Vinogradov's Theorem/Minor Arcs/Lemma 1
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Lemma
For $\beta \in \R$, define:
- $\norm \beta := \min \set {\size {n - \beta}: n \in \Z}$
Then:
- $\ds \forall \alpha \in \R: \size {\sum_{k \mathop = N_1}^{N_2} \map e {\alpha k} } \le \min \set {N_2 - N_1, \frac 1 {2 \norm \alpha} }$
Proof
The bound $N_2 - N_1$ is trivial:
Since $\size {\map e {\alpha k} } = 1$ for all $k$, by the Triangle Inequality:
- $\ds \size {\sum_{k \mathop = N_1}^{N_2} \map e {\alpha k} } \le \sum_{k \mathop = N_1}^{N_2} 1 = N_2 - N_1$
To show the second bound we evaluate the sum as a geometric series.
We have:
- $\map e {\alpha k} = \map e \alpha^k$
So by Sum of Geometric Sequence:
\(\ds \size {\sum_{k \mathop = N_1}^{N_2} \map e {\alpha k} }\) | \(=\) | \(\ds \size {\frac {\map e {\alpha \paren {N_2 + 1} } - \map e {\alpha N_1} } {\map e \alpha - 1} }\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \frac 2 {\size {\map e \alpha - 1} }\) |
By the polar form of a complex number:
- $\map e \alpha - 1 = \map e {\alpha / 2} \paren {\map e {\alpha / 2} - \map e {-\alpha / 2} }$
and:
- $\map e {\alpha / 2} - \map e {-\alpha / 2} = \map \exp {\pi i \alpha} - \map \exp {-\pi i \alpha} = 2 i \map \sin {\pi \alpha}$
Therefore:
- $\ds \size {\sum_{k \mathop = N_1}^{N_2} \map e {\alpha k} } \le \frac 1 {\size {\map \sin {\pi \alpha} } }$
We know that the sine function is concave on $\closedint 0 {\pi / 2}$.
So:
- $\map \sin {\pi \alpha} \ge 2 \alpha$
for $\alpha \in \closedint 0 {1/2}$.
By definition there exists $n \in \Z$ such that:
- $\alpha = n + \norm \alpha$
and:
- $\norm \alpha \in \closedint 0 {\pi/2}$.
So:
\(\ds \size {\map \sin {\pi \alpha} }\) | \(=\) | \(\ds \size {\sin {\pi \norm \alpha} }\) | Sine and Cosine are Periodic on Reals | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \sin {\pi \norm \alpha}\) | as $\sin x > 0$ for $x \in \closedint 0 {\pi/2}$ | |||||||||||
\(\ds \) | \(\ge\) | \(\ds 2 \norm \alpha\) |
This completes the proof.
$\blacksquare$