Prove $2+2\cos(2\pi\theta) \leq 4\exp(-2\|\theta \|^2)$. Let $\theta \in \mathbb{R}$, and let $\|\theta\|$ denote the distance to the integer nearest to $\theta$.  I want to prove  $$2+2\cos(2\pi\theta) \leq 4\exp(-2\|\theta \|^2).$$  From the double angle formula we arrive at the equivalent  $$|\cos(\pi \theta) | \leq e^{-\|\theta\|^2},$$ and since both sides are 1-periodic, we need only check that $$0 \leq e^{-\theta^2} - \cos(\pi \theta) \qquad \text{for $\theta \in \left[0,\frac{1}{2}\right]$},$$ and $$0 \leq e^{-(1-\theta)^2} + \cos(\pi \theta) \qquad \text{for $\theta \in \left[\frac{1}{2},1\right]$}.$$  It seems like these aren't as easy as "take the derivative a few times."  Any suggestions?

In fact, because both sides of $|\cos\pi\theta|\le e^{-\Vert\theta\Vert^2}$ are symmetric around $\frac12$ ($f(\theta)=f(1-\theta)$), we only need to check $0\le e^{-\theta^2}-\cos\pi\theta$ for $\theta\in[0,1/2]$. This can be rewritten as
$$e^{-\theta^2}\ge\cos\pi\theta\tag1$$
The idea of the proof of $(1)$ is to find two polynomials $p(x),q(x)$ with $\cos\pi x\le q(x)\le p(x)\le e^{-x^2}$ with $x\in[0,1/2]$ and then prove $q(x)\le p(x)$ over the same range of $x$. These polynomials are truncations of the Maclaurin series of the two sides of $(1)$, to $x^2$ for the LHS and to $x^4$ for the RHS, so we have
$$e^{-x^2}\ge 1-\frac{x^2}2\ge1-\frac{(\pi x)^2}2+\frac{(\pi x)^4}{24}\ge\cos\pi x$$
$$-\frac{x^2}2\ge-\frac{(\pi x)^2}2+\frac{(\pi x)^4}{24}$$
$$-\frac12\ge-\frac{\pi^2}2+\frac{\pi^4x^2}{24}$$
$$\frac{\pi^2-1}2\cdot\frac{24}{\pi^4}\ge x^2$$
$$x\le\sqrt{\frac{\pi^2-1}2\cdot\frac{24}{\pi^4}}=1.045\dots$$
Since this contains $[0,1/2]$, $q(x)\le p(x)$ over the range in question and $(1)$ is proved.
