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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.
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A '''completely multiplicative function''' (or '''totally multiplicative function''') is an [[arithmetic function]] (that is, a function whose [[Domain (mathematics)|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=30}}</ref>
 
 
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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.
 
 
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==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''<sup>''n''</sup>. Then ''f''(''bc'') = (''bc'')<sup>''n''</sup> = ''b''<sup>''n''</sup>''c''<sup>''n''</sup> = ''f''(''b'')''f''(''c''), and ''f''(1) = 1<sup>''n''</sup> = 1.
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The [[Liouville function]] is a non-trivial example of a completely multiplicative function as are [[Dirichlet character]]s.
  <li>[http://www.cqwkc.net/forum.php?mod=viewthread&tid=137683 http://www.cqwkc.net/forum.php?mod=viewthread&tid=137683]</li>
 
 
==Properties==
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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''<sup>''a''</sup> ''q''<sup>''b''</sup> ..., then ''f''(''n'') = ''f''(''p'')<sup>''a''</sup> ''f''(''q'')<sup>''b''</sup> ...
 
 
</ul>
While the [[Dirichlet convolution]] of two multiplicative functions is multiplicative, the [[Dirichlet convolution]] of two completely multiplicative functions need not be completely multiplicative.
 
There are a variety of statements about a function which are equivalent to it being completely multiplicative. For example, if a function ''f'' multiplicative then is completely multiplicative if and only if the [[Dirichlet inverse]] is <math>\mu f</math> where <math>\mu</math> is the [[Möbius function]].<ref>Apostol, p. 36</ref>
 
Completely multiplicative functions also satisfy a pseudo-associative 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 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 deduced from the latter/above for [both] <math>g = h = 1</math>, where <math>1(n) = 1</math> is well-known [[multiplicative function#Examples|constant function]].
Here <math> \tau</math> is the [[divisor function]].
 
===Proof of pseudo-associative property ===
 
:<math>
\begin{align}
f \cdot \left(g*h \right)(n) &= f(n) \cdot \sum_{d|n} g(d) h \left( \frac{n}{d} \right) = \\
&= \sum_{d|n} f(n) \cdot (g(d) h \left( \frac{n}{d} \right)) = \\
&= \sum_{d|n} (f(d) f \left( \frac{n}{d} \right)) \cdot (g(d) h \left( \frac{n}{d} \right)) \text{ (since } f \text{ is completely multiplicative) } = \\
&= \sum_{d|n} (f(d) g(d)) \cdot (f \left( \frac{n}{d} \right) h \left( \frac{n}{d} \right)) \\
&= (f \cdot g)*(f \cdot h).
\end{align}
</math>
 
===Dirichlet series===
Moreover, The L-function of completely (or totally) [[multiplicative function|multiplicative]] [[Dirichlet series]] a(n) 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==
 
*[[multiplicative function]]
*[[Dirichlet series]]
*[[Dirichlet L-function]]
*[[Arithmetic function]]
 
==References==
<references />
 
[[Category:Multiplicative functions]]

Latest revision as of 13:11, 1 January 2015

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