- First make the substitution
*y*=*x*^{3}. Then*dF*/*dx*= (*dF*/*dy*)(*dy*/*dx*) =*e*^{y2}3*x*^{2}= 3*x*^{2}*e*^{x6}by the chain rule. Therefore*d*^{2}*F*/*dx*^{2}= 6*xe*^{x6}+18*x*^{7}*e*^{x6}. To find the point of inflection, we set*d*^{2}*F*/*dx*^{2}= 0; thus we need to solve 6*x*+ 18*x*^{7}= 0. The only solution is*x*= 0, so this is the point of inflection (perhaps we should note that*d*^{3}*F*/*dx*^{3}is 6≠ 0 at*x*= 0, so*x*= 0 is indeed a point inflection). - The shortest path will first be reflected off the
*x*-axis, then be reflected off the*y*-axis. So we reflect (*x*_{2},*y*_{2}) in the*y*-axis, and then in the*x*-axis, which yields the point (-*x*_{2}, -*y*_{2}). Thus the length of the shortest path is the distance from (*x*_{1},*y*_{1}) to (-*x*_{2}, -*y*_{2}), which is ((*x*_{1}+*x*_{2})^{2}+ (*y*_{1}+*y*_{2})^{2})^{1/2}. - (i)
- We have
*f*(*f*(*x*)) = 1 + sin(*f*(*x*) - 1) = 1 + sin(sin(*x*- 1)), so*f*_{2}(*x*) =*x*if and only if*x*- 1 = sinsin(*x*- 1). If*y*is a real number, then | sin*y*|≤|*y*| with equality if and only if*y*= 0. It follows that*y*= sinsin(*y*) if and only if*y*= 0 and we deduce that there is a unique point*x*_{0}such that*f*_{2}(*x*_{0}) =*x*_{0}, namely*x*_{0}= 1. - (ii)
- From (i), we have
*x*_{n}= 1 for all*n*. Thus we need to find ∑_{n=0}^{∞}1/3^{n}. This is a geometric series with first term 1 and ratio between successive terms 1/3. Therefore this sum is 1/(1 - 1/3) = 3/2.

- Clearly
*t*_{n}≥1 for all*n*. Set*T*= (1 + √5)/2 (the positive root of*x*^{2}-*x*- 1). Note that if 1≤*x*<*T*, then*x*^{2}< 1 +*x*<*T*^{2}. This shows that*t*_{n}<*T*for all*n*, and also that*t*_{n}is an increasing sequence, because*t*_{n+1}^{2}-*t*_{n}^{2}= 1 + +*t*_{n}-*t*_{n}^{2}. Therefore this sequence converges to a number between 1 and*T*. Since the number must satisfy*x*^{2}=*x*+ 1, we deduce that lim_{n--> ∞}*t*_{n}=*T*= (1 + √5)/2. - First we find the eigenvalues and eigenvectors of
*A*. The eigenvalues satisfy*x*(*x*- 3) + 2 = 0, so the eigenvalues are 1,2. To find the eigenvectors corresponding to 1, we need to solve the matrix equation( -1 -2 1 2 ) ( *u**v*) = ( 0 0 ) One solution is

*u*= 2, v= - 1 and we see that( 2 -1 ) is an eigenvector corresponding to 1.

To find the eigenvectors corresponding to 2, we need to solve

( -2 -2 1 1 ) ( *u**v*) = ( 0 0 ) One solution is

*u*= 1,*v*= - 1 and we see that( 1 -1 ) is an eigenvector corresponding to 2.

We now know that if

*T*=( 2 1 -1 -1 ) then

*T*^{-1}*AT*=( 1 0 0 2 ) We deduce that

*T*^{-1}*A*^{100}*T*=( 1 0 0 2 ^{100}) Therefore

*A*^{100}=( 2 - 2 ^{100}2 - 2 ^{101}-1 + 2 ^{100}-1 + 2 ^{101}) - Since
*p*(*r*) = 0, we may write*p*(*x*) =*q*(*x*)(*x*-*r*), where*q*(*x*) is of the form*x*^{2}+*dx*+*e*. Then*p*(*x*)/(*x*-*r*) - 2*p*(*x*+ 1)/(*x*+ 1 -*r*) +*p*(*x*+ 2)/(*x*+ 2 -*r*) =*q*(*x*) - 2*q*(*x*+ 1) +*q*(*x*+ 2) = 2 as required. - Assume that log means natural log. Note that log
*x*is a positive increasing function for*x*> 1. Therefore∫Since ∫_{1}^{n}*x*log*x**dx*≤∑_{t=2}^{n}*t*log*t*__<__∫_{2}^{n+1}*x*log*x**dx*.*x*log*x*= (*x*^{2}log*x*)/2 -*x*^{2}/4, we see that(Now divide by*n*^{2}log*n*)/2 -*n*^{2}/4≤∑_{t=2}^{n}*t*log*t*__<__((*n*+ 1)^{2}log*n*)/2 - (*n*+ 1)^{2}/4 - 2 log 2 + 1.*n*^{2}log*n*and take lim_{n--> ∞}. We obtain1/2≤limWe conclude that the required limit is 1/2._{n--> ∞}(∑_{t=2}^{n}*t*log*t*)/(*n*^{2}log*n*)≤1/2. - For
*G*(3) we have 8 possible rows of goblins, and by writing these out we see that*G*(3) = 17. Similarly for*G*(4) we have 16 possible rows of goblins, and by writing these out we see that*G*(4) = 44.In general, let

*X*be a row of*N*goblins. Then the rows with*N*+ 1 columns are of the form*X*2 or*X*3 (where*X*2 indicates adding a goblin with height 2' to the end of*X*). If*X*ends in 3 or 32 ( 2^{N-1}+2^{N-2}possible rows), then*X*3 has 1 more LGG than*X*; on the other hand if*X*ends in 22, then*X*3 has the same number of LGG's as*X*. If*X*ends in 2 (2^{N-1}possible rows), then*X*2 has 1 more LGG than*X*, while if*X*ends in 3, then*X*2 has the same number of LGG's as*X*. We conclude that for*N*≥2,*G*(*N*+ 1) = 2*G*(*N*) + 2^{N-1}+2^{N-1}+2^{N-2}= 2*G*(*N*) + 5·2^{N-2}.*G*(*N*) =*C*·2^{N}+*aN*2^{N-2}, where*a*is to be determined. Then*G*(*N*+ 1) = 2*G*(*N*) + 5·2^{N-2}yields*a*= 5/2, so*G*(*N*) =*C*·2^{N}+5*N*2^{N-3}. The initial condition*G*(2) = 6 shows that*C*= 1/4 and we conclude that*G*(*N*) = 2^{N-2}+5*N*2^{N-3}for all*N*≥2. We also have*G*(1) = 2.

Peter Linnell 2009-06-12