Difference between revisions of "Dirichlet convolution"

 
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For two functions <math> f, g : \mathbb{N} \rightarrow \mathbb{C} </math>, the '''Dirichlet convolution''' (or simply '''convolution''', when the context is clear) <math> \displaystyle f * g </math> of <math> \displaystyle f </math> and <math> \displaystyle g </math> is defined as
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For two functions <math> f, g : \mathbb{N} \rightarrow \mathbb{C} </math>, the '''Dirichlet convolution''' (or simply '''convolution''', when the context is clear) <math>f * g </math> of <math>f </math> and <math>g </math> is defined as
 
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<math> \sum_{d\mid n} f(d)g\left( \frac{n}{d} \right) </math>.
 
<math> \sum_{d\mid n} f(d)g\left( \frac{n}{d} \right) </math>.
 
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We usually only consider positive divisors of <math> \displaystyle n </math>.  We are often interested in convolutions of weak [[multiplicative function]]s; the set of weak multiplicative functions is closed under convolution.  In general, convolution is commutative and associative; it also has an identity, the function <math> \displaystyle f(n) </math> defined to be 1 if <math> \displaystyle n=1 </math>, and 0 otherwise.  Not all functions have inverses (e.g., the function <math> \displaystyle f(n) : n \mapsto 0 </math> has no inverse, as <math> \displaystyle f*g = f </math>, for all functions <math> g: \mathbb{N} \rightarrow \mathbb{C} </math>), although all functions <math> \displaystyle f </math> such that <math> \displaystyle f(1) \neq 0 </math> have inverses.
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We usually only consider positive divisors of <math>n </math>.  We are often interested in convolutions of weak [[multiplicative function]]s; the set of weak multiplicative functions is closed under convolution.  In general, convolution is commutative and associative; it also has an identity, the function <math>f(n) </math> defined to be 1 if <math>n=1 </math>, and 0 otherwise.  Not all functions have inverses (e.g., the function <math>f(n) : n \mapsto 0 </math> has no inverse, as <math>f*g = f </math>, for all functions <math> g: \mathbb{N} \rightarrow \mathbb{C} </math>), although all functions <math>f </math> such that <math>f(1) \neq 0 </math> have inverses.
  
 
== Closure of Weak Multiplicative Functions Under Convolution ==
 
== Closure of Weak Multiplicative Functions Under Convolution ==
  
'''Theorem.'''  If <math> f,g : \mathbb{N} \rightarrow \mathbb{C} </math> are weak multiplicative functions, then so is <math> \displaystyle f*g </math>.
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'''Theorem.'''  If <math> f,g : \mathbb{N} \rightarrow \mathbb{C} </math> are weak multiplicative functions, then so is <math>f*g </math>.
  
''Proof.''  Let <math> \displaystyle m,n </math> be relatively prime.  We wish to prove that <math> \displaystyle (f*g)(mn) = (f*g)(m)\cdot (f*g)(n) </math>.
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''Proof.''  Let <math>m,n </math> be relatively prime.  We wish to prove that <math>(f*g)(mn) = (f*g)(m)\cdot (f*g)(n) </math>.
  
For <math> a \in \mathbb{N} </math>, let <math> \displaystyle {\rm D}_a </math> be the set of divisors of <math> \displaystyle a </math>.  For relatively prime <math> \displaystyle m,n </math>, we claim that the function <math> p : (d_m,d_n) \mapsto d_md_n </math> is a [[bijection]] from <math> {\rm D}_m \times {\rm D}_n </math> to <math> \displaystyle {\rm D}_{mn} </math>.  Indeed, for any <math> \displaystyle d_m \mid m </math> and <math> \displaystyle d_n \mid n </math>, so <math> p({\rm D}_m \times {\rm D}_n) \subseteq {\rm D}_{mn} </math>.  Furthermore, for each <math> d \mid mn </math>, there exist unique <math> \displaystyle d_m, d_n </math> such that <math> d_m \mid m </math>, <math> d_n \mid n </math>, <math> \displaystyle d_md_n = d </math>.  Thus <math> \displaystyle p </math> is bijective.  As a result of our claim, we have the identity
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For <math> a \in \mathbb{N} </math>, let <math>{\rm D}_a </math> be the set of divisors of <math>a </math>.  For relatively prime <math>m,n </math>, we claim that the function <math> p : (d_m,d_n) \mapsto d_md_n </math> is a [[bijection]] from <math> {\rm D}_m \times {\rm D}_n </math> to <math>{\rm D}_{mn} </math>.  Indeed, for any <math>d_m \mid m </math> and <math>d_n \mid n </math>, so <math> p({\rm D}_m \times {\rm D}_n) \subseteq {\rm D}_{mn} </math>.  Furthermore, for each <math> d \mid mn </math>, there exist unique <math>d_m, d_n </math> such that <math> d_m \mid m </math>, <math> d_n \mid n </math>, <math>d_md_n = d </math>.  Thus <math>p </math> is bijective.  As a result of our claim, we have the identity
 
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<math> \sum_{d_m \mid m, d_n \mid n}a(d_m)b(d_n) = \sum_{d_m \mid m} a(d_m) \cdot \sum_{d_n \mid n}b(d_n) </math>,
 
<math> \sum_{d_m \mid m, d_n \mid n}a(d_m)b(d_n) = \sum_{d_m \mid m} a(d_m) \cdot \sum_{d_n \mid n}b(d_n) </math>,
 
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for any functions <math> \displaystyle a,b </math> mapping subsets of <math> \mathbb{N} </math> into <math> \mathbb{C} </math>.  In particular, we may let the domains of <math> \displaystyle a </math> and <math> \displaystyle b </math> be <math> \displaystyle {\rm D}_m, {\rm D}_n </math>, and define <math> \displaystyle a(d_m) = f(d_m)g(m/d_m) </math> and <math> \displaystyle b(d_n) = f(d_n)g(n/d_n) </math>.  We then have
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for any functions <math>a,b </math> mapping subsets of <math> \mathbb{N} </math> into <math> \mathbb{C} </math>.  In particular, we may let the domains of <math>a </math> and <math>b </math> be <math>{\rm D}_m, {\rm D}_n </math>, and define <math>a(d_m) = f(d_m)g(m/d_m) </math> and <math>b(d_n) = f(d_n)g(n/d_n) </math>.  We then have
 
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<math> \sum_{d_m \mid m}f(d_m)g(m/d_m) \cdot \sum_{d_n \mid n}f(d_n)g(n/d_n) = \sum_{d_m \mid m, d_n \mid n}f(d_m)g(m/d_m)f(d_n)g(n/d_n) </math>.
 
<math> \sum_{d_m \mid m}f(d_m)g(m/d_m) \cdot \sum_{d_n \mid n}f(d_n)g(n/d_n) = \sum_{d_m \mid m, d_n \mid n}f(d_m)g(m/d_m)f(d_n)g(n/d_n) </math>.
 
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But since each divisor of <math> \displaystyle m </math> is relatively prime to every divisor of <math> \displaystyle n </math>, we have
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But since each divisor of <math>m </math> is relatively prime to every divisor of <math>n </math>, we have
 
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<math> \sum_{d_m \mid m, d_n \mid n}f(d_m)g(m/d_m)f(d_n)g(n/d_n) = \sum_{d_m \mid m, d_n \mid n}f(d_md_n)g\left(\frac{mn}{d_md_n}\right) = \sum_{d \mid mn}f(d)g(mn/d) </math>,
 
<math> \sum_{d_m \mid m, d_n \mid n}f(d_m)g(m/d_m)f(d_n)g(n/d_n) = \sum_{d_m \mid m, d_n \mid n}f(d_md_n)g\left(\frac{mn}{d_md_n}\right) = \sum_{d \mid mn}f(d)g(mn/d) </math>,
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[[Category:Definition]]

Latest revision as of 21:40, 21 June 2009

For two functions $f, g : \mathbb{N} \rightarrow \mathbb{C}$, the Dirichlet convolution (or simply convolution, when the context is clear) $f * g$ of $f$ and $g$ is defined as

$\sum_{d\mid n} f(d)g\left( \frac{n}{d} \right)$.

We usually only consider positive divisors of $n$. We are often interested in convolutions of weak multiplicative functions; the set of weak multiplicative functions is closed under convolution. In general, convolution is commutative and associative; it also has an identity, the function $f(n)$ defined to be 1 if $n=1$, and 0 otherwise. Not all functions have inverses (e.g., the function $f(n) : n \mapsto 0$ has no inverse, as $f*g = f$, for all functions $g: \mathbb{N} \rightarrow \mathbb{C}$), although all functions $f$ such that $f(1) \neq 0$ have inverses.

Closure of Weak Multiplicative Functions Under Convolution

Theorem. If $f,g : \mathbb{N} \rightarrow \mathbb{C}$ are weak multiplicative functions, then so is $f*g$.

Proof. Let $m,n$ be relatively prime. We wish to prove that $(f*g)(mn) = (f*g)(m)\cdot (f*g)(n)$.

For $a \in \mathbb{N}$, let ${\rm D}_a$ be the set of divisors of $a$. For relatively prime $m,n$, we claim that the function $p : (d_m,d_n) \mapsto d_md_n$ is a bijection from ${\rm D}_m \times {\rm D}_n$ to ${\rm D}_{mn}$. Indeed, for any $d_m \mid m$ and $d_n \mid n$, so $p({\rm D}_m \times {\rm D}_n) \subseteq {\rm D}_{mn}$. Furthermore, for each $d \mid mn$, there exist unique $d_m, d_n$ such that $d_m \mid m$, $d_n \mid n$, $d_md_n = d$. Thus $p$ is bijective. As a result of our claim, we have the identity

$\sum_{d_m \mid m, d_n \mid n}a(d_m)b(d_n) = \sum_{d_m \mid m} a(d_m) \cdot \sum_{d_n \mid n}b(d_n)$,

for any functions $a,b$ mapping subsets of $\mathbb{N}$ into $\mathbb{C}$. In particular, we may let the domains of $a$ and $b$ be ${\rm D}_m, {\rm D}_n$, and define $a(d_m) = f(d_m)g(m/d_m)$ and $b(d_n) = f(d_n)g(n/d_n)$. We then have

$\sum_{d_m \mid m}f(d_m)g(m/d_m) \cdot \sum_{d_n \mid n}f(d_n)g(n/d_n) = \sum_{d_m \mid m, d_n \mid n}f(d_m)g(m/d_m)f(d_n)g(n/d_n)$.

But since each divisor of $m$ is relatively prime to every divisor of $n$, we have

$\sum_{d_m \mid m, d_n \mid n}f(d_m)g(m/d_m)f(d_n)g(n/d_n) = \sum_{d_m \mid m, d_n \mid n}f(d_md_n)g\left(\frac{mn}{d_md_n}\right) = \sum_{d \mid mn}f(d)g(mn/d)$,

as desired.

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