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How do i get the series expansion $(1-\cos{x})^{-1} = \frac{2}{x^2}+\frac{1}{6}+\frac{x^2}{120}+o(x^4)$ ?

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Do you know the (beginning of the) series expansion of cosine at zero? –  Did Mar 19 '13 at 11:56
    
What is your background, and what do you know already?Thanks in advance still. –  awllower Mar 19 '13 at 12:15
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3 Answers

$$\frac{1}{1 - x} = 1 + x + x^2 + O(x^3)$$

and so

$$\frac{1}{1 - \cos x}= \frac{1}{\frac{x^2}{2} - \frac{x^4}{24} + \frac{x^6}{6!}} = \frac{2}{x^2} \frac{1}{1 - (\frac{x^2}{12} - \frac{x^4}{360} + O(x^6))}$$ $$ = \frac{2}{x^2} \left( 1 + \left(\frac{x^2}{12} - \frac{x^4}{360}\right) + \frac{x^4}{144} + O(x^6) \right) = \frac{2}{x^2} + \frac{1}{6} + \frac{x^2}{120} + O(x^4) $$

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$$ \begin{align} \frac{1}{1 - \cos x} &= \left(\frac{x^2}{2} - \frac{x^4}{24} + \frac{x^6}{720} + o(x^6)\right)^{-1}\\ & = \frac{2}{x^2}\left(1-\frac{x^2}{12} + \frac{x^4}{360} + o(x^4)\right)^{-1}\\ & = \frac{2}{x^2}\left(1 + \frac{x^2}{12} - \frac{x^4}{360} + \frac{x^4}{144} + o(x^4)\right)\\ & = \frac{2}{x^2} + \frac{1}{6} + \frac{x^2}{120} + o(x^2) \end{align} $$

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Can help this: $(1-\cos{x})^{-1} = \dfrac{1}{2\sin^2{\dfrac{x}{2}}}=-\dfrac{d}{dx}\left(\cot{\dfrac{x}{2}}\right)?$

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