Tài liệu Đề thi Olympic sinh viên thế giới năm 2002 ngày 1 doc

... ···+ 1  n 1 n 1   ; then x n +1 = 2 n +1 n + 1 n  k=0 1  n k  = 2 n n + 1  1 + n 1  k=0  1  n k  + 1  n k +1   + 1  = = 2 n n + 1 n 1  k=0 n−k n + k +1 n  n 1 k  + 2 n +1 n + 1 = 2 n n n 1  k=0 1  n 1 k  + 2 n +1 n ... = 1, 2, . . . , n. Show that a 1 − b 1 1 + a 2 − b 2 2 + ···+ a n − b n n = 0. (1) Solution. Since k  n k  = n ...

Ngày tải lên: 26/01/2014, 16:20

... 1 2 . . . 1 1 . . . . . . . . . . . . . . . . . . 1 1 1 . . . 2 1 1 1 1 . . . 1 2              =              1 1 0 0 . . . 0 0 0 1 1 0 . . . 0 0 0 0 1 1 ... the third row to the second one, , adding the nth row to the (n − 1) th, the determinant does not change and we have det(A) =              2 1 +1 . . . 1 1 1 2 1 . . . 1 1...

Ngày tải lên: 26/01/2014, 16:20

... have 1 + 1 n k − 1 < θ k 1 . Then we get α (1 + 1 n 1 ) (1 + 1 n 2 ) . . . = θ k 1 (1 + 1 n k ) (1 + 1 n k +1 ) . . . ≥ θ k 1 (1 + 1 n k ) (1 + 1 n 2 k ) . . . = θ k 1 1 + 1 n k 1 > 1 – a ... 1 . Since θ k 1 ≤ 1 + 1 n k − 1 we have 1 + 1 n k +1 < θ k = θ k 1 1 + 1 n k ≤ 1 + 1 n k 1 1 + 1 n k = 1 + 1 n 2 k − 1 . 3 He...

Ngày tải lên: 21/01/2014, 21:20

... for some θ ∈ (0, 1) we get |f(n +1) −f(n)| = |f  (n+θ)| ≤ 1 + α n α +1 . Since  1 + α n α +1 < +∞ for α > 0 and f(n) −→ n→∞ 0 we get that ∞  n =1 ( 1) n 1 f(n) = ∞  n =1 (f(2n 1) −f(2n)) converges. Now ... some C 2 -functions. 1 Let f (x) = (x + |x| 3/2 ) 2 = x 2 + 2x 2  |x| + |x| 3 , so f is C 2 . For x > 0, g(x) = 1 2  1 1 + 3 2 √ x   = − 1 2 · 1 (1 +...

Ngày tải lên: 21/01/2014, 21:20

... of zero elements in A 1 . b) How many zero elements are there in the inverse of the n × n matrix A =          1 1 1 1 . . . 1 1 2 2 2 . . . 2 1 2 1 1 . . . 1 1 2 1 2 . . . 2 . . . . . ... except b 1, 1 = 2, b n,n = ( 1) n , b i,i +1 = b i +1, i = ( 1) i for i = 1, 2, . . . , n − 1. Problem 2. (13 points) Let f ∈ C 1 (a, b), lim x→a+ f(x) = +∞, lim x→b− f(x) =...

Ngày tải lên: 21/01/2014, 21:20

... Then N  n =1 b n 2 n = N  n =1 (a n − 1) 2 2 n = N  n =1 [a 2 n 2 n − a n 2 n +1 + 2 n ] = N  n =1 [(a n 1 − 1) 2 n − (a n − 1) 2 n +1 ] = (a 0 − 1) 2 1 − (a N − 1) 2 N +1 = 2 − 2 2 2 −N − 1 2 −N . Put ... 1)  dx = 1 2 h + ∆, where |∆| ≤ 1 2  1 + 2 N  n=2  1 n 3 2 − 1 + 1 n 3 2 + 1  ≤ 1 2 + 2 N  n=2 1 n 3 2 − 1 . 10 We use that n 3 2 − 1...

Ngày tải lên: 21/01/2014, 21:20

... 1) . Consequently, (2) ∞  n =1 (a 1 a 2 ···a n ) 1/ n ≤ ∞  n =1 a n c n  ∞  m=n (m(m + 1) ) 1  . Since ∞  m=n (m(m + 1) ) 1 = ∞  m=n  1 m − 1 m + 1  = 1/ n we have ∞  n =1 a n c n  ∞  m=n (m(m + 1) ) 1  = ∞  n =1 a n c n /n = ∞  n =1 a n ((n ... ε) N  n=K +1 1 n +  e − ε 2  1 N  n=K +1 1 n = = 1 e −ε N  n=K +1 1 n ≤ 1 e −ε ∞  n =1 (a 1...

Ngày tải lên: 21/01/2014, 21:20

... − 1 2 ) 2  − 1 2  2 k = 1 2 − 2 2 k 1  −2(x − 1 2 ) 2  2 k = 1 2 − 2 2 k +1 1 (x − 1 2 ) 2 k +1 which is (2) for n = k + 1. Using (1) we can compute the integral,  1 0 f n (x)dx =  1 2 x − 2 2 n 1 2 n + ... n = 1, 2, . . . . For n = 1 this is true, since f(x) = 2x (1 − x) = 1 2 − 2(x − 1 2 ) 2 . If (1) holds for some n = k, then we have f k +1 (x) = f k...

Ngày tải lên: 26/01/2014, 16:20

... thus π (1) + . . . + π(n) ≥ 1 + . . . + n = n(n +1) 2 . By this inequality, ∞  n =1 π(n) n 2 = ∞  n =1  π (1) + . . . + π(n)   1 n 2 − 1 (n + 1) 2  ≥ ≥ ∞  n =1 n(n + 1) 2 · 2n + 1 n 2 (n + 1) 2 = ∞  n =1 2n ... w 1 , . . . , w k such that (ua 1 )w 1 ∼ u, (ua 1 a 2 )w 2 ∼ ua 1 , . . . , (ua 1 . . . a k )w k ∼ ua 1 . . . a k 1 . Then u ∼ ua 1 w 1 ∼...

Ngày tải lên: 26/01/2014, 16:20

... C ∞  i =1 x i . Solution. a) ( n  i =1 x i √ i ) 2 = n  i,j x i x j √ i √ j ≥ n  i =1 x i √ i i  j =1 x i √ j ≥ n  i =1 x i √ i i x i √ i = n  i =1 x 2 i b) ∞  m =1 1 √ m ( ∞  i=m x 2 i ) 1/ 2 ≤ ∞  m =1 1 √ m ∞  i=m x i √ i − m + 1 by a) = ∞  i =1 x i i  m =1 1 √ m √ i − m + 1 You can get a sharp bound on sup i i  m =1 1 √ m √ i − m + 1 by checking that it is a...

Ngày tải lên: 26/01/2014, 16:20