Advanced book on Mathematics Olympiad

570 Real Analysis and this is equivalent to(a+b−c)^2 >0. Hence the conclusion. (Kvant(Quantum), proposed by R.P. Ushakov) 514 ...

Real Analysis 571 D Figure 71 J=− 1 (u^2 +v^2 )^2 . Therefore, ∫∫ D dxdy (x^2 +y^2 )^2 = ∫∫ D 1 dudv= 1 12 . (D. Flondor, N. Don ...

572 Real Analysis (Gh. Bucur, E. Câmpu, S. G ̆aina, ̆ Culegere de Probleme de Calcul Diferen ̧tial ̧si Integral(Collection of Pr ...

Real Analysis 573 Consequently, the integral we are computing is equal toπ 2 32. (American Mathematical Monthly, proposed by M. ...

574 Real Analysis 522.First, note that forx>0, e−sxx−^1 |sinx|<e−sx, so the integral that we are computing is finite. Now ...

Real Analysis 575 Remark.This is a particular case of integrals of the form ∫∞ 0 f(ax)−f(bx) x dx, known as Froullani integrals. ...

576 Real Analysis Using the identity ∑ n≥ 1 1 n^2 = π^2 6 , we obtain ∫∞ 0 F(t)dt= π^3 12 . (Gh. Sire ̧tchi,Calcul Diferen ̧tial ...

Real Analysis 577 while the area element is dσ= 1 cosα dxdy, αbeing the angle formed by the normal to the sphere with thexy-plan ...

578 Real Analysis Proof.For the sake of completeness we will prove Green’s identity. Consider the vector field −→ F =f∇g. Then d ...

Real Analysis 579 532.The condition curl −→ F =0 suggests the use of Stokes’ theorem: ∫∫ S curl −→ F ·−→ndS= ∮ ∂C −→ F ·d −→ R. ...

580 Real Analysis = ∫ D×{b} G(x,y,t) −→ k ·d−→n + ∫ D×{a} G(x,y,t) −→ k ·d−→n + ∫b a ∫b 1 a 1 F 1 (x, a 2 )dx− ∫b a ∫a 1 b 1 F 1 ...

Real Analysis 581 ∮ C Pdx+Qdy+Rdz= ∫∫ S ( ∂Q ∂x − ∂P ∂y ) dxdy+ ( ∂R ∂y − ∂Q ∂z ) dydz + ( ∂P ∂z − ∂R ∂x ) dzdx. Writing the par ...

582 Real Analysis = ∮ C 2 ∂ ∂z′ ((x′−x)^2 +(y′−y)^2 +(z′−z)^2 )−^3 /^2 dz′= 0 , where the last equality is a consequence of the ...

Real Analysis 583 f(eiπ/^3 )=β, f(e−iπ/^3 )= ̄α+ 1 − ̄αβ, whereβis an arbitrary complex parameter. (20th W.L. Putnam Competition ...

584 Real Analysis Proof.Letg(c)=y. Thenc=g(g(c))=g(y); hencey=g(c)=g(g(y)). Thusyis a fixed point ofg◦g.Ify=a, thena=g(a)=g(y)=c ...

Real Analysis 585 whencef(π 2 )=± √ c−1. Substituting in the original equationy=πwe also obtain f(x+π)=−f(x). We then have −f(x) ...

586 Real Analysis g(x)= 1 3 g ( x− 1 2 ) , for allx∈R. Forx=−1 we haveg(− 1 )=^13 g(− 1 ); henceg(− 1 )=0. In general, for an ar ...

Real Analysis 587 544.We should keep in mind thatf(x)=sinxandg(x)=cosxsatisfy the condition. As we proceed with the solution to ...

588 Real Analysis 546.Adding 1 to both sides of the functional equation and factoring, we obtain f(x+y)+ 1 =(f (x)+ 1 )(f (y)+ 1 ...

Real Analysis 589 It now becomes natural to letg(x)=f(x)x , which satisfies the equation g(xy)=g(x)+g(y). The particular casex = ...