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%                     Homework 3:                                        %
%                     Analytic Number Theory                             %
%                                                       GM   18/05/2016  %
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\begin{document}
\begin{center}
{\bf\LARGE{Analytic Number Theory:} Homework 3%\\[.3cm]
%    \large{sheet 1}
    }
\end{center}
\vspace{1cm}

\bigskip
\begin{enumerate}
\item The following steps provide the asymptotic behavior for the sum $\sum_{n\leq X} \frac{\mu(n)^2\sigma^2(n)}{\eul(n)}$.
\begin{Lista}
\item[a)] Let $c_n := \frac{\mu(n)^2\sigma^2(n)}{n\eul(n)}$ and let $F(s):=\sum_n c_n/n^s$. Prove that
          \[
           F(s) = \zeta(s)G(s),
           \quad
           \text{with}
           \quad
           G(s):=\prod_p\Big(1+\frac{3p+1}{p(p-1)p^s}-\frac{(p+1)^2}{p(p-1)p^{2s}}\Big).
          \]
          The Euler product defining $G(s)$ converges absolutely in $\Ree(s)>1/2$, therefore the previous
          equality gives the meromorphic continuation of $F(s)$ in $\Re(s)>1/2$, with a unique, simple,
          pole at $s=1$, coming from the pole of $\zeta$. The residue of $F(s)$ at $s=1$ is $G(1)$.
%
\item[b)] Let $g(n)$ denote the Dirichlet coefficients of $G(s)$, so that
          \[
           c_n = \sum_{k|n} g(k).
          \]
          Deduce that
          \[
           \sum_{n\leq X} c_n = \sum_{n\leq X} \sum_{k|n} g(k)
                              = \sum_{k\leq X} g(k) \intpart{\frac{X}{k}}
                              =X\sum_{k\leq X} g(k)/k + O(\sum_{k\leq X}|g(k)|).
          \]
%
\item[c)] Prove that:
          \begin{align*}
            &\sum_{k\leq X} g(k)/k \to G(1) \quad \text{as $X\to\infty$}, && \\
           %
            &\sum_{k\leq X}|g(k)|
             \leq X^\theta \sum_{k\leq X}|g(k)|k^{-\theta}
             \ll_\theta  X^\theta                                         &&\forall \theta>1/2,\\
           %
            &\sum_{k> X} |g(k)|/k
            \leq X^{\theta-1}\sum_{k> X} |g(k)|/k^\theta
            \ll_\theta X^{\theta-1}
                                                                          &&\forall \theta>1/2,
          \end{align*}
          so that
          \[
            \sum_{n\leq X} c_n = G(1)X + O_\theta(X^\theta),
            \qquad \forall \theta>1/2.
          \]
%
\item[d)] Let $d_n := \frac{\mu(n)^2\sigma^2(n)}{\eul(n)}$, so that $c_n = \frac{d_n}{n}$. By partial
          summation from Step~c deduce that
          \[
           \sum_{n\leq X} d_n = \frac{G(1)}{2}\,X^2 + O_\theta(X^{1+\theta})
           \qquad \forall\theta>1/2.
          \]
%
\item[e)] Prove that $G(1)=\prod_p\big(1 + \frac{2p+1}{p^3}\big) = 2.4748\ldots$.
\end{Lista}
      \bigskip
      \bigskip

\item Let $f$ be the \emph{multiplicative} arithmetical function which is supported on squarefree numbers
      (i.e., such that if $n$ not squarefree then $f(n)=0$), and whose values on prime numbers is:
      \[
       f(p)=
       \begin{cases}
         2            & p=2        \\
       \frac{2p+2}{p} & p=1\pmod{4}\\
       \frac{2p}{p+2} & p=3\pmod{4}.
       \end{cases}
      \]
\begin{Lista}
\item[a)] Prove that
      \[
       \sum_{n\leq X} f(n) = \alpha X\log X + O(X)
      \]
      with
      \[
       \alpha := \prod_{p}\Big(1 + \frac{f(p)}{p}\Big)\Big(1 - \frac{1}{p}\Big)^2 = 0.23504\ldots.
      \]

\item[b)] Are you able to devise a method to \emph{prove} that the first four digits of values in Step a)
      are correct? In other words, you are asked to provide an explicit inequality of type
      \[
       \Big|\prod_{\substack{p=1\!\!\pod{4}\\ p>M}}\Big(1 + \frac{f(p)}{p}\Big)\Big(1 - \frac{1}{p}\Big)^2
       -1\Big|
       \leq \text{something decaying with $M$}
      \]
      in order to be able to select a convenient value of $M$ such that when the product is truncated
      to primes $\leq M$ then its value differs from the true value by a quantity which is for sure
      smaller than $10^{-4}$.

\item[c)] Prove that both products
       \[
         \prod_{p=1\!\!\pod{4}}\Big(1 + \frac{f(p)}{p}\Big)\Big(1 - \frac{1}{p}\Big)^2
         \qquad,\qquad
         \prod_{p=3\!\!\pod{4}}\Big(1 + \frac{f(p)}{p}\Big)\Big(1 - \frac{1}{p}\Big)^2
       \]
       converge: to $0.9328\ldots$ and $0.5039\ldots$, respectively.
%
%G prodeuler(p=1,10000,if(Mod(p,4)==Mod(1,4),(1.+(2*p+2)/p^2)*(1-1/p)^2,1))
%G \\ 0.9328632125264094359803156692
%G prodeuler(p=1,10000,if(Mod(p,4)==Mod(3,4),(1.+2/(p+2))*(1-1/p)^2,1))
%G \\ 0.5039172324011527157262807938
\end{Lista}
      \noindent%
      {\bf Hint:} for Item a), imitate what we have done for the proof of Proposition~3.3 in the Notes to the course.
      \bigskip
      \bigskip

%Repeat the previous exercise for the sum $\sum_{n\leq X} \frac{\mu(n)^2 n^3}{\sigma_2(n)}$, where $\sigma_2(n):= \sum_{d|n}d^2$.\\
%      Deduce that
%      \[
%       \sum_{n\leq X} \mu(n)^2\Big[\frac{n^3}{\sigma_2(n)} - \frac{\eul^2(n)}{\sigma(n)}\Big]
%       > 0
%      \]
%      when $X$ is large enough.
%      \bigskip
%      \bigskip

\item With the notation in Chapter~3 of the notes to the course, suppose that $b(2) = 1$, $b(3) = 2$ and
      $b(p) = 3$ for every $p\geq 5$ (and dividing $P$, of course).
      Imitating the proof of Proposition 3.3, prove that
      \[
       S_{\mathcal{B}}(x,y;P)
        \leq 6 c_3 \frac{y}{\log^3 z}
             + O\Big(\frac{y}{\log^4z}\Big)
             + O\Big(\frac{z^2}{\log^2z}\Big)
      \]
      where $c_3 := \frac{9}{2}\prod_{p\geq 5}\big(1 - \frac{3p-1}{(p-1)^3}\big)= 2.8582\ldots$.
      \bigskip
      \bigskip

\item Let $S_3(x,y)$ be the number of primes $p$ in $(x,x+y]$ such that also $p+2$ and $p+6$ are primes
      (as it happens for example for $p=5,11,17$ and $41$). Prove that
      \[
       S_3(x,y)
        \leq 48 c_3 \frac{y}{\log^3 y}\Big(1+O\Big(\frac{\log\log y}{\log y}\Big)\Big)
      \]
      \noindent%
      {\bf Hint:} imitate the proof of Theorem 3.3 in the Notes to the course.
\end{enumerate}
\end{document}
