- Let
*I*= ∫*d*θ/(2 + tanθ) We make the substitution*y*= tanθ. Then*dy*= sec^{2}θ*d*θ = (1 +*y*^{2})*d*θ and we find that*I*= ∫*dy*/((1 +*y*^{2})(2 +*y*)).*y*^{2})(2 +*y*)) = 1/(2 +*y*) -*y*/(1 +*y*^{2}) + 2/(1 +*y*^{2}), we find that5Therefore 5*I*= ∫*dy*/(*y*+ 2) - ∫*ydy*/(1 +*y*^{2}) + ∫2*dy*/(1 +*y*^{2}) = ln(2 +*y*) - (ln(1 +*y*^{2}))/2 + 2 tan^{-1}*y*.*I*= ln(2 + tanθ)/secθ +2θ = ln(2 cosθ + sinθ) + 2θ and we deduce that*I*= (2θ + ln(2 cosθ + sinθ))/5,hence ∫Plugging in_{0}^{x}*d*θ/(2 + tanθ) = (2*x*+ ln(2 cos*x*+ sin*x*) - ln 2)/5.*x*= π/4, we conclude that∫_{0}^{π/4}*d*θ/(2 + tanθ) = (π +2 ln(3/√2) - 2 ln 2)/10 = (π + ln(9/8))/10. - Let
*A*= 1 + ∑_{n=1}^{∞}(*n*+ 1)/(2*n*+ 1)! and*B*= ∑_{n=1}^{∞}*n*/(2*n*+ 1)!, so*A*and*B*are the values of the sums in (a) and (b) respectively. Now*A*+*B*= ∑ _{n=0}^{∞}1/(2*n*)! = (*e*+*e*^{-1})/2,*A*-*B*= ∑ _{n=0}^{∞}1/(2*n*+ 1)! = (*e*-*e*^{-1})/2.

Therefore*A*=*e*/2 and*B*= 1/(2*e*). - We make the substitution
*y*=*e*^{u}where*u*is a function of*x*to be determined. Then*y'*=*u'e*^{u}and plugging into the given differential equation, we find that*u'e*^{u}=*e*^{u}*u*+*e*^{u}*e*^{x}, hence*u'*-*u*=*e*^{x}. This is a first order linear differential equation which can be solved in several ways, for example one method would be to multiply by the integrating factor*e*^{-x}. We obtain the general solution*u*=*xe*^{x}+*Ce*^{x}, where*C*is an arbitrary constant. We are given*y*= 1 when*x*= 0, and then*u*= 0. Therefore*u*=*xe*^{x}and we conclude that*y*=*e*^{xex}. - Ceva's theorem applied to the triangle
*ABC*shows that (*AR*/*RB*)(*BP*/*PC*)(*CQ*/*QA*) = 1. Since*RP*bisects ∠*BRC*, we see that*BP*/*PC*=*BR*/*RC*. Therefore*AR*/*RC*=*AQ*/*QC*, consequently ∠*ARQ*= ∠*QRC*and the result follows. - Let
*A*= (2 + √5) ^{100}((1 + √2)^{100}+ (1 + √2)^{-100})*B*= (√5 -2) ^{100}((1 + √2)^{100}+ (1 + √2)^{-100})*C*= (√5 +2) ^{100}+ (√5 -2)^{100}*D*= (√2 +1) ^{100}+ (√2 -1)^{100}

First note that*C*and*D*are integers; one way to see this is to use the binomial theorem. Also √2 -1 = (√2 +1)^{-1}. Thus*A*+*B*=*CD*is an integer. Now √5 -2 < 1/4, √2 +1 < 2.5 and √2 -1 < 1. Therefore 0 <*B*< (5/8)^{100}+ (1/4)^{100}< 10^{-4}. We conclude that there is a positive number ε < 10^{-4}such that*A*+ ε is an integer, and the result follows. - Suppose
det(
*A*^{2}+*B*^{2}) = 0. Then*A*^{2}+*B*^{2}is not invertible and hence there exists a nonzero*n*×1 matrix (column vector)*u*with real entries such that (*A*^{2}+*B*^{2})*u*= 0. Then*u'A*^{2}*u*+*u'B*^{2}*u*= 0, where*u'*denotes the transpose of*u*, a 1×*n*matrix. Therefore (*Au*)'(*Au*) + (*Bu*)'(*Bu*) = 0 and we deduce that*u'A*=*u'B*= 0, consequently*u'*(*AX*+*BY*) = 0. This shows that det(*AX*+*BY*) = 0, a contradiction and the result follows. - We claim that
*x*^{1/(ln(ln x))2}> (ln*x*)^{2}for large*x*. Indeed by taking logs, we need (ln*x*)/(ln(ln*x*))^{2}> 2 ln(ln(*x*)), that is ln*x*> 2(ln(ln*x*))^{3}. So by making the substitution*y*= ln*x*, we want*y*> 2(ln*y*)^{3}, which is true for*y*large. It now follows that for large*n*,*n*^{-(1+1/(ln(ln n))2)}= (1/*n*)1/(*n*^{1/(ln(ln n))2}) < 1/(*n*(ln*n*)^{2}).*n*(ln*n*)^{2}) is well known to be convergent, by using the integral test, and it now follows from the basic comparison test that the given series is also convergent.

Peter Linnell 2007-11-07