Algotssleeddynf

2021-12-28

How do we show $1-\mathrm{cos}x\ge \frac{{x}^{2}}{3}$ for $\left|x\right|\le 1$ ? 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.

which is true for all

Ethan Sanders

Beginner2021-12-29Added 35 answers

For $\left|x\right|\le 1$ 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 $n\ge 1$

is an alternating series with terms that decrease in absolute value. It follows that for these x

and therefore

That is an even better estimate since

The same approach can be used to show that

because the

Cheryl King

Beginner2021-12-30Added 36 answers

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

Vasquez

Expert2022-01-09Added 669 answers

Now

Hence

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