đề thi olympic sinh viên toán

Ngày tải lên: 23/03/2014, 08:20

... B j,j and C n,n are relatively prime, we can choose integers C j,n and B j,n such that this equation is satisfied. Doing this step by step for all j = n − 1, n − 2, . . . , 1, we finally get B and C such ... 3. (20 points) Solution. We note that the problem is trivial if A j = λI for some j, so suppose this is not the case. Consider then first the situation where some A j , say A 3 , has two distinct ... minimum of f on [x, y]. Then f([x, y]) = [f(b), f (a)]; hence y − x = f (a) − f(b) ≤ |a − b| ≤ y − x This implies {a, b} = {x, y}, and therefore f is a monotone function. Suppose f is increasing. Then f(x)...

Ngày tải lên: 17/10/2013, 23:15

... using only a translation or a rotation. Does this imply that f(x) = ax + b for some real numbers a and b ? Solution. No. The function f(x) = e x also has this property since ce x = e x+log c . Problem ...  ≥|ξ−b| ≥ m|ξ − b| m−1 ≥ d m−1 m m−2 we get f(ξ) + f ′ (ξ) ≥ f(b) + ε. Together with (2) this shows (3). This finishes the proof of Lemma 2. b a ξ f ′ f f + f ′ 4 For every element X = {x 1 , x 2 , ... 0, since A 2 y = 0. So, A ¯ X annihilates the span of all the v Y with X  Y . This implies that v X does not lie in this span, because A ¯ X v X = v {1,2, ,k} = 0. Therefore, the vectors v X (with...

Ngày tải lên: 20/10/2013, 18:15

... required properties. For an arbitrary rational q, consider the function g q (x) = f(x+q)−f(x). This is a continuous function which attains only rational values, therefore g q is constant. Set ... b)) we can assume that P (X) = 0. Now we are going to prove that P (X k ) = 0 for all k ≥ 1. Suppose this is true for all k < n. We know that P (X n + e) = 0 for e = −P (X n ). From the induction ... . 0 b 2k−2,2k−1 0 b 2k−1,2 0 . . . b 2k−1,2k−2 0          . Note that every matrix of this form has determinant zero, because it has k columns spanning a vector space of dimension at...

Ngày tải lên: 20/10/2013, 18:15

... VnMath.Com vnMath.com Dịch vụ Toán học info@vnmath.com Sách Đại số Giải tích Hình học Các loại khác Chuyên đề Toán Luyện thi Đại học Bồi dưỡng HSG Đề thi Đáp án Đại học Cao học Thi lớp 10 Olympic Giáo án các ... nghịch và vì vậy I = (A + B)A −1 (I − C) = A −1 (A + B)(I − C), 4 Dịch Vụ Toán Học Đáp án Đề thi Olympic Toán Sinh viên năm 2010 Đại số và Giải tích WWW.VNMATH.COM ... HỘI TOÁN HỌC VIỆT NAM BỘ GIÁO DỤC VÀ ĐÀO TẠO ĐÁP ÁN OLYMPIC TOÁN SINH VIÊN LẦN THỨ XVIII Môn : Đại số Câu 1. Cho A, B là các ma trận...

Ngày tải lên: 15/12/2013, 01:16

... 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,i+1 = ... y) such that f(c) = 0 and f (x) > 0 for x ∈ (c, y]. For x ∈ (c, y] we have |f  (x)| ≤ λf(x). This implies that the function g(x) = ln f (x) − λx is not increasing in (c, y] because of g  (x) ... The number of indices (i, j) for which a ii = a jj = c m for some m = 1, 2, . . . , k is d 2 m . This gives the desired result. Problem 5. (18 points) Let x 1 , x 2 , . . . , x k be vectors of...

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| p + |y| p = 2}. Since D δ is compact it is enough to show that f is continuous on D δ . For this we show that the denominator of f is different from zero. Assume the contrary. Then |x + y|...

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, ... and T  is its reflexion about the x-axis, then C(E) = 8 > K(E). Remarks: All distances used in this problem are Euclidian. Diameter of a set E is diam(E) = sup{dist(x, y) : x, y ∈ E}. Contraction ... < ε. So the sequence cannot come into the interval (x − δ, x + δ), but also cannot jump over this interval. Then all cluster points have to be at most x − δ (a contradiction with L being a...

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 = ... number and det ω(BA − AB) = ω n det(BA − AB) and det(BA − AB) = 0, then ω n is a real number. This is possible only when n is divisible by 3. 2 For each k we write θ k = p k q k as a fraction ... 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 FOURTH INTERNATIONAL...

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 ... that x is uniquely determined by the triple u, v, w; since the set of such triples is countable, this will finish the proof. To prove the claim, suppose, that from some x  ∈ M we arrived to the ... induction that (1) f n (x) = 1 2 − 2 2 n −1  x − 1 2  2 n holds for 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...

Ngày tải lên: 26/01/2014, 16:20

... 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 ... contains at most one marked point, delete it. This decreases n + k by 1 and the number of the marked points by at most 1, so the condition remains true. Repeat this step until each row and column contains ... was bcdedef = xyzyxzxyzxyxzxzyxyzyzxyzxyxzxzyxyzyzyxyxzyxzxzxzxyxyzxyz, which is of length 46. This is not the shortest way: reducing the length of word a can be done for example by the following...

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 ... c 0 I + CA. It is well-known that the characteristic polynomials of AC and CA are the same; denote this polynomial by f(x). Then the characteristic polynomials of matrices q(e AB ) and q(e BA ) are ... = 0 is x m , so the same holds for the matrix q(e BA ). By the theorem of Cayley and Hamilton, this implies that  q(e BA )  m =  p(e BA )  km = 0. Thus the matrix q(e BA ) is nilpotent, too. 4 ...

Ngày tải lên: 26/01/2014, 16:20

... 0, we have lim x→∞  f(x) ·  g(x)  A     g(x)  A+1   = B A + 1 . By l’Hospital’s rule this implies lim x→∞ f(x) g(x) = lim x→∞ f(x) ·  g(x)  A  g(x)  A+1 = B A + 1 . 4 8 th IMC ... 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 ... 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...

Ngày tải lên: 26/01/2014, 16:20

Ngày tải lên: 26/01/2014, 16:20

Ngày tải lên: 26/01/2014, 16:20

Ngày tải lên: 26/01/2014, 16:20

Ngày tải lên: 26/01/2014, 16:20

Ngày tải lên: 26/01/2014, 16:20

Ngày tải lên: 12/07/2014, 07:00