đề thi olympic sinh viên toán 2013

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

Đề thi Olympic sinh viên thế giới năm 2006

... Bj,jand Cn,nare relatively prime, we can choose integers Cj,nand Bj,nsuch that this equation issatisfied. 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 Aj= λI for some j, so suppose this is not the case.Consider then first the situation where some Aj, say A3, has two distinct ... minimum of f on [x, y]. Then f([x, y]) = [f(b), f (a)]; hencey − x = f (a) − f(b) ≤ |a − b| ≤ y − xThis implies {a, b} = {x, y}, and therefore f is a monotone function. Suppose f is increasing. Thenf(x)...

Đề thi Olympic sinh viên thế giới năm 2007

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... using only a translation or a rotation. Does this imply thatf(x) = ax + b for some real numbers a and b ?Solution. No. The function f(x) = exalso has this property since cex= ex+log c.Problem ... ≥|ξ−b|≥ m|ξ − b|m−1≥dm−1mm−2we getf(ξ) + f′(ξ) ≥ f(b) + ε.Together with (2) this shows (3). This ﬁnishes the proof of Lemma 2.baξf′ff + f′4 For every element X = {x1, x2, ... 0,since A2y= 0. So, A¯Xannihilates the span of all the vYwith X  Y . This implies that vXdoes notlie in this span, because A¯XvX= v{1,2, ,k}= 0. Therefore, the vectors vX(with...

Đề thi Olympic sinh viên thế giới năm 2008

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... required properties. For an arbitrary rational q, consider thefunction gq(x) = f(x+q)−f(x). This is a continuous function which attains only rational values, thereforegqis constant.Set ... b))we can assume that P (X) = 0.Now we are going to prove that P (Xk) = 0 for all k ≥ 1. Suppose this is true for all k < n. We knowthat P (Xn+ e) = 0 for e = −P (Xn). From the induction ... . 0 b2k−2,2k−10 b2k−1,20 . . . b2k−1,2k−20.Note that every matrix of this form has determinant zero, because it has k columns spanning a vectorspace of dimension at...

Tài liệu Đáp án đề thi Olympic sinh viên 2010 doc

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... VnMath.ComvnMath.comDịch vụ Toán họcinfo@vnmath.comSáchĐại sốGiải tíchHình họcCác loạikhácChuyên đề Toán Luyện thi Đại họcBồi dưỡngHSG Đề thi Đáp ánĐại họcCao học Thi lớp 10 Olympic Giáo áncác ... nghịch và vì vậyI = (A + B)A−1(I − C) = A−1(A + B)(I − C),4 Dịch Vụ Toán HọcĐáp án Đề thi Olympic Toán Sinh viên năm 2010Đại số và Giải tíchWWW.VNMATH.COM ... HỘI TOÁN HỌC VIỆT NAM BỘ GIÁO DỤC VÀ ĐÀO TẠOĐÁP ÁN OLYMPIC TOÁN SINH VIÊN LẦN THỨ XVIIIMôn : Đại sốCâu 1. Cho A, B là các ma trận...

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

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... at leastone is negative. Hence we have at least two non-zero elements in everycolumn of A−1. This proves part a). For part b) all bijare zero exceptb1,1= 2, bn,n= (−1)n, bi,i+1= ... y) such that f(c) = 0 and f (x) > 0 for x ∈ (c, y]. For x ∈ (c, y] wehave |f(x)| ≤ λf(x). This implies that the function g(x) = ln f (x) − λx isnot increasing in (c, y] because of g(x) ... The number of indices (i, j)for which aii= ajj= cmfor some m = 1, 2, . . . , k is d2m. This gives thedesired result.Problem 5. (18 points)Let x1, x2, . . . , xkbe vectors of...

Tài liệu Đề thi Olympic sinh viên thế giới năm 1995 docx

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... −13.From the hypotheses we have101xf(t)dtdx ≥101 − x22dx or10tf(t)dt ≥13. This completes the proof.Problem 3. (15 points)Let f be twice continuously diﬀerentiable on (0, ... |x|p+ |y|p= 2}.Since Dδis compact it is enough to show that f is continuous on Dδ.For this we show that the denominator of f is diﬀerent from zero. Assumethe contrary. Then |x + y|...

Tài liệu Đề thi Olympic sinh viên thế giới năm 1996 doc

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... 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 −cosh2(m + 1)θ − 1.√cosh2mθ − 1Set 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. Diameterof a set E is diam(E) = sup{dist(x, y) : x, y ∈ E}. Contraction ... < ε. So the sequencecannot come into the interval (x − δ, x + δ), but also cannot jump over thisinterval. Then all cluster points have to be at most x − δ (a contradictionwith L being a...

Tài liệu Đề thi Olympic sinh viên thế giới năm 1997 ngày 1 pptx

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... suﬃcesto show that this sequence is strictly decreasing. Now,pk− qk− (pk−1− qk−1) = nkpk−1− (nk+ 1)qk−1− pk−1+ qk−1= (nk− 1)pk−1− nkqk−1and this is negative becausepk−1qk−1= ... number anddet ω(BA − AB) = ωndet(BA − AB) and det(BA − AB) = 0, then ωnis areal number. This is possible only when n is divisible by 3.2 For each k we writeθk=pkqkas a fraction ... is the rational numberpq. Our aim is to showthat for some m,θm−1=nmnm− 1.Suppose this is not the case, so that for every m,(3) θm−1<nmnm− 1.4 FOURTH INTERNATIONAL...

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

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... then thereare axes making k2πnangle). If A is infinite then we can think that A = Zand f (m) = m + 1 for every m ∈ Z. In this case we define g1as a symmetryrelative to12, g2as 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 regularn polygon and that f is rotation by2πn....

Tài liệu Đề thi Olympic sinh viên thế giới năm 1998 ppt

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... containsonly powers of ¯y = yK, i.e., S4/K is cyclic. It is easy to see that this factor-group is not comutative(something more this group is not isomorphic to S3).3) n = 5a) If x = (12), then for ... thatx is uniquely determined by the triple u, v, w; since the set of such triples is countable, this will ﬁnishthe proof.To prove the claim, suppose, that from some x∈ M we arrived to the ... induction that(1) fn(x) =12− 22n−1x −122nholds for n = 1, 2, . . . . For n = 1 this is true, since f(x) = 2x(1 − x) =12− 2(x −12)2. If (1) holds forsome n = k, then we...

Tài liệu Đề thi Olympic sinh viên thế giới năm 1999 doc

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... opposite inequality holds ∀m  1. Thiscontradiction 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 thenumber of the marked points by at most 1, so the condition remains true. Repeat this step until each rowand column contains ... wasbcdedef = xyzyxzxyzxyxzxzyxyzyzxyzxyxzxzyxyzyzyxyxzyxzxzxzxyxyzxyz,which is of length 46. This is not the shortest way: reducing the length of word a can be done for exampleby the following...

Tài liệu Đề thi Olympic sinh viên thế giới năm 2000 doc

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... 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 ... c0I + CA. It is well-known that the characteristicpolynomials of AC and CA are the same; denote this polynomial by f(x). Then thecharacteristic polynomials of matrices q(eAB) and q(eBA) are ... = 0 is xm,so the same holds for the matrix q(eBA). By the theorem of Cayley and Hamilton, thisimplies thatq(eBA)m=p(eBA)km= 0. Thus the matrix q(eBA) is nilpotent, too.4...

Tài liệu Đề thi Olympic sinh viên thế giới năm 2001 ppt

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... 0, we havelimx→∞f(x) ·g(x)Ag(x)A+1=BA + 1.By l’Hospital’s rule this implieslimx→∞f(x)g(x)= limx→∞f(x) ·g(x)Ag(x)A+1=BA + 1.4 8thIMC ... 4/x2. It is a matter of simple manipulation to prove that2f(x) > x for all x ∈ (0, 2), this implies that the sequence (2nan) is strictly1 increasing. The inequality 2g(x) < x ... integers x and y,b = bxr+ys= (br)x(bs)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 S7of order...

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

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... it converges to zero. Therefore an≤4c2ln+1for n ≥ 2l+1,meaning that c2l+1≤ 4c2l. This implies that a sequence ((4cl)2−l)l≥0is non-increasing and therefore bounded from above ... 1n−1k=0n−kn+k+1nn−1k+2n+1n + 1=2nnn−1k=01n−1k+2n+1n + 1= xn+2n+1n + 1.This implies (2) for n + 1.Problem 4. Let f : [a, b] → [a, b] be a continuous function and let...

Tài liệu Đề thi Olympic sinh viên thế giới năm 2002 ngày 2 doc

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... bnis an integer.Solution. We prove by induction on n that an/e and bne are integers, weprove this for n = 0 as well. (For n = 0, the term 00in the deﬁnition of thesequences must be replaced ... subsets of slice OBC withoutcommon interior point, and they do not cover the whole slice OBC; thisimplies the statement. In cases (b) and (c) where at least one of the signsis negative, projections ... x2.Prove that∀x1, x2∈ Rn∇f(x1) − ∇f(x2)2≤ L∇f(x1) − ∇f(x2), x1− x2. (1)In this formula a, b denotes the scalar product of the vectors a and b.Solution. Let g(x) = f(x)−f(x1)−∇f...