- Let
*I*= ∫_{0}^{1}∫_{0}^{x}∫_{0}^{1-x2}*e*^{(1-z)2}*dzdydx*. We change the order of integration, so we write*I*= ∫∫∫_{V}*e*^{(1-z)2}*dV*, where*V*is the region of integration.It can be described as the cylinder with axis parallel to the

*z*-axis and cross-section*A*, bounded below by*z*= 0 and bounded above by*z*= 1 -*x*^{2}. This region can also be described as the cylinder with axis parallel to the*y*-axis and cross-section*B*, bounded on the left by*y*= 0 and on the right by*y*=*x*. Therefore*I*= ∫ _{0}^{1}∫_{0}^{√1-z}∫_{0}^{x}*e*^{(1-z)2}*dydxdz*= ∫ _{0}^{1}∫_{0}^{√1-z}*xe*^{(1-z)2}*dxdz*= ∫_{0}^{1}(1 -*z*)*e*^{(1-z)2}/2*dz*= [- *e*^{(1-z)2}/4]_{0}^{1}= (*e*- 1)/4.

- We need to prove that
*pq*≤∫_{0}^{p}*f*(*t*)*dt*+ ∫_{0}^{q}*g*(*t*)*dt*. Either*q*≤*f*(*p*) or*q*≥*f*(*p*) and without loss of generality we may assume that*q*≥*f*(*p*) (if*q*≤*f*(*p*), then we interchange*x*and*y*; alternatively just follow a similar argument to what is given below). Then we have the following diagram.

We now interpret the quantities in terms of areas:

*pq*is the area of*A*∪*C*, ∫_{0}^{p}*f*(*t*)*dt*is the area of*C*, and ∫_{0}^{q}*g*(*t*)*dt*is the area of*A*∪*B*. The result follows. - Differentiating both sides with respect to
*x*, we obtain 2*ff'*=*f*^{2}-*f*^{4}+ (*f'*)^{2}. Thus*f*^{4}= (*f*-*f'*)^{2}, hence*f*-*f'*= ±*f*^{2}and we deduce that*dx*/*df*= 1/(*f*±*f*^{2}). We have two cases to consider; first we consider the + sign, that is*dx*/*df*= 1/*f*- 1/(*f*+ 1) and we obtain*x*= ln|*f*| - ln|*f*+ 1| +*C*, where*C*is an arbitrary constant. Now we have the initial condition*f*(0) = ±10. If*f*(0) = 10, we find that*C*= ln(11/10) and consequently*x*= ln(11/10) - ln|(*f*+ 1)|/*f*|. Solving this for*x*, we see that*f*(*x*) = 10/(11*e*^{-x}- 10). On the other hand if*f*(0) = - 10, then*C*= ln(9/10), consequently*x*= ln(9/10) - ln|(*f*+ 1)/*f*|. Solving this for*x*, we conclude that*f*(*x*) = 10/(9*e*^{-x}- 10).Now we consider the - sign, that is

*dx*/*df*= 1/*f*- 1/(*f*- 1) and we obtain*x*= ln|*f*| - ln|*f*- 1| +*D*, where*D*is an arbitrary constant. If the initial condition*f*(0) = 10, we find that*D*= ln(9/10) and consequently*x*= ln|*f*/(*f*- 1)| + ln(9/10). Solving this for*x*, we deduce that*f*(*x*) = 10/(10 - 9*e*^{-x}). On the other hand if the initial condition is*f*(0) = - 10, then*D*= ln(11/10) and hence*x*= ln|*f*/(*f*- 1)| + ln(11/10). Solving for*x*, we conclude that*f*(*x*) = 10/(10 - 11*e*^{-x}).Summing up, we have

*f*(*x*) = ±10/(10 - 9*e*^{-x}) or ±10/(10 - 11*e*^{-x}). - Set
*f*(*x*) =*ax*^{4}+*bx*^{3}+*x*^{2}+*bx*+*a*= 0. We will show that the maximum value of*a*+*b*is -1/2; certainly -1/2 can be obtained, e.g. with*a*= 1 and*b*= - 3/2, because in this case 1 is a root of*f*. Furthermore*f*(1) = 2(*a*+*b*) + 1, hence we may assume that 1 is not a root of*f*. Note that if α is a root of*f*, then so is 1/α. Thus we may assume that two of the roots of*f*are 1/α,α, where 1 < α is a real number.We claim that the other two roots of

*f*cannot be positive. Indeed suppose β is another positive root of*f*, which we may suppose is not 1 (though could be α or 1/α). Since the product of the roots of*f*is*a*/*a*= 1, we see that the fourth root of*f*is 1/β. It now follows that the sum of the products of roots of*f*two at a time is greater than α(1/α) + β(1/β) = 2; however this sum is also 1/*a*and we have contradicted the hypothesis*a*> 1/2.Thus

*f*has exactly two positive roots, namely 1/α,α, and by considering the graph of*f*, we see that*f*(1) < 0. Thus*a*+*b*+ 1 +*b*+*a*< 0 and the result follows. - We follow the hint given, so suppose
β is another eigenvector
corresponding to 1 with components
β
_{i}( 1≤*i*≤*n*). Choose*t*such that α_{t}/|β_{t}|≤α_{i}/|β_{i}| for all*i*(if β_{i}= 0, then we interpret α_{i}/|β_{i}| = + ∞; note that β_{t}≠ 0). Multiplying β by α_{t}/β_{t}, we may assume that β_{t}= α_{t}, and then we have β_{i}≤α_{i}for all*i*. Since∑we see that β_{i=1}^{n}*a*_{ti}α_{i}= α_{t}= β_{t}= ∑_{i=1}^{n}*a*_{ti}β_{i},_{i}= α_{i}for all*i*and the result follows. - Set
*a*_{n}=*f*(*n*)^{2}. Then squaring*f*(*n*+ 1) = 2√*f*(*n*)^{2}+*n*, we obtain*a*_{n+1}= 4*a*_{n}+ 4*n*. We solve this recurrence relation in a similar way to solving the corresponding first order differential equation*y'*= 4*y*+ 4*t*. The solution to*a*_{n+1}= 4*a*_{n}is*a*_{n}=*C*4^{n}for some constant*n*. Then we look for a solution to*a*_{n+1}= 4*a*_{n}+ 4*n*in the form*a*_{n}=*An*+*B*, where*A*,*B*are constants to be determined. Plugging this into the recurrence relation, we obtain*A*(*n*+ 1) +*B*= 4*An*+ 4*B*+ 4*n*, and then equating the coefficients of*n*and the constant term, we find that*A*= - 4/3,*B*= - 4/9. Therefore*a*_{n}=*C*4^{n}- 4*n*/3 - 4/9, and then plugging in*a*_{1}= 1, we see that*C*= 25/36 and we conclude that*a*_{n}= 25·4^{n}/36 - 4*n*/3 - 4/9. We now need to calculate ∑_{n=1}^{N}*a*_{n}. This is25(4^{N}-1)/27 - 2*N*^{2}/3 - 10*N*/9. - We have
*x*_{n+3}= 19*x*_{n+2}/(94_{n+1}) = 19^{2}/(94^{2}*x*_{n})*x*_{n+6}=*x*_{n}for all nonnegative integers*n*. It follows that ∑_{n=0}^{∞}*x*_{6n}/2^{n}= ∑_{n=0}^{∞}10/2^{n}= 20.

Peter Linnell 2007-06-16