In case is of the form
for some arithmetic function , one has,
| | (1) |
Generalized identities of the previous form are found here. This identity often provides a practical way to calculate the mean value in terms of the Riemann zeta function. This is illustrated in the following example.
The density of the kth-power-free integers in ℕ
For an integer the set of kth-power-free integers is
We calculate the natural density of these numbers in ℕ, that is, the average value of , denoted by , in terms of the zeta function.
The function is multiplicative, and since it is bounded by 1, its Dirichlet series converges absolutely in the half-plane , and there has Euler product
By the Möbius inversion formula, we get
where stands for the Möbius function. Equivalently,
where
and hence,
By comparing the coefficients, we get
Using (1), we get
We conclude that,
where for this we used the relation
which follows from the Möbius inversion formula.
In particular, the density of the square-free integers is .
Visibility of lattice points
We say that two lattice points are visible from one another if there is no lattice point on the open line segment joining them.
Now, if gcd(a, b) = d > 1, then writing a = da2, b = db2 one observes that the point (a2, b2) is on the line segment which joins (0,0) to (a, b) and hence (a, b) is not visible from the origin. Thus (a, b) is visible from the origin implies that (a, b) = 1. Conversely, it is also easy to see that gcd(a, b) = 1 implies that there is no other integer lattice point in the segment joining (0,0) to (a,b).
Thus, (a, b) is visible from (0,0) if and only if gcd(a, b) = 1.
Notice that is the probability of a random point on the square :\max(|r|,|s|)=n\}}
to be visible from the origin.
Thus, one can show that the natural density of the points which are visible from the origin is given by the average,
is also the natural density of the square-free numbers in ℕ. In fact, this is not a coincidence. Consider the k-dimensional lattice, . The natural density of the points which are visible from the origin is , which is also the natural density of the k-th free integers in ℕ.
Definition
Let h(x) be a function on the set of monic polynomials over Fq. For we define
This is the mean value (average value) of h on the set of monic polynomials of degree n. We say that g(n) is an average order of h if
as n tends to infinity.
In cases where the limit,
exists, it is said that h has a mean value (average value) c.
Zeta function and Dirichlet series in Fq[X]
Let Fq[X] = A be the ring of polynomials over the finite field Fq.
Let h be a polynomial arithmetic function (i.e. a function on set of monic polynomials over A). Its corresponding Dirichlet series define to be
where for , set if , and otherwise.
The polynomial zeta function is then
Similar to the situation in N, every Dirichlet series of a multiplicative function h has a product representation (Euler product):
where the product runs over all monic irreducible polynomials P.
For example, the product representation of the zeta function is as for the integers: .
Unlike the classical zeta function, is a simple rational function:
In a similar way, If f and g are two polynomial arithmetic functions, one defines f * g, the Dirichlet convolution of f and g, by
where the sum extends over all monic divisors d of m, or equivalently over all pairs (a, b) of monic polynomials whose product is m. The identity still holds. Thus, like in the elementary theory, the polynomial Dirichlet series and the zeta function has a connection with the notion of mean values in the context of polynomials. The following examples illustrate it.
Examples
Polynomial Divisor functions
In Fq[X], we define
We will compute for .
First, notice that
where and .
Therefore,
Substitute we get,
and by Cauchy product we get,
Finally we get that,
Notice that
Thus, if we set then the above result reads
which resembles the analogous result for the integers:
Polynomial von Mangoldt function
The Polynomial von Mangoldt function is defined by:
where the logarithm is taken on the basis of q.
Proposition. The mean value of is exactly 1.
Proof.
Let m be a monic polynomial, and let be the prime decomposition of m.
We have,
Hence,
and we get that,
Now,
Thus,
We got that:
Now,
Hence,
and by dividing by we get that,