# Help with Calculus?

May 9, 2018

Use the taylor series for $\cos x$, which is 1 - x^2/(2!) + x^4/(4!) + ...

We now rewrite the inequality.

0 ≤ 1- x^2/(2!) + x^4/(4!) - 1 + x^2/(2!) ≤ 1/24

0 ≤ x^4/(4!) ≤ 1/24

The maximum of this will be at $x = 1 \mathmr{and} - 1$, or 1/(4!) = 1/24. The minimum will be at $x = 0$ or $0$.

If we add more terms it makes no difference.

0 ≤ x^4/(4!) - x^6/(6!) ≤ 1/24

The minimum will ALWAYS be $0$ no matter the value of $x$. The maximum will be at most $\frac{1}{24}$, because the terms after x^4/(4!) converge to $0$ when $x = 1$.

We have therefore proved the required statement.

Hopefully this helps!