\begin{equation}
\sum_{d \le \sqrt{x}}\!\!\phantom{}^{\prime}\ M\left( \frac{x}{d} \right) + \sum_{d \le \sqrt{x}} \mu(d) \omega_P\left( \frac{x}{d} \right) - \omega_P(\sqrt{x})M(\sqrt{x}) =0
\end{equation}
where
I calculate an example for \( x = 50 \) and \( P = \{ 2,3 \} \). The first term is
\begin{equation}
\sum_{d \le 7}\!\phantom{}^{\prime}\ M\left( \frac{50}{d} \right) = M(50) + M(10) + M(7)
\end{equation}
The second term is
\begin{equation}
\sum_{d \le 7} \mu(d) \omega_P\left( \frac{50}{d} \right)
\end{equation}
If I use
\begin{equation}
\omega_P(x) = \lfloor x \rfloor - \left\lfloor \frac{x}{2} \right\rfloor - \left\lfloor \frac{x}{3}\right\rfloor + \left\lfloor \frac{x}{6} \right\rfloor
\end{equation}
then it can be calculated that the second term is zero. The third term is \( \omega(7) M(7) = 3 M(7) \). All together, von Sterneck's formula gives the recursive relation
\begin{equation}
M(50) + M(10) +M(7) + 0 - 3 M(7) =0
\end{equation}
Because \( M(10) = -1 \) and \( M(7) = -2 \) this gives \( M(50) = -3 \).
Sketch of the proof of the formula
The proof proceeds as follows. Define the function \( e_P \)
\begin{equation}
e_P(m) = \begin{cases} 1 \quad\text{if}\quad p_1 \nmid m \text{ and } \cdots \text{ and } p_j \nmid m\\ 0 \quad\text{otherwise} \end{cases}
\end{equation}
then calculate the Dirichlet convolution
\begin{equation} e_P \ast\mu = 1_{\langle P \rangle} \cdot \mu- \( \mu \) the Möbius function
- \( M(x) = \sum_{ n \le x} \mu(n) \) the Mertens function
- \( \omega_P(x) = \left\{ k \in\mathbb{N} \middle| 1 \le k \le x \text{ and } p_1 \nmid k \text{ and } \cdots \text{ and } p_j \nmid k\right\} \) with \(P = \left\{ p_1, \ldots p_j \right\}\) a non-empty set of primes
- \(\sum_{k \le x}^{\prime}\) the sum over integers \(k\) with \( p_1 \nmid k \text{ and } \cdots \text{ and } p_j \nmid k\)
- \( x \ge p_1 p_2 \cdots\ p_j \)
I calculate an example for \( x = 50 \) and \( P = \{ 2,3 \} \). The first term is
\begin{equation}
\sum_{d \le 7}\!\phantom{}^{\prime}\ M\left( \frac{50}{d} \right) = M(50) + M(10) + M(7)
\end{equation}
The second term is
\begin{equation}
\sum_{d \le 7} \mu(d) \omega_P\left( \frac{50}{d} \right)
\end{equation}
If I use
\begin{equation}
\omega_P(x) = \lfloor x \rfloor - \left\lfloor \frac{x}{2} \right\rfloor - \left\lfloor \frac{x}{3}\right\rfloor + \left\lfloor \frac{x}{6} \right\rfloor
\end{equation}
then it can be calculated that the second term is zero. The third term is \( \omega(7) M(7) = 3 M(7) \). All together, von Sterneck's formula gives the recursive relation
\begin{equation}
M(50) + M(10) +M(7) + 0 - 3 M(7) =0
\end{equation}
Because \( M(10) = -1 \) and \( M(7) = -2 \) this gives \( M(50) = -3 \).
Sketch of the proof of the formula
The proof proceeds as follows. Define the function \( e_P \)
\begin{equation}
e_P(m) = \begin{cases} 1 \quad\text{if}\quad p_1 \nmid m \text{ and } \cdots \text{ and } p_j \nmid m\\ 0 \quad\text{otherwise} \end{cases}
\end{equation}
then calculate the Dirichlet convolution
\end{equation} where \( 1_{\langle P \rangle}\) is the characteristic function on the set \( \langle P \rangle = \left\{ n \in \mathbb{N} \middle|\quad \forall p: p | n \implies p \in P \right\} \). Because \( \sum_{ k \le x}e_P(k) = \omega_P(x) \), Dirichlet's hyperbola method gives
\begin{equation}
\sum_{n \le x} 1_{\langle P \rangle} (n) \mu(n) = \sum_{d \le \sqrt{x}}e_P(d) \ M\left( \frac{x}{d} \right) + \sum_{d \le \sqrt{x}} \mu(d) \omega_P\left( \frac{x}{d} \right) - \omega_P(\sqrt{x})M(\sqrt{x})
\end{equation}
The result then follows because the left hand side is zero for \( x \ge p_1 p_2 \cdots\ p_j \)
Von Sterneck used this formula with \( P = \{ 2,3,5,7 \} \). In that way he could calculate values of \(M(n) \) for \( n \) up to 5,000,000 based on a pre-computed table of \(M(n) \) for \( n \) up to 500,000.
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