1
$\begingroup$

Suppose we have $$f(x) = a_n(x-x_1)(x-x_2)\cdots(x-x_n) = a_n \prod_{k=1}^n \left(x-x_k\right)$$

Then by Newton's identities we can write:

$$k e_k(x_1,\cdots,x_n) = \sum_{j=1}^{k-1} (-1)^{j-1} e_{k-j} (x_1,\cdots,x_n) p_{j}(x_1,\cdots,x_n) $$

where $e$ is an elementary symmetric function of the roots and $p$ is a power sum of the roots.

I've also stumbled across this link:

https://artofproblemsolving.com/wiki/index.php/Newton%27s_Sums

Which claims the $p_j$ can be expressed in terms of lower order power sums and coefficients $a_l$ of the original polynomial.

e.g. $$ \begin{split} a_n p_1 + a_{n-1} &= 0\\ a_n p_2 + a_{n-1}p_1 + 2 a_{n-2} &= 0 \end{split} $$

Is it possible to express a given power sum $p_j$ in terms of only coefficients of the polynomial and perhaps elementary symmetric functions such that no power sum terms are used?

Thank you.

$\endgroup$
0

0

You must log in to answer this question.

Browse other questions tagged .