Page 114 - MATINF Nr. 13-14

114 M.N. Popescu

15. Avem

card {σ ∈ S n : |X n (σ) − ln n| > ε ln (n)}
P (|X n − ln n| > ε ln (n)) =
card S n
® 2 ´
1 |ω (σ) − ln n|
= card σ ∈ S n : > 1
n! ε (ln (n)) 2
2
1 X |ω (σ) − ln n| 2
≤ ,
2
n! ε (ln (n)) 2
σ∈S n
deci, cu punctul 14.b, ∃C 0 > 0 s , i n 0 ∈ N astfel ca

1 c 1
P (|X n − ln n| > ε ln (n)) ≤ + + C 0 · , ∀n ≥ n 0 .
2
2
ε ln (n) ε (ln (n)) 2 n ln (n)
Dintre fract , iile din dreapta, cea mai mare este prima, deci exist˘ o constant˘a C > 0 astfel ca
a
C
P (|X n − ln n| > ε ln (n)) ≤ , ∀n ≥ 2.
2
ε ln (n)

Partea a doua

16. Explicit˘am funct , ia A (t):


a 2 pentru t ∈ [2, 3)


 pentru t ∈ [3, 4)
 a 2 + a 3

 . .

 . . . .
A (t) = .
 a 2 + a 3 + . . . + a n−1 pentru t ∈ [n − 1, n)

 a 2 + a 3 + . . . + a n−1 + a n pentru t ∈ [n, n + 1)


 . .

. . . .

Avem
n 3 4 n
Z Z Z Z
0
0
0
0
b (t) A (t) dt = b (t) A (t) dt + b (t) A (t) dt + . . . + b (t) A (t) dt
2 2 3 n−1
3 4 n
Z Z Z
0
0
0
= b (t) a 2 dt + b (t) (a 2 + a 3 ) dt + . . . + b (t) (a 2 + . . . + a n−1 ) dt
2 3 n−1
3 4 n

= a 2 b (t) + (a 2 + a 3 ) b (t) + . . . + (a 2 + . . . + a n−1 ) b (t)
2 3 n−1
= a 2 [b (3) − b (2)] + (a 2 + a 3 ) [b (4) − b (3)] + . . .
+ (a 2 + . . . + a n−1 ) [b (n) − b (n − 1)]
= a 2 {[b (3) − b (2)] + [b (4) − b (3)] + . . . + [b (n) − b (n − 1)]}
+a 3 {[b (4) − b (3)] + . . . + [b (n) − b (n − 1)]}
. . .
+a n−1 {[b (n) − b (n − 1)]}