International Competition in Mathematics for Universtiy Students in Plovdiv, Bulgaria 1994

Transcription

1 International Competition in Mathematics for Universtiy Students in Plovdiv, Bulgaria 1994

2 1 PROBLEMS AND SOLUTIONS First day July 29, 1994 Problem points a Let A be a n n, n 2, symmetric, invertible matrix with real positive elements. Show that z n n 2 2n, where z n is the number of zero elements in A 1. b How many zero elements are there in the inverse of the n n matrix A = ? Solution. Denote by a ij and b ij the elements of A and A 1, respectively. Then for k m we have n i= a ki b im = and from the positivity of a ij we conclude that at least one of {b im : i = 1, 2,..., n} is positive and 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 = b i+1,i = 1 i for i = 1, 2,..., n 1. Problem points Let f C 1 a, b, lim x a+ fx = +, lim x b fx = and f x + f 2 x 1 for x a, b. Prove that b a π and give an example where b a = π. Solution. From the inequality we get d arctg fx + x = f dx x 1 + f 2 x + 1 for x a, b. Thus arctg fx+x is non-decreasing in the interval and using the limits we get π 2 + a π + b. Hence b a π. One has equality for 2 fx = cotg x, a =, b = π. Problem points

3 2 Given a set S of 2n 1, n N, different irrational numbers. Prove that there are n different elements x 1, x 2,..., x n S such that for all nonnegative rational numbers a 1, a 2,..., a n with a 1 + a a n > we have that a 1 x 1 + a 2 x a n x n is an irrational number. Solution. Let I be the set of irrational numbers, Q the set of rational numbers, Q + = Q [,. We work by induction. For n = 1 the statement is trivial. Let it be true for n 1. We start to prove it for n. From the induction argument there are n 1 different elements x 1, x 2,..., x n 1 S such that 1 a 1 x 1 + a 2 x a n 1 x n 1 I for all a 1, a 2,..., a n Q + with a 1 + a a n 1 >. Denote the other elements of S by x n, x n+1,..., x 2n 1. Assume the statement is not true for n. Then for k =, 1,..., n 1 there are r k Q such that 2 Also b ik x i + c k x n+k = r k for some b ik, c k Q +, b ik + c k >. 3 k= d k x n+k = R for some d k Q +, k= d k >, R Q. If in 2 c k = then 2 contradicts 1. Thus c k and without loss of generality one may take c k = 1. In 2 also b ik > in view of x n+k I. Replacing 2 in 3 we get k= d k b ik x i + r k = R or k= d k b ik x i Q, which contradicts 1 because of the conditions on b s and d s. Problem points Let α R \ {} and suppose that F and G are linear maps operators from R n into R n satisfying F G G F = αf. a Show that for all k N one has F k G G F k = αkf k. b Show that there exists k 1 such that F k =.

4 3 Solution. For a using the assumptions we have F k G G F k = = = k F k i+1 G F i 1 F k i G F i = k F k i F G G F F i 1 = k F k i αf F i 1 = αkf k. b Consider the linear operator LF = F G G F acting over all n n matrices F. It may have at most n 2 different eigenvalues. Assuming that F k for every k we get that L has infinitely many different eigenvalues αk in view of a a contradiction. that Problem points a Let f C[, b], g CR and let g be periodic with period b. Prove b Find fxgnxdx has a limit as n and lim fxgnxdx = 1 fxdx gxdx. n b Solution. Set g 1 = π sin x lim n 1 + 3cos 2 nx dx. gx dx and ωf, t = sup { fx fy : x, y [, b], x y t}. In view of the uniform continuity of f we have ωf, t as t. Using the periodicity of g we get k/n fxgnxdx = fxgnxdx bk 1/n k/n k/n = fbk/n gnxdx + {fx fbk/n}gnxdx bk 1/n bk 1/n = 1 fbk/n gxdx + Oωf, b/n g 1 n

5 4 = 1 b + 1 b = 1 b k/n bk 1/n fxdx gxdx b n fbk/n k/n fxdx fxdx gxdx + Oωf, b/n g 1 bk 1/n gxdx + Oωf, b/n g 1. This proves a. For b we set b = π, fx = sin x, gx = 1 + 3cos 2 x 1. From a and π sin xdx = 2, π 1 + 3cos 2 x 1 dx = π 2 we get π sin x lim n 1 + 3cos 2 dx = 1. nx Problem points Let f C 2 [, N] and f x < 1, f x > for every x [, N]. Let m < m 1 < < m k N be integers such that n i = fm i are also integers for i =, 1,..., k. Denote b i = n i n i 1 and a i = m i m i 1 for i = 1, 2,..., k. a Prove that 1 < b 1 a 1 < b 2 a 2 < < b k a k < 1. b Prove that for every choice of A > 1 there are no more than N/A indices j such that a j > A. c Prove that k 3N 2/3 i.e. there are no more than 3N 2/3 integer points on the curve y = fx, x [, N]. Solution. a For i = 1, 2,..., k we have b i = fm i fm i 1 = m i m i 1 f x i for some x i m i 1, m i. Hence b i a i = f x i and so 1 < b i a i < 1. From the convexity of f we have that f is increasing and b i a i = f x i < f x i+1 = b i+1 a i+1 because of x i < m i < x i+1.

6 5 b Set S A = {j {, 1,..., k} : a j > A}. Then k N m k m = a i a j > A S A j S A and hence S A < N/A. c All different fractions in 1, 1 with denominators less or equal A are no more 2A 2. Using b we get k < N/A + 2A 2. Put A = N 1/3 in the above estimate and get k < 3N 2/3. Second day July 3, 1994 Problem points Let f C 1 [a, b], fa = and suppose that λ R, λ >, is such that f x λ fx for all x [a, b]. Is it true that fx = for all x [a, b]? Solution. Assume that there is y a, b] such that fy. Without loss of generality we have fy >. In view of the continuity of f there exists c [a, y such that fc = and fx > for x c, y]. For x c, y] we have f x λfx. This implies that the function gx = ln fx λx is not increasing in c, y] because of g x = f x λ. Thus ln fx λx fx ln fy λy and fx e λx λy fy for x c, y]. Thus = fc = fc + e λc λy fy > a contradiction. Hence one has fx = for all x [a, b]. Problem points Let f : R 2 R be given by fx, y = x 2 y 2 e x2 y 2. a Prove that f attains its minimum and its maximum. b Determine all points x, y such that f f x, y = x, y = and x y determine for which of them f has global or local minimum or maximum. Solution. We have f1, = e 1, f, 1 = e 1 and te t 2e 2 for t 2. Therefore fx, y x 2 + y 2 e x2 y 2 2e 2 < e 1 for x, y / M = {u, v : u 2 + v 2 2} and f cannot attain its minimum and its

7 6 maximum outside M. Part a follows from the compactness of M and the continuity of f. Let x, y be a point from part b. From f x, y = x 2x1 x 2 + y 2 e x2 y 2 we get 1 x1 x 2 + y 2 =. Similarly 2 y1 + x 2 y 2 =. All solutions x, y of the system 1, 2 are,,, 1,, 1, 1, and 1,. One has f1, = f 1, = e 1 and f has global maximum at the points 1, and 1,. One has f, 1 = f, 1 = e 1 and f has global minimum at the points, 1 and, 1. The point, is not an extrema point because of fx, = x 2 e x2 > if x and fy, = y 2 e y2 < if y. Problem points Let f be a real-valued function with n + 1 derivatives at each point of R. Show that for each pair of real numbers a, b, a < b, such that fb + f b + + f n b ln fa + f a + + f n = b a a there is a number c in the open interval a, b for which f n+1 c = fc. Note that ln denotes the natural logarithm. Solution. Set gx = fx + f x + + f n x e x. From the assumption one get ga = gb. Then there exists c a, b such that g c =. Replacing in the last equality g x = f n+1 x fx e x we finish the proof. Problem points Let A be a n n diagonal matrix with characteristic polynomial x c 1 d 1 x c 2 d 2... x c k d k, where c 1, c 2,..., c k are distinct which means that c 1 appears d 1 times on the diagonal, c 2 appears d 2 times on the diagonal, etc. and d 1 +d 2 + +d k = n.

8 7 Let V be the space of all n n matrices B such that AB = BA. Prove that the dimension of V is d d d 2 k. Solution. Set A = a ij n i,j=1, B = b ij n i,j=1, AB = x ij n i,j=1 and BA = y ij n i,j=1. Then x ij = a ii b ij and y ij = a jj b ij. Thus AB = BA is equivalent to a ii a jj b ij = for i, j = 1, 2,..., n. Therefore b ij = if a ii a jj and b ij may be arbitrary if a ii = a jj. 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 points Let x 1, x 2,..., x k be vectors of m-dimensional Euclidian space, such that x 1 +x 2 + +x k =. Show that there exists a permutation π of the integers {1, 2,..., k} such that k 1/2 x πi x i 2 Note that denotes the Euclidian norm. for each n = 1, 2,..., k. Solution. We define π inductively. Set π1 = 1. Assume π is defined for i = 1, 2,..., n and also 1 2 x πi x πi 2. Note 1 is true for n = 1. We choose πn + 1 in a way that 1 is fulfilled with n + 1 instead of n. Set y = n x πi and A = {1, 2,..., k} \ {πi : i = 1, 2,..., n}. Assume that y, x r > for all r A. Then y, x r > r A and in view of y + x r = one gets y, y >, which is impossible. r A Therefore there is r A such that 2 y, x r. Put πn + 1 = r. Then using 2 and 1 we have n+1 2 x πi = y + x r 2 = y 2 + 2y, x r + x r 2 y 2 + x r 2

9 8 n+1 x πi 2 + x r 2 = x πi 2, which verifies 1 for n + 1. Thus we define π for every n = 1, 2,..., k. Finally from 1 we get 2 k x πi x πi 2 x i 2. Problem points ln 2 N 2 N 1 Find lim. Note that ln denotes the natural N N ln k lnn k k=2 logarithm. Solution. Obviously 1 A N = ln2 N N A N = ln2 N N N 2 k=2 1 ln k lnn k ln2 N N N 3 ln 2 N = 1 3 N. 1 Take M, 2 M < N/2. Then using that is decreasing in ln k lnn k [2, N/2] and the symmetry with respect to N/2 one get N 2 Choose M = 2 A N ln2 N N M k=2 N M 1 + k=m+1 + k=n M 1 ln k lnn k { M 1 2 ln 2 lnn 2 + N 2M 1 } ln M lnn M 1 2M ln N 1 N ln M + O. ln N 2 ln 2 M ln N N + [ ] N ln to get N 1 2 N ln 2 N Estimates 1 and 2 give ln N 1 ln N 2 ln ln N +O 1+O ln N ln 2 N 2 N 1 lim N N ln k lnn k = 1. k=2 ln ln N. ln N