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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.

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2 Answers 2

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Maclaurin series can be treated as infinite polynomials (most of the time), so we have $$\cos2x-1=- \frac{4x^2}{2!} + \frac{16x^4}{4!} - \frac{64x^6}{6!}+\dots$$ $$\frac{\cos2x-1}{x^2}=- \frac{4}{2!} + \frac{16x^2}{4!} - \frac{64x^4}{6!}+\dots$$ $$=-2+\frac23x^2-\frac4{45}x^4+\dots$$

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  • $\begingroup$ So if I am understanding correctly, you divided the maclaurin series of $cos(2x)-1$ by $\frac{1}{x^2}$ $\endgroup$
    – Ludwig
    Commented Mar 12, 2019 at 2:50
  • $\begingroup$ @Ludwig Yes, after subtracting $1$. $\endgroup$ Commented Mar 12, 2019 at 2:50
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Since \begin{eqnarray} \cos(2x) = 1 - \frac{4x^2}{2!} + \frac{16x^4}{4!} - \frac{64x^6}{6!} \cdots \end{eqnarray} $$\cos (2x) - 1 = \frac{4x^2}{2!} + \frac{16x^4}{4!} - \frac{64x^6}{6!} \cdots$$ $$\frac{\cos(2x)-1}{x^2} = \frac{4}{2!} + \frac{16x^2}{4!} - \frac{64x^4}{6!} \cdots$$ $$=\boxed{2 + \frac{16x^2}{4!} - \frac{64x^4}{6!} \cdots}$$

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