Concrete mathematics : a foundation for computer science

A ANSWERS TO EXERCISES 575
9.53 By induction, the 0 term is (m - l)!--’ s,” tmP’f(“‘)(x - t) dt. Since
f(ln+‘) has the opposite sign to fcm), the absolute value of this integral is
bounded by If(“‘(O) 1 J,” tm-’ dt; so the error is bounded by the absolute value
of the first discarded term.
9.54 Let g(x) =~f(x)/xrx. Then g’(x) N -oLg(x)/x as x t 00. By the mean
Sounds like a nasty value theorem, g(x - i) - g(x + i) = -g’(y) - ag(y)/y for some y between
theorem. x - i and x + i. Now g(y) = g(x)(l +0(1/x)), so g(x - i) - g(x + i) -
ag(x)/x = af(x)/xlta. Therefore
x f(k) ~ =
k3n k’+”
(J(t(g&- :I - g(k+ iI)) = o(g(n- :I).
k3n
9.55 The estimate of (n + k + i) ln(l + k/n) + (n - k + i) ln(1 - k/n) is
extended to k2/n + k4/6n3 + O(nP3/2+5E), so we apparently want to have an
extra factor ePk4/6n3 in bk(n), and ck(n) = 22nn-2+5eePk*/n. But it turns
out to be better to leave bk(n) untouched and to let
ck(n) = 22nTL-2+5ce-kZ/n + 22nn-5+5~,&-kz/~,
thereby replacing e-1c4/6n3 by 1 + 0 ( k4/n3 ) . The sum 1 k k4 eP k2/n is 0 ( n512 ) ,
as shown in exercise 30.
9.56 If k < n’/‘+’ we have ln(nk/nk) = -gk’/n + ik/n - ik3/n2 +
0 (n- 1+4E) by Stirling’s app roximation, hence
nk/nk = ePkzi2n(l + k/2n - $k3/(2n)2 + O(nP”4’)) .
Summing with the identity in exercise 30, and remembering to omit the term
for k = 0, gives -1 + 01~ + O:‘,’ - $G:“,’ + O(nP1/2+4’) = m - 5 +
O(n- 1/2+4e) .
9.57 Using the hini;, the given sum becomes J,” ueCU