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I need to solve the below mentioned equation and try to find a unique solution for $\epsilon$ for the range between (-1,1) in terms of $n$.

$$\begin{equation} \sum_{j=1}^{n-1} \frac{(2{\alpha_j} + 1)( \alpha_j -1)}{{\bigg[(\alpha_j - 1) \bigg((2{\alpha_j}+3) - \epsilon(2{\alpha_j}+1)\bigg) \bigg]\,\,\,}^{2}} = 0 \end{equation} $$

where

$\alpha_j=\cos\frac{2{\pi}j}{n}$.

With simulations, I can see that there's a unique solution for $\epsilon$ when $\epsilon$ is in (-1,1) and I need to find that as an expression in terms of $n$. Any suggestions

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I removed the (soft-question) tag, which is for questions that don't admit a definitive answer. Please read the tag summaries that are offered in the tag selection menu when selecting tags. –  joriki Oct 27 '11 at 8:00
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1 Answer

Your sum has following closed form:

$$\frac{\left( n-1 \right) \left( 2\,\cosh \left( 2\,{\frac {\pi }{n}} \right) +1 \right)} {\left( \cosh \left( 2\,{\frac {\pi }{n}} \right) -1 \right) \left( 2\,\cosh \left( 2\,{\frac {\pi }{n}} \right) + 3-\epsilon \, \left( 2\,\cosh \left( 2\,{\frac {\pi }{n}} \right) +1 \right) \right) ^{2}}$$

So when this expression is equal to $0$ , $\epsilon$ may be an arbitrary number.

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Since $j$ is not an imaginary unit here, can we use the above mentioned closed form for this problem ? Is there any other way to solve this problem? –  Udara Oct 30 '11 at 9:28
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