Math 331, Section 201—Homework #6
due in class Tuesday, April 8, 2003
Warm-Up Questions—do not hand in
Graham, Knuth, and Patashnik, Chapter 9, pp. 489–490, #2, #6, #7, #12, #13, #20
Homework Questions—turn in solutions to these problems
−2
I. Fix an integer k ≥ 2. Find an asymptotic formula for (kn
n ) with relative error n . If
ε is a positive number, which expression is larger when n is large, (kn)n or (n!)1+ε ?
2
4
4n
3n
−2
II. Find asymptotic formulas for (9n
n )( n ) and ( n ) , both with relative error n .
Which of these expressions is larger when n is large? Find real numbers α and
β, with β 6= 0, such that
lim
n→∞
4n
(9n
n )( n )
4
2
α
(3n
n) n
= β.
The last two problems involve the Möbius mu-function µ (n), which is defined recursively
for n ∈ N by
∑ [d\n]µ (d) = [n = 1].
d≥1
(The notation d\n means that d is a divisor of n; see page 102 of Graham, Knuth, and
Patashnik.) The first few values are
µ (1) = 1, µ (2) = −1, µ (3) = −1, µ (4) = 0, µ (5) = −1, µ (6) = 1, . . . .
An explicit formula for µ (n) is known in terms of the factorization of n, but for these
problems you will only need to use the recursive definition above and the fact (you don’t
have to prove it) that |µ (n)| ≤ 1 for all n ∈ N.
III. Fix a real number s > 1. Prove that
µ (n)
1
∑ ns = ζ (s) .
n≥1
(Hint: find an asymptotic formula for the partial sum ∑n≤ N
∑n≤ N n1s in class. Then analyze
µ (n)
1
∑ s
∑ s
n≤ N n
n≤ N n
µ (n)
ns
like we did for
in two ways, first by multiplying the sums together and simplifying, then by multiplying the asymptotic formulas for the sums together.)
IV. It is an oft-quoted “fact” that if two positive integers are chosen at random, the
probability that they have no common factor is π62 ; consider the following quantitative formulation of this statement. Fix a positive integer N. Let X and Y be two
independent random variables, each of which is uniform on the set {1, 2, . . . , N }.
Find an asymptotic formula (in terms of N) for the probability that X and Y have
no common factors. (Hint: notice that
[m and n have no common factors] = [gcd(m, n) = 1]
=
∑ [d\ gcd(m, n)]µ (d)
d≥1
=
∑ [d\m][d\n]µ (d)
d≥1
by the recursive definition of µ (n) above and the properties of gcd. Form a sum
of this expression over all possible values X = m and Y = n, then make the sum
over d the outermost sum.)