The Prime Number Theorem
We study the distribution of primes via the function π(x) = the number of primes x 6 5 4 3 2 2 3 4 5 6 7 8 9 0 2 3 4 5 2
It s easier to draw this way: π(x) = the number of primes x 6 5 4 3 2 2 3 4 5 6 7 8 9 0 2 3 4 5 3
Clearly, π(x) < x (not every number is prime) and lim x π(x) = (there are infinitely many primes) 80 60 40 20 00 200 300 400 500 How many primes are there among the first 000 000 000 integers? By brute force, π( 000 000 000) = 50 847 534. Can we approximate π(x) using elementary functions? 4
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) 0 2 25 2.7 3.3 0.320 0 3 68 44.8 23.2 0.38 0 4 229 085.7 43.3 0.66 0 5 9592 8685.9 906. 0.0945 0 6 78 498 72 382.4 65.6 0.0779 0 7 664 579 620 42.0 44 58.3 0.0664 5
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
( 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
The Prime Number Theorem 40 lim x π(x) ( ) =. x log(x) 30 20 0 50 00 50 200 The convergence is very slow: At x = 0 6, π(x)/(x/ log(x)).08449 At x = 0 9, π(x)/(x/ log(x)).05373 At x = 0 2, π(x)/(x/ log(x)).0395. 8
It s also messy. y = π(x) ( ) x log(x).09040.09035.09030.09025.09020.0905 500 500 50 000 50 500 502 000.078.077.076.05582.05582.05582.05582.05582.05582.00005 0 7.0000 0 7.0005 0 7.00020 0 7 5.0000 0 8 5.0000 0 8 5.00002 0 8 5.00002 0 8 9
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() = + 2 + 3 + ζ(2) = 2 + 2 2 + 3 2 + ζ(4) = 4 + 2 4 + 3 4 + diverges π2 converges to 6 π4 converges to 90 Hadamard 0
The function ζ(s) is defined for 2 all s = x + iy in C, 0.0 0 with a simple pole at s =. 0.5 20.0 0.5 What s the connection with primes? 2.0-20 80 60 40 20 00 200 300 400 500 Riemann
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
Recall that P 8 = {2, 3, 5, 7}. Consider the product ( 2 + 0 2 + ) ( 2 + 2 3 + 0 3 + ) 3 + ( 2 5 + 0 5 + ) ( 5 + 2 7 + 0 7 + ) 7 + 2 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
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 ( ) ( ) ( ) ( ) = 2 3 5 7 = 05 24. + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 0 + 2 + 4 + 5 + 6 + 8 + 20 + 2 + 4
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
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, 2 2 3 4 5 6 7 0 2 3 6 7 8 9 22 23 28 29 can be made arbitrarily small. 6
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
Proof sketch Because a n converges, we can find N so that Write S = n= a n = a + a 2 + + 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
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
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
Old Stuff: n is divergent. New Stuff: p is divergent. + 2 + 3 + 4 + 5 + 6 + 7 + = 2 + 3 + 5 + 7 + + 3 + = 2 3 4 5 6 7 8 9 0 2 3 4 5 6 7 8 9 20 The harmonic series diverges slowly: It takes about a million terms to get to a sum of 2. 2 3 4 5 6 7 8 9 0 2 3 4 5 6 7 8 9 20 The /p series diverges very slowly: A million terms gets you only to about 3.07. But with either series, if you add up enough terms, you can make the sum as large as you want. 2
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
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 5 000 000 n 3.6398 π(5 000 00) π(5 000 000) = 4. [M, M + M]. 23
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
Heuristic + Heuristic 2: There are approximately M M n n= in the interval primes [M, M + M]. Example and 00 ( 2 000 000 n= ) 6.6287 n π(2 000 00) π(2 000 000) = 6. 25
Heuristic 3: 0.8 M n log(m + ). 0.6 0.4 0.2 0 2 3 4 5 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..2.0 5 In fact, we have [ M ] lim log(m + ) M n n= where γ 0.5772. γ So for large M, M n= n log(m + ) + γ. 26
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
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 30 20 0 dt log(t + ) + γ. 50 00 50 200 28
Comparing the approximations 50 π(x) + x 2 x log(x) log(t + ) + γ dt 00 50 200 400 600 800 000 29
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
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
Some numbers x x log(x) + x 2 (log(x + ) + γ) dt π(x) 0 6 72 382 75 82 78 498 0 9 48 254 942 49 394 762 50 847 532 0 2 36 9 206 825 36 807 68 22 37 607 92 08 0 5 28 952 965 460 27 29 338 656 062 654??? 32