Tài liệu Đề thi Olympic sinh viên thế giới năm 1994 pptx

... Competition in Mathematics for Universtiy Students in Plovdiv, Bulgaria 1994 1 PROBLEMS AND SOLUTIONS First day — July 29, 1994 Problem 1. (13 points) a) Let A be a n × n, n ≥ 2, symmetric, invertible ... at least one is negative. Hence we have at least two non-zero elements in every column of A −1 . This proves part a). For part b) all b ij are zero except b 1,1 = 2, b n,n = (−1) n , b i...

Ngày tải lên: 21/01/2014, 21:20

... − 1 3 . From the hypotheses we have  1 0  1 x f(t)dtdx ≥  1 0 1 − x 2 2 dx or  1 0 tf(t)dt ≥ 1 3 . This completes the proof. Problem 3. (15 points) Let f be twice continuously differentiable on (0, ... x tends to 0+ we obtain −x 0 ≤ lim inf x→0+ f(x) f  (x) ≤ lim sup x→0+ f(x) f  (x) ≤ 0. Since this happens for all x 0 ∈ (0, r) we deduce that lim x→0+ f(x) f  (x) exists and lim x→0+ f(x...

Ngày tải lên: 21/01/2014, 21:20

... consider k ≥ 2. For any m we have (2) cosh θ = cosh ((m + 1)θ − mθ) = = cosh (m + 1)θ.cosh mθ − sinh (m + 1)θ .sinh mθ = cosh (m + 1)θ.cosh mθ −  cosh 2 (m + 1)θ − 1. √ cosh 2 mθ − 1 Set cosh kθ = a, ... the first and the third integral tend to − 1 f  (0) as n → ∞, hence so does the second. Also n  1 A (f(x)) n dx ≤ n(f(A)) n −→ n→∞ 0 (f (A) < 1). We get L = − 1 f  (0) in this case....

Ngày tải lên: 21/01/2014, 21:20

... suffices to show that this sequence is strictly decreasing. Now, p k − q k − (p k−1 − q k−1 ) = n k p k−1 − (n k + 1)q k−1 − p k−1 + q k−1 = (n k − 1)p k−1 − n k q k−1 and this is negative because p k−1 q k−1 = ... is the rational number p q . Our aim is to show that for some m, θ m−1 = n m n m − 1 . Suppose this is not the case, so that for every m, (3) θ m−1 < n m n m − 1 . 4 For each k we...

Ngày tải lên: 21/01/2014, 21:20

... then there are axes making k 2π n angle). If A is infinite then we can think that A = Z and f (m) = m + 1 for every m ∈ Z. In this case we define g 1 as a symmetry relative to 1 2 , g 2 as a symmetry ... It is enough to prove the theorem for every such set. Let A = T (x). If A is finite, then we can think that A is the set of all vertices of a regular n polygon and that f is rotation by 2π n .

Ngày tải lên: 21/01/2014, 21:20

... contains only powers of ¯y = yK, i.e., S 4 /K is cyclic. It is easy to see that this factor-group is not comutative (something more this group is not isomorphic to S 3 ). 3) n = 5 a) If x = (12), then for ... vector space. Solution First choose a basis {v 1 , v 2 , v 3 } of U 1 . It is possible to extend this basis with vectors v 4 ,v 5 and v 6 to get a basis of U 2 . In the same way we can e...

Ngày tải lên: 26/01/2014, 16:20

... (1) From this recursion we can compute the probabilities for small values of n and can conjecture that p (r) n = 1 5 + 4 5·6 n if n ≡ r (mod )5 and p (r) n = 1 5 − 1 5·6 n otherwise. From (1), this ... opposite inequality holds ∀m  1. This contradiction shows that g is a constant, i.e. f(x) = Cx, C > 0. Conversely, it is easy to check that the functions of this type verify the conditions...

Ngày tải lên: 26/01/2014, 16:20

... e, this yields e + 2ef = 0, and similarly f + 2ef = 0, so that f = −2ef = e, hence e = f = g by symmetry. Hence, finaly, 3e = e + f + g = 0, i.e. e = f = g = 0. For part (i) just omit some of this. Problem ... into n smaller cubes may be refined to give a dissection into n + (a d − 1) cubes, for any a ≥ 1. This refinement is achieved by picking an arbitrary cube in the dissection, and cutting it...

Ngày tải lên: 26/01/2014, 16:20

... integers x and y, b = b xr+ys = (b r ) x (b s ) y = e. It follows similarly that a = e as well. 2. This is not true. Let a = (123) and b = (34567) be cycles of the permu- tation group S 7 of order ... 4/x 2   . It is a matter of simple manipulation to prove that 2f(x) > x for all x ∈ (0, 2), this implies that the sequence (2 n a n ) is strictly 1 increasing. The inequality 2g(x) < x ....

Ngày tải lên: 26/01/2014, 16:20

... are pairwise disjoint, so there are at most countably many of the third type. b. Let f be such a map. Then for each value y of this map there is an x 0 such that y = f(x) and f  (x) = 0, because ... δ n centered at p n such that there are there are infinitely many k’s such that p k /∈ n  j=0 I j , this can be done by induction. Let n 0 = 0 and n m+1 be the smallest integer k > n m such th...

Ngày tải lên: 26/01/2014, 16:20