Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Find the sum of the thirteenth powers of the roots of $x^{13} + x - 2\geq 0$.

Any solution for this question would be greatly appreciated.

share|cite|improve this question
up vote 2 down vote accepted

Any root $r_i$ of $x^{13} + x - 2 = 0$ satisfies $r_i^{13} + r_i - 2 = 0,$ or $r_i^{13} = 2 - r_i.$ A polynomial of degree $13$ has $13$ roots (counting repititons). See here and here. Sum them up: $$ \sum_{i = 1}^{13} r_i^{13} = 26 - \sum_{i = 1}^{13} r_{i} .$$ Also observe that the $x^{n-k}$th coefficient of a polynomial is the $k$th symmetric polynomials in the roots (see this), with $$\text{ceoff of } x^{12} = r_1 + r_2 + \ldots + r_{13} = 0.$$ (to convince yourself of the latter fact, expand a smaller example: $(x-r_1)(x-r_2)(x-r_3),$ and observe the coefficient of $x^2$.)

share|cite|improve this answer

If $r_i$ is a root, then $r_i^{13}=2-r_i$. Add up, $i=1$ to $13$. Note that $\sum r_i=0$, because the coefficient of $x^{12}$ is $0$.

share|cite|improve this answer
sorry i did not get you.. – mgh Jul 14 '12 at 4:23
If $r$ is a root, then $r^{13}+r-2=0$, so $r^{13}=2-r$. So – André Nicolas Jul 14 '12 at 4:25
so the final answer is? – mgh Jul 14 '12 at 4:32
We want $\sum_{i=1}^{13}(2-r_i)$. This is $\sum_{i=1}^{13} 2-\sum_{i=1}^{13}r_i$. The first term is $26$. The second term is $0$, since the sum of the roots is the negative of the coefficient of $x^{12}$, divided by the coefficient of $x^{13}$. So the answer is $26$. – André Nicolas Jul 14 '12 at 4:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.