對狄利克雷卷積的深刻刻畫

對於數論函式 \(f\)，我們定義其對應的狄利克雷生成函式（\(\rm DGF\)）為：

\[\overline{f}(x) = \sum\limits_{i \ge 1} = \frac{f(i)}{i ^ x} \]

• \((1)\) 狄利克雷生成函式的卷積對應數論函式狄利克雷卷積後的生成函式：

\[\overline{f}(x) \overline{g}(x) = \sum\limits_{i \ge 1, j \ge 1} \frac{f(i)g(j)}{(ij) ^ x} = \sum\limits_{i \ge 1} \frac{\sum\limits_{d \mid i} f(d)g(\frac{i}{d})}{i ^ x} = \overline{f * g}(x) \]

• \((2)\)
  積性函式迪利克雷生成函式更本質的形式：

\[\overline{f}(x) = \prod\limits_{p \in \mathbb{P}} \left(\sum\limits_i \frac{f(p ^ i)}{p ^ {ix}}\right) \]

• \((3)\) 常見數論函式的生成函式。

在此之前，我們先給出黎曼函式 \(\zeta(x) = \sum\limits_{i \ge 1} \frac{1}{i ^ x} = \prod\limits_{p \in \mathbb{P}} \left(\sum\limits_{i \ge 0} \frac{1}{p ^ {ix}}\right) = \prod\limits_{p \in \mathbb{P}} (1 - p ^ {-x}) ^ {-1}\)

1. 單位函式 \(\epsilon(n)\)：\(\epsilon(x) = 1\).

2. 恆等函式 \(1(n)\)：\(\overline{1}(x) = \zeta(x)\).

3. 標誌函式 \(id(n)\)：\(\overline{id}(x) = \zeta(x - 1)\)

\[\overline{id}(x) = \sum\limits_{i \ge 1} \frac{i}{i ^ x} = \sum\limits_{i \ge 1} \frac{1}{i ^ {x - 1}} = \zeta(x - 1) \]

4. 標誌函式 \(k\) 次冪 \(id_k(n)\)：\(\overline{id}(x) = \zeta(x - k)\)

   .

5. 莫比烏斯函式 \(\mu(n)\)：\(\overline{\mu}(x) = \zeta ^ {-1}(x)\).

\[\overline{\mu}(x) = \prod\limits_{p \in \mathbb{P}} (1 - p ^ {-x}) = \zeta ^ {-1}(x) \]

由此可以直觀感受得知：\(\mu * 1 = \epsilon\).

6. 劉維爾函式 \(\lambda(n)\)：\(\overline{\lambda}(x) = \zeta(2x)\zeta ^ {-1}(x)\).

\[\overline{\lambda}(x) = \prod\limits_{p \in \mathbb{P}} \left(\sum\limits_i \frac{(-1) ^ i}{p ^ {ix}}\right) = \prod\limits_{p \in \mathbb{P}} \frac{1}{1 + p ^ {-x}} =\prod\limits_{p \in \mathbb{P}}\frac{1 - p ^ {-x}}{1 - p ^ {-2x}} = \zeta(2x)\zeta ^ {-1}(x) \]

因此有 \(\lambda * 1 = \zeta(2x) = \sum\limits_{i \ge 1} \frac{1}{(i ^ 2) ^ x}= 1_{Sq}(1_{Sq}(n) = [\exist i, i ^ 2 = n])\)。

根據莫比烏斯反演有：\(\lambda = \mu * 1_{Sq}\).

7. \(f(n) = \mu ^ 2(n)\)：\(\overline{f}(x) = \zeta ^ {-1}(2x)\zeta(x)\).

\[\overline{f}(x) = \prod\limits_{p \in \mathbb{P}} (1 + p ^ {-x}) = \overline{\lambda} ^ {-1}(x) = \zeta ^ {-1}(2x)\zeta(x) \]

因此有 \(\lambda\) 與 \(\mu ^ 2\)（點積）互為逆元。

8. 尤拉函式 \(\varphi\)：\(\overline{\varphi}(x) = \zeta(x - 1)\zeta ^ {-1}(x)\).

\[\begin{aligned} \overline{\varphi}(x) &= \prod\limits_{p \in \mathbb{P}} \left(1 + (p - 1)\sum\limits_i \frac{p ^ {i}}{p ^ {(i + 1)x}}\right)\\ &= \prod\limits_{p \in \mathbb{P}} \left(1 + \frac{p - 1}{p ^ x}\sum\limits_i \frac{1}{p ^ {i(x - 1)}}\right)\\ &= \prod\limits_{p \in \mathbb{P}} \frac{p ^ x - 1}{p ^ x(1 - p ^ {1 - x})}\\ &= \prod\limits_{p \in \mathbb{P}} \frac{1 - p ^ {-x}}{1 - p ^ {1 - x}} = \zeta(x - 1)\zeta ^ {-1}(x)\\ \end{aligned} \]

因此有：\(\varphi * 1 = id\)，同時由莫比烏斯反演 \(\varphi = \mu * id\).

9. 約數個數函式 \(d(n)\)：\(\overline{d}(x) = \zeta ^ 2(x)\).

\[\overline{d}(x) = \prod\limits_{p \in \mathbb{P}} \left(\sum\limits_i \frac{i + 1}{p ^ {ix}}\right) \]

\(\forall p\)，令 \(S = \sum\limits_i \frac{i + 1}{p ^ {ix}}\)，那麼有：

\[p ^ xS = \sum\limits_i \frac{i + 1}{p ^ {(i + 1)x}} = \sum\limits_{i \ge 1} \frac{i}{p ^ {ix}} \]

則有：\(p ^ {-x}S + \sum\limits_i \frac{1}{p ^ {ix}} = p ^ {-x}S + (1 - p ^ {-x}) ^ {-1} = S \Rightarrow S = (1 - p ^ {-x}) ^ {-2}\)，那麼有：

\[\overline{d}(x) = \prod\limits_{p \in \mathbb{P}} (1 - p ^ {-x}) ^ {-2} = \zeta ^ 2(x) \]

10. 約數 \(k\) 次方和函式 \(\sigma_k(n)\)：\(\overline{\sigma_k}(x) = \zeta(x - k)\zeta(x)\)

\[\overline{\sigma_k}(x) = \prod\limits_{p \in \mathbb{P}} \left(\sum\limits_i \frac{\sum\limits_j ^ i p ^ {jk}}{p ^ {ix}}\right) = \prod\limits_{p \in \mathbb{P}} \left(\sum\limits_i \frac{1 - p ^ {(i + 1)k}}{(1 - p ^ k)p ^ {ix}}\right) = \prod\limits_{p \in \mathbb{P}} \frac{1}{1 - p ^ k}\sum\limits_i \left(\frac{1}{p ^ {ix}} - \frac{(p ^ k) ^ {i + 1}}{(p ^ x) ^ i}\right) \]

一方面，令 \(S_1 = \frac{1}{1 - p ^ k}\sum\limits_i \frac{1}{p ^ {ix}} = \frac{1}{(1 - p ^ k)(1 - p ^ {-x})}\).

另一方面，令 \(S_2 = \frac{1}{1 - p ^ k}\sum\limits_i \frac{(p ^ k) ^ {i + 1}}{(p ^ x) ^ i} = \frac{p ^ k}{(1 - p ^ k)(1 - p ^ {k - x})}\).

則有 \(S_1 - S_2 = (1 - p ^ {-x}) ^ {-1}(1 - p ^ {k - x}) ^ {-1}\)，因此有：

\[\overline{\sigma_k}(x) = \prod\limits_{p \in \mathbb{P}} (1 - p ^ {-x}) ^ {-1}(1 - p ^ {k - x}) ^ {-1} = \zeta(x - k)\zeta(x) \]

由此我們可知 \(\sigma_k = id_k * 1\).

• \((4)\) 狄利克雷生成函式基本運算

1. 函式平移：\(\overline{f \times id_k}(x) = \sum\limits_{i \ge 1} \frac{f(i)}{i ^ {x - k}} = \overline{f}(x - k)\).

因此有：\(f * g = h \Rightarrow (f \times id_k) * (g \times id_k) = h \times id_k\).

常見的運用：\(\mu * 1 = \epsilon \Rightarrow (\mu \times id_k) * id_k = \epsilon/\varphi * 1 = id \Rightarrow (\varphi \times id_k) * id_k = id_{k + 1}\)

2. 乘法：\(f * g = h \Rightarrow \overline{f}(x) \times \overline{g}(x) = \overline{h}(x)\).

證明在本文開始已經提及。

3. 除法：\(f / g = f * g ^ {-1} \Rightarrow \frac{\overline{f}(x)}{\overline{g}(x)} = \overline{f * g ^ {-1}}(x)\)

4. 求導：\(\overline{f}'(x) = \left(\sum\limits_{i \ge 1} \frac{f(i)}{i ^ x}\right)' = \sum\limits_{i \ge 1} \frac{-\ln i f(i)}{i ^ x}\).

5. 積分：\(\int \overline{f}(x) = \int \sum\limits_{i \ge 1} \frac{f(i)}{i ^ x} = \sum\limits_{i \ge 1} \frac{-f(i)}{\ln i \times i ^ x}\).

以下部分暫時只記結論，一段時間後來補具體證明。

6. 求 \(\ln\)

首先 \(\ln n\) 在 \(n\) 為正整數的時候值域為實數域，無法很好的儲存，因此使用另一個函式 \(\Omega(n)\) 為 \(n\) 所有素因子次數和來代替，其滿足 \(\Omega(ab) = \Omega(a) + \Omega(b)\).

這樣一來求導和積分的結果就為整數了。

於是求 \(\overline{g}(x) = \ln \overline{f}(x)\) 有：\(\overline{g}'(x) = \frac{\overline{f}'(x)}{\overline{f}(x)}\)，通過求導和求逆即可得到 \(\overline{g}'(x)\) 最後積分還原。

7. 求 \(\exp\)

有 \(\overline{g}(x) = \exp \overline{f}(x) \Rightarrow \ln \overline{g}(x) = \overline{f}(x) \Rightarrow \overline{f}'(x) = \frac{\overline{g}'(x)}{\overline{g}(x)} \Rightarrow \overline{g}'(x) = \overline{f}'(x)\overline{g}(x)\).

於是有等式：

\[\begin{aligned} g'(n) &= \sum\limits_{d \mid n} f'(d)g(\frac{n}{d})\\ &= \sum\limits_{d \mid n} -\ln d \cdot f(d)g(\frac{n}{d})\\ &= -\ln 1 f(1)g(n) + \sum\limits_{d \mid n, d \ne 1} -\ln d \cdot f(d)g(\frac{n}{d})\\ &= \sum\limits_{d \mid n, d \ne 1} -\ln d \cdot f(d)g(\frac{n}{d})\\ \end{aligned} \]

另一方面：\(g'(n) = -\ln n \cdot g(n)\) 那麼就可以得到遞推式：\(g(n) = \frac{1}{\ln n}\sum\limits_{d \mid n, d \ne 1} \ln d \cdot f(d)g(\frac{n}{d})\).

GO!