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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.

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  • $\begingroup$ Is it 1/cos, as in the title, or 1/(1-cos) as in the first sentence? $\endgroup$
    – coffeemath
    Mar 7, 2014 at 20:55
  • $\begingroup$ Sorry, corrected it - but you beat me to it. It's 1/(1-cos). $\endgroup$
    – user133846
    Mar 7, 2014 at 20:56
  • $\begingroup$ Also, if you want the expansion around the (arbitrary) point $x=c$ it will be very complicated. Are you sure you don't mean to expand around $x=0$? [Looks by your attempts you do mean expand around $0$.] $\endgroup$
    – coffeemath
    Mar 7, 2014 at 20:59
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    $\begingroup$ $ 1 - \cos x = 2 \sin^2 \frac{x}{2} $ $\endgroup$
    – hjpotter92
    Mar 7, 2014 at 20:59
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    $\begingroup$ @coffeemath I wish it was indeed expand around $x=0$. But the intent is for the point of expansion to be arbitrary, which understandably makes the problem harder. My attempts may have been flawed in that sense, but I wanted something to start with, and finding it around $x=0$ was one of those "give it a shot." moments. $\endgroup$
    – user133846
    Mar 7, 2014 at 21:01

3 Answers 3

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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.

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  • $\begingroup$ This works for $c=0$. I don't get anything that doesn't look messy for general $c$. Do you get anything that looks decent? $\endgroup$
    – robjohn ♦
    Mar 7, 2014 at 21:53
  • $\begingroup$ actually this series expansion holds $\forall x \in (0,2 \pi)$, because the expansion of $\cot(x)$ holds $\forall x \in(0,\pi)$, and when you substitute $x$ with $x/2$ the $\pi$ becomes $2\pi$ $\endgroup$
    – Paglia
    Mar 7, 2014 at 21:58
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    $\begingroup$ What the OP means by "around $x=c$" is expanding in a power series in $x-c$; i.e. $$\sum_{k=0}^\infty a_k(x-c)^k$$ $\endgroup$
    – robjohn ♦
    Mar 7, 2014 at 22:01
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    $\begingroup$ Was an initial term $1/x^2$ dropped, since the expansion for $\cot(x)$ begins with $x^{-1}$ which after sign change and derivative would give $1/x^2$? $\endgroup$
    – coffeemath
    Mar 7, 2014 at 22:03
  • $\begingroup$ I think that in this sense the question is probably not answerable. Anyway, way do you need an expansion of this form, if you already have a series that holds $\forall x$ inside your domain? $\endgroup$
    – Paglia
    Mar 7, 2014 at 22:06
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$\frac{1}{1-\cos(x)}=\frac{1+\cos(x)}{1-\cos^2(x)}=\frac{1+\cos(x)}{\sin^2(x)}=\csc^2(x)+\csc(x)\cot(x)$

That last one can be easily noticed to be the derivative of $-(\cot(x)+\csc(x))$. Now the question is a matter of finding the power series expansions of $\cot(x)$ and $\csc(x)$, which are more easily found, and differentiating.

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integration of 1/1-cos(x)=
= 1/1-cos(x)(1+cos(x)/1+cos(x) = 1+cos(x)/[1-cos(x)][1+cos(x)] = 1+cos(x)/[sin(x)][sin(x) = 1/sin(x)[sin(x)]+cos(x)/sin(x)[sin(x)] = cosec(x)[cosec(x)+cot(x)cosec(x) = -cot(x)+cosec(x)+c

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    $\begingroup$ Welcome to MSE. Please use MathJax. $\endgroup$
    – José Carlos Santos
    Aug 4, 2017 at 6:53
  • $\begingroup$ mst ho to frnds me share kro $\endgroup$
    – Shivam Pandey
    Aug 4, 2017 at 6:53
  • $\begingroup$ very good answer $\endgroup$
    – Shivam Pandey
    Aug 4, 2017 at 7:34

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