Verifying the bound $\cos^2x\leq 2 e^{-x^2/4}-1$

A CAS plot suggests that for $x\in[0,\pi/2]$:

$$\cos^2x\leq 2\, e^{-x^2/4}-1.$$

The difference is not increasing in this range so I don't think differentiation alone will do.

Any suggestions welcome.

Note $\cos x\leq e^{-x^2/2}$ in this range.

We can actually prove that

$$2e^{-x^2/4}-1\ge\cos x,\quad\forall x\in[0,\pi/2].\tag{1}$$

It is a tighter inequality because $\cos x\ge\cos^2x$ in the interval. We have $$(1)\iff e^{-x^2/4}\ge \frac{1+\cos x}{2}=\cos^2(x/2)\quad\forall x\in[0,\pi/2].\tag{2}$$ Now denote $t=x/2\in[0,\pi/4]$ and continue $$(2)\iff e^{-t^2}\ge\cos^2t\iff e^{-t^2/2}\ge\cos t,\quad\forall t\in[0,\pi/4].\tag{3}$$ The last inequality in (3) can be estimated from the Taylor expansions using at the second step that $|t|<1$, hence, $t^6\le t^4$ $$e^{-t^2/2}\ge 1-\frac{t^2}{2}+\frac{t^4}{8}-\frac{t^6}{48}\ge 1-\frac{t^2}{2}+t^4\Big(\underbrace{\frac{1}{8}-\frac{1}{48}}_{=\frac{5}{48}>\frac{1}{24}}\Big)\ge 1-\frac{t^2}{2}+\frac{t^4}{24}\ge\cos t.$$

P.S. Explanations of inequalities from Taylor expansions with Lagrange remainders: \begin{eqnarray} e^{-x}&=&1-x+\frac{x^2}{2}-\frac{x^3}{3!}+\frac{e^{-\xi}}{4!}x^4\ge1-x+\frac{x^2}{2}-\frac{x^3}{3!},\\ \cos x&=&1-\frac{x^2}{2}+\frac{x^4}{4!}-\frac{\cos\xi}{6!}x^6\le1-\frac{x^2}{2}+\frac{x^4}{4!}. \end{eqnarray}

EDIT: Thanks to Marty Cohen sharing the link to a really elementary proof of (3) by differentiation.

• How do you know that you can truncate the series? For some valid proofs, see math.stackexchange.com/questions/2769766/… – marty cohen Jun 7 '18 at 21:01
• @martycohen Truncations can be validated by the fact that the Lagrange remainder is non-negative for $\exp$ and non-positive for $\cos$. – A.Γ. Jun 7 '18 at 21:03
• Then it should be stated with a link to a proof. – marty cohen Jun 8 '18 at 2:31

The given inequality is a consequence of $$f(x)=\frac{\sin^2 x}{x^2} > \frac{2-2e^{-x^2/4}}{x^2}=g(x) \tag{1}$$ for any $x\in\left(0,\frac{\pi}{2}\right)$. $(1)$ is a pretty loose inequality, and both $f(x)$ and $g(x)$ are decreasing and concave functions over $I=\left(0,\frac{\pi}{2}\right)$. In particular $$\forall x\in I,\quad f(x)\geq 1-\frac{2}{\pi}\left(1-\frac{4}{\pi^2}\right)x>\begin{smallmatrix}\text{the equation of the tangent line}\\\text{to the graph of }g(x)\text{ at }x=\pi/2\end{smallmatrix}\geq g(x)\tag{2}$$ and $(1)$ is proved.

Hint. Split the interval $[0,\pi/2]$ in two halves.

1) Show that for $x\in [0,\pi/4]$, $$\cos^2(x)\leq 1-\frac{x^2}{2}\leq 2 e^{-x^2/4}-1.$$ Note that the Taylor expansions at $x=0$ of the two functions are $$\cos^2(x)=1-\frac{x^2}{2} +O(x^4)\quad\mbox{and}\quad 2 e^{-x^2/4}-1=1-x^2+O(x^4).$$

2) Show that for $x\in [\pi/4,\pi/2]$, $$\cos^2(x)\leq 1-\frac{2x}{\pi}\leq 2 e^{-x^2/4}-1.$$ Note that $\cos^2(x)$ is convex and $2 e^{-x^2/4}-1$ is concave in such interval.