Average Values of
Arithmetic Functions
Rose Wong
December 5, 2006
18.104
1
Outline
1. Introduction
2. Average number of representations of n
as a sum of two squares
3. Average number of representations of n
as a sum of three squares
4. Average number of primitive Pythagorean
triangles with hypotenuse n
5. Average number of divisors of n
2
Introduction
• Recall: Arithmetic Functions
f ( n) : Ν → C
• Examples:
Π (x)
d (n)
» Number of primes less than or equal to x
» Number of divisors of n
• The average of an arithmetic function is defined to be:
f (1)+ ... + f (n)
, n = 1,2,3,...
n
• In particular, we want to look at the average as nÆ ∞ .
f (1)+ ... + f (n)
lim
n →∞
n
3
Outline
1. Introduction
2. Average number of representations of
n as a sum of two squares
3. Average number of representations of n
as a sum of three squares
4. Average number of primitive Pythagorean
triangles with hypotenuse n
5. Average number of divisors of n
4
Average number of representations of n as a
sum of two squares
• Let r(n) be the number of representations
2
2
n
=
x
+
y
such that
• Theorem: The average number of
representations of a natural number as a
sum of two squares is π . That is
r (1)+ ... + r (n)
=π
lim
n →∞
n
5
Average number of representations of n as a
sum of two squares (continued)
•
•
•
Proof
R (n) = r (1)+ ... + r ( n)
Let
and r(n) be the number of
lattice points with integer
coordinates on the circle
x2 + y2 = n
R(n) = r (1)+ ... + r (n)
Hence, 1+R(n) is the number
of lattice points inside the
2
2
circle x + y ≤ n
Idea- Place a unit square on
each lattice point inside the
circle and count the number of
squares to find the area.
6
Average number of representations of n as a
sum of two squares (continued)
•
•
•
•
The total area of unit squares is 1+R(n)
⎛
1 ⎞
Upper bound of area is π ⎜⎜⎝ n + 2 ⎟⎟⎠
Lower bound of area is π ⎛⎜⎜ n − 1 ⎞⎟⎟
2⎠
⎝
2
2
So,
⎛
⎛
1 ⎞
1 ⎞
2
2
π ⎜⎜ n −
⎝
•
⎟⎟ < 1 + R (n) < π ⎜⎜ n +
2⎠
⎝
Using fact that π
2 <5
and that
πn − 6 n < R (n) < πn + 6 n
0<
π
2
⎟⎟
2⎠
−1< n
n = 1,2,3...
R(n) − πn < 6 n
R ( n)
6
−π <
n
n
R ( n)
→π
• Taking limit as nÆ ∞ , we have
n
7
Outline
1. Introduction
2. Average number of representations of n
as a sum of two squares
3. Average number of representations of
n as a sum of three squares
4. Average number of primitive Pythagorean
triangles with hypotenuse n
5. Average number of divisors of n
8
Average number of representations of n as a
sum of three squares
• Let f(n) be the number of representations
2
2
2
n
=
x
+
y
+
z
such that
• Theorem: The average number of
representations of a natural number as a
4
sum of three squares is 3 π .That is
f (1)+ ... + f (n) 4
lim
= π
n →∞
n
3
9
Average number of representations of n as a
sum of three squares (continued)
Proof
• Let
•
F ( n) = f (1)+ ... + f ( n)
Let f(n) be the number of lattice points with integer coordinates on
the sphere
2
2
2
x + y +z =n
•
Hence, 1+F(n) is the number of lattice points inside the sphere
x2 + y2 + z2 ≤ n
•
Idea- Place a unit cube on each lattice point inside the sphere and
count the number of cubes to find the volume. To estimate volume,
we find volume of an outer sphere that encloses all cubes and an
inner sphere that is inscribed inside the cubes.
10
Average number of representations of n as a
sum of three squares (continued)
•
•
•
The total area of unit cubes is
Radius of outer sphere is
Radius of inner sphere is
•
Comparing volumes, we get
•
Using fact that 2 3π < 11 and
that 3 π < 3 , and rearranging,
2
we get
•
Dividing by n
3/ 2
•
Taking limit as nÆ ∞, we
have
1 + F ( n)
3
2
3
n−
2
n+
3
4 ⎛
3⎞
4 ⎛
3⎞
⎟ < 1 + R ( n) < π ⎜ n +
⎟
π ⎜⎜ n −
⎟
⎜
3 ⎝
2 ⎠
3 ⎝
2 ⎟⎠
3
4
R(n) − πn 3 / 2 − 3π n + 1 < 11n − 3
3
R (n) 4
3π
1
11
3
−
π
−
+
<
−
3/ 2
n n3/ 2
n3/ 2 3
n n
R ( n)
4
→
π
3/ 2
3
n
11
Outline
1. Introduction
2. Average number of representations of n
as a sum of two squares
3. Average number of representations of n
as a sum of three squares
4. Average number of primitive
Pythagorean triangles with
hypotenuse n
5. Average number of divisors of n
12
Average number of primitive Pythagorean
triangles with hypotenuse n
• Let P(n) be the number of primitive Pythagorean
triangles with hypotenuse equal to n.
3
5
10
6
4
Primitive
8
Not Primitive
• Example: P(5)=1 and P(65)=2
32 + 4 2 = 52
33 2 + 56 2 = 63 2 + 16 2 = 65 2
• We want to look at:
P(1)+ ... + P(n)
=K
n →∞
n
lim
13
Average number of primitive Pythagorean
triangles with hypotenuse n (continued)
•
•
We want to examine the average of P(n) for large values of n.
Approach (computer): For each value of n from 1 to 1000, we generate
the average values and plot them on a graph. We want to look at the
behavior as n grows large.
Plot of
P(1)+ ... + P(n)
n
vs n
14
Average number of primitive Pythagorean
triangles with hypotenuse n (continued)
• From the figure, we see that the average values
(measured on the vertical axis) oscillate, but that the
average levels level off quickly.
• We can conjecture that
P (1)+ ... + P(n)
lim
≈ 0.159
n →∞
n
15
Outline
1. Introduction
2. Average number of representations of n
as a sum of two squares
3. Average number of representations of n
as a sum of three squares
4. Average number of primitive Pythagorean
triangles with hypotenuse n
5. Average number of divisors of n
16
Average number of divisors
• Let d(n) be the number of divisors of the natural number n.
• Example: d(8)=4
• Theorem: The average value of the number of divisors
of natural numbers grows like log n.
d (1)+ ... + d (n)
~ log n
n
17
Average number of divisors (continued)
Proof
• Let k be a fixed integer. Listing multiples of k less than or equal to n,
we have
⎡n⎤
k ,2k ,3k ,...⎢ ⎥ k
⎣k ⎦
•
⎡n⎤
There are ⎢⎣ k ⎥⎦ multiples, where [ ] denotes the floor function. Each
of these multiples contributes 1 to the sum d (1) + ... + d ( n)
•
Examining multiples of all integers
k≤n
⎡n⎤
∑
⎢ k ⎥ = d (1) + ... + d ( n)
k =1 ⎣ ⎦
n
18
Average number of divisors (continued)
⎡n⎤
∑
⎢ ⎥
d (1) + ... + d (n)
k =1 ⎣ k ⎦
lim
= lim
=1
n →∞
n → ∞ n log n
n log n
n
•
Now, we want to prove that
•
First we establish the relationship:
•
Summing over k gives:
n
⎡n⎤ n
−1< ⎢ ⎥ <
k
⎣k ⎦ k
⎛ n ⎞ n ⎡n⎤ n n
⎜ − 1⎟ < ∑ ⎢ ⎥ < ∑
∑
⎠ k =1 ⎣ k ⎦ k =1 k
k =1 ⎝ k
n
n
1
⎛ 1 1 ⎞ n ⎡n⎤
n∑ ⎜ − ⎟ < ∑ ⎢ ⎥ < n∑
n ⎠ k =1 ⎣ k ⎦
k =1 ⎝ k
k =1 k
n
•
Factoring out n gives
•
The first and last term are
rewritten in integral form
n
n⎛ 1
n⎛ 1 ⎞
1⎞
⎡n⎤
n ∫ ⎜ − ⎟dk < ∑ ⎢ ⎥ < n ∫ ⎜ ⎟dk
1
1
⎝k n⎠
⎝k⎠
k =1 ⎣ k ⎦
19
Average number of divisors (continued)
Integrating gives
⎡n⎤
n log n − n + 1 < ∑ ⎢ ⎥ < n log n
k =1 ⎣ k ⎦
•
So taking nÆ∞ , we have
⎡n⎤
∑
⎢ ⎥
k =1 ⎣ k ⎦
lim
=1
n → ∞ n log n
•
And so,
d (1)+ ... + d (n)
~ log n
n
n
•
n
20