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.
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.
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 $x^2$.)
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 $0$.