First day August 1, Problems and Solutions

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1 FOURTH INTERNATIONAL COMPETITION FOR UNIVERSITY STUDENTS IN MATHEMATICS July 30 August 4, 997, Plovdv, BULGARIA Frst day August, 997 Problems and Solutons Problem. Let {ε n } n= be a sequence of postve real numbers, such that lm ε n = n 0. Fnd n lm n n n + ε n, where denotes the natural logarthm. Soluton. It s well known that = (Remman s sums). Then n 0 n n + ε n xdx = lm n n n n ( k n) n n. n Gven ε > 0 there exst n 0 such that 0 < ε n ε for all n n 0. Then n n n + ε n n n n + ε. Snce lm n n n n + ε = 0 (x + ε)dx = +ε ε xdx

2 we obtan the result when ε goes to 0 and so lm n n n n + ε n =. Problem 2. Suppose a n converges. Do the followng sums have to converge as n= well? a) a + a 2 + a 4 + a 3 + a 8 + a 7 + a 6 + a 5 + a 6 + a a 9 + a 32 + b) a + a 2 + a 3 + a 4 + a 5 + a 7 + a 6 + a 8 + a 9 + a + a 3 + a 5 + a 0 + a 2 + a 4 + a 6 + a 7 + a 9 + Justfy your answers. Soluton. a) Yes. Let S = n= a n, S n = n a k. Fx ε > 0 and a number n 0 such that S n S < ε for n > n 0. The partal sums of the permuted seres have the form L 2 n +k = S 2 n + S 2 n S 2 n k, 0 k < 2 n and for 2 n > n 0 we have L 2 n +k S < 3ε,.e. the permuted seres converges. b) No. Take a n = ( )n+.then L n 3.2 n 2 = S 2 n + 2n and L 3.2 n 2 S 2 n 2 n 2, so L 2 n n 3.2 n 2. n 2k + k=2 n 2 Problem 3. Let A and B be real n n matrces such that A 2 +B 2 =AB. Prove that f BA AB s an nvertble matrx then n s dvsble by 3. Soluton. Set S = A + ωb, where ω = We have 2 SS = (A + ωb)(a + ωb) = A 2 + ωba + ωab + B 2 = AB + ωba + ωab = ω(ba AB), because ω + = ω. Snce det(ss) = det S. det S s a real number and det ω(ba AB) = ω n det(ba AB) and det(ba AB) 0, then ω n s a real number. Ths s possble only when n s dvsble by 3. 2

3 Problem 4. Let α be a real number, < α < 2. a) Show that α has a unque representaton as an nfnte product α = ( + ) ( + )... n n2 where each n s a postve nteger satsfyng n 2 n +. b) Show that α s ratonal f and only f ts nfnte product has the followng property: For some m and all k m, n k+ = n 2 k. Soluton. a) We construct nductvely the sequence {n } and the ratos θ k = α k ( + n ) so that θ k > for all k. Choose n k to be the least n for whch (θ 0 = α) so that for each k, + n < θ k () + n k < θ k + n k. Snce we have + n k+ < θ k = θ k + n k θ k + + n k n k + = + n 2 n k k. 3

4 Hence, for each k, n k+ n 2 k. Snce n 2, n k so that θ k. Hence α = ( + nk ). The unquness of the nfnte product wll follow from the fact that on every step n k has to be determne by (). Indeed, f for some k we have + n k θ k then θ k, θ k+ < and hence {θ k } does not converge to. Now observe that for M >, (2) ( + ) ( + ) ( M M 2 + ) M 4 = + M + Assume that for some k we have Then we get + n k < θ k. M 2 + M 3 + = + M. α ( + n )( + n 2 )... = θ k ( + n k )( + n k+ )... θ k ( + n k )( + )... = θ k + n 2 n k k > a contradcton. b) From (2) α s ratonal f ts product ends n the stated way. Conversely, suppose α s the ratonal number p. Our am s to show q that for some m, θ m = n m n m. Suppose ths s not the case, so that for every m, (3) θ m < n m n m. 4

5 For each k we wrte θ k = p k q k as a fracton (not necessarly n lowest terms) where p 0 = p, q 0 = q and n general p k = p k n k, q k = q k (n k + ). The numbers p k q k are postve ntegers: to obtan a contradcton t suffces to show that ths sequence s strctly decreasng. Now, p k q k (p k q k ) = n k p k (n k + )q k p k + q k and ths s negatve because p k q k = θ k < by nequalty (3). = (n k )p k n k q k n k n k Problem 5. For a natural n consder the hyperplane } n R0 {x n = = (x, x 2,..., x n ) R n : x = 0 and the lattce Z0 n = {y Rn 0 : all y are ntegers}. Defne the (quas )norm ( n ) /p n R n by x p = x p f 0 < p <, and x = max x. = a) Let x R0 n be such that max x mn x. For every p [, ] and for every y Z n 0 max x p x + y p. = prove that b) For every p (0, ), show that there s an n and an x R0 n wth x and an y Z0 n such that x mn x p > x + y p. 5

6 mn sets: Soluton. a) For x = 0 the statement s trval. Let x 0. Then max x > 0 and x < 0. Hence x <. From the hypothess on x t follows that: ) If x j 0 then max x x j +. ) If x j 0 then mn x x j. Consder y Z0 n, y 0. We splt the ndces {, 2,..., n} nto fve I(0) = { : y = 0}, I(+, +) = { : y > 0, x 0}, I(+, ) = { : y > 0, x < 0}, I(, +) = { : y < 0, x > 0}, I(, ) = { : y < 0, x 0}. As least one of the last four ndex sets s not empty. If I(+, +) Ø or I(, ) Ø then x + y > x. If I(+, +) = I(, ) = Ø then y = 0 mples I(+, ) Ø and I(, +) Ø. Therefore ) and ) gve x + y x whch completes the case p =. Now let p <. Then usng ) for every j I(+, ) we get x j + y j = y j + x j + y j + max x. Hence x j + y j p y j + x k p for every k I(, +) and j I(+, ). Smlarly x j + y j p y j + x k p for every k I(+, ) and j I(, +); x j + y j p y j + x j p for every j I(+, +) I(, ). Assume that. Then = j I(+, ) x + y p p x p p j I(+,+) I(, ) + j I(,+) j I(+,+) I(, ) j I(,+) ( x j + y j p x j p ) + x j + y j p y j + k I(+, ) j I(+, ) j I(+, ) x k p ( y j ) 6 x j + y j p k I(,+) x k p

7 = + The case ( y j ) + j I(,+) j I(+, ) j I(,+) n y 2 = 2 (y j ) + 2 y j 0. = j I(+, ) j I(+, ) j I(+,+) j I(+, ) j I(,+) s smlar. Ths proves the statement. b) Fx p (0, ) and a ratonal t ( 2, ). Choose a par of postve ntegers m and l such that mt = l( t) and set n = m + l. Let x = t, =, 2,..., m; x = t, = m +, m + 2,..., n; y =, =, 2,..., m; y m+ = m; y = 0, = m + 2,..., n. Then x R n 0, max x mn x =, y Z0 n and x p p x + y p p = m(t p ( t) p ) + ( t) p (m + t) p, whch s posstve for m bg enough. Problem 6. Suppose that F s a famly of fnte subsets of N and for any two sets A, B F we have A B Ø. a) Is t true that there s a fnte subset Y of N such that for any A, B F we have A B Y Ø? b) Is the statement a) true f we suppose n addton that all of the members of F have the same sze? Justfy your answers. Soluton. a) No. Consder F = {A, B,..., A n, B n,...}, where A n = {, 3, 5,..., 2n, 2n}, B n = {2, 4, 6,..., 2n, 2n + }. b) Yes. We wll prove nductvely a stronger statement: Suppose F, G are two famles of fnte subsets of N such that: ) For every A F and B G we have A B Ø; 2) All the elements of F have the same sze r, and elements of G sze s. (we shall wrte #(F ) = r, #(G) = s). 7

8 Then there s a fnte set Y such that A B Y Ø for every A F and B G. The problem b) follows f we take F = G. Proof of the statement: The statement s obvous for r = s =. Fx the numbers r, s and suppose the statement s proved for all pars F, G wth #(F ) < r, #(G ) < s. Fx A 0 F, B 0 G. For any subset C A 0 B 0, denote F (C) = {A F : A (A 0 B 0 ) = C}. Then F = F (C). It s enough to prove that for any par of nonempty sets C, D A 0 B 0 the famles F (C) and G(D) satsfy the statement. Ø C A 0 B 0 Indeed, f we denote by Y C,D the correspondng fnte set, then the fnte set Y C,D wll satsfy the statement for F and G. The proof C,D A 0 B 0 for F (C) and G(D). If C D Ø, t s trval. If C D = Ø, then any two sets A F (C), B G(D) must meet outsde A 0 B 0. Then f we denote F (C) = {A \ C : A F (C)}, G(D) = {B \ D : B G(D)}, then F (C) and G(D) satsfy the condtons ) and 2) above, wth #( F (C)) = #(F ) #C < r, #( G(D)) = #(G) #D < s, and the nductve assumpton works. 8