ĐỀ THI OLYMPIC SINH VIÊN TOÁN TOÀN QUỐC MÔN ĐẠI SỐ NĂM 2007 potx

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 ... is not divisible by 3. Now add triangles P 1 P k P k+2 , P k P k+1 P k+2 and P 1 P k+2 P k+3 . This way we introduce two new triangles at vertices P 1 and P k so parity is preserved. The ve...

Ngày tải lên: 17/10/2013, 23:15

... IMC2007, Blagoevgrad, Bulgaria Day 1, August 5, 2007 Problem 1. Let f be a polynomial of degree 2 with integer coefficients. ... the first columns of A 1 , . . . , A k are linearly dependent, the first 1 IMC2007, Blagoevgrad, Bulgaria Day 2, August 6, 2007 Problem 1. Let f : R → R be a continuous function. Suppose that for ... using only a translation or a rotation. Does this imply that f(x)...

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...

Ngày tải lên: 20/10/2013, 18:15

... dưỡng HSG Đề thi Đáp án Đại học Cao học Thi lớp 10 Olympic Giáo án các môn 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 ... bài toán. —————————————————— 5 About 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 th...

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 = ... −  bk/n b(k−1)/n f(x)dx   b 0 g(x)dx + O(ω(f, b/n)g 1 ) = 1 b  b 0 f(x)dx  b 0 g(x)dx + O(ω(f, b/n)g 1 ). This proves a). For b) we set b = π, f(x) = sin x, g(x) = (1 + 3cos 2 x) −1 . From...

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