狄利克雷卷积
积性函数\(f\)和\(g\),
狄利克雷卷积的形式:
或者
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它满足
- 结合律 \((f \ast g) \ast h = f \ast ( g \ast h )\)
证明
等同于
这个式子满足轮换对称,\(i,j,k\)怎么换都一样
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\([a]\)表示 \(a\) 成立为 \(1\) ,否则为 \(0\)
有一些基本的积性函数:
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\[\epsilon(n) = [n = 1] \]
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\[1(n) = 1 \]
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\[id(n) = n \]
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\[d(n) = \sum\limits_{d|n}1 \]
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\[\sigma(n) = \sum\limits_{d|n}d \]
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\[d_w(n) =\sum\limits_{d|n}1^w \]
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\[\sigma_w(n) = \sum\limits_{d|n}d^w \]
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\[g(n) = \sum\limits_{i=1}^{n}gcd(n,i) \]
常用的卷积:
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\[f \ast \epsilon = f \]
证明:
当且仅当\(d=n\)时,\(\epsilon(1) = 1\)
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\[\mu \ast 1 = \epsilon \]
证明:
特别的,\(n=1\)时,\(\mu(1)=1\),任意积性函数的\(1\)都是\(1\)
当\(n \ne 1\)时,
令\(d = p_{1}^{x_1}p_2^{x_2}...p_k^{x_k}\)
根据\(\mu\)的定义,
\(\mu(i) = \left\{ \begin{aligned} 1\quad \quad \quad ,n=1 \\ (-1)^k \quad ,\prod_{i=1}^{k}x_i=1 \\ 0 \quad,\prod_{i=1}^{k}x_i\ne1\\ \end{aligned} \right. \)
当\(i > 1\)时,只有中间的一项有贡献
意义是从\(k\)个中选有\(i\)个质因子的数的个数,\(\mu\)如果有奇数个质因子,则为\(-1\),\(i = 0\)意味着不选,也就是\(\mu(1)\)的贡献。
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\[\varphi\ast 1 = id \]
证明:
根据\(\varphi(i)\)的定义,
\(\varphi(i)=\) \(1\)~\(i\) 与 \(i\) 互质的个数
考虑构造有\(n\)个元素的集合\(A=\left\{\frac{i}{n}|i\in Z,i\in [1,n] \right\}\)
有\(\frac{x}{n}\in A\),把它化解到最简\(\frac{a}{b}\),显然每个 \(\frac{a}{b}\) 仅会对应一个\(\frac{x}{n}\)
显然\(b|n\),因为同时除以了公因数。
由于是最简形式,对于每个分母是\(b\)的分数, 显然\(a\)都与\(b\)互质,
所以\(a\)的个数为\(\varphi(b)\)
所以
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\[\varphi = id \ast \mu \]
证明:
同乘\(1\)
化简得
得
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\[1 \ast 1 = d \]
证明:
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\[1 = d \ast \mu \]
证明:
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\[id \ast 1 = \sigma \]
证明:
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\[id = \sigma \ast \mu \]
证明:
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\[id_w \ast 1 = \sigma_{w} \]
证明:
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\[id_w = \sigma_w \ast \mu \]
证明:
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\[\sigma = \varphi \ast d \]
证明:
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\[g = \varphi \ast id \]