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\begin{fcthmstar}[Euler product]
  \label{thm:eulprod}
  Assuming:
  - $f : \Nbb \to \Cbb$ be bounded ($|f(n)| \vll
  1$) and \gls{mult}
  Then:
  \[
    \Df_f(s) = \prod_p \left( 1 + \frac{f(p)}{p^2}
    + \frac{f(p^2)}{p^{2s}} + \cdots \right)
  ,\]
  for $\Re(s) > 1$. Furthermore, if $f$ is
  \gls{cmult}, then
  \[
    \Df_f(s) = \prod_p \left( 1 - \frac{f(p)}{p^s}
    \right)^{-1}
  ,\]
  for $\Re(s) > 1$.
\end{fcthmstar}

\begin{proof}
  Let $N \in \Nbb$, $\Re(s) = \sigma > 1$. Let
  \[
    D_f(s, N) = \prod_{p \le N} \left( 1 +
    \frac{f(p)}{p^s} + \frac{f(p^2)}{p^{2s}} +
    \cdots \right)
  .\]
  Note that the series defining the factors are
  absolutely convergent, since
  \[
    \sum_{k = 1}^\infty \frac{|f(p^k)|}{|p^{ks}|}
    \vll \sum_{k = 1}^\infty \frac{1}{p^{ks}}
    < \infty
  \]
  (geometric series).

  Therefore, multiplying out,
  \[
    D_f(s, N) = \sum_{n = 1}^\infty a(n, N)
    f(n)n^{-s}
  \]
  where
  \[
    a(n, N) = \#\{\text{ways to write $n$ as a
    product of prime powers, where the primes are
    $\le N$}\}
  .\]
  The fundamental theorem of arithmetic tells us
  that $a(n, N) \in \{0, 1\}$ and $a(n, N) = 1$
  for $n \le N$.

  Now,
  \begin{align*}
    |\Df_f(S) - D_f(s, N)|
    &\le \sum_{n = N + 1}^\infty |f(n)| |n^{-s}|
    \\
    &\vll \sum_{n = N + 1}^\infty n^{-\sigma} \\
    &\stackrel{N \to \infty}{\longrightarrow} 0
  \end{align*}
  (since $\sum_{n = 1}^\infty n^{-\sigma} <
  \infty$). Hence $\Df_f(s) = \lim_{N \to \infty} \Df_f(s,
  N)$.

  Finally, for $f$ \gls{cmult}, use geometric
  formula:
  \[
    \sum_{k = 1}^\infty \frac{f(p)^k}{p^{ks}} =
    \frac{1}{1 - \frac{f(p)}{p^s}}
    .
    \qedhere
  \]
\end{proof}

\begin{fclemmastar}[]
  Assuming:
  - $f, g : \Nbb \to \Cbb$
  - $|f(n)|, |g(n)| \le n^{o(1)}$
  Then:
  \[
    \Df_{f \conv g}(s) = \Df_f(s) \Df_g(s)
  \]
  for $\Re(s) > 1$.
\end{fclemmastar}

\begin{proof}
  We know $\Df_f(s)$ and $\Df_g(s)$ are absolutely
  convergent, so can expand out the product.
  \begin{align*}
    \Df_f(s) \Df_g(s)
    &= \sum_{a, b = 1}^\infty f(a) g(b) (ab)^{-s}
    &= \sum_{n = 1}^\infty \sum_{n = ab} f(a)g(b)
    n^{-s} \\
    &= \sum_{n =  1}^\infty (f \conv g)(n) n^{-s} \\
    &= \Df_{f \conv g}(s)
    \qedhere
  \end{align*}
\end{proof}

\begin{fcdefnstar}[Riemann zeta function]
  \glssymboldefn{riemannzeta}%
  For $\Re(s) > 1$, define
  \[
    \zeta(s) = \sum_{n = 1}^\infty n^{-s}
  .\]
\end{fcdefnstar}

\begin{example*}
  \phantom{}
  \begin{itemize}
    \item $\sum_{n = 1}^\infty \frac{\tauf(n)}{n^s}
      = \rzeta(s)^2$ for $\Re(s) > 1$ (since $\tauf
      = 1 \conv 1$).
    \item $\sum_{n = 1}^\infty \frac{\mob(n)}{n^s}
      = \frac{1}{\rzeta(s)}$ for $\Re(s) > 1$
      (since $\mob$ is the Dirichlet inverse of
      $1$, so $\mob \conv 1 = \dId$).
    \item $\rzeta'(s) = -\sum_{n = 1}^\infty
      \frac{\log n}{n^s}$ for $\Re(s) > 1$, since
      $\frac{\dd}{\dd s} n^{-s} = -(\log n)
      n^{-s}$. Can differentiate termwise, since
      if $F_n$ analytic and $F_n \to F$ uniformly,
      then $F$ is analytic and $F_n' \to F'$. We
      know that $\sum_{n = 1}^\infty f(n) n^{-s}$
      converges uniformly for $\Re(s) > 1$ if
      $|f(n)| \le n^{o(1)}$.
    \item $\frac{\zeta'(s)}{\zeta(s)} = -\sum_{n =
      1}^\infty \frac{\vman(n)}{n^s}$ for $\Re(s)
      > 1$. This is because $-\log = -1 \conv
      \vman$ -- see the definition of the
      \gls{vman}.
  \end{itemize}
\end{example*}

\subsubsection*{Dirichlet hyperbola method}

Problem: How many lattice points $(a, b) \in
\Nbb^2$ satisfy $ab \le x$?
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/6a87e90233c04622.png}
\end{center}
Note that this number is $\sum_{n \le x} \tauf(n)
= \sum_{ab \le x} 1$.

Dirichlet proved that for $x \ge 2$,
\[
  \sum_{n \le x} \tauf(n)
  = \mathcloze{x \log x + (2\gamma - 1)x +
  \On(x^{1 / 2})}
\]
where $\eulgamma$ is \gls{eulc}.

We will see a proof of this shortly.

Conjecture: Can have $\On_\eps(x^{\frac{1}{4} +
\eps})$. Current best exponent is $0.314$.

First, we prove a lemma:

\begin{fclemmastar}[Dirichlet hyperbola method]
  \label{dhypm}
  Assuming:
  - $f, g : \Nbb \to \Cbb$
  - $x \ge y \ge 1$
  Then:
  \[
    \sum_{n \le x} (f \conv g)(n) = \sum_{d \le y}
    f(d) \sum_{m \le \frac{x}{d}} g(m)
    + \sum_{m \le \frac{x}{y}} g(m) \sum_{y < d
    \le \frac{x}{m}} f(d)
  .\]
\end{fclemmastar}

\begin{proof}
  $\sum_{n \le x} (f \conv g)(n) = \sum_{dm \le x}
  f(d) g(m)$. Split this sum into parts with $d
  \le y$ and $d > y$ to get the conclusion.
\end{proof}
