Exercise 11.2.8

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Prove that Gauss formula (11.10) is equivalent to the formula given in Theorem 11.2.4.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
Let $n = p_1^{\alpha_1}\cdots p_s^{\alpha_s}$ the decomposition of $n$ in primes.
If $m\mid n$, then $m = p_1^{\beta_1}\cdots p_s^{\beta_s^s}$, $0\leq \beta_k \leq \alpha_k$.

If $\beta_k>1$ for some $k=1,\ldots,s$, then by definition $\mu(m)=0$, so $\beta_k = 0$ or $1$ if $\mu(m) 
eq 0$.

The nonzero terms in the sum $$N_n = \frac{1}{n} \sum_{m\mid n} \mu(m) p^{\frac{n}{m}}$$ are those which satisfy $m = p_{i_1}\cdots p_{i_r}$, with $1\leq i_1<\cdots < i_r \leq s$, so $\mu(m) = (-1)^r$.

Thus
$$N_n = \frac{1}{n} \sum_{m\mid n} \mu(m)\,  p^{\frac{n}{m}} =  \frac{1}{n} \ \sum_{1\leq i_1<\cdots < i_r \leq s} (-1)^r\,  p^{\frac{n}{p_{i_1}\cdots p_{i_r}}}.$$
With a less formal writing, we obtain
$$N_n = \frac{1}{n} \sum_{m\mid n} \mu(m)\,  p^{\frac{n}{m}} = \frac{1}{n}\left(p^n - \sum_{a} p^{\frac{n}{a}} + \sum_{a,b} p^{\frac{n}{ab}} - \sum_{a,b,c} p^{\frac{n}{abc}}+\cdots ight),$$
where $\sum_a$ is the sum over all distinct primes $a$ dividing $n$, $\sum_{a,b}$ is the sum over all products of distinct primes $a,b$ dividing $n$, and so on.
\end{proof}

Exercise 11.2.8

Answers

Comments

Exercise 11.2.8

Answers

Comments

Add answer