... 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)...
... 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...
... 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...
... VnMath.ComvnMath.comDịch vụ Toán học info@vnmath.comSáchĐại sốGiải tíchHình học Các loạikhácChuyên đề ToánLuyện thi Đại học Bồi dưỡngHSG Đề thi Đáp ánĐại học Cao 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 Học Đáp án ĐềthiOlympic 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 SINHVIÊN LẦN THỨ XVIIIMôn : Đại sốCâu 1. Cho A, B là các ma trận vuông...
... 3300/1,2 0,54IO I IE V E V Nên 46H IOhoặc 3IOđều có thể oxi hóa Ithành 3I. Như vậy Ichỉ bị oxi hóa thành 3I. +1,20 +1,7 +1,23 +1,51 Lấy một nửa lượng ... 2MnOkhông thể tồn tại vì 420/MnO MnOEvà 220/MnO MnEđều lớn hơn 30/IIE nên 4MnOvà 2MnOđều có thể oxi hóa I thành 3I. Như vậy 4MnO bị khử hoàn toàn thành ... 4MnOsẽ oxi hóa 2Mnthành 2MnO. Khi 2MnOdư thì 3I và I cũng không thể tồn tại vì: 4 2 3 3 30 0 0/ / /,MnO MnO I I IO IE E E nên 4MnOoxi hóa là 3I...
... Phản ứng: x x Tuyến chọn một số bài từ sách TUYỂN TẬP 10 NĂM ĐỀTHIOLYMPIC 30/4 HÓAHỌC 10- NXB GIÁO DỤC PHÂN I: HALOGEN Câu 4: (đề 1996 trang 7) Xét phản ứng tổng hợp hiđro iođua: H2(khí) ... 119,8 – 230 = - 484,4 kcal/mol Câu 6: chuyên đề phản ứng oxi hóa khử trang 147 1. Viết các phản ứng hóahọc trong các trường hợp sau: a, Ozon oxi hóa I- trong môi trường trung tính. b, Sục ... lượng khí sinh ra khi cho cùng lượng kim loại đó tác dụng hoàn toàn với axit clohiđric dư trong cùng điều kiện. Khối lượng muối clo sinh ra trong phản ứng với clo gấp 1,2886 lần lượng sinh ra...
... 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...
... −13.From the hypotheses we have101xf(t)dtdx ≥101 − x22dx or10tf(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|...
... 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 Tis 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...
... 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...
... 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....
... 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−1x −122nholds 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...
... 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...
... 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 thatq(eBA)m=p(eBA)km= 0. Thus the matrix q(eBA) is nilpotent, too.4...