Completely multiplicative function: Difference between revisions

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{{Short description|Arithmetic function}}

In [[number theory]], functions of [[positive integer]]s which respect products are important and are called '''completely multiplicative functions''' or '''totally multiplicative functions'''. A weaker condition is also important, respecting only products of [[coprime]] numbers, and such functions are called [[multiplicative function]]s. Outside of number theory, the term "multiplicative function" is often taken to be synonymous with "completely multiplicative function" as defined in this article.

==Definition==

A '''completely multiplicative function''' (or '''totally multiplicative function''') is an [[arithmetic function]] (that is, a function whose [[Domain (mathematics)of a function|___domain]] is the [[natural number]]s), such that ''f''(1) = 1 and ''f''(''ab'') = ''f''(''a'') ''f''(''b'') holds ''for all'' positive integers ''a'' and ''b''.<ref>{{cite book|last=Apostol|first=Tom|title=Introduction to Analytic Number Theory|year=1976|publisher=Springer|isbn=0-387-90163-9|pages=[https://archive.org/details/introductiontoan00apos_0/page/30 30]|url-access=registration|url=https://archive.org/details/introductiontoan00apos_0/page/30}}</ref>

In logic notation: <math>f(1) = 1</math> and <math>\forall a, b \in \text{___domain}(f), f(ab) = f(a)f(b)</math>.

Without the requirement that ''f''(1) = 1, one could still have ''f''(1) = 0, but then ''f''(''a'') = 0 for all positive integers ''a'', so this is not a very strong restriction.

Without the requirement that ''f''(1) = 1, one could still have ''f''(1) = 0, but then ''f''(''a'') = 0 for all positive integers ''a'', so this is not a very strong restriction. If one did not fix <math>f(1) = 1</math>, one can see that both <math>0</math> and <math>1</math> are possibilities for the value of <math>f(1)</math> in the following way:

<math display="block">

\begin{align}

f(1) = f(1 \cdot 1) &\iff f(1) = f(1)f(1) \\

&\iff f(1) = f(1)^2 \\

&\iff f(1)^2 - f(1) = 0 \\

&\iff f(1)\left(f(1) - 1\right) = 0 \\

&\iff f(1) = 0 \lor f(1) = 1.

\end{align} </math>

The definition above can be rephrased using the language of algebra: A completely multiplicative function is a [[homomorphism]] from the [[monoid]] <math>(\mathbb Z^+,\cdot)</math> (that is, the positive integers under multiplication) to some other monoid.

==Examples==

The easiest example of a completely multiplicative function is a [[monomial]] with leading coefficient 1: For any particular positive integer ''n'', define ''f''(''a'') = ''a''^''n''. Then ''f''(''bc'') = (''bc'')^''n'' = ''b''^''n''''c''^''n'' = ''f''(''b'')''f''(''c''), and ''f''(1) = 1^''n'' = 1.

The [[Liouville function]] is a non-trivial example of a completely multiplicative function as are [[Dirichlet character]]s, the [[Jacobi symbol]] and the [[Legendre symbol]].

==Properties==

A completely multiplicative function is completely determined by its values at the prime numbers, a consequence of the [[fundamental theorem of arithmetic]]. Thus, if ''n'' is a product of powers of distinct primes, say ''n'' = ''p''^''a'' ''q''^''b'' ..., then ''f''(''n'') = ''f''(''p'')^''a'' ''f''(''q'')^''b'' ...

While the [[Dirichlet convolution]] of two multiplicative functions is multiplicative, the [[Dirichlet convolution]] of two completely multiplicative functions need not be completely multiplicative. Arithmetic functions which can be written as the Dirichlet convolution of two completely multiplicative functions are said to be quadratics or specially multiplicative multiplicative functions. They are rational arithmetic functions of order (2, 0) and obey the Busche-Ramanujan identity.

There are a variety of statements about a function which are equivalent to it being completely multiplicative. For example, if a function ''f'' is multiplicative then it is completely multiplicative if and only if theits [[Dirichlet inverse]] is <math>\mu\cdot f</math> where <math>\mu</math> is the [[MobiusMöbius function]].<ref>Apostol, p. 36</ref>

Completely multiplicative functions also satisfy a pseudo-associativedistributive law. If ''f'' is completely multiplicative then

<math>f \cdot (g*h)=(f \cdot g)*(f \cdot h)</math>

where ''*'' represents the [[Dirichlet product]] and <math>\cdot</math> represents [[Hadamard product (matrices)|pointwise multiplication]].<ref>Apostol pg. 49</ref>. One consequence of this is that for any completely multiplicative function ''f'' one has

<math>f*f = \tau \cdot f.</math>

which can be deduced from the above by putting both <math>g = h = 1</math>, where <math>1(n) = 1</math> is the [[multiplicative function#Examples|constant function]].

Here <math> \tau</math> is the [[divisor function]].

===Proof of pseudo-associativedistributive property ===

:<math>

<math> f \cdot \left(g*h \right)(n) = f(n) \sum_{d|n} g(d) h \left( \frac{n}{d} \right) </math>

\begin{align}

<math> f \cdot \left(g*h \right)(n) &= f(n) \cdot \sum_{d|n} g(d) h \left( \frac{n}{d} \right) </math>= \\

<math> f \cdot \left(g*h \right)(n) &= \sum_{d|n} f(n) \cdot (g(d) h \left( \frac{n}{d} \right)) </math>= \\

<math> f \cdot \left(g*h \right)(n) &= \sum_{d|n} (f(d) f \left( \frac{n}{d} \right)) \cdot (g(d) h \left( \frac{n}{d} \right)) </math>\text{ (since ''} f'' \text{ is completely multiplicative) } = \\

<math> f \cdot \left(g*h \right)(n) &= \sum_{d|n} (f(d) g(d)) \cdot (f \left( \frac{n}{d} \right) h \left( \frac{n}{d} \right)) </math>\\

&= (f \cdot g)*(f \cdot h).

\end{align}

</math>

===Dirichlet series===

<math> f \cdot \left(g*h \right)(n) = \sum_{d|n} f(n) g(d) h \left( \frac{n}{d} \right) </math>

The L-function of completely (or totally) [[multiplicative function|multiplicative]] [[Dirichlet series]] <math>a(n)</math> satisfies

:<math>L(s,a)=\sum^\infty_{n=1}\frac{a(n)}{n^s}=\prod_p\biggl(1-\frac{a(p)}{p^s}\biggr)^{-1},</math>

which means that the sum all over the natural numbers is equal to the product all over the prime numbers.

==See also==

<math> f \cdot \left(g*h \right)(n) = \sum_{d|n} f(d) f \left( \frac{n}{d} \right) g(d) h \left( \frac{n}{d} \right) </math> (since ''f'' is completely multiplicative)

*[[Arithmetic function]]

<math> f \cdot \left(g*h \right)(n) = \sum_{d|n} f(d) g(d) f \left( \frac{n}{d} \right) h \left( \frac{n}{d} \right) </math>

*[[Dirichlet L-function]]

*[[Dirichlet series]]

*[[Multiplicative function]]

==References==

<math> f \cdot \left(g*h \right)(n) = (f \cdot g)*(f \cdot h).</math>

<references />

* T. M. Apostol, Some properties of completely multiplicative arithmetical functions, Amer. Math. Monthly 78 (1971) 266-271.

==See also==

* P. Haukkanen, On characterizations of completely multiplicative arithmetical functions, in Number theory, Turku, de Gruyter, 2001, pp. 115–123.

* E. Langford, Distributivity over the Dirichlet product and completely multiplicative arithmetical functions, Amer. Math. Monthly 80 (1973) 411–414.

* V. Laohakosol, Logarithmic operators and characterizations of completely multiplicative functions, Southeast Asian Bull. Math. 25 (2001) no. 2, 273–281.

==References==

<references />

* K. L. Yocom, Totally multiplicative functions in regular convolution rings, Canad. Math. Bull. 16 (1973) 119–128.

[[Category:Multiplicative functions]]