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I have to solve the following exercise:

Give $a$, $b$, $c \in \mathbb R$ such that $$\frac{1}{1-\cos x} = \frac{a}{x^2} + b +cx^2 + o(x^2)$$ for $x\to 0$.

Here's my attempt:
I know that $$\cos x = \frac{1}{2}\left( e^{ix} + e^{-ix}\right) = \frac{1}{2} \sum_{n=0}^\infty \frac{(ix)^{2n}}{(2n)!} = \frac{1}{2}\sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{(2n)!} = \frac{1}{2}\left(1-\frac{x^2}{2}+o(x^2)\right)$$ for $x\to 0$. The question now is how to resolve $\frac{1}{1-\cos x}$. This much looks like an application of the geometric series here. I assume I can use it since $\cos x < 1$ for all sufficiently small $x \neq 0$. I wouldn't know how to solve this exercise otherwise. Proceeding, it follows that $$\frac{1}{1-\cos x} = \frac{1}{1-\left(\frac{1}{2} - \frac{x^2}{4} + o(x^2)\right)} = \sum_{n=0}^\infty\left(\frac{1}{2} - \frac{1}{4}x^2+o(x^2)\right)^n = 1+ \frac{1}{2}-\frac{1}{4}x^2+o(x^2).$$ Choosing $a=0$, $b = 1,5$ and $c=-\frac{1}{4}$ establishes the case.

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    Your expansion is wrong. Personally, I just look up the expansion for trig functions and similar well-known functions via google. – 2017-01-08
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    Where is the mistake? – 2017-01-08
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    When you add the expansions for $e^{ix}$ and $e^{-ix}$, you forget there are twice as many terms, which cancels with the $1/2$ in the front. – 2017-01-08
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    Damn, I see. I was already wondering why that $1/2$ didn't vanish. But is the rest of the attempt fine? – 2017-01-08
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    Looks good. Right ideas everywhere. – 2017-01-08
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    Is the application of the geometric series allowed, even though $\cos 0 = 1$? – 2017-01-08
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    @lappen68 If you look carefully, you should see that one uses the geometric series expansion for $$g(x)=1-\frac2{x^2}(1-\cos x)$$ not for $\cos x$ itself. And $g(0)=0$, as desired... (Excellent comment, by the way.) – 2017-01-08
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    Sure. But we only achieve that by factoring out $\frac{2}{x^2}$ which again we only get when we expand at order $6$ right? So just using the expansion up to order $2$ or $4$ would not suffice I assume. – 2017-01-08

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Your expansion of $\cos(x)$ at $0$ is obviously wrong, since it implies $\lim_{x\to 0}\cos(x)=\frac 12$.

You needn't use the power series for $e^{ix}$. I'd suggest that you rather use Taylor expansion at order $6$ for $\cos$, since the derivatives at $0$ are very simple. This method yields $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+o(x^6)$$

Hence $$\begin{align} \frac{1}{1-\cos x} &=\frac{1}{1-\left(1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+o(x^6) \right)}\\ &=\frac{2}{x^2}\frac{1}{1-\left(\frac{x^2}{12}-\frac{x^4}{360}+o(x^4)\right)}\\ &=\frac{2}{x^2} \left( 1+\left(\frac{x^2}{12}-\frac{x^4}{360}+o(x^4)\right)+ \left(\frac{x^2}{12}-\frac{x^4}{360}+o(x^4)\right)^2+ o\left(\left(\frac{x^2}{12}-\frac{x^4}{360}+o(x^4)\right)^2\right)\right)\\ &=\frac{2}{x^2}\left( 1+\frac{x^2}{12}-\frac{x^4}{360}+\frac{x^4}{144}+o\left(x^4\right)\right)\\ &=\frac{2}{x^2} + \frac 16 + \frac{x^2}{120}+o(x^2) \end{align}$$

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    That is a smooth answer, making sure the term for the geometric series is $< 1$. – 2017-01-08
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    @lappen68 it is, whenever $|x|$ is smaller than some $\epsilon$. – 2017-01-08
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hint

$$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+o(x^6)$$ and after factoring out by $x^2$ ,

$$\frac{1}{1-X}=1+X+X^2+o(X^2).$$

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    Why passing by the square of the sine of the half angle any simpler than expanding the cosine one step further? – 2017-01-08
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    The exercise is taken from a course exam, I doubt they expect you to come up with this kind of solution. Thanks for the hint though! – 2017-01-08
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    And now the expansion of cosine you suggest does not allow to compute $c$. Sorry but I must downvote this. – 2017-01-08
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    @Did Now you will remember it. – 2017-01-08
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    "@Did Now you will remember it." Sorry but what do you mean by that? This sounds exciting but please be explicit. – 2017-01-08
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    @Did It is in the positive sense. – 2017-01-08
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    Note for your future use of the site: When you modify a post in reaction to a comment, it is customary (and polite) to signal the modification in a new comment. – 2017-01-08
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    I will remember it. thanks. – 2017-01-08