Concrete mathematics : a foundation for computer science

562 ANSWERS TO EXERCISES
And since gk,O = 91, mr we must have gk,,, = m+n+2mn. (c) The recurrence
is satisfied when mn > 0, because
sin(2mf l)e = &% (
sin(2m- l)tl
4
+ sin(2mt l)G + sin(2m+3)8 .
2 4 1 ’
this is a consequence of the identity sinjx - y) i- sinjx + y) = 2sin x cosy. So
all that remains is to check t.he boundary conditions.
8.50 (a) Using the hint, we get
3(1 42; (‘f) (;z)Yl -Z)2 k
= 3(1 -z)2F (‘f) (i)kF (ktjp3)zj+k;
now look at the coefficient of z3+‘. (b) H(z) == $ + &z + i J& c~+~z~+~.
(c) Let r = J(1 -2)(9-z). 0 ne can show that (z-3+r)(z-3-r) ~42,
and hence that (r/(1 -2)+2)’ = (13-5z+4r)/(l-z) = (9-H(z))/(l -H(z)).
(d) Evaluating the first derivative at z = 1 shows that Mean(H) = 1. The
second derivative diverges at z = 1, so the variance is infinite.
8.51 (a) Let H,(z) be the pgf for your holdings after n rounds of play, with
Ho(z) = z. The distribution for n rounds is
H,,+liz) = H,(W) ,
so the result is true by induction (using the amazing identity of the preceding
problem). (b) g,, =H,(O)-H, l(O) =4/n(n$-l)(n+2) =4(n-l)A. The
mean is 2, and the variance is infinite. (c) The expected number of tickets you
buy on the nth round is Mean(H,,) = 1, by exercise 15. So the total expected
number of tickets is infinite. (Thus, you almost surely lose eventually, and you
expect to lose after the second game, yet you also expect to buy an infinite
number of tickets.) (d) Now the pgf after n games is H,(z)‘, and the method
of part (b) yields a mean of 16 - 4x2 z 2.8. (The sum t,,, l/k2 = 7rL/6
shows up here.)
8.52 If w and w’ are events with Pr(w) > Pr(w’), then a sequence of
n independent experiments .will encounter cu more often than w’, with high
probability, because w will occur very nearly nPr(w) times. Consequently,
as n + co, the probability approaches 1 that the median or mode of the