mairie0708zl

2022-01-26

Proof that $|\mathrm{arctan}\left(x\right)-\frac{\pi }{4}-\frac{\left(x-1\right)}{2}|\le \frac{{\left(x-1\right)}^{2}}{2}$
I'm trying to prove that, for every $x\ge 1$ :
$|\mathrm{arctan}\left(x\right)-\frac{\pi }{4}-\frac{\left(x-1\right)}{2}|\le \frac{{\left(x-1\right)}^{2}}{2}$
I could do it graphically on $\mathbb{R}$, but how to make a formal algebraic proof?

chaloideq1

Expert

We have that for $x\ge 1$
$\mathrm{arctan}x-\frac{\pi }{4}-\frac{\left(x-1\right)}{2}\le 0⇒|\mathrm{arctan}x-\frac{\pi }{4}-\frac{\left(x-1\right)}{2}|=\frac{\pi }{4}+\frac{\left(x-1\right)}{2}-\mathrm{arctan}x$
then consider
$f\left(x\right)=\frac{{\left(x-1\right)}^{2}}{2}+\mathrm{arctan}x-\frac{\pi }{4}-\frac{\left(x-1\right)}{2}⇒{f}^{\prime }\left(x\right)=\frac{1}{{x}^{2}+1}+x-\frac{3}{2}\ge 0$
and since f(1)=0 we have that $f\left(x\right)\ge 0$ and the inequality is proved.

euromillionsna

Expert

Hint: For each $x>1$, Taylor gives
$\mathrm{arctan}\left(x\right)=\frac{\pi }{4}+\frac{\left(x-1\right)}{2}+\frac{\mathrm{arctan}\left({c}_{x}\right)}{2}{\left(x-1\right)}^{2}$
where $1<{c}_{x} Thus all you need to show is that $|\mathrm{arctan}\left(c\right)|\le 1$ for all $c\ge 1$

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