Page 124 - MATINF Nr. 13-14

124 M.N. Popescu

s , i
(ln 2 (x)) 2 X 2 E (x) 2
1 = (ln 2 (x)) · = (ln 2 (x)) .
x x x→∞
n≤x
Prin urmare
1 X 2 1 X 2 2
(ω (n) − ln 2 (x)) = (ω (n)) − (ln 2 (x)) + O (ln 2 (x)) .
x x→∞ x
n≤x n≤x

22. b. Avem

Å ãÅ ã Å ã
X 2 X X X X X X
(ω (n)) = 1 1 = 1
n≤x n≤x p 1 |n p 2 |n n≤x p 1 |n p 2 |n
p 1 prim p 2 prim p 1 prim p 2 prim
Å ã
X X X
= 1
p 1 ≤x p 2 ≤x n≤x
p 1 prim p 2 prim p 1 |n, p 2 |n
X X
∗
= card {n ∈ N : n ≤ x, p 1 |n, p 2 |n}.
p 1 ≤x p 2 ≤x
p 1 prim p 2 prim

22. c. Dac˘a p 1 6= p 2 sunt prime s , i divid pe n, atunci s , i p 1 · p 2 divide n. Astfel

Å ã
x
∗
card {n ∈ N : n ≤ x, p 1 |n s , i p 2 |n} = E ,
p 1 p 2
deci
Å ã
X X x
∗
card {n ∈ N : n ≤ x, p 1 |n s , i p 2 |n} = E .
p 1 p 2
p 1 ,p 2 ≤x p 1 6=p 2 prime
p 1 6=p 2 prime p 1 p 2 ≤x
Acum, din
Å ã
X x X X X x
x
E − ≤ 1 ≤ ,

p
p 1 p 2 p 1 p 2
p 1 6=p 2 prime p 1 6=p 2 prime p 1 6=p 2 prime p prim
p 1 p 2 ≤x p 1 p 2 ≤x p 1 p 2 ≤x p≤x
cu punctul 21.b, avem
Å ã
X x X x
E = + O (xln 2 (x)) .
x→∞
p 1 p 2 p 1 p 2
p 1 6=p 2 prime p 1 6=p 2 prime
p 1 p 2 ≤x p 1 p 2 ≤x
Apoi, din
x x x
X X X
= −
p 1 p 2 p 1 p 2 p 1 p 2
p 1 6=p 2 prime p 1 6=p 2 prime p 1 6=p 2 prime
p 1 p 2 ≤x p 1 ,p 2 ≤x p 1 ,p 2 ≤x
p 1 p 2 >x
√ √
s , i, din p 1 p 2 > x ⇒ p 1 > x sau p 2 > x, avem
X x X x
≤ √ ,
p 1 p 2 p x
p 1 6=p 2 prime p prim
p 1 ,p 2 ≤x p≤x
p 1 p 2 >x