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If one sees the simplification done in equation $5.3$ (bottom of page 29) of this paper it seems that a trigonometric identity has been invoked of the kind,

$$\ln(2) + \sum _ {n=1} ^{\infty} \frac{\cos(n\theta)}{n} = - \ln \left\vert \sin\left(\frac{\theta}{2}\right)\right\vert $$

Is the above true and if yes then can someone help me prove it?

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    Are you familiar with writing $\cos$ as a sum of complex exponentials? Are you familiar with the Taylor series expansion of $-\log(1-x)$?2012-07-24

1 Answers 1

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Hint 1: Use that $$ \log(1-z)=-\sum\limits_{n=1}^\infty\frac{z^n}{n} $$

Hint 2: Set $$ z=r e^{i\theta} $$

Hint 3: Take a real part

Hint 4: Take a limit $r\to 1-0$ and use Abel's summation formula

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    Just a small remark : in the second step, some work is required to justify convergence (one might use Abel's summation formula for example).2012-07-24
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    ...finally: $1-\cos\,\theta=2\sin^2\frac{\theta}{2}$2012-07-24
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    @Joel, yes, since $|\exp\,i\theta|=1$, which is exactly the boundary of the convergence region of the series being used...2012-07-24
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    MSE users are always on alert, thanks for your remarks!2012-07-24
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    For people not used to Norbert's notation in Hint 4: what he wrote is the same as $r\to1^{-}$ (that is, approach from the left).2012-07-24
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    @J.M. Is this notation is not common?2012-07-24
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    It's not as common as the one I was talking about, I've found, so I mentioned it for completeness.2012-07-24
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    @J.M.The same notation used in the reference I've posted in my answer2012-07-24
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    Thanks for the help! The identity is very curious to me inspite of having been doing advanced mathematics for years now. Is it part of some general thinking..which seems I have missed out on! @J.M I initially did try by writing as you suggest $sin (\frac{\theta}{2})$ in terms of $cos(\theta)$ and then taking the log of it but that doesn't seem to produce the answer..is there such a direct way to get there?2012-07-25
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    @6818, you still need to take the series route...2012-07-25