# 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.

Maclaurin series can be treated as infinite polynomials (most of the time), so we have $$\cos2x-1=- \frac{4x^2}{2!} + \frac{16x^4}{4!} - \frac{64x^6}{6!}+\dots$$ $$\frac{\cos2x-1}{x^2}=- \frac{4}{2!} + \frac{16x^2}{4!} - \frac{64x^4}{6!}+\dots$$ $$=-2+\frac23x^2-\frac4{45}x^4+\dots$$
• So if I am understanding correctly, you divided the maclaurin series of $cos(2x)-1$ by $\frac{1}{x^2}$ Commented Mar 12, 2019 at 2:50
• @Ludwig Yes, after subtracting $1$. Commented Mar 12, 2019 at 2:50
Since $$\begin{eqnarray} \cos(2x) = 1 - \frac{4x^2}{2!} + \frac{16x^4}{4!} - \frac{64x^6}{6!} \cdots \end{eqnarray}$$ $$\cos (2x) - 1 = \frac{4x^2}{2!} + \frac{16x^4}{4!} - \frac{64x^6}{6!} \cdots$$ $$\frac{\cos(2x)-1}{x^2} = \frac{4}{2!} + \frac{16x^2}{4!} - \frac{64x^4}{6!} \cdots$$ $$=\boxed{2 + \frac{16x^2}{4!} - \frac{64x^4}{6!} \cdots}$$