Concrete mathematics : a foundation for computer science

A ANSWERS TO EXERCISES 567
9.9 (For completeness, we assume that there is a side condition n + 00,
so that two constants are implied by each 0.) Every function on the left has
the form a(n) + b(n), where there exist constants Q, B, no, C such that
la(n)/ 6 Blf(n)[ for n 3 mc and [b(n)1 6 Clg(n)l for n 3 no. Therefore the
left-handfunctionisatmostmax(B,C)(lf(n)l+Ig(n)l),forn3max(~,no),
so it is a member of the right side.
9.10 If g(x) belongs to the left, so that g(x) = cosy for some y, where
Iy/ < Clxl for some C, then 0 6 1 - g(x) = 2sin2(y/2) < $y2 6 iC2x2; hence
the set on the left is contained in the set on the right, and the formula is true.
9.11 The proposition is true. For if, say, 1x1 < /yI, we have (x + Y)~ 6 4y2.
Thus (x+Y)~ = 0(x2) +O(y’). Thus O(x+y)’ = O((x+y)‘) = 0(0(x2) +
O(y2)) = 0(0(x2)) -t O(O(y2)) = 0(x2) + O(y2).
9.12 1 +2/n + O(nP2) = (1 + 2/n)(l + O(nP2)/(1 +2/n)) by (g.26), and
l/(1 +2/n) = O(1); now use (9.26).
9.13 n”(1 + 2nP’ + O(nP2))” = nnexp(n(2n-’ + O(nP2))) = e2nn +
O(n”-‘).
9.14 It is nn+Pexp((n+ @)(ol/n- ta2/n2 +O(ne3)))
9.15 In (n2n) = 3nln3 - 1nn+tln3-ln2n+
(+f)n-’ +O(nP3), so
the answer is‘
=(I - 5n-l + 82jnp2 + o(n-3)).
9.16 If 1 is any integer in the range a 6 1 < b we have
1
1
B(x)f(l+x) dx = B(x)f(l+x) dx-
0 l/2 s0
=s
1
l/2
l/2
B(l -x)f(l+x)dx
B(x)(f(l+x) -f(l+ 1 -x)) dx.
Since 1 + x > 1 + 1 - x when x 3 i, this integral is positive when f(x) is
nondecreasing.
9.17 L>O B,(i)z."'/m! = ~e~'~/(e~-l) = z/(eZ/2-1)-z/(e"-1)
9.18 The text’s derivation for the case OL = 1 generalizes to give
2(2n+1/2)a
bk(n) = - - e -k’a/n
(27rn)"/2 '
the answer is 22na(~n)i’~a1’20L~1’2(1 + O(n-1/2+36)).
ck(n) = 22nan -(l+cx)/2+3ykb./n.
I