Artur o problema que vc esta querendo resolver é feito utilizando as
identidades de Newton
procure no google: newton´s identities ou symmetric polynomial
Consider the polynomial
[image: p(\lambda) = \prod_{\alpha=1}^n \left( \lambda - x_\alpha \right) =
\sum_{j=0}^n (-1)^{j} a_j \lambda^{n-j}]
where the *x*α are the roots and the *a**j* are the
coefficients<http://en.wikipedia.org/wiki/Coefficients>.
Define the *power sums*
[image: t_j = \sum_{\alpha=1}^n x_\alpha^j \ {\rm \ for\ } j \geq 0 \,.]
Then the original form of Newton's identities is the recurrence
[image: t_1 = a_1 \,,][image: t_2 = a_1 t_1 - 2 a_2 \,,][image: t_3 = a_1
t_2 - a_2 t_1 + 3 a_3 \,,][image: t_4 = a_1 t_3 - a_2 t_2 + a_3 t_1 - 4 a_4
\,,][image: t_5 = a_1 t_4 - a_2 t_3 + a_3 t_2 - a_4 t_1 + 5 a_5 \,,]
where the pattern is obvious. For an elementary proof see the book by Tignol
cited below.
[edit<http://en.wikipedia.org/w/index.php?title=Newton%27s_identities&action=edit§ion=2>
] Computing power sums
From these formulae we can readily obtain more useful formulae, expressing
the power sums in terms of the coefficients:
[image: t_1 = a_1\,,][image: t_2 = a_1^2 - 2 a_2\,,][image: t_3 = a_1^3 - 3
a_1 a_2 + 3 a_3\,,][image: t_4 = a_1^4 - 4 a_1^2 a_2 + 4 a_1 a_3 + 2 a_2^2 -
4 a_4\,.]Espero que isto te ajude!
Jones
2008/2/12 Artur Costa Steiner <[EMAIL PROTECTED]>:
> Se q é um inteiro positivo, existe alguma forma relativamente fácil de se
> determinar a soma das potências q das raÃzes de um polinômio? Algo, por
> exemplo, baseado nas reações de Girard?
>
> Obrigado
> Artur
>
