Concrete mathematics : a foundation for computer science

568 ANSWERS TO EXERCISES
9.19 Hlo = 2.928968254 z 2.928968256; lo! =I 3628800 z 3628712.4; B,,., =
0.075757576 z 0.075757494; n( 10) = 4 z 10.0017845; e".' = 1.10517092 z
1.10517083;ln1.1 = 0.0953102 z 0.0953083; 1.1111111 z 1.1111000~ l.l@.' =
1.00957658 z 1.00957643. (The approximation to n(n) gives more significant
figures when n is larger; for example, rc( 1 09) = 50847534 zz 50840742.)
9.20 (a) Yes; the left side is o(n) while the right side is equivalent to O(n).
(b) Yes; the left side is e. eoi’/ni. (c) No; the left side is about J;; times the
bound on the right.
9.21 WehaveP,=m=n(lnm-1 -l/lnm+O(l/logn)2), where
lnm = lnn+lnlnm- l/lnn+lnlnn/(lnn)2 +O(l/logn)2;
lnlnn (lnlnn)’ lnlnn
lnlnm = 1nlnn-t -In -
2(lnn)2 +- (lnn)2
It follows that
P, = n ( lnn+lnlnn-1
lnlnn-2 t(lnlnn)’ - 31nlnn
+ - - -
hi n (lnn)2
+ O(l/logn)‘.
+ O(l/logn)’ .
)
(A slightly better approximation replaces this 0( l/logn)’ by the quantity
-5/(lnn)’ + O(loglogn/logn)3; then we estimate P~OOOOOO z 15483612.4.)
9.22 Replace O(nzk) by --&npLk + O(n 4k) in the expansion of H,r; this
replaces O(t3(n2)) by -h.E3(n2) + O(E:3(n4)) in (9.53). We have
,X3(n) = ii-i- ’ + &n,F2 + O(np3),
hence the term O(n2) in ($1.54) can be replaced by -gnp2 + O(n 3).
g.23 nhn = toskcn hk/(n~-k) +ZcH,/(n+ l)(n+2). Choose c = enL/6 =
tkaogk so that tka0 hk := 0 and h, = O(log n)/n3. The expansion of
t OSk