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Extra info for Elementary Number Theory (Math 780 instructors notes)
Example text
It is known that pn cn log n for some constant c. Using this information and Theorem 31, prove that c = 1. R R The Number of Prime Divisors of n: Notation. (n). De nition. Let f : + ! + and g : + ! + . Then f (n) is said to have normal order g(n) if for every " > 0, the number of positive integers n x satisfying (1 ? , for almost all positive integers n, f (n) 2 ((1 ? ")g(n); (1+ ")g(n))). TheoremX 32. (n) has normal order log log n. Lemma. (n) ? log log x 2 x log log x. Z R Z R n x Proof. We examine each term on the right-hand side of the equation X X X?
Iii) 1 ? 1p log x p x Proof of parts (i) and (iii): For (i), we use integration by parts and Chebyshev's Theorem to obtain (ii) X 1 = Z x 1 d (t) = (x) + Z x (t) dt = O 1 + Z x 1 dt + C (x) 1 x log x 1 t 1 t2 2 t log t p xp X where C1 (x) = Z x 1 2 t2 (t) ? logt t dt: By a previous homework problem and Theorem 34, x 1 Z 2 t2 It follows that 11 Z exists. Also, observe that Z 11 t2 x t2 2 x Z (t) ? logt t dt 2 1 dt t log2 t 1: (t) ? 1 C1 (x) (t) ? logt t dt Z 1 1 dt x t log2 t 1 : log x Hence, C1 (x) = 11 Z t2 11 = 2 t2 2 Z Z 1 t 1 (t) ?
N)2 = = X X n x pjn XX n x p= 6 q pqjn 1 1+ 2 = X XX n x pjn qjn XX n x pjn 1= 1 XX n x p6=q pqjn 1 + x log log x + O(x): 37 We proceed by observing that XX n x p6=q pqjn 1= X X p6=q n x pq x pqjn 1= x = X x + O(x) = X x ? X x + O(x): 2 pq x pq p px p p6=q pq p6=q pq pq x pq x X Theorem 31 imlies that each of the sums (log log x)2 +O(log log x) = Also, X (1=p2 ) = O(1) p px deduce that 1 p p px X since X n x X p X (1=p) and p px 2 (1=p2 ) 1 pq x pq X X (1=p) is log log x + O(1) so that p x 1 2= (log log x)2 +O(log log x): p xp X converges (by comparison with 1 X (1=n2 )).
Elementary Number Theory (Math 780 instructors notes)
by George
4.4