Concrete mathematics : a foundation for computer science

576 ANSWERS TO EXERCISES
9.58 Let 0 < 8 6 1 and f(z) = e2xiro/( eZnir - 1). We have
when xmod 1 = 4;
when lyl 3 c.
Therefore /f(z)1 is bounded on the contour, and the integral is O(Mlmm).
The residue of 2nif(z)/zm at z = k # 0 is eznike/km; the residue at z = 0 is
the coefficient of 2-l in
e2niz0 2rriz
Zm+l (Bo + B1 T $- . . >
27riz
= &,(Wi +W+ +-.) ,
namely (2ti)“‘B,(O)/m!. Therefore the sum of residues inside the contour is
(27ri)m enim/2 COS (2nk6 -- nm/2)
m,B,(B) + 2F
kz=l
km
This equals the contour integral O(Mlpm), so it approaches zero as M -+ 00.
9.59 If F(x) is sufficiently well behaved, we have the general identity
x F(k + t) = t G(2rm.)eZRint ,
k n
where G(y) = ST,” eciyXF(x) dx. (This is “Poisson’s summation formula:
which can be found in standard texts such as Henrici (151, Theorem 10.6e].)
9.60 The stated formula is equivalent to
5 21
___-
+ 1024n3 32768n4
+ O(C5)
by exercise 5.22. Hence the result follows from exercises 6.64 and 9.44.
9.61 The idea is to make cr “almost” rational. Let ok = 22zk be the kth
partial quotient of 01, and let n = ;a,,,+, qm, where qm = K(al,. . . , a,) and
m is even. Then 0 < {q,,,K} < l/Q(al,...,a,+.,) < 1/(2n), and if we take
v = a,,,+1 /(4n) we get a discrepancy 3 :a,+, . If this were less than n’-’ we
would have
E
%+1 = WlAy),
but in fact a,+1 > 42,"