<s>[INST] Maclaurin series for $\frac{\cos2x-1}{x^2}$ using Maclaurin series for $\cos2x$ I am having some trouble finding the first three nonzero terms for the maclaurin series of $\frac{\cos2x-1}{x^2}$ using the maclaurin series for $\cos2x$.
So far I have the maclaurin series for $\cos2x$ being:
$$\cos2x=1-\frac{4x^2}{2!} + \frac{16x^4}{4!} - \frac{64x^6}{6!}\dots$$
However I am not sure how to proceed in finding the macluarin series for $\frac{\cos2x-1}{x^2}$ now. 
 [/INST] Since \begin{eqnarray}
cos(2x) = 1 - \frac{4x^2}{2!} + \frac{16x^4}{4!} - \frac{64x^6}{6!} ...
\end{eqnarray}
$$\cos (2x) -  1 = \frac{4x^2}{2!} + \frac{16x^4}{4!} - \frac{64x^6}{6!} ...$$
$$\frac{\cos(2x)}{x^2} = \frac{4}{2!} + \frac{16x^2}{4!} - \frac{64x^4}{6!} ...$$
$$=2 + \frac{16x^2}{4!} - \frac{64x^4}{6!} ...$$
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