Advanced book on Mathematics Olympiad

530 Real Analysis which is, in fact, the condition forfto be convex. (P.N. de Souza, J.N. Silva,Berkeley Problems in Mathematics ...

Real Analysis 531 (a^21 +a 22 +a^23 )(a 14 +a^42 +a 34 )≥(a^31 +a^32 +a 33 )^2. This is just the Cauchy–Schwarz inequality appli ...

532 Real Analysis 444.Split the integral as ∫ ex 2 dx+ ∫ 2 x^2 ex 2 dx. Denote the first integral byI 1. Then use integration by ...

Real Analysis 533 With the substitutionx−^1 x=twe have( 1 +x^12 )dx=dt, and the integral takes the form ∫ 1 t^2 + 1 dt=arctant+C ...

534 Real Analysis In= ∫ n!(fn(x)−fn′(x)) fn(x) dx=n! ∫ ( 1 − fn′(x) fn(x) ) dx=n!x−n!lnfn(x)+C =n!x−n!ln ( 1 +x+ x^2 2! +···+ xn ...

Real Analysis 535 =arctanx+ 1 3 arctanx^3. To write the answer in the required form we should have 3 arctanx+arctanx^3 =arctan P ...

536 Real Analysis 455.Denote the value of the integral byI. With the substitutiont=abx we have I= ∫b a e bt −e ta ab t · −ab t^2 ...

Real Analysis 537 This we already computed in the previous problem. (“Happiness is longing for repeti- tion,’’ says M. Kundera.) ...

538 Real Analysis arctan ( 1 −cosα sinα ) −arctan ( − 1 −cosα sinα ) , where the angles are to be taken between−π 2 andπ 2. But ...

Real Analysis 539 =−e−xP(x)|t 0 −e−xP′(x)|t 0 −···−e−xP(n)(x)|t 0. Because limt→∞e−tP(k)(t)=0,k= 0 , 1 ,...,n, when passing to t ...

540 Real Analysis In= ⎧ ⎪⎪⎨ ⎪⎪⎩ 1 · 3 · 5 ···( 2 k− 1 ) 2 · 4 · 6 ···( 2 k) · π 2 , ifn= 2 k, 2 · 4 · 6 ···( 2 k) 1 · 3 · 5 ···( ...

Real Analysis 541 468.We have sn= 1 √ 4 n^2 − 12 + 1 √ 4 n^2 − 22 +···+ 1 √ 4 n^2 −n^2 = 1 n ⎡ ⎣√^1 4 − ( 1 n ) 2 + 1 √ 4 − ( 2 ...

542 Real Analysis and, using the inequalityex> 1 +x, 2 i/n 1 +ni^1 = 2 (i−^1 )/n 21 /n 1 +ni^1 = 2 (i−^1 )/n eln 2/n 1 +ni^1 ...

Real Analysis 543 We are left to compute the limit of this expression asngoes to infinity. Ifa≤1, this limit is equal to 0. Ifa& ...

544 Real Analysis We now recognize a perfect square and write this as ∫ 1 0 ( f(x)− x 2 ) 2 dx= 0. The integral of the nonnegati ...

Real Analysis 545 We have 0 ≤ ∫ 1 0 (f (x)−g(x))^2 dx= ∫ 1 0 f (x)(f (x)−g(x))dx− ∫ 1 0 g(x)(f (x)−g(x))dx = ∫ 1 0 f^2 (x)−f (x) ...

546 Real Analysis Ifχ[ 0 ,ai]is the characteristic function of the interval[ 0 ,ai](equal to 1 on the interval and to 0 outside) ...

Real Analysis 547 481.Suppose thatx>y. Transform the inequality successively into mn(x−y)(xm+n−^1 −ym+n−^1 )≥(m+n− 1 )(xm−ym) ...

548 Real Analysis Becausef′( 1 )>0 and arctany≤yfory≥0, we further obtain ∫ 1 0 dx f^2 (x)+ 1 ≤ arctanf( 1 ) f′( 1 ) ≤ f( 1 ) ...

Real Analysis 549 2 m(A)IA+ 2 m(B)IB≥IA+IB. Sincem(A)+m(B)=1, this inequality translates into ( m(A)− 1 2 ) (IA−IB)≥ 0 , which w ...