Tài liệu Đề thi Olympic sinh viên thế giới năm 1997 ngày 2 docx

... 4c 2 x 2 + O(x 3 ), 2f(x)f  (x) = 4c 2 x 2 + O(x 3 ) and 2( f  (x)) 2  f(x) = 2( 4c 2 x 2 + O(x 3 ))|x|  c + O(x). g is bounded because 2( f  (x)) 2  f(x) |x| 3 −→ x→0 8c 5 /2 = 0 and f  (x) 2 − ... C 2 (R)? Solution. Let c = 1 2 f  (0). We have g = (f  ) 2 − 2ff  2( f  ) 2 √ f , where f(x) = cx 2 + O(x 3 ), f  (x) = 2cx + O(x 2 ), f  (x) =...

Ngày tải lên: 21/01/2014, 21:20

... L 3 .2 n 2 = S 2 n−1 + 2 n−1 −1  k =2 n 2 1 √ 2k + 1 and L 3 .2 n 2 − S 2 n−1 ≥ 2 n 2 1 √ 2 n −→ n→∞ ∞, so L 3 .2 n 2 −→ n→∞ ∞. Problem 3. Let A and B be real n×n matrices such that A 2 +B 2 =AB. ... partial sums of the permuted series have the form L 2 n−1 +k = S 2 n−1 + S 2 n − S 2 n −k , 0 ≤ k < 2 n−1 and for 2 n−1 > n 0 we have |L 2 n−1 +k − S|...

Ngày tải lên: 21/01/2014, 21:20

... University Students Warsaw, July 19 - July 25 , 20 02 Second Day Problem 1. Compute the determinant of the n × n matrix A = [a ij ], a ij =  (−1) |i−j| , if i = j, 2, if i = j. Solution. Adding the second ... domains AOB, BOC and COA; the intersec- tions are “slices” and their areas are 1 2 γ, 1 2 α and 1 2 β, respectively. 2 Now project the slices AOC and COB to the plane OAB....

Ngày tải lên: 26/01/2014, 16:20

... a 2 k ≤ c 2 l (k+1) 2 ≤ 4c 2 l 2k+1 − 4c 2 l 2k +2 , whereas in the second case there is a 2k+1 − a 2k +2 ≤ a k a k+1 ≤ c 2 l (k+1)(k +2) ≤ 4c 2 l 2k +2 − 4c 2 l 2k+3 . Hence a sequence (a n − 4c 2 l n+1 ) n 2 l+1 is non-decreasing ... 4c 2 l . Indeed, for any integer n ≥ 2 l+1 there exists an integer k ≥ 2 l such that n = 2k or n = 2k + 1. In the first case there...

Ngày tải lên: 26/01/2014, 16:20

... e −1 , f(0, 1) = −e −1 and te −t ≤ 2e 2 for t ≥ 2. Therefore |f(x, y)| ≤ (x 2 + y 2 )e −x 2 −y 2 ≤ 2e 2 < e −1 for (x, y) /∈ M = {(u, v) : u 2 + v 2 ≤ 2} and f cannot attain its minimum ... b). From ∂f ∂x (x, y) = 2x(1 − x 2 + y 2 )e −x 2 −y 2 we get (1) x(1 − x 2 + y 2 ) = 0. Similarly (2) y(1 + x 2 − y 2 ) = 0. All solutions (x, y) of the system (1), (2...

Ngày tải lên: 21/01/2014, 21:20

... o(t 2 )  + 1 2p  1 p − 1  (−pt + o(t)) 2 + o(t 2 ) = 1 − t − p − 1 2 t 2 + o(t 2 ) − p −1 2 t 2 + o(t 2 ) = 1 − t − (p − 1)t 2 + o(t 2 ). We have (x − y) 2 = (2t + o(t)) 2 = 4t 2 + o(t 2 ) and 4−(x+y) 2 =4− (2 (p−1)t 2 +o(t 2 )) 2 =4−4+4(p−1)t 2 +o(t 2 )=4(p−1)t 2 +o(t 2 ). So ... Then N  n=1 b n 2 n = N  n=1 (a n − 1) 2 2 n = N  n=1 [a...

Ngày tải lên: 21/01/2014, 21:20

... O(h 2 ). From (1), (2) and (3) we get      ∞  n=1 nx (n 2 + x) 2 − 1 2      ≤ ∞  n=1 h 2  nh+ h 2 nh− h 2 |f  (t)|dt + O(h 2 ) = = h 2  ∞ h 2 |f  (t)|dt + O(h 2 ) = O(h 2 ) = O(x −1 ). 13 Problem ... One has a 2 = 1 2 >  2 3  2 = q 2 . Going by induction we have for N ≥ 3 a N = 2 N a N−1 + 1 N N 2  k =2 a k a N−k ≥ 2 N q N−1 + N...

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 + ... 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) =...

Ngày tải lên: 26/01/2014, 16:20

... . + 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 + 1 2n(n + 1) ≥ ∞  n=1 1 n ... , a 2k , such that a 1 and a 2 are in the same row, a 2 and a 3 are in the same column, . . . , a 2k−1 and a 2k are in the same row, and a 2k and a 1 are in the same column...

Ngày tải lên: 26/01/2014, 16:20

... b) 2 x 2 dx = a 7 21 0 (7k 2 + 7k + 2) and b  0 (x + a) 2 (x − b) 2 x 2 dx = a 7 21 0 (2k 2 + 7k + 7) Then K = k 5 2k 2 + 7k + 7 7k 2 + 7k + 2 . The derivative of the function f(k) = 2k 2 +7k+7 7k 2 +7k +2 is ... most  i+1 0 1 √ x √ i + 1 − x dx = π 2 Alternatively you can observe that i  m=1 1 √ m √ i + 1 − m = 2 i /2  m=1 1 √ m √ i + 1 − m ≤ ≤ 2 1  i...

Ngày tải lên: 26/01/2014, 16:20