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
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:
so |g(z)| has a global maximum of 3 at 0.
so in order to ensure choose
(there might be an easier way....feel free, people).