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Homemultiplicative function
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multiplicative function
In number theory, a multiplicative function is an arithmetic function $f\colon\mathbb{N}\to\mathbb{C}$ such that $f(1)=1$ and, for all $a,b\in\mathbb{N}$ with $\gcd(a,b)=1$, we have $f(ab)=f(a)f(b)$.
An arithmetic function $f(n)$ is said to be completely multiplicative if $f(1)=1$ and $f(ab)=f(a)f(b)$ holds for all positive integers $a$ and $b$, even when they are not relatively prime. In this case, the function is a homomorphism of monoids and, because of the fundamental theorem of arithmetic, is completely determined by its restriction to prime numbers. Every completely multiplicative function is multiplicative.
Outside of number theory, the term multiplicative is usually used for all functions with the property $f(ab)=f(a)f(b)$ for all arguments $a$ and $b$. This entry discusses number theoretic multiplicative functions.
Examples
Examples of multiplicative functions include many important functions in number theory, such as:

$\varphi(n)$: the Euler totient function (also denoted $\phi(n)$), counting the totatives of $n$;

$\mu(n)$: the Möbius function, which determines the parity of the prime factors of $n$ if $n$ is squarefree;

$\tau(n)$: the divisor function (also denoted $d(n)$), counting the positive divisors of $n$;

$\sigma(n)$: the sum of divisors function (also denoted $\sigma_{1}(n)$), summing the positive divisors of $n$;

$\sigma_{{k}}(n)$: the sum of the $k$th powers of all the positive divisors of $n$ for any complex number $k$ (typically a natural number);

$\hbox{id}(n)$: the identity function, defined by $\hbox{id}(n)=n$;

$\hbox{id}^{{k}}(n)$: the power functions, defined by $\hbox{id}^{{k}}(n)=n^{{k}}$ for any complex number $k$ (typically a natural number);

$1(n)$: the constant function, defined by $1(n)=1$;

$\varepsilon(n)$: the convolution identity function, defined by:
$\varepsilon(n)=\sum_{{dn}}\mu(d)=\begin{cases}1&\text{if }n=1\\ 0&\text{if }n\neq 1\end{cases}$ where $d$ runs through the positive divisors of $n$.
Properties
A multiplicative function is completely determined by its values at the powers of prime numbers, a consequence of the fundamental theorem of arithmetic. Thus, if $n$ is a product of powers of distinct prime numbers, say $n=p^{a}q^{b}\cdots$, then $f(n)=f(p^{a})f(q^{b})\cdots$. This property of multiplicative functions significantly reduces the need for computation, as in the following examples for $n=144=2^{4}\cdot 3^{2}$:
$\displaystyle\sigma(144)$  $\displaystyle=$  $\displaystyle\sigma_{{1}}(144)=\sigma_{{1}}(2^{4})\sigma_{{1}}(3^{2})=(1^{1}+2% ^{1}+4^{1}+8^{1}+16^{1})(1^{1}+3^{1}+9^{1})=31\cdot 13=403$  
$\displaystyle\sigma_{{2}}(144)$  $\displaystyle=$  $\displaystyle\sigma_{{2}}(2^{4})\sigma_{{2}}(3^{2})=(1^{2}+2^{2}+4^{2}+8^{2}+1% 6^{2})(1^{2}+3^{2}+9^{2})=341\cdot 91=31031$  
$\displaystyle\sigma_{{3}}(144)$  $\displaystyle=$  $\displaystyle\sigma_{{3}}(2^{{4}})\sigma_{{3}}(3^{{2}})=(1^{3}+2^{3}+4^{3}+8^{% 3}+16^{3})(1^{3}+3^{3}+9^{3})=4681\cdot 757=3543517$ 
Similarly, we have:
$\displaystyle\tau(144)$  $\displaystyle=$  $\displaystyle\tau(2^{4})\tau(3^{2})=(4+1)(2+1)=5\cdot 3=15$  
$\displaystyle\varphi(144)$  $\displaystyle=$  $\displaystyle\varphi(2^{4})\varphi(3^{2})=2^{3}(21)3^{1}(31)=8\cdot 1\cdot 3% \cdot 2=48$ 
Convolution
Recall that, if $f$ and $g$ are two arithmetic functions, one defines a new arithmetic function $f*g$, the Dirichlet convolution (or simply convolution) of $f$ and $g$, by
$(f*g)(n)=\sum_{{dn}}f(d)g\left({n\over d}\right),$ 
where the sum extends over all positive divisors $d$ of $n$. Some general properties of this operation with respect to multiplicative functions include (here the argument $n$ is omitted in all functions):

If both $f$ and $g$ are multiplicative, then so is $f*g$ (proven here);

$f*g=g*f$ (proven here);

$(f*g)*h=f*(g*h)$ (proven here);

$f*\varepsilon=\varepsilon*f=f$ (proven here);

If $f$ is multiplicative, there exists a multiplicative function $g$ such that $f*g=\varepsilon$ (proven here). In other words, every multiplicative function has a convolution inverse that is also multiplicative.
This shows that, with respect to convolution, the multiplicative functions form an abelian group with identity element $\varepsilon$. Relations among the multiplicative functions discussed above include:

$1*1=\tau$

$\hbox{id}*1=\sigma$

$\hbox{id}^{{k}}*1=\sigma^{{k}}$

$\phi*1=\hbox{id}$
Given a completely multiplicative function $f$, its convolution inverse is $f\mu$. See this entry for a proof.
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Attached Articles
proof that Euler $\varphi$ function is multiplicative by Wkbj79
criterion for a multiplicative function to be completely multiplicative by Wkbj79
elementary results about multiplicative functions and convolution by Wkbj79
composition of multiplicative functions by Wkbj79
Comments
What are the units in the monoid?
Why not mention, in passing, that the monoid of functions under Dirichlet convolution contains a submonoid, in fact a group, namely the functions that take at 1 a value different from 0 ?
Re: What are the units in the monoid?
This directly goes to monoids. I do not quite understand what to correct in entry "multiplicative function". I guess your message is just mediate reference. Let me know. Best regard.
Re: What are the units in the monoid?
>This shows that the multiplicative functions
>with convolution form a commutative monoid with
>identity element . relations among the
>multiplicative functions discussed above include
In reference to this. I often see the exclusion of the zero function from the multiplicative functions, so any multiplicative function F that is not identically zero must have F(1) = 1 (easy to show). Moreover, under Dirichlet convolution, a function G has a Dirichlet inverse G^1 whenever G(1) \not= 0, so that the monoid you mention, given appropriate modifications to the definition of a multiplicative functions, is in fact a group.
Re: What are the units in the monoid?
Oops, your definition of multiplicative already specifies f(1) = 1. This is equivalent to specifying that a multiplicative function cannot be identically zero. So ignore the previous comment concerning changing the definition of multiplicative.
Re: What are the units in the monoid?
Okay. Thank you anyway for your effort. We all miss something along the way, don't we. So I believe that a subject is explained enough understandable. Best regard.
Re: What are the units in the monoid?
Ok, I think I'm not being clear.
Change
> this shows that the multiplicative functions with the convolution form a commutative monoid
into
> his shows that the multiplicative functions with the convolution form a commutative group
And I'll be happy.
totient
Nice article. Maybe this is the object in which we should define "totient". Anyway, I wondered for years why Sylvester coined that strange word in 1882.
An arithmetic function f is called a "totient" if g*f=h for some two completely multiplicative functions g and h, where * is the convolution product.
Maybe the word "totient" is a hybrid of "total" (from "totally multiplicative") and "quotient", since we can indeed write f=h/g.
Re: totient
Yes, dear Sylvester, the Master of coining. For sure. Just mail me your propositions and I'll try to put them in. Best regards.
the tau function
Overall, I think that this entry is very informative and interesting. One comment that I'd like to make is that I have only seen the function that counts the positive divisors of a positive integer as the tau function. Again, a very good entry. :)