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.
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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 $0$.
André Nicolas
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sorry i did not get you.. – mgh Jul 14 '12 at 04:23
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2If $r$ is a root, then $r^{13}+r-2=0$, so $r^{13}=2-r$. So – André Nicolas Jul 14 '12 at 04:25
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so the final answer is? – mgh Jul 14 '12 at 04:32
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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 04:42
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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 $x^2$.)
Glorfindel
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