Algotssleeddynf

2021-12-28

How do we show $1-\mathrm{cos}x\ge \frac{{x}^{2}}{3}$ for $|x|\le 1$? My first idea was to write
$1-\mathrm{cos}x=\frac{1}{2}{|{e}^{ix}-1|}^{2}$
which is true for all $x\in \mathbb{R}$, but I don't have a suitable lower bound for the right-hand side at hand.

Ethan Sanders

Expert

For $|x|\le 1$ the Taylor series
$\mathrm{cos}\left(x\right)=1-\frac{{x}^{2}}{2!}+\frac{{x}^{4}}{4!}-\frac{{x}^{6}}{6!}+\dots$
is an alternating series with terms that decrease in absolute value. It follows that for these x
$\mathrm{cos}\left(x\right)\le 1-\frac{{x}^{2}}{2!}+\frac{{x}^{4}}{4!}\le 1-\frac{{x}^{2}}{2!}+\frac{{x}^{2}}{4!}=1-\frac{11}{24}{x}^{2}$
and therefore
$1-\mathrm{cos}\left(x\right)\ge \frac{11}{24}{x}^{2}$
That is an even better estimate since $\frac{11}{24}>\frac{13}{}$
The same approach can be used to show that $1-\mathrm{cos}\left(x\right)\ge \frac{{x}^{2}}{3}$ holds on the larger interval [−2,2]:
$\mathrm{cos}\left(x\right)\le 1-\frac{{x}^{2}}{2!}+\frac{{x}^{4}}{4!}\le 1-\frac{{x}^{2}}{2!}+\frac{4{x}^{2}}{4!}=1-\frac{1}{3}{x}^{2}$
because the $\frac{{x}^{2n}}{\left(2n\right)!}$ terms decrease in absolute value for $n\ge 1$

Cheryl King

Expert

Since sinx is concave for $x\in \left[0,\frac{\pi }{2}\right]$, if $x\in \left[0,1\right]$ then $\mathrm{sin}x\le x\mathrm{sin}1⇒1-\mathrm{cos}x\ge {x}^{2}\frac{12}{\mathrm{sin}1}$. The ${x}^{2}$ coefficient approximates 0.42. The generalization $x\in \left[-1,1\right]$ follows from $1-\mathrm{cos}x,c{x}^{2}$ being even.

Vasquez

Expert

$f\left(x\right)=\frac{1-\mathrm{cos}x}{{x}^{2}}$ is an even function, so we only look at $x\in \left[0,1\right]$. It's easy to prove f(x) is continuous and differentiable at x=0, and
$f\prime \left(x\right)=\frac{x\mathrm{sin}x+2\mathrm{cos}x-2}{{x}^{3}}$
Now
$x\mathrm{sin}x+2\mathrm{cos}x-2\le 2\mathrm{tan}\frac{x}{2}\cdot \mathrm{sin}x-4{\mathrm{sin}}^{2}\frac{x}{2}$
$=2\mathrm{tan}\frac{x}{2}\cdot 2\mathrm{sin}\frac{x}{2}\mathrm{cos}\frac{x}{2}-4{\mathrm{sin}}^{2}\frac{x}{2}=0,\mathrm{\forall }x\in \left[0,1\right]$
Hence

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