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\begin{document}

\title{Arithmetic Combinatorics Homework}
\author{Dakota Blair}
\maketitle

\noindent  Let $\primes$ be the set of primes and for $p \in \primes$ let $\nu_p(n)$ be the $p$-adic valuation of $n$.

\begin{lem}
\label{lem1}
If $f(n)$ is multiplicative then
\[
\sum_{n = 1}^\infty \frac{f(n)}{n^{-x}} = \prod_{p \in \primes} (1 + f(p)p^{-x} + f(p^2)p^{-2x} + \cdots )
\]
\end{lem}

\begin{lem}
\label{lem2}
Let $a_n$ and $b_n$ be two sequences then
\[
\left(\sum_{n=1}^\infty \frac{a_n}{n^s}\right)
\left(\sum_{n=1}^\infty \frac{b_n}{n^s}\right)
=
\sum_{n=1}^\infty \frac{\sum_{d \div n} a_d b_{\frac{n}{d}}}{n^s}
\]
\end{lem}

\begin{hw}
\[
\frac{1}{\zeta(s)} = \prod_{p \in \primes} (1 - p^{-s}) = \sum_{m=1}^\infty \frac{\mu(m)}{m^s}
\]
\end{hw}
\begin{proof}
From Lemma \ref{lem1} we see that
\begin{align*}
\sum_{m=1}^\infty \frac{\mu(m)}{m^s}
& = \prod_{p \in \primes} (1 + \mu(p)p^{-s} + \mu(p^2)p^{-2s} + \mu(p^3)p^{-3s} + \cdots ) \\
& = \prod_{p \in \primes} (1 + -p^{-s} + 0 p^{-2s} + 0 p^{-3s} + \cdots )\\
\sum_{m=1}^\infty \frac{\mu(m)}{m^s}
& = \prod_{p \in \primes} (1 - p^{-s}) = \frac{1}{\zeta(s)}
\end{align*}
\end{proof}

\begin{hw}
Let $\delta(m) = [m=1]$.
\[
\delta(m) = \sum_{d \div m} \mu(d)
\]
\end{hw}
\begin{proof}
From Lemma \ref{lem2} we see that
\[
\sum_{n=1}^\infty \frac{\delta(n)}{n^s}
= 1 = \zeta(s) \frac{1}{\zeta(s)} =
\left(\sum_{n=1}^\infty \frac{1}{n^s}\right)
\left(\sum_{n=1}^\infty \frac{\mu(n)}{n^s}\right)
= \sum_{n=1}^\infty \frac{\sum_{d \div n} \mu(d)}{n^s}
\]
and hence the assertion.
\end{proof}

\begin{hw}
\label{mob-inv}
For all $f, g : \natural \to \complex$
\[
\tag{3}
g(n) = \sum_{d \div n} f(d)
 \iff
f(n) = \sum_{d \div n} \mu(d) g\left(\frac{n}{d}\right)
\]
\end{hw}
\begin{proof}
Let 
\[
F(s) = \left(\sum_{n=1}^\infty \frac{f(s)}{n^s}\right)
\eqand
G(s) = \left(\sum_{n=1}^\infty \frac{1}{n^s}\right).
\]
Then by Lemma \ref{lem2}
\[
F(s)\zeta(s)
= \left(\sum_{n=1}^\infty \frac{f(n)}{n^s}\right)
  \left(\sum_{n=1}^\infty \frac{1}{n^s}\right)
= \sum_{n=1}^\infty \frac{\sum_{d \div n} f(d)}{n^s}
= \sum_{n=1}^\infty \frac{g(n)}{n^s}
= G(s)
\]
therefore, again by Lemma \ref{lem2},
\[
F(s) = \frac{1}{\zeta(s)}G(s)
= \left(\sum_{n=1}^\infty \frac{\mu(n)}{n^s}\right)
  \left(\sum_{n=1}^\infty \frac{g(n)}{n^s}\right)
= \sum_{n=1}^\infty \frac{
    \sum_{d \div n} \mu(d)g\left(\frac{n}{d}\right)
  }{n^s}
= \sum_{n=1}^\infty \frac{f(n)}{n^s}
.
\]
Hence
\[
g(n) = \sum_{d \div n} f(d)
 \iff
f(n) = \sum_{d \div n} \mu(d) g\left(\frac{n}{d}\right).
\]
\end{proof}

\begin{hw}
Let $\Lambda(n) = [\exists (p, \alpha) n = p^\alpha]\log p$,
\begin{align*}
  \log n
 & = \sum_{d \div n} \Lambda(d)\\
  \Lambda(n)
 & = \sum_{d \div n} \mu(d)\log \frac{n}{d}
   = -\sum_{d \div n} \mu(d) \log d\\
\end{align*}
\end{hw}
\begin{proof}
Note that $\log 1 = \Lambda(1) = 0$.
If $n > 1$ then
\[
n = \prod_{p \in \primes}p^{\nu_p(n)}
\]
therefore
\[
\log n = \sum_{p \in \primes} \nu_p(n) \log p
= \sum_{p \in \primes} \nu_p(n) \Lambda(p)
= \sum_{p \in \primes} \sum_{p^i \div n} \Lambda(p)
= \sum_{d \div n} \Lambda(d)
\]
hence by M\"{o}bius inversion (\ref{mob-inv})
\[
\Lambda(n)
 = \sum_{d \div n} \mu(d)\log \frac{n}{d}
 = -\sum_{d \div n} \mu(d) \log d.\\
\]
\end{proof}

\end{document}
