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[SOLVED] Summation Challenge

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,812
Let $x_1,\,x_2,\,\cdots,\,x_{2014}$ be the roots of the equation $x^{2014}+x^{2013}+\cdots+x+1=0$. Evaluate $\displaystyle \sum_{k=1}^{2014} \dfrac{1}{1-x_k}$.
 

lfdahl

Well-known member
Nov 26, 2013
733
Since the geometric series equals zero, alle the $2014$ roots, $x_k$, must obey the equation: $\frac{1-x_k^{2015}}{1-x_k} = 0 \;\;\;\;(1)$.

There is no real solution to the equation, so we are looking for complex solutions: $x_k = r_k \cdot e^{i \theta_k}$.

But, from the condition: $x_k^{2015} = 1$, we must require: $|x_k| = 1$ which implies: $r_k = 1$ for all $k$.

Thus all the roots have the form: $x_k = e^{i \theta_k}$, and the $\theta_k$-angles are readily found:

\[\left ( e^{i \theta_k } \right )^{2015} = 1 \Rightarrow \cos (2015 \; \theta_k)+i \sin (2015 \; \theta_k) = 1 \Rightarrow 2015 \; \theta_k = k\; 2\pi \Rightarrow \theta_k = \frac{2\pi}{2015}k,\;\;\; k = 1,2, ... , 2014.\]

Note, that the cases $k = 0$ and $k = 2015$ are not allowed, because of the singularity in $(1)$.

Rewriting the $k$th term in the sum:

\[\frac{1}{1-x_k}= \frac{1}{1-e^{i\theta_k}} = \frac{1-e^{i\theta_k}}{2-(e^{i\theta_k}+e^{-i\theta_k})}= \frac{1-\cos \theta_k - i \sin \theta_k}{2(1-\cos \theta_k)} = \frac{1}{2}\left ( 1- i\frac{\sin \theta_k}{1-\cos \theta_k} \right ) = \frac{1}{2}\left ( 1-i \cot\left ( \frac{\theta_k}{2} \right ) \right )\]

Finally, the sum can be evaluated:

\[\sum_{k=1}^{2014} \frac{1}{2}\left ( 1-i \cot\left ( \frac{\theta_k}{2} \right ) \right ) = 1007 - \frac{i}{2}\sum_{k=1}^{2014}\cot\left ( \frac{\pi}{2015}k \right )\]

The imaginary part is a telescoping sum:

\[\sum_{k=1}^{2014}\cot\left ( \frac{\pi}{2015}k \right ) = \sum_{k=1}^{1007}\left ( \cot\left ( \frac{\pi}{2015}k \right ) + \cot\left ( \frac{\pi}{2015} (2015-k) \right ) \right ) = \sum_{k=1}^{1007}\left ( \cot\left ( \frac{\pi}{2015}k \right ) + \cot\left (- \frac{\pi}{2015}k \right ) \right ) = 0\]

Thus we end up with the answer: \[\sum_{k=1}^{2014}\frac{1}{1-x_k} = 1007.\]
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,740
The numbers $x_1,x_2,\ldots,x_{2014}$, together with $1$, are the solutions of $x^{2015}-1=0$.

Replacing $x$ by $1-x$, the numbers $1-x_1,1-x_2\ldots,1-x_{2014}$, together with $0$, are the solutions of $(1-x)^{2015}-1=0$.

Replacing $x$ in that equation by $\dfrac1x$, the numbers $\dfrac1{1-x_1},\dfrac1{1-x_2}\ldots,\dfrac1{1-x_{2014}}$, are the solutions of $\left(1-\dfrac1x\right)^{2015}-1=0$, or $(x-1)^{2015} - x^{2015} = 0$. (The extra solution from the previous equations has now disappeared because the coefficient of $x^{2015}$ in that last equation is zero, so in fact there are only 2014 solutions.)

Using the binomial expansion of $(x-1)^{2015}$, that last equation becomes $$-{2015\choose1}x^{2014} + {2015\choose2}x^{2013} - \ldots = 0.$$ The sum of the roots is given by Vieta's formula as $\dfrac{2015\choose2}{2015\choose1} = \frac12(2014) = 1007.$