where the first sum is over positive divisors of \(n\) and the second sum is over pairs \((d,e)\) of positive integers whose product is \(n\text{.}\)
Example2.3.2.
\(\phi * 1 = N \text{:}\) Here \(\phi\) is the Euler totient function, \(1\) is the function \(1(n) = 1\) for all \(n\text{,}\) and \(N\) is the function \(N(n) = n\) for all \(n\text{.}\) We have shown that \(\sum_{d \mid n} \phi(d) = n\text{.}\) We can rewrite this as \(\sum_{d \mid n} \phi(d) 1(n/d) = N(n)\text{,}\) where the left hand side is \((\phi * 1)(n)\text{.}\) This shows that \(\phi * 1 = N\) as claimed.
Other notable operations on arithmetic functions include:
The sum: \((f+g)(n) = f(n)+g(n)\)
The pointwise product: \((fg)(n) = f(n) g(n)\)
The composition: \((f \circ g)(n) = f(g(n))\) (assuming \(g\) has values in \(\bbN\text{,}\) the domain of \(f\))
The Cauchy product, defined by \(\sum_{d=0}^n f(d)g(n-d) \text{.}\) (I’m not sure if there is a common notation for the Cauchy product.)
Theorem2.3.3.
The Dirichlet product is associative, commutative, and distributes over the sum. The function \(e\) (defined by \(e(n)=1\) if \(n=1\text{,}\) otherwise \(e(n)=0\)) is an identity for the Dirichlet product, that is, \(f*e = f\) for any arithmetic function \(f\text{.}\)
since \(e(n/d) = 0\) for all \(d\) other than \(d=n\text{.}\) So \(f*e = f\text{.}\)
Theorem2.3.4.
Let \(f\) be an arithmetic function. There exists a Dirichlet inverse \(g\) such that \(f*g = e\) if and only if \(f(1) \neq 0\text{.}\) If \(f(1) \neq 0\text{,}\) then \(g\) is defined recursively by \(g(1) = \frac{1}{f(1)}\) and for \(n \gt 1\text{,}\)
If \(f*g = e\) then \(f(1)g(1) = e(1) = 1\text{,}\) so \(f(1) \neq 0\text{.}\) The converse direction is left as an exercise.
Subsection2.3.2Dirichlet product of multiplicative functions
Theorem2.3.5.
If \(f,g\) are both multiplicative, then so is \(f * g \text{.}\)
Proof.
Suppose \((n_1,n_2) = 1\text{.}\) Observe that if \(de = n_1 n_2\) then there exist factorizations \(d = d_1 d_2\text{,}\)\(e = e_1 e_2\text{,}\) such that \(d_1 e_1 = n_1\) and \(d_2 e_2 = n_2\text{.}\) (Simply, \(d_1\) is the product of prime factors that are in \(d\) and \(n_1\text{,}\)\(d_2\) is the product of prime factors that are in \(d\) and \(n_2\text{,}\) and so on.) Observe that in this factorization, \((d_1,d_2) = (e_1,e_2) = 1\text{.}\)
For all \(s\text{,}\)\(\sigma_s\) is multiplicative.
Proof.
Let \(g(n) = n^s\text{,}\) so \(g\) is multiplicative (in fact, completely multiplicative). Then \(\sigma_s = g * 1\) where \(1\) denotes the arithmetic function \(1(n) = 1\) for all \(n\text{.}\) Here \(1\) is completely multiplicative. Since \(g\) and \(1\) are multiplicative, so is \(\sigma_s\text{.}\)
Remark2.3.7.
Even if both \(f\) and \(g\) are completely multiplicative, in general \(f*g\) is only multiplicative, not necessarily completely multiplicative. For example let \(f = g = 1\text{,}\) then \((f*g)(n) = \sum_{d \mid n} 1 = d(n) = \sigma_0(n)\text{,}\) the number of positive divisors of \(n\text{.}\) This is multiplicative but not completely multiplicative, for example \(d(4) = 3 \neq d(2)^2\text{.}\)
Corollary2.3.8.
Let \(f\) be a multiplicative function. For all \(n \geq 1\) we have
The left hand side is \((f * 1)(n)\text{.}\) Since \(f\) and \(1\) are multiplicative, then so is \(f * 1\text{.}\) Let \(n\) have prime factorization \(n = p_1^{a_1} \dotsb p_k^{a_k}\text{.}\) It follows that
Let \(f,g\) be arithmetic functions. If \(g\) and \(f*g\) are multiplicative, then so is \(f\text{.}\)
Proof.
First of all we have \((f*g)(1) = 1\) and \(g(1) = 1\) since \(f*g\) and \(g\)are multiplicative. By definition of Dirichlet product, \((f*g)(1) = f(1)g(1)\text{.}\) Therefore \(f(1) = 1\text{.}\) We have \(f(1 \cdot 1) = 1\text{,}\) that is, \(f(mn) = 1\) for \(m=n=1\text{.}\)
We will prove the following statement by induction on \(N\text{:}\)\(f(mn) = f(m)f(n)\) for all \(m,n\) such that \((m,n)=1\) and \(mn \leq N\text{.}\) We have seen that this holds for \(N=1\text{.}\) Suppose inductively that it holds for \(N-1\text{,}\) and suppose \(m,n\) are positive integers such that \((m,n)=1\) and \(mn=N\text{.}\) Since \(f*g\) is multiplicative we have
\begin{align*}
(f*g)(mn) \amp = \sum_{d \mid mn} f(d) g(\frac{mn}{d}) \\
\amp = \sum_{d \mid m, e \mid n} f(de) g(\frac{mn}{de}) \\
\amp = \sum_{d \mid m, e \mid n, de \lt mn} f(de) g(\frac{mn}{de}) + f(mn)g(1) \\
\amp = \sum_{d \mid m, e \mid n, de \lt mn} f(d)f(e)g(\frac{m}{d})g(\frac{n}{d}) + f(mn) \\
\amp = \left( \sum_{d \mid m} f(d)g(\frac{m}{d}) \right)
\left( \sum_{e \mid n} f(e)g(\frac{n}{e}) \right) - f(m)f(n) + f(mn) \\
\amp = (f*g)(m) (f*g)(n) - f(m)f(n) + f(mn) \\
\amp = (f*g)(mn) - f(m)f(n) + f(mn) \text{.}
\end{align*}
Here we use that \(f(de) = f(d)f(e)\) for \(de \lt mn\text{,}\) where \((d,e)=1\) since \(d \mid m\) and \(e \mid n\text{;}\) we use that \((m/d,n/e) = 1\) and that \(g\) is multiplicative; at the end we use that \(f*g\) is multiplicative.
It follows that \(f(mn) = f(m)f(n)\text{.}\)
Corollary2.3.10.
If \(f\) is multiplicative then its Dirichlet inverse \(f^{-1}\) is also multiplicative.
Proof.
Both \(f\) and \(f*f^{-1} = e\) are multiplicative.
Corollary2.3.11.
The set of arithmetic functions \(f\) with \(f(1) \neq 0\) is an abelian group under Dirichlet product. The subset consisting of multiplicative functions is a subgroup.
Subsection2.3.3Explanation of Dirichlet product
Definition2.3.12.
A power series centered at \(c\) is an infinite sum
Again, the \(a_n\) are coefficients of the Dirichlet series. The sum defines a function of \(s\) provided that it converges.
Multiplication of power series corresponds to the Cauchy product of coefficients, and multiplication of Dirichlet series corresponds to the Dirichlet product of coefficients.
This shows that the product of power series with sequences of coefficients \((a_i)\) and \((b_j)\) is given by the power series with coefficients given by the Cauchy product of \((a_i)\) and \((b_j)\text{.}\)
Similarly, the Dirichlet product corresponds to multiplication of Dirichlet series. Let us multiply two Dirichlet series:
