ASYMPTOTIC ANALYSIS OF ARITHMETIC FUNCTIONS
(MA2316, ELEVENTH WEEK)
VLADIMIR DOTSENKO
Theorem 1. We have
Mτ (n) = ln n + O(1).
Proof. Note that
X
XX
nMτ (n) =
τ (k) =
1=
k≤n
=
k≤n ab=n
X X
1+
√
a≤ n b≤n/a
X
1=
a,b≤n : ab≤n
X X
X
1−
√
a≤ n b≤n/a
1=2
√
√
a≤ n,b≤ n
X jnk
√
− b nc2 =
a
√
a≤ n
X n
√
√
=2
+ O( n) + O(n) = 2(n ln n + O(n)) + O(n) = n ln n + O(n).
√ a
a≤ n
Here, we use two estimates for the sum of inverse integers:
Z n
n
X
1
dx
≤1+
= 1 + ln n
k
x
1
k=1
and
Z n+1
n
X
dx
1
≥
= ln(n + 1),
k
x
1
k=1
which together imply that
n
X
1
= ln n + O(1).
k
k=1
Dividing nMτ (n) = n ln n + O(n) by n, we get the required statement.
Theorem 2. We have
3n
+ O(ln n).
π2
P
Proof. Let us examine the function Φ(n) = nMφ (n) =
φ(k). Extend it to all nonnegative real
Mφ (n) =
k≤n
numbers, putting Φ(x) = Φ(bxc) when x is not an integer. Let us first show that
X
bxcbx + 1c
Φ(x/d) =
.
2
d≥1
Indeed, the right hand side is equal to the number of pairs (m, n) with 0 ≤ m < n ≤ x. The
number of such pairs with gcd(m, n) = d is equal to Φ(x/d), since factoring out d reduces counting
these pairs to counting pairs 0 ≤ m0 ≤ n0 ≤ x/d, and for each n0 ≤ x/d the number of allowed m0
is φ(n0 ).
To make use of the formula we just proved, we shall invoke the following generalisation of Möbius
inversion which you will prove in the next tutorial:
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Suppose f, g are P
two functions with complex values defined on [0, +∞), and assume
in addition that k,d≥1 |f (x/(kd))| < +∞ (for instance, that happens when f (x) =
0 for x < 1). Show that if
X
g(x) =
f (x/d),
d≥1
then we have
f (x) =
X
µ(d)g(x/d).
d≥1
Thanks to this statement, we have
nMφ (n) =
1X
µ(k)bn/kcb1 + n/kc.
2
k≥1
Let us play around with this formula a little bit. Clearly,
bn/kcb1 + n/kc = (n/k + O(1))(n/k + O(1)) = (n2 /k 2 + O(n/k)),
so
n
nMφ (n) = Φ(n) =
1X
µ(k)(n2 /k 2 + O(n/k)) =
2
k=1
=
n
X
n2 X µ(k) n2 X µ(k)
−
+
n
(O(1/k)) =
2
k2
2
k2
k≥1
=
k>n
n2 X
2
k≥1
k=1
µ(k)
3n2
+
O(n)
+
O(n
ln
n)
=
+ O(n ln n).
k2
π2
Here we use obvious estimates
X µ(k) X 1
X
1
1
≤
<
=
2
2
k k
k(k − 1)
n
k>n
and
k>n
k>n
n
X
1
= O(ln n)
k
k=1
which we proved above already, as well as the formula
X µ(k)
Y 1
1
=
1− 2 = Q 2
k
p
p prime
1−
k≥1
p prime
1
p2
1
−1 = P
m≥1
1
m2
=
1
π2
6
=
6
.
π2
Dividing by n, we get
Mφ (n) =
3n
+ O(ln n),
π2
as required.
This theorem easily implies that the “probability” for two randomly chosen numbers to be
coprime is π62 . (If we agree that the probability in question is the (limit as N → ∞ of the)
proportion of pairs (m, n) with coprime m, n among all pairs m, n with 0 ≤ m, n ≤ N ).
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