- #1
mmzaj
- 107
- 0
We have the rational function :
$$f(x)=\frac{(1+ix)^{n}-1}{(1-ix)^{n}-1}\left(\frac{1-ix}{1+ix}\right)^{n/2}\;\;\;,\;\;n\in \mathbb{Z}^{+}$$
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:
$$\log f(x)=-in\tan^{-1}(x)-i\pi+2i\pi n+2i\sum_{k=1}^{n-1}\tan^{-1}\left(\frac{x}{1-\xi_{n}^{k}}\right)$$
But it looks nothing like what wolframalpha returns. What am i doing wrong here ?
$$f(x)=\frac{(1+ix)^{n}-1}{(1-ix)^{n}-1}\left(\frac{1-ix}{1+ix}\right)^{n/2}\;\;\;,\;\;n\in \mathbb{Z}^{+}$$
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:
$$\log f(x)=-in\tan^{-1}(x)-i\pi+2i\pi n+2i\sum_{k=1}^{n-1}\tan^{-1}\left(\frac{x}{1-\xi_{n}^{k}}\right)$$
But it looks nothing like what wolframalpha returns. What am i doing wrong here ?
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