An important binary operation on arithmetic functions is the Dirichlet convolution: for two artithmetic functions *f*(*n*) and *g*(*n*), the Dirichlet convolution gives a new arithmetic function, denoted *f***g*, defined by

.

Noting, first of all, that if *d* is a divisor of *n*, then *n*/*d* is also a divisor, we see that

,

and so the Dirichlet convolution is commutative.

Another property (but more complicated to prove, so I present without proof) is that the Dirichlet convolution is also associative: *f**(*g***h*)=(*f***g*)**h*.

Now, suppose that *f* and *g* are both multiplicative functions. Then let us consider (*f***g*)(*mn*), where m and n are coprime:

. Let *d*_{1} be any divisor of *m*, and *d*_{2} be any divisor of *n*. Then *d*_{1}*d*_{1} divides *mn*. Further, since gcd(*m*,*n*)=1, any divisor *d* of *mn* may be uniquely expressed as such a product *d*_{1}*d*_{2}. This allows us to rewrite the sum in the Dirichlet convolution above:

,

and, since gcd(*d*_{1},*d*_{2})=1, and thus (since respective factors of coprime integers must also be coprime), we apply the fact that *f* and *g* are both multiplicative:

,

and so the Dirichlet convolution of multiplicative functions is also multiplicative. Since a multiplicative function can be defined entirely by its values on the powers of primes, this makes finding the Dirichlet convolution of multiplicative functions much easier.

For example, consider the Dirichlet convolution of the Möbius function *μ*(*n*), and the multiplicative function 2^{ω(n)} appearing in this post. Since both are multiplicative, *μ**2^{ω} will be multiplicative, so we need only examine its values over the primes.

Now, from the definition of the Dirichlet convolution,

.

Now, for *p* prime, the divisors of *p ^{k}* are just 1,

*p*,

*p*

^{2}, …,

*p*. Thus,

^{k}.

Now, we recall that for powers of primes,

and

.

Thus, we see that for

*k*=0, we get

,

as we expect of a multiplicative function; for

*k*=1:

,

Now, for

*k*≥2, we see that any terms in the sum with

*i*≥2 will be zero, as the Möbius function for

*p*is zero for i≥2. Similarly, for

^{i}*k*≥2, both and both equal 2, since both

*k*and

*k*-1 are greater than zero.

Thus, we find that for

*k*≥2,

Thus,

*μ**2

^{ω}is the multiplicative function with

,

and so we see that

Similarly, we can see that the Dirichlet convolution of the constant function 1(

*n*) with itself is

,

the divisor function, and the Dirichlet convolution of the constant function 1(

*n*) with the identity function

*id*(

*n*) is

,

the sum-of-divisors function.

Lastly, consider the “unit function”

*ε*(n) defined by

. We see that this function is completely multiplicative; for any positive integers

*m*and

*n*.

Now, consider the Dirichlet convolution of this function with any arithmetic function

*f*(

*n*); all terms of the sum will be zero except that where

*d*=1, and thus

*n*/

*d*=

*n*:

.

Thus, we see that the unit function serves as the identity element for Dirichlet convolution.

Tags: Dirichlet Convolution, Divisors, Math, Monday Math, Multiplicative Function

October 12, 2009 at 1:16 am |

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