How to see $\cos x \leq \exp(-x^2/2)$ on $x \in [0,\pi/2]$?

Solution 1:

Without any special treatment for $x=\frac{\pi}{2}$, which can be verified manually, we have $\forall x \in \left [0, \frac{\pi}{2} \right )$:

$$0<\cos(x)\leq e^{-\frac{x^2}{2}} \Leftrightarrow \ln(\cos(x)) \leq -\frac{x^2}{2}$$ just because $\ln(x)$ is ascending, strictly.

As a result, let's look at this function $f(x)=-\frac{x^2}{2}-\ln(\cos(x))$. $$f'(x)=\tan(x)-x$$ $$f''(x) = (\tan(x))^2$$

Second derivative says that the first one is ascending, thus for $x\geq0$: $$f'(x)=\tan(x)-x \geq f'(0)=0$$ So $f(x)$ is also ascending, which means $x\geq0$: $$f(x)=-\frac{x^2}{2}-\ln(\cos(x)) \geq f(0)=0$$

Solution 2:

You may use the Weierstrass product for the cosine function: $$ \cos x = \prod_{n\geq 0}\left(1-\frac{4x^2}{(2n+1)^2\pi^2}\right) \tag{1}$$ and by taking $\log$s: $$\begin{eqnarray*}\log\cos x&=&\sum_{n\geq 0}\log\left(1-\frac{4x^2}{(2n+1)^2\pi^2}\right)\\&=&-\sum_{n\geq 0}\sum_{m\geq 1}\frac{4^m x^{2m}}{m(2n+1)^{2m}\pi^{2m}}\\&=&-\sum_{m\geq 1}(4^m-1)\frac{\zeta(2m)}{m\pi^{2m}}\,x^{2m}\tag{2}\end{eqnarray*} $$ so: $$ \log\cos x\leq -3\frac{\zeta(2)}{\pi^2}x^2 = -\frac{x^2}{2}\tag{3}$$ as wanted. The same approach also proves $\cos(x)\leq \exp\left(-\frac{x^2}{2}-\frac{x^4}{12}-\frac{x^6}{45}\right)$.
An alternative approach is to notice that $\tan(x)\geq x$ for any $x\in\left(0,\frac{\pi}{2}\right)$ holds by convexity, hence by integrating both sides over the interval $(0,\theta)$ we get $\log\cos\theta\leq-\frac{\theta^2}{2}$ for any $\theta\in\left(0,\frac{\pi}{2}\right)$.

Solution 3:

Here is another option to prove the inequality by Taylor expansions as the OP suggested. So we are to prove that

$$ \exp(-x^2/2)\ge\cos x,\quad x\in[0,\pi/2]. $$

By doing Taylor expansions at zero with the Lagrange remainders, we can get the estimations \begin{eqnarray} \exp(-t)&=&1-t+\frac{t^2}{2!}-\frac{t^3}{3!}+\underbrace{\frac{\exp(-\xi)}{4!}t^4}_{\ge 0}\ge 1-t+\frac{t^2}{2!}-\frac{t^3}{3!},\\ \cos x&=&1-\frac{x^2}{2!}+\frac{x^4}{4!}-\underbrace{\frac{\cos\xi}{6!}x^6}_{\ge 0}\le 1-\frac{x^2}{2!}+\frac{x^4}{4!}. \end{eqnarray} Now we can write with $t=x^2/2$ $$ \exp(-x^2/2)\ge 1-\frac{x^2}{2}+\frac{x^4}{8}\color{red}{-\frac{x^6}{48}}\ge 1-\frac{x^2}{2}+\frac{x^4}{8}\color{red}{-\frac{x^4}{12}}= 1-\frac{x^2}{2}+\frac{x^4}{24}\ge\cos x,\quad x\in[0,\pi/2]. $$ Here at the second step we have used the bound $$ \color{red}{-\frac{x^6}{48}}\ge\color{red}{-\frac{x^4}{12}}\iff x^4(2+x)(2-x)\ge 0 $$ which is true for $0\le x\le\pi/2<2$.