you have your original series incorrect, btw, it should be alternating signs.

your 4th derivative is correct, however.

so you should wind up with:

there is a reason i brought |x^4| = x^4 out of the remainder term. if we can find some global maximum for the rest, then the remainder will only depend on the value of |x|.

we can simplify a bit:

now, let's look at the expression inside the absolute value sign, we would like to find an upper bound for it. if

then

note that for g'(z) = 0, either z = 0, or z is a solution to , which is a quadratic in .

the solutions to this quadratic are:

as z goes to plus or minus infinity, eventually the exponential term dominates, so:

so the global maximum for |g(z)| is one of the roots of g'(z). g(z) is an even function, so it suffices to check the 3 non-negative roots of g'(z).

g(0) = 3 (well, that was easy).

at , we get:

at ,

so |g(z)| has a global maximum of 3 at 0.

so:

so in order to ensure choose

(there might be an easier way....feel free, people).