Vinogradov's Theorem/Major Arcs/Lemma 1
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Lemma
Let $\phi$ be the Euler $\phi$ function.
Let $\mu$ be the Möbius function.
Let $c_q$ be the Ramanujan sum modulo $q$.
Let $P, N \ge 1$.
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Let:
- $\ds \map {\SS_P} N := \sum_{q \mathop \le P} \frac {\map \mu q \map {c_q} N} {\map \phi q^3}$
- $\ds \map \SS N := \lim_{P \mathop \to \infty} \map {\SS_P} N$
Then:
- $\map \SS N = \map {\SS_P} N + \map \OO {P^{\epsilon -1} }$
and $\SS$ has the Euler product:
- $\ds \map \SS N = \prod_{p \mathop \nmid N} \paren {1 + \frac 1 {\paren {p - 1}^3} } \prod_{p \mathop \divides N} \paren {1 - \frac 1 {\paren {p - 1}^2} }$
where:
- $p \nmid N$ denotes that $p$ is not a divisor of $N$
- $p \divides N$ denotes that $p$ is a divisor of $N$.
Proof
We have:
\(\ds \size {\map {c_q} N}\) | \(\le\) | \(\ds \sum_{\substack {1 \mathop \le a \mathop \le q \\ \map \gcd {a, q} \mathop = 1} } 1\) | Triangle Inequality | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi q\) | Definition of Euler $\phi$ Function |
Trivially we have $\size {\map \mu q} \le 1$.
Therefore:
\(\ds \size {\map \SS N - \map {\SS_P} N}\) | \(=\) | \(\ds \sum_{q \mathop > P} \frac {\size {\map \mu q \map {c_q} N} } {\map \phi q^3}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \sum_{q \mathop > P} \frac 1 {\map \phi q^2}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \sum_{q \mathop > P} q^{\epsilon - 2}\) | for large $P$ and $\epsilon \in \R_{>0}$, by Asymptotic Growth of Euler Phi Function | |||||||||||
\(\ds \) | \(\le\) | \(\ds P^{\delta - 1} \sum_{q \mathop > P} q^{\epsilon - \delta - 1}\) | for any $\delta > \epsilon$ |
Therefore by Convergence of Powers of Reciprocals:
- $\map \SS N = \map {\SS_P} N + \map \OO {P^{\epsilon - 1} }$
as claimed.
By definition of Euler product, and because $\map \mu {p^k} = 0$ for $k > 1$:
- $\ds \map \SS N = \prod_p \paren {1 + \frac {\map \mu q \map {c_q} N} {\map \phi q^3} }$
Now for a prime $p$ we have:
- $\map \mu p = -1$
- $\map \phi p = p - 1$
We also have Kluyver's Formula for Ramanujan's Sum:
- $\ds \map {c_p} n = \sum_{d \mathop \divides \map \gcd {p, n} } \rd \map \mu {\frac p d}$
Let $p \divides N$.
Then: $\map \gcd {p, N} = p$
which gives:
- $\map {c_p} N = p \map \mu 1 + \map \mu p = p - 1$
Let $p \nmid N$.
Then:
- $\map \gcd {p, N} = 1$
and so:
- $\map {c_p} N = -1$
Therefore:
- $\ds \map \SS N = \prod_{p \mathop \divides N} \paren {1 - \frac 1 {\paren {p - 1}^2} } \prod_{p \mathop \nmid N} \paren {1 + \frac 1 {\paren {p - 1}^3} }$
This completes the proof.
$\blacksquare$