Second day August 2, Problems and Solutions

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1 FOURTH INTERNATIONAL COMPETITION FOR UNIVERSITY STUDENTS IN MATHEMATICS July 30 August 4, 1997, Plovdiv, BULGARIA Secod day August, 1997 Problems ad Solutios Let Problem 1. Let f be a C 3 (R) o-egative fuctio, f(0)=f (0)=0, 0 < f (0). g(x) = ( ) f(x) f (x) for x 0 ad g(0) = 0. Show that g is bouded i some eighbourhood of 0. Does the theorem hold for f C (R)? where Let c = 1 f (0). We have g = (f ) ff (f ) f, f(x) = cx + O(x 3 ), f (x) = cx + O(x ), f (x) = c + O(x). Therefore (f (x)) = 4c x + O(x 3 ), f(x)f (x) = 4c x + O(x 3 ) ad (f (x)) f(x) = (4c x + O(x 3 )) x c + O(x). g is bouded because (f (x)) f(x) x 3 8c 5/ 0 x 0 ad f (x) f(x)f (x) = O(x 3 ). The theorem does ot hold for some C -fuctios. 1

2 Let f(x) = (x + x 3/ ) = x + x x + x 3, so f is C. For x > 0, g(x) = 1 ( ) = 1 x 1 (1 + 3 x) x x 0 Problem. Let M be a ivertible matrix of dimesio, represeted i block form as [ ] [ ] A B M = ad M 1 E F =. C D G H Show that det M. det H = det A. Let I deote the idetity matrix. The [ ] [ ] [ A B I F A 0 det M. det H = det det = det C D 0 H C I ] = det A. Problem 3. Show that Set f(t) = =1 ( 1) 1 si (log ) α coverges if ad oly if α > 0. si (log t) t α. We have f (t) = α cos (log t) si (log t) + tα+1 t α+1. So f (t) 1 + α t α+1 for α > 0. The from Mea value theorem for some θ (0, 1) we get f(+1) f() = f (+θ) 1 + α 1 + α. Sice < + α+1 α+1 for α > 0 ad f() 0 we get that ( 1) 1 f() = (f( 1) f()) =1 =1 coverges. si (log ) Now we have to prove that α does ot coverge to 0 for α 0. It suffices to cosider α = 0. We show that a = si (log ) [ does ot ted to zero. Assume the cotrary. There exist k N ad λ 1, 1 ] for > e such that log = k + λ. The a = si λ. Sice a 0 we get λ 0.

3 = We have k +1 k = log( + 1) log (λ +1 λ ) = 1 ( log ) (λ +1 λ ). The k +1 k < 1 for all big eough. Hece there exists 0 so that k = k 0 for > 0. So log = k 0 + λ for > 0. Sice λ 0 we get cotradictio with log. Problem 4. a) Let the mappig f : M R from the space M = R of matrices with real etries to reals be liear, i.e.: (1) f(a + B) = f(a) + f(b), f(ca) = cf(a) for ay A, B M, c R. Prove that there exists a uique matrix C M such that f(a) = tr(ac) for ay A M. (If A = {a ij } i, the tr(a) = a ii ). i=1 b) Suppose i additio to (1) that () f(a.b) = f(b.a) for ay A, B M. Prove that there exists λ R such that f(a) = λ.tr(a). a) If we deote by E ij the stadard basis of M cosistig of elemetary matrix (with etry 1 at the place (i, j) ad zero elsewhere), the the etries c ij of C ca be defied by c ij = f(e ji ). b) Deote by L the 1-dimesioal liear subspace of M cosistig of all matrices with zero trace. The elemets E ij with i j ad the elemets E ii E, i = 1,..., 1 form a liear basis for L. Sice E ij = E ij.e jj E jj.e ij, i j E ii E = E i.e i E i.e i, i = 1,..., 1, the the property () shows that f is vaishig idetically o L. Now, for ay A M we have A 1 tr(a).e L, where E is the idetity matrix, ad therefore f(a) = 1 f(e).tr(a). 3

4 Problem 5. Let X be a arbitrary set, let f be a oe-to-oe fuctio mappig X oto itself. Prove that there exist mappigs g 1, g : X X such that f = g 1 g ad g 1 g 1 = id = g g, where id deotes the idetity mappig o X. Let f = f f f, f 0 = id, f = (f 1 ) for every atural }{{} times umber. Let T (x) = {f (x) : Z} for every x X. The sets T (x) for differet x s either coiside or do ot itersect. Each of them is mapped by f oto itself. It is eough to prove the theorem for every such set. Let A = T (x). If A is fiite, the we ca thik that A is the set of all vertices of a regular polygo ad that f is rotatio by. Such rotatio ca be obtaied as a compositio of symmetries mappig the polygo oto itself (if is eve the there are axes of symmetry makig agle; if = k + 1 the there are axes makig k agle). If A is ifiite the we ca thik that A = Z ad f(m) = m + 1 for every m Z. I this case we defie g 1 as a symmetry relative to 1, g as a symmetry relative to 0. Problem 6. Let f : [0, 1] R be a cotiuous fuctio. Say that f crosses the axis at x if f(x) = 0 but i ay eighbourhood of x there are y, z with f(y) < 0 ad f(z) > 0. a) Give a example of a cotiuous fuctio that crosses the axis ifiiteley ofte. b) Ca a cotiuous fuctio cross the axis ucoutably ofte? Justify your aswer. a) f(x) = x si 1 x. b) Yes. The Cator set is give by C = {x [0, 1) : x = b j 3 j, b j {0, }}. There is a oe-to-oe mappig f : [0, 1) C. Ideed, for x = a j j, a j {0, 1} we set f(x) = (a j )3 j. Hece C is ucoutable. 4

5 For k = 1,,... ad i = 0, 1,,..., k 1 1 we set k a k,i = 3 k k 6 a j 3 j + 1, b k,i = 3 k 6 a j 3 j +, where i = k a j j, a j {0, 1}. The k 1 1 [0, 1) \ C = (a k,i, b k,i ), k=1 i=0 i.e. the Cator set cosists of all poits which have a triary represetatio with 0 ad as digits ad the poits of its complimet have some 1 s i their triary represetatio. Thus, k 1 1 (a k,i, b k,i ) are all poits (exept a k,i ) which i=0 have 1 o k-th place ad 0 or o the j-th (j < k) places. Noticig that the poits with at least oe digit equals to 1 are everywhere dece i [0,1] we set f(x) = ( 1) k g k (x). k=1 where g k is ( a piece-wise ) liear cotiuous fuctios with values at the kots ak,i + b k,i g k = k, g k (0) = g k (1) = g k (a k,i ) = g k (b k,i ) = 0, i = 0, 1,..., k 1 1. The f is cotiuous ad f crosses the axis at every poit of the Cator set. 5