The problem is, as the title suggests, to find the Power Series Expansion of $\frac{1}{1- \cos x}$ around $x=c$.
What I've tried:
- Direct Computation: Derivatives get very ugly quickly, and don't yield a nice formula that I can recognize as a "series."
- Tried finding the integral of $\frac{1}{1- \cos x}$, finding it's series and then differentiating it to get the new series.
- Tried reverse of the above, differentiating and finding it's series, then integrating (very messy).
- Then I tried some substitution "tricks", like using the series of $\frac{1}{1-x}$ and then plugging in the series expansion for $\cos x$, but that's a double sum that I struggled to produce anything useful from
:$\displaystyle \sum_{k=0}^\infty\left(\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n)!}\right)^k$
I am literally at my witts end with this problem. I have spent perhaps a day or two trying to figure it out, because I feel that I am so close – but just barely missing something. I do not want the solution posted – now it's personal and I have to figure it out, but I would greatly appreciate a hint in the right direction, or to point out a mistake that I may be overlooking.
Best Answer
As @hjpotter92 suggest, you have $$\frac{1}{1-\cos(x)} = \frac{1}{2\sin^2(x/2)} = -\frac{\text{d}}{\text{d}x}(\cot(x/2)).$$ Now you can exploit the series expansion of $$\cot(x)=\sum_{n=0}^\infty\frac{(-1)^n 2^{2n}B_{2n}}{(2n)!}x^{2n-1}, \quad \forall 0<\left|x\right|<\pi.$$ Now, by evaluating in $x/2$, differentiating each coefficient and changing the sign, you get $$\frac{1}{1-\cos(x)} =\sum_{n=0}^\infty\frac{(-1)^{n+1} 2(2n-1)B_{2n}}{(2n)!}x^{2n-2}, \quad \forall 0<\left|x\right|<2\pi,$$ where $B_n$ are the Bernoulli numbers.