Concrete mathematics : a foundation for computer science

574 ANSWERS TO EXERCISIES
length of the interval of numbers that round to 0. We have /y -01 6 id; also
y-e < Q--i6 andy+c’ > Q-t-:8. Therefore e+c’ > 6. Andif 6 < min(e, E’),
the rounding does distinguish ij from both y - e and y + 6’. Hence 10Ph” <
l/(k-l)+l/kand 10IPbk 3 l/k; we have bk = log k+O(l). Finally, therefore,
Et=, dk = ,& (logk+loglogk+O(l)), which is nlogn+nloglogn+O(n)
by Euler’s summation formula.
9.49 We have H, > lnn+y+ in-’ - &nP2 = f(n), where f(x) is increasing
for all x > 0; hence if n 3 ea Y we have H, 3 f(e”-Y) > K. Also H,-, <
Inn + y - in--’ = g(n), where g(x) is increasing for all x > 0; hence if
n 6 eaPy we have H,-l $ g(e”--Y) < 01. Therefore H,-r < OL 6 H, implies
that eaPv+l >n>ea+Y-l. (Sharper results have been obtained by Boas
and Wrench [27].)
9.50 (a) The expected return is ,YlsksN k/(k’HE’) = HN/H~‘, and we
want the asymptotic value to O(N-’ ):
1nN +y+O(N-‘) 6lnlO 6y 361n10 n
n2/6-N-l+O(N-2) = ~n+~~+~~+o(lo-n)*
The coefficient (6 In 1 O)/n2 = 1.3998 says that we expect about 40% profit.
(b) The probability o:f profit is x,, mk-1 such that
3 f(k+ 1)‘.
Now let A(z) = tk>, (z/k)mk. This power series converges for all z, because
the terms for k > Iz/ are bounded by a geometric series. Also A(n + 1) 3
((n+ l)/n)“‘n 3 f(n+l)‘, hence lim,,,f(n)/A(n) =O.