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Transcript
Chebychev’s Theorem on the Density of Prime Numbers Yaron Heger and Tal Kaminker So far we only proved that there are ∞ primes. But can we say more than this? We will prove in this lecture a good estimation on the density of primes Density of primes Let us define = number of primes till x Compare this with: For large numbers… x pi(x) x/ln x x/(ln x -1) 1000 168 145 169 10000 1229 1086 1218 100000 9592 8686 9512 1000000 78498 72382 78030 10000000 664579 620420 661459 100000000 5761455 5428681 5740304 Basically it is possible to prove that: The numbers of primes till n is asymptotically the same as n/ln(n) Meaning: This was proven in 1896 by Hadamard and de la Vallée Poussin But we can’t prove it right now… So we will prove less powerful but still good estimation We will prove that: π(x) = Θ(x/ log x) This is a result of a work by Chebychev made in 1852. Chebychev-type estimates We will now prove that: Lower bound The heart of the proof is understanding some facts of the binomial coefficient 2n (2n)! 2n (2n 1) (2n 2) (n 1) n(n 1)( n 2) 2 1 n n!n! Let us define: 100 3 4 2 3 1113 17 19 29 31 53 59 61 67 71 83 89 97 50 Overview First we will prove that: 2n 2 2 n n 2n (1) Then we will prove that: For each i p 2n ki i Thus, we get that: ( 2) N ( n) 2 log( N ) Assuming (1) and (2) we conclude… For N = 2n (N is even): N ( n) 1 log( N ) For N = 2n+1 (N is odd): N ( n) 2 log( N ) Proof: (n 1) (n) n n 1 1 n 1 1 n 1 1 1 1 2 log n log n log n log n log( n 1) 2n 2 2 n n 2n Proof of Using the Newton's binomial: 1 1 2n Thus 2n i 2 n i 2 n 2n 1 1 i 0 i i 0 i 2n 2n 2 i 0 i i 2 n 2n … 2n n So, if we manage to prove that is the largest coefficient we will get that: 2n 2n 2 n 2n 1 Using simple arguments (for example that the first and the last coefficients are 1) we can get that: 2n 2 2 n n 2n Proof that 2n n is the largest coefficient 2n 2n (2n 1) (n 1) 2n (2n 1) (2n k 1) 2n ? n (n 1) 2 1 k (k 1) 2 1 n k (2n k ) (2n k 1) (n 1) 1 ? n (n 1) (k 1) 1 Both the numerator and the denominator have n-k elements. But any of the numerator’s elements is larger than any of the denominator elements, thus: 2n 2n n k 0 n k This is enough since 2n 2n n k n k A graph: Now we know that 2n 2 n 2 2n 2 n 2n Now we will prove that: For each Where: i p 2n ki i Proof that: For each i p 2n ki i First define: For n N and prime p, define p (n ) = the power to which p appears in the factorization of n. Thus, p (n ) is the largest k≥0 such that pk | n And we also get: np p (n) p| n For example, 3 (162) 3 (3 2) 4 4 Proof that: For each i p 2n ki i Next, define the floor function: a b = a div b For example: 16 5 3 Also, this equals to the number of multiplies b,2b,3b,4b,…mb that do not exceed a Proof that: For each i p 2n ki i Legendre’s lemma n p (n!) k k 1 p Proof: First define: Rp,n = {(i, k) | 1 ≤ i ≤ n and p k divides i }. Legendre’s lemma Rp,n = {(i, k) | 1 ≤ i ≤ n and k pdivides i }. Example for p=2 and n=20: i=1 2 k=1 2 3 ● 3 4 ● ● 5 6 ● 7 8 ● 9 10 11 12 13 14 15 16 17 18 19 20 ● ● ● 4 For every ● the pair (i,k) is in Rp,n ● ● ● ● ● ● ● ● ● ● Legendre’s lemma According to what we have seen on the blackboard, since both sums represents |Rp,n| we have: n p (n!) k k 1 p And this proves the Legendre’s lemma. So far we have seen… We proved ● 2n 2 2 n n 2n n p (n!) k k 1 p ● And we defined: And now we will prove that: for each i : piki 2n Proof that: For each i p 2n ki i 2n If we denote, l p n According to the proof on the blackboard, we get 2n n l k 2 k k 1 p p And since, y R 2 y 2 y {0,1} We conclude that: p l 2n Summary : We first proved that: ● Then we proved: ● 2n 2 2 n n 2n n p (n!) k k 1 p Finally we proved: For each i ki i p 2n And from here we concluded the lower bound theorem.