Find $\sum_{k=1}^{14} \frac{1}{\left(\omega^{k}-1\right)^{3}}$

Question

For $\displaystyle\omega = \exp\left({2\pi \over 15}\,\mathrm{i}\right),\quad$ find $\displaystyle\ \sum_{k=1}^{14} \frac{1}{\left(\omega^{k}-1\right)^{3}}$.

I tried to write $x^{15}-1$=$(x-1) (x-\omega).....(x-\omega^{14})$ And took log and differentiate thrice but it's very lenghty.

Answer 1

Let $n = 15$ and $P(x) = \frac{x^n-1}{x-1} = \sum\limits_{k=0}^{n-1} x^k$. Last part of your question suggest you already know:

$$\mathcal{S} \stackrel{def}{=} \sum_{k=1}^{n-1} \frac{1}{(\omega^k - 1)^3} = - \frac12 \left.\frac{d^2}{dx^2} \frac{P'(x)}{P(x)}\right|_{x=1} = -\frac12\left.\frac{d^3}{dx^3}\log P(x)\right|_{x=1}$$ To evaluate the derivative, change variable to $t = \log x$ and let $D$ be the operator $\frac{d}{dt}$, we have

$$-2\mathcal{S} = \left.\left(x^3\frac{d^3}{d x^3}\right)\log P(x)\right|_{x=1} = D(D-1)(D-2)\left.\log P(e^t)\right|_{t=0}\tag{*1} $$ Notice $$\log P(e^t) = \log\frac{e^{nt} - 1}{e^t-1} = \log n + f(nt) - f(t) \quad\text{ where }\quad f(t) = \log\frac{e^t - 1}{t}$$ We just need to figure out the Taylor expansion of $f(t)$ up to $O(t^4)$. Since

$$f(t) = \log\left( e^{\frac{t}{2}} \frac{\sinh(\frac{t}{2})}{\frac{t}{2}}\right) = \frac{t}{2} + \log\left( 1 + \frac{t^2}{3! 2^2} + O(t^4)\right) = \frac{t}{2} + \frac{t^2}{24} + O(t^4)$$ We have $$Df(t) = \frac12 + \frac{t}{12} + O(t^3)$$ and hence $$Df(t)|_{t=0} = \frac12,\quad D^2f(t)|_{t=0} = \frac{1}{12}\quad\text{ and }\quad D^3f(t)|_{t=0} = 0$$ Substitute this in $(*1)$, we get

$$-2\mathcal{S} = \bigg[(D^2 - 3D + 2)D(f(nt) - f(t))\bigg]_{t=0} = -\frac{3}{12}(n^2-1) + \frac{2}{2}(n-1)$$ Similplify this give us

$$\sum_{k=1}^{n-1} \frac{1}{(\omega^k - 1)^3} = \mathcal{S} = \frac{(n-3)(n-1)}{8}$$

For $n = 15$, this reduces to $\displaystyle\;\frac{(15-3)(15-1)}{8} = 21$ as first pointed out by @user64494 in comment.

Answer 2

$w_k$ are the roots of $$y^{15}-1=0$$

Now let $p_k=\dfrac1{(w_k-1)^3}, w_k-1=\sqrt[3]{\dfrac1{p_k}}$

Writing $p_k$ as $z$ and $w_k$ is a root of $y^{15}=1$

$$1=\left(1+\sqrt[3]{\dfrac1z}\right)^{15}$$

$$z^5=(1+\sqrt[3]z)^{15}$$

$$\iff - \sum_{r=0}^4z^r\binom{15}{3r}=z^{1/3}\sum_{r=0}^4 z^r\binom{15}{3r+1}+z^{2/3}\sum_{r=0}^4 z^r\binom{15}{3r+2} $$

Now to rationalize take cube in both sides,

$$-\left(\sum_{r=0}^4z^r\binom{15}{3r}\right)^3=z\left(\sum_{r=0}^4 z^r\binom{15}{3r+1}\right)^3+z^2\left(\sum_{r=0}^4 z^r\binom{15}{3r+2}\right)^3+3z\left(- \sum_{r=0}^4z^r\binom{15}{3r}\right)$$

$$\left(\binom{15}{3\cdot4+2}\right)^3z^{4\cdot3+2}+z^{4\cdot3+1}\left(\left(\binom{15}{3\cdot4+1}\right)^3-3\binom{15}{12}\right)+\cdots=0$$

$$\implies\sum_{k=1}^{14}p_k=-\dfrac{\left(\binom{15}2\right)^3-3\binom{15}{12}}{\left(\binom{15}1\right)^3}$$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ With $\ds{\omega \equiv \exp\pars{2\pi\ic/15}}$: \begin{align} \sum_{k = 1}^{14}{1 \over \pars{\omega^{k} - 1}^{3}} & = 2\,\Re\sum_{k = 1}^{7}{1 \over \pars{\expo{2k\pi\ic/15} - 1}^{3}} \\[5mm] & = 2\,\Re\sum_{k = 1}^{7}{1 \over \expo{k\pi\ic/5}\pars{\expo{k\pi\ic/15} - \expo{-k\pi\ic/15}}^{3}} \\[5mm] & = 2\,\Re\sum_{k = 1}^{7}{\expo{-k\pi\ic/5} \over \bracks{2\ic\sin\pars{{k\pi/15}}}^{3}} = {1 \over 4}\sum_{k = 1}^{7}\underbrace{{\sin\pars{k\pi/5} \over \sin^{3}\pars{{k\pi/15}}}}_{\ds{3\cot^{2}\pars{k\pi \over 15} - 1}} \\ & = -\,{7 \over 4} + {3 \over 4} \underbrace{\sum_{k = 1}^{7}\cot^{2}\pars{k\pi \over 15}} _{\ds{\color{red}{\Large\S}\quad{91 \over 3}}} = \bbx{21} \end{align} $\ds{\color{red}{\Large\S}}$: I'm still trying to work out the sum !!!.

Answer 4

Strigthforward approach:
First we combine $\frac{1}{(\omega^{k}-1)^3}$ with $\frac{1}{(\omega^{15-k}-1)^3}$ to get rid of $i$ in the denominator.
$$\sum\limits_{k=1}^{14}\frac{1}{(\omega^k-1)^3}= \sum\limits_{k=1}^{7}\frac{(\omega^{15-k}-1)^3+(\omega^{k}-1)^3} {\left((\omega^k-1)(\omega^{15-k}-1)\right)^3}=$$ $$\sum\limits_{k=1}^{7}\frac{ \left((\omega^{15-k}-1)+(\omega^{k}-1)\right) \left( (\omega^{15-k}-1)^2-(\omega^{15-k}-1)(\omega^{k}-1)+(\omega^{k}-1)^2 \right) } {\left(2-\omega^k-\omega^{15-k}\right)^3}=$$ $$-\sum\limits_{k=1}^{7}\frac{ (\omega^{15-k}-1)^2-(\omega^{15-k}-1)(\omega^{k}-1)+(\omega^{k}-1)^2 } {\left(2-\omega^k-\omega^{15-k}\right)^2}=$$ $$-\sum\limits_{k=1}^{7}\frac{ \omega^{30-2k}-2\omega^{15-k}+1-2+\omega^k+\omega^{15-k}+\omega^{2k}-2\omega^{k}+1 } {\left(2-\omega^k-\omega^{15-k}\right)^2}=$$ $\displaystyle-\sum\limits_{k=1}^{7}\frac{ \omega^{30-2k}+\omega^{2k}-\omega^{15-k}-\omega^{k} } {\left(2-\omega^k-\omega^{15-k}\right)^2}=$ $\displaystyle-\sum\limits_{k=1}^{7}\frac{\omega^{2k} + \omega^{k} + 1}{\omega^{2k} - 2 \omega^{k} + 1} $ $=\displaystyle-7-3\sum\limits_{k=1}^{7}\frac{1}{\omega^{k} - 2 + \omega^{-k}}=$ $$-7-\frac32\sum\limits_{k=1}^{7}\frac{1}{\cos\frac{2k\pi}{15}-1}= -7+\frac34\sum\limits_{k=1}^{7}\frac{1}{\sin^2\frac{k\pi}{15}}=$$ with a large overkill $\displaystyle -7+\frac34\cdot\frac23\cdot 7 \cdot 8=21$