狄利克雷卷積總結

狄利克雷卷積

約定

$\ n=p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{r}^{a_{r}}$

$\ a \mid b$：$\ a$整除$\ b$

$\ a \nmid b$：$\ a$不整除$\ b$

$\ a^{k} \parallel b$：$\ a^{k} \mid b$且$\ a^{k+1} \nmid b$

$\ (a,b)$最大公約數

$\ [a,b]$最小公倍數

定義

$h(n)= \sum_{d \mid n} f(n) g(\frac{n}{d})=\sum_{d_{1} d_{2} =n} f(d_{1})g(d_{2})$

也可寫作：

$h=f \circ g$

這相當於一種運算。

性質

結合律

$(f \circ g) \circ h=f \circ(g \circ h)$

證明

$\begin{array}{crl} & &\,\,\,\,\,\,\,(f \circ g) \circ h\\& &=\sum_{d_{1} \mid n}\{ \sum_{d_{2} \mid d_{1}}f(d_{2})g(\frac{d_{1}}{d_{2}})\} g(\frac{n}{d_{1}}) \\\,\\ & &=\sum_{d_{1} d_{2} d_{3}= n} (f(d_{1})g(d_{2}))g(d_{3}) \\\,\\ & &=\sum_{d_{1} d_{2} d_{3}= n} f(d_{1})(g(d_{2})g(d_{3}))\\\,\\ & &=f \circ(g \circ h)\\\,\\ \end{array}$
證畢

交換律

$f \circ g=g \circ f$

證明

$f \circ g=\sum_{d_{1} d_{2} =n} f(d_{1})g(d_{2})=\sum_{d_{1} d_{2} =n} g(d_{1})f(d_{2})=g \circ f$

證畢

逆元

對於$\ f$存在$\ g$使
$f \circ g = \epsilon(n)$
其中
$\epsilon(n)= [n=1]$
證明

$g(n)=\frac{1}{f(1)} \left( \left[ n=1 \right] - \sum_{i \mid n,i \neq n}f(i) g(\frac{n}{i}) \right)$

如此一來

$\sum_{d \mid n}f(d)g(\frac{n}{d})=f(1)g(n)-\sum_{d \mid n,d \neq 1}f(d)g(\frac{n}{d})=\left[ n=1 \right]$

證畢

分配率

$f \circ (g + h)=f \circ g+f \circ h$

證明

$f \circ (g + h)=\sum_{d_{1} d_{2} =n} f(d_{1})(g(d_{2})+h(d_{2}))=\sum_{d_{1} d_{2} =n} f(d_{1})g(d_{2})+\sum_{d_{1} d_{2} =n} f(d_{1})h(d_{2})=f \circ g+f \circ h$

數乘結合律

$(s \cdot f) \circ g=s \cdot (f \circ g)$

證明

$(s \cdot f) \circ g=\sum_{d_{1} d_{2} =n} (s\cdot f(d_{1}))g(d_{2})=s\sum_{d_{1} d_{2} =n}f(d_{1})g(d_{2})$

證畢

積性的傳遞性

若$\ f(n),g(n)$爲積性$\ h=f \circ g$也是積性的

$\begin{array}{rcl} & &\,\,\,\,\,\,\,h(nm)\\\,\\ & &=\sum_{d \mid nm} f(d) g(\frac{nm}{d}) \\\,\\ & &=\sum_{a \mid n,b \mid m} f(ab)g(\frac{nm}{ab}) \\\,\\ & &=\sum_{a \mid n,b \mid m} f(a)g(\frac{n}{a}) f(b)g(\frac{m}{b}) \\\,\\ & &=\{ \sum_{a \mid n} f(a)g(\frac{n}{a}) \}\{ \sum_{b \mid m} f(b)g(\frac{m}{b}) \} \\\,\\ & &=h(n)h(m) \end{array}$

證畢

$\ f$積性則$\ f^{-1}$積性

證明

設$\ nm>1$時有$\ n'm' < nm,g(n'm')=g(n')g(m')$成立
$\ f(n)$的逆元$\ g(n)=\frac{1}{f(1)} \left( \left[ n=1 \right] - \sum_{i \mid n,i \neq n}f(i) g(\frac{n}{i}) \right)$
$\begin{array}{rcl} & &\,\,\,\,\,\,\,g(nm) \\\,\\ & &=- \sum_{d \mid nm,d \neq 1}f(d)g(\frac{nm}{d}) \\\,\\ & &=- \sum_{a \mid n,b \mid m,ab \neq 1} f(ab) g(\frac{nm}{ab}) \\\,\\ & &=f(1)f(1)g(n)g(m)- \sum_{a \mid n,b \mid m,ab \neq 1} f(a) f(b) g(\frac{n}{a}) g(\frac{m}{b}) \\\,\\ & &=g(n)g(m)- \{ \sum_{a \mid n,a \neq 1} f(a) g(\frac{n}{a}) \} \{ \sum_{b \mid m,b \neq 1} f(b) g(\frac{m}{b}) \}\\\,\\ & &=g(n)g(m)-\epsilon(n) \epsilon(m) \\\,\\ & &=g(n)g(m) \end{array}$

證畢

應用

我們總可以證明求一個狄利克雷卷積的複雜度是$\ O(n \ln n)$的。