Confusion on applying Newton formulae (not Newton's method) technique in finding roots

Question

In this post, Alexandre Eremenko outlines a method to find solutions to systems of equations of the form:

$$\begin{cases} \sum_i x_i = c_1 \\ \;\vdots \\ \sum_i x_i^n = c_n \end{cases}$$

I tried this method, but ended up with a wrong result. Suppose I have:

$$\begin{cases} x_1 + x_2 = 1 \\ x_1^2 + x_2^2 = 3\end{cases}$$

In the language of the Wikipedia article linked in the post, this is $p_1 = 1$ and $p_2 = 3$. Thus if $e_i$ is the $i$th elementary symmetric polynomial, we know that $e_1 = 1$ and $2e_2 = p_1^2 - p_2 = -2$, thus:

$$\begin{cases} e_1 = 1 \\ e_2 = -1 \end{cases}$$

According to Alexandre, the solutions $(x_1,x_2)$ are roots of the polynomial:

$$x^2 + x - 1$$

but these roots are in fact $(-\varphi,-\phi)$ (where $\varphi$ is the golden ratio and $\phi$ its conjugate), when the answer should be $(\varphi,\phi)$. What went wrong?

Answer 1

If you expand a factored (monic) polynomial, you can see the coefficients in terms of the roots. In degree $2$, $$ (x - r_1)(x - r_2) = x^2 - (r_1 + r_2)x + (r_1r_2) = x^2 + a_1x + a_0. $$ Notice that $$ -a_1 = r_1 + r_2 = e_1(r_1, r_2) \qquad \text{and} \qquad +\!a_0 = r_1r_2 = e_2(r_1, r_2). $$

Note the alternating sign! The general formula, often attributed to Viète, looks like $$ (-1)^k a_{n-k} = e_k(r_1, \dots, r_n) \qquad \text{for } k = 1, \dots, n. $$

In your case, your polynomial turns out to be $$ x^2 - x - 1 $$ with roots $\{\varphi, \phi\}$.