is, can be made arbitrarily small. 6

17 A small digression: Claim: all primes ( ) p = 0 implies p prime p is divergent Remark: This gives another proof of the fact that there are infinitely many primes. The proof of the claim itself follows from a Lemma: Let > a > a 2 > a 3 > be a decreasing sequence of positive numbers such that a n converges. Then. ( a n ) 0 n= 7

18 Proof sketch Because a n converges, we can find N so that Write S = n= a n = a + a a N }{{} >S 2 ( a n ) = n= N ( a n ) n= a n < 2. n=n + a N + a N+ + }{{} < 2 ( a n ). n=n It will suffice to show that the second factor is non-zero. 8

19 Write ( a n ) = ( a N )( a N+ )( a N+2 ) n=n = i a i + i<j a i a j a i a j a k i<j<k + where all the indices start at N. 9

20 We can then show that a i < 2, a i a j < 2 2, i and i i<j That is, the right side of a i > i<j n=n( a n ) = i i<j<k a i a j > a i + i<j a i a j a k < 2 3, i<j<k a i a j a k > a i a j i<j<k satisfies the hypotheses of the Alternating Series Test. and so on, a i a j a k + By the AST, the series (and therefore the product) converges to a number between and i non-zero. a i. Since i a i <, we know that this number is 2 20

21 Old Stuff: n is divergent. New Stuff: p is divergent = = The harmonic series diverges slowly: It takes about a million terms to get to a sum of The /p series diverges very slowly: A million terms gets you only to about But with either series, if you add up enough terms, you can make the sum as large as you want. 2

22 Estimating π(x) Using Probability Pick a random X with X > M and X near M. What s the probability that you get a prime? Pr(X is not divisible by 2) = 2 = ( 2 Pr(X is not divisible by 3) = 2 3 = ( ) 3 Pr(X is not divisible by 5) = 4 5 = ( ) 5. Pr(X is not divisible by any prime in P M ) )?? M?????? = p P M ( ) p =. n n Λ M 22

23 Heuristic : Among the numbers between M and M + M, the proportion of primes is about. n n Λ M There are approximately M n Λ M n primes in the interval Example and 00 n Λ n π( ) π( ) = 4. [M, M + M]. 23

24 Heuristic 2: We can approximate In fact, n with M n Λ M n= n Λ M n = M n= n + n. n Λ M n > M n. What happens to the second summation on the right as M? The number M n= n is called the M th harmonic number. 24

25 Heuristic + Heuristic 2: There are approximately M M n n= in the interval primes [M, M + M]. Example and 00 ( n= ) n π( ) π( ) = 6. 25

26 Heuristic 3: 0.8 M n log(m + ) n= 4 The picture shows that is equal to dx, which is log(5), plus the n n= x dark red steps, whose total area is less than In fact, we have [ M ] lim log(m + ) M n n= where γ γ So for large M, M n= n log(m + ) + γ. 26

27 Heuristic + Heuristic 2 + Heuristic 3: There are approximately [M, M + M]. M log(m + ) + γ primes in the interval That is, π(m + M) π(m) π(m + M) π(m) M or M log(m + ) + γ log(m + ) + γ or d dm [π(m)] log(m + ) + γ. 27

28 Calculus? In a Number Theory talk? Now that we know (approximately) the derivative of π(x), we can use the Fundamental Theorem of Calculus to find π(x) itself. Using the fact that π(2) =, we get 40 π(x) π(2) so that π(x) + x 2 x 2 dt log(t + ) + γ dt log(t + ) + γ

29 Comparing the approximations 50 π(x) + x 2 x log(x) log(t + ) + γ dt

30 Our approximation can be simplified to π(x) + x 2 log(t + ) + γ dt e γ [Ei(log(x + ) + γ) Ei(log(3) + γ)], where Ei is the exponential integral function, Ei(z) = z e t t dt. But it can t be written in terms of elementary functions. 30

31 The good news is that you can use l Hôpital s rule to show that ( x ) + 2 log(t + ) + γ dt lim ( ) =, x x log(x) so by the real Prime Number Theorem, our heuristic approximation also works. That is, lim x ( x + 2 π(x) log(t + ) + γ dt ) =. 3

32 Some numbers x x log(x) + x 2 (log(x + ) + γ) dt π(x) ??? 32