TITLE: Show that $\bigl| e^x + e^{-x}-2-x^2\bigr| \le {x^4 \over 6} $ for $|x| \leq 1$
QUESTION [5 upvotes]: My try at it $$ \left| e^x + e^{-x}-2-x^2\right| \iff | f(x) - p_2(x)| = |R_3(x)| $$
where $ f(x) = e^x + e^{-x} $ and $ |x| \le 1 $ 
This gets me
$$ |R_3(x)| \le (e-e^{-1}) {x^3 \over 6} $$
This is as far as I've gotten. Where did I mess up?

REPLY [2 votes]: The user OohAah pointed me in the right direction.
$$ |f(x)-p_3(x)| = |R_4(x)| $$
$$ |R_4(\phi x)| = (e^{\phi x}+e^{-\phi x}) {x^4 \over 24} $$
where $ 0 \le \phi \le 1 $. Since $ |x| \le 1 $ it follows that
$$ \phi x \le 1 $$
Ergo 
$$ |R_4(x)| \le (e+e^{-1}){x^4 \over 24} \le {x^4 \over 6} $$
If my mathematical writing is sloppy, please do point it out. I'm trying to improve.