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let $x_1, x_2, ... , x_{2014} $ be the roots for $x^{2014} + x^{2013} + ... +x +1 = 0 $ find $\sum_{k=1}^{2014} 1/(1-x_k) $.

So What I tried to do is to consider x^3 function and get the elementary symmetric functions. the find the sum of $\sum_{k=1}^{3} 1/(1-x_k) $. I lost and did not work.

How I can approach this problem?

Jyrki Lahtonen
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Syrian Inshad Tarab
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    Let $P(x)=x^{2014}+...+x+1$. Then the sum that you are trying to compute is $\frac{P'(1)}{P(1)}$. Note that if $P(x)=\prod_{k=1}^{2014}(x-x_k)$, then $P'(x)=\sum_{k=1}^{n}\prod_{l=1,l\neq k}^{n}(x-x_l)$. Therefore, when you divide by $P(x)$ you get $\sum_{k=1}^{n}\frac{1}{x-x_k}$. – conditionalMethod Nov 16 '19 at 19:25
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    For further reference, $\frac{P'(1)}{P(1)}=\frac{2014+2013+...+2+1}{1+1+...+1}=\frac{\frac{2014\cdot 2015}{2}}{2015}=\frac{2014}{2}=1007$. – conditionalMethod Nov 16 '19 at 19:37
  • See also: https://math.stackexchange.com/questions/2209034/finding-sum-k-0n-1-frac-alpha-k2-alpha-k-where-alpha-k-are-the – lab bhattacharjee Nov 16 '19 at 19:42
  • Stil not accepting the answers – nonuser Nov 18 '19 at 19:53

2 Answers2

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Let $y_k=\dfrac1{1-x_k}\implies x_k=\dfrac{y_k-1}{y_k}$ and we need

$$\sum_{k=1}^{2014}y_k$$

Replacing $x$ with $\dfrac{y-1}y$ in the given equation to get

$$0=\sum_{r=0}^{2014}y^r(y-1)^{2014-r}$$

$$2015y^{2014}-y^{2013}\sum_{k=0}^{2014}\binom k1+\cdots+1=0$$

Apply Vieta's formula

lab bhattacharjee
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We notice that the equation: $$x^{2014} + x^{2013} + ... +x +1 = 0 $$ is a geometric sequence and can be rewritten as: $$x^{2014} + x^{2013} + ... +x +1 = \sum_{n=0}^{2014}x^n = \frac{1-x^{2015}}{1-x} = 0 $$ From the denominator $x-1 $ we conclude that $x\ne1$, but we can rewrite $ 1 = e^{2\pi k} $. So from: $$ 1-x^{2015} = 0 \rightarrow x^{2015} = e^{2\pi k} $$ $$ x_k = e^{2\pi k/2015} $$ Where $k \ne 0$ and goes from $1$ to $2014$. Because $k\ne 0$ then $x\ne1$ and there is no problem with the denominator anymore. The sum becomes: $$ \sum_{k=1}^{2014} \frac{1}{1-x_k} = \sum_{k=1}^{2014} \frac{1}{1-e^{2\pi k/2015}} = \sum_{k=1}^{2014} \frac{e^{-\pi k/2015}}{e^{-\pi k/2015}-e^{\pi k/2015}} = \sum_{k=1}^{2014} \frac{\cos(\pi k/2015)-i\sin(\pi k/2015)}{-2i\sin(\pi k/2015)} = $$ $$ \sum_{k=1}^{2014} \frac{1}{2} + i\frac{\cot(\pi k/2015)}{2} = 1007 + \frac{i}{2}\sum_{k=1}^{2014}\cot(\pi k/2015) = 1007 $$ From Wolfram alpha $ \sum_{k=1}^{2014}\cot(\pi k/2015) = 0 $ but you can prove it easily. An example: $$ \cot(\frac{2014\pi}{2015}) = \cot(\frac{2015\pi-\pi}{2015}) = \cot(\pi- \frac{\pi}{2015}) = -\cot(\frac{\pi}{2015}) $$ which will cancel out with the first term and so on with other terms.

Dinno Koluh
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