%! TEX root = NT.tex
% vim: tw=50
% 08/11/2023 10AM

\gls{d_series} are interesting when $a_n$ is an
arithmetically interesting sequence, and then
$\sum_{n = 1}^\infty a_n n^{-s}$ is a kind of
generating function.

\begin{flashcard}[Dirichlet-convolution-defn]
\begin{definition*}[Dirichlet convolution]
  \glsnoundefn{d_conv}{convolution}{convolutions}
  \glsnoundefn[d_conv]{dirich_conv}{convolution}{convolutions}
  \glssymboldefn{d_conv}{$*$}{$*$}
  \cloze{
  The \emph{Dirichlet convolution} of functions
  $f, g : \NN \to \CC$ is defined by
  \[ (f * g)(n) = \sum_{d \divides n} f(d) g \left(
  \frac{n}{d} \right) .\]
  This satisfies the property that
  \[ \left( \sum_{n = 1}^\infty f(n)n^{-s} \right)
  \left( \sum_{m = 1}^\infty g(m) m^{-s} \right)
  = \sum_{n, m \ge 1} g(n) g(m) (nm)^{-s}
  = \sum_{n = 1}^\infty h(n) n^{-s} .\]
  }
\end{definition*}
\end{flashcard}

\begin{lemma}
  Let $f, g, h : \NN \to \CC$. Then:
  \begin{enumerate}[(1)]
    \item $f \conv g = g \conv f$ as functions
      $\NN \to \CC$.
    \item $(f \conv g) \conv h = f \conv (g \conv
      h)$.
    \item If $f, g$ are mutliplicative (i.e.
      $f(mn) = f(m)f(n)$, $\gcdbrack(m, n) = 1$)
      then $f \conv g$ is also multiplicative.
  \end{enumerate}
\end{lemma}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(1)]
    \item \[ (f \conv g)(n) = \sum_{d \divides}
      f(d) g \left( \frac{n}{d} \right) = \sum_{
      \substack{a, b \in \NN \\ ab = n}} f(a) g(b)
      .\]
      This is symmetric in $f$ and $g$.
    \item \begin{align*}
        ((f \conv g) \conv h)(n)
        &= \sum_{d_1 d_2 = n} (f \conv g)(d_1)
        h(d_2) \\
        &= \sum_{d_1 d_2 = n} \sum_{e_1 e_2 = d_1}
        f(e_1)g(e_2)h(d_2) \\
        &= \sum_{\substack{a, b, c \in \NN \\ abc
        = n}} f(a) g(b) h(c)
      \end{align*}
      A computation shows this is equal to $(f
      \conv (g \conv h))(n)$.
    \item Let $m, n \in \NN$, $\gcdbrack(m, n) =
      1$. Then
      \begin{align*}
        (f \conv g)(mn)
        &= \sum_{d \divides mn} f(d) g \left(
        \frac{mn}{d} \right) \\
        &= \sum_{\substack{d_1 \divides m \\ d_2
        \divides n}} f(d_1 d_2) g \left(
        \frac{mn}{d_1 d_2} \right) \\
        &= \sum_{\substack{d_1 \divides m \\ d_2
        \divides n}} f(d_1) f(d_2) g \left(
        \frac{m}{d_1} \right) g \left(
        \frac{m}{d_2} \right) \\
        &= \left( \sum_{d_1 \divides m} f(d_1) g
        \left( \frac{m}{d_1} \right) \right)
        \left( \sum_{d_2 \divides n} f(d_2) g
        \left( \frac{n}{d_2} \right) \right) \\
        &= (f \conv g)(m) (f \conv g)(n) \qedhere
      \end{align*}
  \end{enumerate}
\end{proof}

\begin{example*}
  \begin{align*}
    \zetafn(s - 1) \zetafn(s)
    &= \sum_{n = 1}^\infty n^{1 - s} \sum_{m =
    1}^\infty m^{-s} \\
    &= \sum_{n = 1}^\infty n \cdot n^{-s} \sum_{m
    = 1}^\infty m^{-s} \\
    &= \sum_{n = 1}^\infty (f \conv g)(n)n^{-s} \\
    &= \sum_{n = 1}^\infty \sigmafactorsum(n) n^{-s}
  \end{align*}
  where we use $f(n) = n$, $g(n) = 1$. Then $(f
  \conv g)(n) = \sum_{d \divides n} d =
  \sigmafactorsum(n)$.
\end{example*}

\begin{flashcard}[mobius-function-defn]
\begin{definition}[M\"obius function]
  \glssymboldefn{mobius_function}{$\mu$}{$\mu$}
  \glsnoundefn{mobius_function}{M\"obius function}{N/A}
  \cloze{The M\"obius function $\mu : \NN \to \CC$ is
  defined by
  \[ \mu(n) = \begin{cases}
    0 & \text{$n$ is not squarefree} \\
    (-1)^k & \text{$n = p_1 \cdots p_k$, $p_i$
    distinct primes}
  \end{cases} \]
  In particular, $\mu(1) = (-1)^0 = 1$.}
\end{definition}
\end{flashcard}

\begin{flashcard}[mobius-one-delta-lemma]
\begin{lemma}
  Let $\allone : \NN \to \CC$ be $\allone(n) = 1
  ~\forall n \in \NN$, and $\delta : \NN \to \CC$
  be $\delta(n) = 1$ if $n = 1$, $\delta(n) = 0$
  if $n > 1$. Then:
  \begin{enumerate}[(1)]
    \item $\delta$ is an identity for
      \gls{dirich_conv}: $\delta \conv f = f$,
      $\forall f : \NN \to \CC$.
    \item TODO
  \end{enumerate}
\end{lemma}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(1)]
    \item TODO
    \item \cloze{TODO
      So it's enough to show $(\mobius \conv
      \allone)(p^k) = \ddelta(p^k)$ if $p$ is
      prime, $k \ge 0$. For $k = 0$:
      \[ (\mobius \conv \allone)(p^k) = \sum_{d
      \divides 1} \mobius(d) = 1 \]
      For $k \ge 1$,
      \[ (\mobius \conv \allone)(p^k) = \sum_{i =
      0}^k \mobius(p^i) = \mu(1) + \mu(p) + \cdots
      + \mu(p^k) = 1 - 1 + 0 \cdots + 0 = 0 =
      \ddelta(p^k). \qedhere \]}
  \end{enumerate}
\end{proof}
\end{flashcard}

\begin{flashcard}[mobius-inversion-formula]
\begin{proposition}[M\"obius inversion formula]
  \label{mob_inversion_formula}
  \cloze{
  Suppose $f, g : \NN \to \CC$ are such that
  \[ f(n) = \sum_{d \divides n} g(d) \qquad
  \forall n \in \NN \]
  Then
  \[ g(n) = \sum_{d \divides n} \mobius(d) f
  \left( \frac{n}{d} \right) \qquad \forall n \in
  \NN \]
  }
\end{proposition}

\begin{proof}
  \cloze{By definition, we have $f = g \conv \allone$,
  and we need to show $g = \mobius \conv f$. But
  $\mobius \conv f = \mobius \conv g \conv \allone
  = g \conv (\allone \conv \mobius) = g \conv
  \ddelta = g$.}
\end{proof}
\end{flashcard}

\begin{flashcard}[Mongoldt-function-defn]
\begin{definition}[von Mongoldt function]
  \glssymboldefn{von_mongoldt}{$\Lambda$}{$\Lambda$}
  \cloze{
  The von Mongoldt function $\Lambda : \NN \to
  \CC$ is defined by
  \[ \Lambda(n) = \begin{cases}
    0 & \text{if $n$ is not a \gls{prime} power} \\
    \log p & \text{if $n = p^k$, $p$ \gls{prime},
    $k\ge 1$}
  \end{cases} \]
  ``Weighted indicator function'' of prime powers.
  }
\end{definition}
\end{flashcard}
\vspace{-1em}

The Chebyshev function $\psi : [1, \infty) \to
\CC$ is defined by $\psi(x) = \sum_{1 \le n \le x}
\vonmongoldt(n) = \sum_{p^k \le x} \log p$.
One can show using elementary methods that
\[ \psi(x) \sim \pc(x) \log (x) \]
where $\pc(x)$ is the prime counting function as
usual. Recall \cref{pnt} (\nameref{pnt}) says that
$\pc(x) \sim \frac{x}{\log x}$ as $x \to \infty$.
This is equivalent to saying that
\[ \psi(x) \sim x \]
as $x \to \infty$ (i.e. $\lim_{x \to \infty}
\frac{\psi(x)}{x} = 1$).

\begin{theorem}
  If $s \in \CC$, $\sigmareal = \Re(s) > 1$, then
  \[ \frac{-\zetafn'(s)}{\zetafn(s)} = \sum_{n =
  1}^\infty \vonmongoldt(n) n^{-s} \]
\end{theorem}

\begin{proof}
  Both $\LHS$ and $\RHS$ are holomorphic, so it's
  enough to show equality when $s = \sigmareal$ is
  real (identity principle for holomorphic
  functions).
  \begin{align*}
    -\frac{\zetafn'(s)}{\zetafn(s)}
    &= -\frac{\dd}{\dd \sigmareal} \log
    \zetafn(\sigmareal) \\
    &= -\frac{\dd}{\dd \sigmareal} \log \prod_p (1
    - p^{-\sigmareal})^{-1} \\
    &= -\frac{\dd}{\dd \sigmareal} \sum_p -\log(1
    - p^{-\sigmareal}) \\
    &= -\frac{\dd}{\dd \sigmareal} \sum_p \sum_{k
    \ge 1} \frac{p^{-k\sigmareal}}{k}
  \end{align*}
  Using $-\log(1 - x) = \sum_{k \ge 1}
  \frac{x^k}{k}$, $|x| < 1$. We can interchange
  order of differentiation and summation, using
  uniform convergence. So
  \begin{align*}
    -\frac{\zetafn'(\sigmareal)}{\zetafn(\sigmareal)}
    \\
    &= -\sum_{\substack{\text{$p$ \gls{prime}} \\
    k \ge 1}} \frac{\dd}{\dd \sigmareal}
    \frac{p^{-k\sigmareal}}{k} \\
    &= -\sum_p \frac{\dd}{\dd \sigmareal}
    \frac{\exp(-k\sigmareal \log p)}{k} \\
    &= \sum_{\substack{p \\ k \ge 1}} (\log p)
    p^{-k\sigmareal} \\
    &= \sum_{n = 1}^\infty \vonmongoldt(n)
    n^{-\sigmareal}
  \end{align*}
\end{proof}

What happens next? If $\zetafn(s)$ has a zero of
order $k$ at $s = s_0$, then
$-\frac{\zetafn'(s)}{\zetafn(s)}$ will have a
simple pole at $s = s_0$ of residue $-k$. You can
consider a contour integral of
$-\frac{\zetafn'(s)}{\zetafn(s)} \frac{x^s}{x}$
and evaluate using Cauchy's residue theorem to
prove a formula
\[ \psi(x) = x - \sum_\rho \frac{x^\rho}{\rho} -
\frac{\zetafn'(0)}{\zetafn(0)} \]
valid when $x > 2$ is not a prime power, where the
sum $\sum_\rho$ is over zeroes $\rho$ of the
Riemann $\zetafn$-function. ``Riemann's explicit
relation''.

We now turn to elementary techniques to study the
distribution of \glspl{prime}. Main goal:
\nameref{chebyshevs_thm}:
\[ c_1 \frac{x}{\log x} \le \pc(x) \le c_2
\frac{x}{\log x} \]
Main tool: \gls{prime} factorisation of binomial
coefficients ${2n \choose n}$, $n \in \NN$.