- Let
*I*= ∫_{0}^{1}∫_{√y-y2}^{√1-y2}*xe*^{(x4+2x2y2+y4)}*dxdy*. We change to polar coordinates to obtain*I*= ∫_{0}^{π/2}∫_{sinθ}^{1}*r*cosθ*e*^{r4}*rdrd*θ = ∫_{0}^{π/2}∫_{sinθ}^{1}*r*^{2}*e*^{r4}cosθ*drd*θ.*t*= θ. This yields*I*= ∫ _{0}^{1}∫_{0}^{sin-1r}*r*^{2}*e*^{r4}cos*t**dtdr*= ∫_{0}^{1}[*r*^{2}*e*^{r4}sin*t*]_{0}^{sin-1r}*dr*= ∫ _{0}^{1}*r*^{3}*e*^{r4}*dr*= [*e*^{r4}/4]_{0}^{1}= (*e*- 1)/4.

- Write
*r*_{1}=*m*_{1}/*n*_{1}and*r*_{2}=*m*_{2}/*n*_{2}, where*m*_{1},*n*_{1},*m*_{2},*n*_{2}are positive integers and gcd(*m*_{1},*n*_{1}) = 1 = gcd(*m*_{2},*n*_{2}). Set*Q*= ((*m*_{1}+*m*_{2})/(*n*_{1}+*n*_{2}), 1/(*n*_{1}+*n*_{2})). We note that*Q*is on the line joining (*r*_{1}, 0) with*P*(*r*_{2}), that is the line joining (*m*_{1}/*n*_{1}, 0) with (*m*_{2}/*n*_{2}, 1/*n*_{2}). This is because(Similarly*m*_{1}+*m*_{2})/(*n*_{1}+*n*_{2}) =*m*_{1}/*n*_{1}+ (*m*_{2}/*n*_{2}-*m*_{1}/*n*_{1})(*n*_{2}/(*n*_{1}+*n*_{2})).*Q*is on the line joining*P*(*r*_{1}) with (*r*_{2}, 0). It follows that (*m*_{1}+*m*_{2})/(*n*_{1}+*n*_{2}), 1/(*n*_{1}+*n*_{2}) is the intersection of the line joining (*r*_{1}, 0) to*P*(*r*_{2}) and the line joining*P*(*r*_{1}) and (*r*_{2}, 0). Set*P*=*P*((*r*_{1}*f*(*r*_{1}) +*r*_{2}*f*(*r*_{2}))/(*f*(*r*_{1}) +*f*(*r*_{2}))).*P*= ((*m*_{1}+*m*_{2})/(*n*_{1}+*n*_{2}),/*f*((*m*_{1}+*m*_{2})/(*n*_{1}+*n*_{2}))),*P*is the point of intersection of the two given lines if and only if*f*((*m*_{1}+*m*_{2})/(*n*_{1}+*n*_{2})) =*n*_{1}+*n*_{2}. We conclude that the necessary and sufficient condition required is that gcd(*m*_{1}+*m*_{2},*n*_{1}+*n*_{2}) = 1. - Taking logs, we get
*dy*/*dx*=*y*ln*y*, hence*dx*/*dy*= 1/(*y*ln*y*). Integrating both sides, we obtain*x*= ln(ln*y*) +*C*where*C*is an arbitrary constant. Plugging in the initial condition*y*=*e*when*x*= 1, we find that*C*= 1. Thus ln(ln*y*) =*x*- 1 and we conclude that*y*=*e*^{(ex-1)}. - Set
*g*(*x*) =*x*^{2}*f*(*x*). Then the given limit says lim_{x--> ∞}*g''*(*x*) = 1. Therefore lim_{x--> ∞}*g'*(*x*) = lim_{x--> ∞}*g*(*x*) = ∞. Thus by l'Hôpital's rule,limWe deduce that lim_{x--> ∞}*g*(*x*)/*x*^{2}= lim_{x--> ∞}*g'*(*x*)/(2*x*) = lim_{x--> ∞}*g*(*x*)/2 = 1/2._{x--> ∞}*f*(*x*) = 1/2 and lim_{x--> ∞}(*xf'*(*x*)/2 +*f*(*x*)) = 1/2, and the result follows. - Set
*f*(*x*)= *a*_{1}+*b*_{1}*x*+ 3*a*_{2}*x*^{2}+*b*_{2}*x*^{3}+5*a*_{3}*x*^{4}+*b*_{3}*x*^{5}+7*a*_{4}*x*^{6},*g*(*x*)= *a*_{1}*x*+*b*_{1}*x*^{2}/2 +*a*_{2}*x*^{3}+*b*_{2}*x*^{4}/4 +*a*_{3}*x*^{5}+*b*_{3}*x*^{6}/6 +*a*_{4}*x*^{7}.

Then*g*(1) =*g*(- 1) because*a*_{1}+*a*_{2}+*a*_{3}+*a*_{4}= 0, hence there exists*t*∈(- 1, 1) such that*g'*(*t*) = 0. But*g'*(*x*) =*f*(*x*) and the result follows. - We choose the
*n*line segments so that the sum of their lengths is as small as possible. We claim that no two line segments intersect. Indeed suppose*A*,*B*are red balls and*C*,*D*are green balls, and*AC*intersects*BD*at the point*P*. Since the length of one side of a triangle is less than the sum of the lengths of the two other sides, we have*AD*<*AP*+*PD*and*BC*<*BP*+*PC*, consequently*AD*+*BC*<*AP*+*PC*+*BP*+*PD*=*AC*+*BD*, - We have
*f*_{n, j+1}(*x*) -*f*_{n, j}(*x*) = √*x*/*n*, hence*f*_{n, j}(*x*) =*f*_{0, j}(*x*) +*j*√*x*/*n*=*x*+ (*j*+ 1)√*x*/*n*.*f*_{n, n}(*x*) =*x*+ (*n*+ 1)√*x*/*n*and we see that lim_{n--> ∞}*f*_{n, n}(*x*) =*x*+ √*x*.

Peter Linnell 2007-05-20