Show that $\bigl| e^x + e^{-x}-2-x^2\bigr| \le {x^4 \over 6} $ for $|x| \leq 1$

Question

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?

Answer 1

Look at the series of $2\cosh x$: $$2\cosh x = 2 + x^2+\frac{x^4}{12}+ \ldots= \sum_{k=0}^\infty 2\frac{x^{2k}}{(2k)!}$$

Thus, when $|x|<1$: $$e^x+e^{-x}-2-x^2=\sum_{k=2}^\infty 2\frac{x^{2k}}{(2k)!} < x^4\sum_{k=2}^\infty\frac{2}{(2k)!}= (2\cosh 1 -3)x^4$$ $$\approx 0.086\ x^4 <x^4/6$$

Answer 2

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.