We have the rational function : $$f(x)=\frac{(1+ix)^{n}-1}{(1-ix)^{n}-1}\;\;\;,\;\;n\in \mathbb{Z}^{+}$$ It's not hard to prove that : $$\frac{(1+ix)^{n}-1}{(1-ix)^{n}-1}=(-1)^{n}\prod_{k=1}^{n-1}\frac{x+i(\xi_{n}^{k}-1)}{x-i(\xi_{n}^{k}-1)}\;\;\;,\;\;\xi_{n}^{k}=e^{2\pi i k/n}$$ Now we want to compute $\log f(x)$ for $x>0$. The logarithm of the individual factors can be written as :

$$\log\left(\frac{x+i(\xi_{n}^{k}-1)}{x-i(\xi_{n}^{k}-1)}\right)=2i\tan^{-1}\left(\frac{x}{1-\xi_{n}^{k}}\right)+i\pi;\;\;\;\;x>0$$ So, one would expect: $$\log f(x)=-i\pi+2i\pi n+2i\sum_{k=1}^{n-1}\tan^{-1}\left(\frac{x}{1-\xi_{n}^{k}}\right)$$ But it looks nothing like what wolframalpha returns. What am i doing wrong here ?

  • $\begingroup$ what does WA return? $\endgroup$ – tired Jul 26 '16 at 20:51
  • $\begingroup$ Could you provide what wolframalpha returned? Your answer might actually be the same, but we don't know what we're comparing to, to answer the question. $\endgroup$ – Simply Beautiful Art Jul 26 '16 at 20:52
  • $\begingroup$ Does $\tan^{-1}=\cot\text{ or }\arctan$? $\endgroup$ – Simply Beautiful Art Jul 26 '16 at 20:53
  • $\begingroup$ $\tan^{-1}$ is $\arctan$ $\endgroup$ – Mohammad Al Jamal Jul 27 '16 at 5:29
  • $\begingroup$ WF returns the difference as a sum of step functions linkl $\endgroup$ – Mohammad Al Jamal Jul 27 '16 at 6:30

I think that, if you use the standard branch of $\log$ used by Wolfram Alpha, your third identity should read:

$$\log\left(\frac{x+i(\xi_{n}^{k}-1)}{x-i(\xi_{n}^{k}-1)}\right)=2i\tan^{-1}\left(\frac{x}{1-\xi_{n}^{k}}\right)-\pi i,\quad x>0.$$

Compare for exampe: arctan and log.

You are using a branch of the logarithm, so in general it is not true that $\log (ab) = \log(a) + \log(b)$. You can check that the difference between the corrected expression $$\log f(x)=-\pi(n-1)i + 2i\sum_{k=1}^{n-1}\tan^{-1}\left(\frac{x}{1-\xi_{n}^{k}}\right),$$ and the original $\log f(x)$ is a multiple of $\pi$.

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  • $\begingroup$ i still don't understand why WA returns the above + some sort of a step function !! here is an example $\endgroup$ – Mohammad Al Jamal Jul 31 '16 at 6:59
  • $\begingroup$ Like I said, $\log(ab) \ne \log(a) + \log(b)$, for example $\log(e^{\pi i} \cdot e^{\pi i}) = 0$, and $\log(e^{\pi i}) = \pi i$. So the difference in general is going to be a multiple of $\pi i$. $\endgroup$ – sometempname Jul 31 '16 at 9:08
  • $\begingroup$ i understand that. but what are the points at which there is a step difference by a multiple of $\pi i$ $\endgroup$ – Mohammad Al Jamal Jul 31 '16 at 9:52
  • $\begingroup$ The jumps are at the points where the sum is equal to a multiple of $\pi i$. $\endgroup$ – sometempname Aug 2 '16 at 21:30

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