In his Mémoire Chebyshev uses bounds he found on the function \( \theta(x) =\sum_{p\le x} \log p\) to calculate numerically sums over primes
\begin{equation*}
\sum_{p =2 }^{\infty} f(p)
\end{equation*} I explain his method in this post; to simplify the presentation, I use less sophisticated bounds on \( \theta(x) \) than Chebyshev. As an application, I calculate numerically the sum
\begin{equation*}
\sum_{p =2 }^{\infty} \frac{1}{p \log p}
\end{equation*}
with explicit bound on the obtained error.
Wednesday, September 16, 2015
Thursday, September 10, 2015
Sign patterns of the Möbius function
I test some of the statements in the paper "Sign patterns of the Liouville and Möbius function". On page 6 in the paper one can find "the three events \( \mu(n) = +1, \mu(n) = 0, \mu(n) = -1 \) occur with asymptotic probability \( \frac{1}{2 \zeta(2)}, 1 - \frac{1}{\zeta(2)}, \frac{1}{2 \zeta(2)} \) respectively". Here, \( \mu \) is the Möbius function and \(\zeta \) is Riemann's zeta function. I interpret the concept of asymptotic probability in a naive way and just count these events in Mathematica for \( n \le 100 \).
Saturday, September 5, 2015
Variations on a theme by Chebyshev
In his paper "Mémoire sur les nombres premiers", Chebyshev calculates a lower and upper bound on the function
\begin{equation*}
\psi(x) = \sum_{n \le x } \Lambda(n)
\end{equation*}
from an analysis of the expression
\begin{equation}\label{eq:2}
T(x) - T\left( \frac{x}{2} \right)- T\left( \frac{x}{3} \right)- T\left( \frac{x}{5} \right)+ T\left( \frac{x}{30} \right)
\end{equation}
with \( T(x) = \sum_{ 1 \le n \le x} \log n\ \). In this post, I quickly review Chebyshev's argument and then calculate which bounds on \(\psi(x)\) one can get from expressions that are similar to \eqref{eq:2}.
Wednesday, August 26, 2015
Lower and upper bound for Chebyshev's psi function
In this post, I prove the inequalities
\begin{equation}\label{eq:1}
\log 2 \cdot x - 3 \log x \le \psi(x) \le 2 \log 2 \cdot x + \frac{3}{\log 2} \log^2 x
\end{equation}
for all real \( x \ge 2 \). Hereby is \(
\psi(x) = \sum_{n \le x } \Lambda(n)
\)
with \( \Lambda \) the von Mangoldt function. This is a second post motivated by Chebyshev's paper "Mémoire sur les nombres premiers" from 1852. In this paper, Chebyshev uses a more complex variant of the calculation below to obtain stronger inequalities. Similar inequalities can also be found on page 50 in Murty's book.
Sunday, August 23, 2015
How not to prove Bertrand's postulate
This is a first post motivated by Chebyshev's paper "Mémoire sur les nombres premiers" from 1852. In this paper Chebyshev proves Bertrand's postulate that there is always a prime between \( a \) and \( 2a \) for all \( a \ge 2 \). Chebyshev bases his proof on inequalities for the function
\begin{equation}\label{HNTPeq:1}
T(x) - T\left( \frac{x}{2} \right)- T\left( \frac{x}{3} \right)- T\left( \frac{x}{5} \right)+ T\left( \frac{x}{30} \right)
\end{equation}
with
\begin{equation*}
T(x) = \sum_{ 1 \le n \le x} \log n
\end{equation*}
Chebyshev's paper is quite easy to read and can be found at this link. In this post I will follow Chebyshev's reasoning on an easier version of \eqref{HNTPeq:1}.
Saturday, July 25, 2015
Neue Empirische Daten Über Die Zahlentheoretische Funktion sigma(n), von Sterneck, 1913. Part II
This post is about a formula that von Sterneck used to calculate values of \( M(n) \) for \( n \) up to 5,000,000. The formula is
\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.
\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.
Sunday, July 19, 2015
Neue Empirische Daten Über Die Zahlentheoretische Funktion sigma(n), von Sterneck, 1913. Part I
In the paper in the title, von Sterneck performs numerical tests on the inequality
\begin{equation}\label{Neueeq:1}
| M (n) | \le \sqrt{n}
\end{equation}
where \( M(n) = \displaystyle\sum_{k=1}^n \mu(n) \) is the Mertens function. The inequality (1) is of historical interest, because it implies the correctness of the Riemann hypothesis. In 1985 however, Odlyzko and te Riele, showed that (1) is incorrect, although an explicit number \( n \) for which (1) fails is still not known. Nevertheless, I find von Sterneck's paper still interesting because I want to see how \( M(n) \) was calculated for large values of \( n \) before the advent of (electronic) computers.
\begin{equation}\label{Neueeq:1}
| M (n) | \le \sqrt{n}
\end{equation}
where \( M(n) = \displaystyle\sum_{k=1}^n \mu(n) \) is the Mertens function. The inequality (1) is of historical interest, because it implies the correctness of the Riemann hypothesis. In 1985 however, Odlyzko and te Riele, showed that (1) is incorrect, although an explicit number \( n \) for which (1) fails is still not known. Nevertheless, I find von Sterneck's paper still interesting because I want to see how \( M(n) \) was calculated for large values of \( n \) before the advent of (electronic) computers.
Subscribe to:
Posts (Atom)