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?

  • 3
    $\begingroup$ Are you familiar with writing $\cos$ as a sum of complex exponentials? Are you familiar with the Taylor series expansion of $-\log(1-x)$? $\endgroup$ – anon Jul 24 '12 at 21:30

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

  • 2
    $\begingroup$ Just a small remark : in the second step, some work is required to justify convergence (one might use Abel's summation formula for example). $\endgroup$ – Joel Cohen Jul 24 '12 at 21:40
  • $\begingroup$ ...finally: $1-\cos\,\theta=2\sin^2\frac{\theta}{2}$ $\endgroup$ – J. M. is a poor mathematician Jul 24 '12 at 21:40
  • $\begingroup$ @Joel, yes, since $|\exp\,i\theta|=1$, which is exactly the boundary of the convergence region of the series being used... $\endgroup$ – J. M. is a poor mathematician Jul 24 '12 at 21:41
  • $\begingroup$ MSE users are always on alert, thanks for your remarks! $\endgroup$ – Norbert Jul 24 '12 at 21:50
  • $\begingroup$ 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). $\endgroup$ – J. M. is a poor mathematician Jul 24 '12 at 21:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.