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I ran across an interesting series involving the square of sine. It was solved in a clever manner I do not quite get.

$\sum_{k=0}^{n-1}\frac{1}{1+8\sin^{2}(\frac{k\pi}{n})}$

The solution involved the nth roots of unity.

$\sin^{2}\left(\frac{k\pi}{n}\right)=\frac{2-L^{k}-L^{-k}}{4},$ where $L=\displaystyle e^{\frac{2\pi i}{n}}$

Which becomes $\frac{1}{1+8\sin^{2}(\frac{k\pi}{n})}=\frac{1}{3}\left[\frac{1}{2L^{k}-1}-\frac{2}{L^{k}-2}\right]$

Noting that $L^{k}, \;\ (0\leq k\leq n-1)$ are the nth roots of unity, the following series are used:

$\sum_{k=0}^{n-1}\frac{1}{2L^{k}-1}=\frac{n}{2^{n}-1}$ and

$\sum_{k=0}^{n-1}\frac{1}{L^{k}-2}=\frac{n2^{n-1}}{1-2^{n}}.$

My question is, how are those last two closed form for the series evaluated?

This much is explained: By noting that $z^{n}=1$ and using $y=\frac{1}{2z-1}, \;\ z=\frac{y+1}{2y},$

we can construct $\left(\frac{y+1}{2y}\right)^{n}=1.$

Then, $(y+1)^{n}=2^{n}y^{n}.\tag1$ Then, $(1-2^{n})y^{n}+ny^{n-1}+\cdot\cdot\cdot +1=0.\tag2$

So, from this last equation we get $\sum_{k=0}^{n-1}y_{k}=\sum_{k=0}^{n-1}\frac{1}{2L^{k}-1}=\frac{n}{2^{n}-1},$ where $y_{k}$ are the roots of the equation in $(2).$

My problem is, I still do not quite understand that last line and how that closed form is found. I get down to $(1)$ and after that get lost. Is that a binomial series, and how is it derived from the previous step?.

and how does $\sum_{k=0}^{n-1}y_{k}=\sum_{k=0}^{n-1}\frac{1}{2L^{k}-1}=\frac{n}{2^{n}-1}$

Apparently, $y_{k}=\frac{1}{2L^{k}-1}$ somehow. Thanks very much for any input. I always like to learn something new and there is probably something here to learn.

I suppose there is something about nth roots of unity I do not get as of yet.

1 Answers 1

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The explanation is written in a recondite manner, but its essence is not so hard to grasp. Note that

$ y_k = \frac{1}{2L^k - 1}$

form $n$ solutions of the equation

$ \left( \frac{y+1}{2y} \right)^n = 1 \quad \Longleftrightarrow \quad 2^n y^n - (y+1)^n = 0. $

Since $L^k \neq L^l$ if $k$ and $l$ are different mod $n$, we find that $y_0, \cdots, y_{n-1}$ are all different. But the last equation if of degree $n$, so it has at most $n$ zeros. That is, we must have

$ 2^n y^n - (y+1)^n = (2^n - 1)(y - y_0) \cdots (y - y_{n-1}), $

where the proceeding term $2^n - 1$ in the RHS arises since the leading coefficient of the LHS is $2^n - 1$. Expanding the LHS by binomial theorem, we have

$ 2^n y^n - (y+1)^n = 2^n y^n - (y^n + ny^{n-1} + \cdots) = (2^n - 1)y^n - n y^{n-1} + \cdots ,$

and expanding the RHS yields

$ (2^n - 1)(y - y_0) \cdots (y - y_{n-1}) = (2^n - 1)y^n - (2^n - 1)(y_0 + \cdots + y_{n-1})y^{n-1} + \cdots.$

Therefore we obtain the desired relation

$y_0 + \cdots + y_{n-1} = \frac{n}{2^n - 1}.$

This is quite ingenious and simple, but you can also obtain this formula more naturally as follows: Since

$ \sum_{k=0}^{n-1} \frac{1}{2L^k - 1} = \sum_{k=0}^{n-1} \frac{\frac{L^{-k}}{2}}{1 - \frac{L^{-k}}{2}} = \sum_{k=0}^{n-1} \sum_{l=1}^{\infty} \left( \frac{L^{-k}}{2} \right)^{l} = \sum_{l=1}^{\infty} \sum_{k=0}^{n-1} \left( \frac{L^{-k}}{2} \right)^{l} $

and

$ \sum_{k=0}^{n-1} L^{km} = \begin{cases} n & m \equiv 0 \ (\mathrm{mod} \ n) \\ 0 & \mathrm{otherwise} \end{cases},$

we must have

\sum_{k=0}^{n-1} \frac{1}{2L^k - 1} = \sum_{l'=1}^{\infty} \frac{n}{2^{nl'}} = \frac{n}{2^n - 1},

where l = nl'. Similar consideration also works for the second one.

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    Thanks a lot sos. I appreciate the explanation. I can see there were details left out.2012-03-31