Concrete mathematics : a foundation for computer science

570 ANSWERS TO EXERCIS:ES
9.30 Let g(x) = xLePxL and. f(x) = g(x/fi). Then n “’ ,Yk>O k’ePkz”’ is
,
.I
Oc’ h&4)
cc f(x) dx - f %‘kP”(q - (-1 )-I ,fl"'(x) dx
0 k=l k! 0
= n l/2 g(x) dx - c E!Lnlk~l)i2gik-l1(0) + 0(~-m/2).
k=, k!
Since g(x) = x1 - x2+‘/l ! + x4 ‘l/2! - x6+‘/3! +. . , the derivatives g imi (x) obey
a simple pattern, and the answer is
1,it+l)/2 r 1 - + ’ -
( >
Bt+l b+3np’ Bt+6 2
2 2 (1+1)!0! + (l+3)!1! - (1+5)!2! +Obp3)
9.31 The somewhat surprising identity l/(cmmmk + cm) + l/(~"'+~ + cm) =
1 /cm makes the terms for 0 < k 6 2m sum to (m + +)/cm. The remaining
terms are
1 1
=- C2m+l _ C2m - C3m+2 _ C3m +... )
and this series can be truncated at any desired point, with an error not exceeding
the first omitted term.
9.32 H:) = x2/6 - l/n + O(nP2) by Euler’s summation formula, since we
know the constant; and H, is given by (9.89). So the answer is The world’s top
ney+nL’6 1 - in-’ + O(n-‘)) .
(
9.33 Wehavenk/n’= l-k.(k-l)nP’+~k2(k--l)2n~2+0(k6nP3); dividing
by k! and summing over k 3 0 yields e - en-’ $- I en- 2 + 0 ( nP3 ) .
9.34 A = ey; B = 0; C = -.ie’; D = ieY(l -y); E = :eY; F = &eY(3v+l).
9.35 Since l/k(lnk+ O(l]) = l/kink+ O(l/k(logk)2), the given sum
is Et==, 1 /kink + 0( 1). The remaining sum is In Inn + 0( 1) by Euler's
summation formula.
9.36 This works out beautifully with Euler’s summation formula:
dx
+L--
1 n B2 -2x n
+ O(nm5)
n2 + x2 2 n2 + x2 o +?(n2+x2)2 o
three constants,
(e, n, y), all appear
in this answer.