klamytewoc

2022-03-01

Prove that $x-\frac{{x}^{3}}{3!}<\mathrm{sin}\left(x\right) for all x>0

OxicegeWheegeb0k

Recursive integration We know that
$0\le \mathrm{cos}a\le 1⇒\mathrm{sin}t={\int }_{0}^{t}\mathrm{cos}sds
for $0. Integrating over $\left(0,z\right)$ we get
$1-\mathrm{cos}z={\int }_{0}^{z}\mathrm{sin}tdt<{\int }_{0}^{z}tdt=\frac{{z}^{2}}{2}$
that is for all $0 we have,
$1-\frac{{z}^{2}}{2}<\mathrm{cos}z\le 1$
integrating again over $\left(0,x\right)$ we get
$x-\frac{{x}^{3}}{6}={\int }_{0}^{x}1-\frac{{z}^{2}}{2}dz<{\int }_{0}^{x}\mathrm{cos}zdz=\mathrm{sin}x$
that is
$x-\frac{{x}^{3}}{6}<\mathrm{sin}x
continuing with this process you get,
$1-\frac{{x}^{2}}{2}<\mathrm{cos}x<1-\frac{{x}^{2}}{2}+\frac{{x}^{4}}{24}$
$x-\frac{{x}^{3}}{6}<\mathrm{sin}x
More generally for $n\ge 1$, by induction we get
$\sum _{k=0}^{2n-1}{\left(-1\right)}^{k}\frac{{x}^{2k}}{\left(2k\right)!}<\mathrm{cos}x<\sum _{k=0}^{2n-1}{\left(-1\right)}^{k}\frac{{x}^{2k}}{\left(2k\right)!}+\frac{{x}^{4n}}{\left(4n\right)!}$

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