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Unformatted text preview: Bertrand’s Postulate Jonathan L.F. King University of Florida, Gainesville FL 326112082, USA [email protected] Webpage http://www.math.ufl.edu/ ∼ squash/ 9 September, 2009 (at 18:17 ) Background. Proofs are from Shoup, from Wikipedia and from my notes. The superscript ‘ ⊕ ’. An inequality OTForm ∀ large n : f ( n ) 6 5 ⊕ · h ( n ) means ∀ U > 5 , ∀ large n : f ( n ) 6 U · h ( n ) . Similarly, ∀ large n : f ( n ) > [ 1 3 ] · h ( n ) means ∀ positive L < 1 3 , ∀ large n : f ( n ) > L · h ( n ) . Notice that U and L are quantified before n . Clumps. For p prime, let Divlog p (1500) denote the maximum natnum L st. p L • 1500. Another no tation for this is p L • 1500. So Divlog 5 (1500) = 3. For a nonzero integer B , the “ pclump of B ” , Clm p ( B ), is the largest power of p which divides B . So Clm 5 (1500) is 125, and Clm 2 (1500) = 4. Evidently Clm p ( B ) = p Divlog p ( B ) ; and B ’s clumps multipliedtogether make B . 1 : Lemma. Fix a prime p and natnum K . Then Divlog p ( K ! ) = X ∞ j =1 K p j . ( Exercise ) ♦ 2 : Prop’n. ∀ α ∈ R : b 2 α c  2 b α c is zero or one. ♦ We denote the set of prime numbers by P . Below, “ p ” ranges over the prime numbers. All following definitions are for real x , although usually x will be an integer. First the “ Pr oduct O f P rimes ” , PrOP( N ) := Y p : p 6 x p . Its logarithm is the famous Chebyshev theta fnc : ϑ ( x ) := log ( PrOP( N ) ) = X p : p 6 x log( p ) . Generalizing PrOP. When S is a set of reals, let PrOP( S ) mean the product of the primes in S . 3 : PowFour Lemma. For each x > 1 : PrOP( x ) < 4 x . in other words: ϑ ( x ) < log(4) · x , ♦ Proof. WLOG, x is an integer N . Case: N = 1 : PrOP(1) = 1 < 4 1 . Case: N = 2 : PrOP(2) = 2 < 4 2 . Case: N > 2 and N is even PrOP( N ) = PrOP( N 1) , since N isn’t prime, < 4 N 1 , by induction, which is less than 4 N . N > 2 and N is odd Write N := [2 H + 1]. Induc tion gives ( since 1+ H < N ) that PrOP [ 1 .. 1+ H ] < 4 1+ H , so our goal is to show that PrOP ( 1+ H .. N ] ? 6 4 H . 3 : Flipping a coin N times, the number of coinflip se quences is ( letting j,k range over N ) [1 + 1] N = X j + k = N N j, k ! > ( N H, 1+ H ) + ( N 1+ H, H ) = 2 · ( N H ) . Divide by 2, then exchange sides, to get ( N H ) 6 4 H . Each prime in ( 1+ H .. N ] divides ( N H ) , so PrOP ( 1+ H .. N ] • ( N H ) . Since ( N H ) is positive, PrOP ( 1+ H .. N ] 6 ( N H ) ....
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This note was uploaded on 01/26/2012 for the course MAC 3472 taught by Professor Jury during the Fall '07 term at University of Florida.
 Fall '07
 JURY
 Math, Calculus

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