Mathematical Competition for Students of the Department of Mathematics and Informatics of Vilnius University Problems and Solutions Paulius Drungilas, Artūras Dubickas and Jonas Jankauskas 3 6--6 PROBLEMS Problem. Let S be the set of all real numbers x satisfying x 4 + 36 3x. Find the largest and the smallest values of x 3 3x when x S. Problem. Prove that for every matrix A M (R there exist two matrices B, C M (R such that A B 3 + C 3. (Here, M (R denotes the set of matrices with real coefficients. Problem 3. Let P (x x 6 + a x 5 + a x 4 + + a 5 x + a 6 be a polynomial of degree 6 whose coefficients a j belong to the set {, } for each j,,..., 6. Prove that P has less than 6 distinct real roots. Problem 4. Find the value of the it (i + j + i!. Each problem is worth points. Vilnius University, Department of Mathematics and Informatics, Naugarduko 4, Vilnius LT-35, Lithuania, http://www.mif.vu.lt/ drungilas/ Vilnius University, Department of Mathematics and Informatics, Naugarduko 4, Vilnius LT-35, Lithuania, http://www.mif.vu.lt/ dubickas/ 3 Vilnius University, Department of Mathematics and Informatics, Naugarduko 4, Vilnius LT-35, Lithuania, http://www.mif.vu.lt/ jonakank/ and Montanuniversität Leoben, Lehrstuhl für Mathematik und Statistik, Franz-Josef-Strasse 8/HP, 87 Leoben, Austria
PROBLEMS WITH SOLUTIONS Problem. Let S be the set of all real numbers x satisfying x 4 + 36 3x. Find the largest and the smallest values of x 3 3x when x S. Answer: The largest value is 8; the smallest value is 8. Solution. Since x 4 + 36 3x (x 4(x 9 (x 3(x (x + (x + 3, we find that S [ 3, ] [, 3]. The function f(x : x 3 3x is increasing on [ 3, ] and also increasing on [, 3], since its derivative f (x 3x 3 3(x is positive in S. Thus, the largest value of f(x for x S equals max{f(, f(3} max{, 8} 8. The smallest value of f(x for x S is min{f( 3, f(} min{ 8, } 8. Problem. Prove that for every matrix A M (R there exist two matrices B, C M (R such that A B 3 + C 3. (Here, M (R denotes the set of matrices with real coefficients. a b Solution. We claim that every upper triangular matrix is the cube of some d matrix from M (R, provided that ad. Indeed, using the equality 3 ( x y x 3 y(x + xz + z, ( z z 3 which holds for all x, y, z R, we find that a b d 3 x y z a x 3, b y(x + xz + z, d z 3. For ad we can solve it as follows: x 3 a, z 3 d and y b 3 a + 3 ad + 3 d. 3 (Note that the denominator is positive, i.e., a + 3 ad + 3 d >, if ad. Similarly, by considering the transpose matrices in (, we see that 3 ( x x 3. y z y(x + xz + z z 3
a Hence, every lower triangular matrix is also the cube of some matrix from c d M (R, provided that ad. Finally, note that every matrix from M (R is the sum of an upper and a lower triangular matrices with the required property ad. Indeed, writing a b a + t b t A + ( c d d + t c t with some sufficiently large t R (e.g., for t > a + d, we have (a + t(d + t and t. Hence, by the above, both matrices on the right hand side of ( are cubes of some matrices B and C from M (R, that is, A B 3 + C 3. Solution. Let A M (R. By the Cayley-Hamilton theorem, A aa + di O, where a is the trace of A (i.e., the sum of the elements on the main diagonal of A, d is the determinant of A, O M (R is the zero matrix, and I M (R is the identity matrix. Hence, A aa di. Using this equality, for any t R we deduce that (A + ti 3 A 3 + 3tA + 3t A + t 3 I A(aA di + 3t(aA di + 3t A + t 3 I aa + (3t + 3ta da + (t 3 3tdI a(aa di + (3t + 3ta da + (t 3 3tdI (3t + 3ta + a da + (t 3 3td adi. Thus, denoting u : 3t + 3ta + a d and v : t 3 3td ad, we obtain (A + ti 3 ua + vi. Now, choose sufficiently large t for which u >. Then, ( A u (A + ti3 v 3 ( 3 u I v 3 (A + ti + 3 u u I, 3 whence the result. Problem 3. Let P (x x 6 + a x 5 + a x 4 + + a 5 x + a 6 be a polynomial of degree 6 whose coefficients a j belong to the set {, } for each j,,..., 6. Prove that P has less than 6 distinct real roots.
Solution. Set, for brevity, n 6 and assume that for some choice of the coefficients ± the polynomial P has n distinct real roots α,..., α n. Then, P (x x n + a x n + a x n + + a n x + a n (x α (x α... (x α n. Consider the sum of squares S : α + α + + α n. By Vieta s formulas, S (α + α + + α n (α α + α α 3 +... α n α n a a a. Hence, S + 3. On the other hand, the arithmetic mean vs geometric mean inequality implies S α + α + + α n n(α α... α n /n n ( a n /n n. 4 Consequently, n S 3, which is impossible for each n 4. impossible for n 6. (In particular, it is Problem 4. Find the value of the it Answer: (e /. (i + j + i!. Solution. Fix x and n N. By the Taylor formula for exponential series with the Lagrange remainder form, we have e x n x k k + r k! n (x xn+, where r (n+! n(x e x. Thus, ( e x e x e x x i i! + r x n+ ( x j n(x (n +! + r x n+ n(x (n +! Set ( x i i! ( x j x n+ + r n (x (n +! x i+j i! + r x n+ ( n(x e x + (n +! x n+ ( ϕ n (x : r n (x e x + (n +! x j x n+ + ex r n (x (n +! x j. x j. Observe that for each x [, ] and each n N we have x n+ ( ϕ n (x r n (x e x x j + e x (n +! (n +! (ex + e x e (n +!. Hence, ϕ n(xdx.
Integrating from to and using the equality xi+j dx, we find that i+j+ e x dx Therefore, whence the result. x i+j i! dx + (i + j + i! + (i + j + i! ϕ n (xdx ϕ n (xdx. e x dx xi+j dx dx + i! ϕ n (xdx ex ϕ n (xdx e, 5