Alduccii2

2022-07-22

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

Reinfarktq6

Beginner2022-07-23Added 18 answers

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). 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 kth symmetric polynomials in the roots, with

$\text{ceoff of}{x}^{12}={r}_{1}+{r}_{2}+\dots +{r}_{13}=0.$

(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 ${x}^{2}$.)

$\sum _{i=1}^{13}{r}_{i}^{13}=26-\sum _{i=1}^{13}{r}_{i}.$

Also observe that the ${x}^{n-k}$th coefficient of a polynomial is the kth symmetric polynomials in the roots, with

$\text{ceoff of}{x}^{12}={r}_{1}+{r}_{2}+\dots +{r}_{13}=0.$

(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 ${x}^{2}$.)

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