Using Zeta Zeros to Tally Sigma Times Tau

​
start x axis at (from 0 to 100)
0
length of x axis (from 10 to 50)
30
pairs of zeta zeros (from 0 to 100)
0
always show estimate of T(x) using no zeta zeros
display T(x) values in a 'tooltip'
In number theory, the number of divisors of an integer
n
is usually denoted by
τ(n)
. (
τ
is the lowercase Greek letter tau.) For example, 4 has three divisors (namely, 1, 2, and 4), so
τ(4)=3
. The sum of the divisors of
n
is denoted by
σ(n)
. (
σ
is the lowercase Greek letter sigma.) So,
σ(4)=1+2+4=7
.
Suppose
x≥1
. The sum of
σ(n)τ(n)
for
1≤n≤x
is an irregular step function that jumps up at every integer
x
. For example, for
x=3
, this sum is
σ(1)τ(1)+σ(2)τ(2)+σ(3)τ(3)
=
1×1+3×2+4×2=15
. For
x=4
, this sum is
15+σ(4)τ(4)
=
15+7×3=36
.
This Demonstration shows how we can approximate this step function with a sum that involves zeros of the Riemann zeta (
ζ
) function.

Details

Snapshot 1: the graphs of the step function
T(x)=σ(1)τ(1)+σ(2)τ(2)+σ(3)τ(3)+…+σ(⌊x⌋)τ(⌊x⌋)
and the formula using no zeta zeros
Snapshot 2: the graphs of
T(x)
and the formula using 100 pairs of zeta zeros
After you use the slider to choose
N
(the number of pairs of zeta zeros to use), this Demonstration uses the following formula to calculate
T(x)
:
(1)
T(x)≃
a
1
+
a
2
x+
a
3
2
x
+
a
4
2
x
log(x)+2Re
N
∑
k=1
ρ
k
+1
2
x
ζ
ρ
k
+1
2

2
ζ
ρ
k
-1
2

(
ρ
k
+1)
′
ζ
(
ρ
k
)
+
∞
∑
k=1
(-2k+1)
2
x
ζ
-2k+1
2

2
ζ
-2k-1
2

(-2k+1)
′
ζ
(-2k)
.
In this formula,
a
1
=-
1
48
≃0.020833
,
a
2
=
1
2
=0.5
,
a
3
=
24
2
π
ζ(3)
′
ζ
(2)+
4
π
(-ζ(3)+4ζ(3)-4
′
ζ
(3))
144
2
ζ(3)
≃-0.175402
,
and
a
4
=
4
π
72ζ(3)
≃1.12549
,
where
γ≃0.577216
is Euler's constant.
In equation (1),
ρ
k
is the
th
k
complex zero of the Riemann zeta function with positive real part. The first three complex zeros of the zeta function are approximately
1/2+14.135i
,
1/2+21.022i
, and
1/2+25.011i
. These zeros occur in conjugate pairs, so if
a+bi
is a zero, then so is
a-bi
.
If you use the slider to choose, say,
N=1
(one pair of zeta zeros), then the first sum in equation (1) adds the two terms that correspond to the first pair of conjugate zeros,
1/2+14.135i
and
1/2-14.135i
. These terms are conjugates of each other. When these terms are added, their imaginary parts cancel while their real parts add. So, the
2Re(…)
applied to the first sum is merely an efficient way to combine the two terms for each pair of zeta zeros.
Notice that the second sum has the same form as the first, except that the second sum extends over the real zeros of the zeta function, namely,
s=-2k=-2,-4,-6,…
. However, the second sum is too small to visibly affect the graphs, so this sum is not computed here.
If you plot a graph using no zeta zeros, then the graph is computed with only the terms
a
1
through
a
4
.
​
Where Does Equation (1) Come From?
To prove equation (1), we start with the following identity, which holds for
s>2
(see[1], equation D-58, and[2], Theorem 305):
(2)
∞
∑
n=1
σ(n)τ(n)
s
n
=
2
ζ(s)
2
ζ(s-1)
ζ(2s-1)
.
Perron's formula (see reference[3]) takes an identity like equation (2) and gives a formula for the sum of the numerators as a function of
x
, in this case,
x
∑
n=1
σ(n)τ(n)
.
When we apply Perron's formula to equation (2), we get equation (1). To apply Perron's formula, we integrate this integrand
2
ζ(s)
2
ζ(s-1)
ζ(2s-1)
s
x
s
around a contour in the complex plane. Each part of equation (1) is the residue at one of the poles of this integrand. The residue at the pole at
s=0
is
a
1
. At
s=1
,
ζ(s)
has a pole (of order 1), so the integrand has a pole of order 2 at
s=1
due to the
2
ζ(s)
. The residue at
s=1
is
a
2
x
. Similarly, the integrand also has a pole of order 2 at
s=2
. The expression
a
3
2
x
+
a
4
2
x
log(x)
is the residue at
s=2
.
Mathematica can compute these residues. For example, this calculation Residue[Zeta[s]^2 Zeta[s-1]^2 / Zeta[2 s - 1] x^s/s, {s, 2}] gives
a
3
2
x
+
a
4
2
x
log(x)
, where
a
3
and
a
4
have the values given above.
Finally, the integrand has a pole at each complex zero of
ζ(2s-1)=0
. The first sum in equation (1) is just the sum of the residues at these complex zeros of zeta. Each complex zero gives rise to one term in the sum. (
ρ
k
is the
th
k
complex zero of
ζ(s)=0
, so
ρ
k
+1
2
is the
th
k
complex zero of
ζ(2s-1)=0
).
In the same way, the second sum arises from the real zeros (
-2k=-2,-4,-6,…
) of
ζ(2s-1)=0
. (
-2k
is the
th
k
real zero of
ζ(s)=0
, so
-2k+1
2
is the
th
k
real zero of
ζ(2s-1)=0
).
The Demonstration "Using Zeta Zeros to Tally Squarefree Divisors" gives additional details on Perron's formula.

References

[1] H. W. Gould and T. Shonhiwa, "A Catalog of Interesting Dirichlet Series."
[2] G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, 4th ed., Oxford: Oxford University Press, 1965, p. 256.
[3] H. L. Montgomery and R. C. Vaughan, Multiplicative Number Theory: I. Classical Theory, Cambridge: Cambridge University Press, 2007, p. 397.

External Links

Riemann Zeta Function Zeros (Wolfram MathWorld)
Divisor Function (Wolfram MathWorld)
How the Zeros of the Zeta Function Predict the Distribution of Primes
Using Zeta Zeros to Compute the Chebyshev Psi Function
Using Zeta Zeros to Compute the Mertens Function
Using Zeta Zeros to Compute a Summatory Liouville Function
Using Zeta Zeros to Count the Squarefree Integers
Using Zeta Zeros to Tally the Euler Phi Function
Using Zeta Zeros to Tally Squarefree Divisors
Using Zeta Zeros to Compute Sigma Sums
Using Zeta Zeros to Compute Tau Sums

Permanent Citation

Robert Baillie
​
​"Using Zeta Zeros to Tally Sigma Times Tau"​
​http://demonstrations.wolfram.com/UsingZetaZerosToTallySigmaTimesTau/​
​Wolfram Demonstrations Project​
​Published: March 7, 2011