Sums of powers of roots

It is known since Newton's time that sums of the form \(\sigma_k:=\sum x_i^k\) are polynomial expressions in the coefficients of \(f=\prod(z-x_i)\). What follows is my best understanding of this. We use a formal power series approach, and therefore define \(N_f:=\sum\sigma_kz^k\). I want to prove that \[z^nf(\frac1z)N_f(z) = z^{n-1}f'(\frac1z).\label{th1}\tag{1}\] (It is written this way since \(z^nf(\frac1z)\) and \(z^{n-1}f'(\frac1z)\) are polynomials.)

The proof uses logarithmic differentiation and geometric series expansion only. \begin{align*} \frac{z^{n-1}f'(\frac1z)}{z^nf(\frac1z)} &= \frac1z(\log f)'\left(\frac1z\right) &&= \frac1z\sum\frac1{\frac1z-x_i} \\ &= \sum\frac1{1-x_iz} &&= \sum x_i^kz^k \\ &= \sum\sigma_kz^k &&= N_f(z) \end{align*} For example suppose \(f=(z-x_0)(z-x_1)=(z-2)(z-3)=z^2-5z+6\). The first few sums are \[\sigma_0 = 2,\quad\sigma_1=5,\quad\sigma_2=13,\quad\sigma_3=35,\quad\sigma_4=97\] so \(N_f\approx2+5z+13z^2+35z^3+97z^4\). Furthermore \(z^nf(\frac1z)=1-5z+6z^2\) and \(z^{n-1}f'(\frac1z)=2-5z\). We may test \((\ref{th1})\) by computing its left- and right-hand sides up to fourth order: \[(2+5z+13z^2+35z^3+97z^4)(1-5z+6z^2) = (2-5z) \mod z^5.\] Conversely, we can compute more terms \(\sigma_k\) by polynomial division: \[N_f = \frac{2-5z}{1-5z+6z^2} = \sum_{0\leq k\lt m}\sigma_kz^k + z^m\frac{\textrm{remainder polynomial}}{1-5z+6z^2}\] In python:

>>> z = sympy.var('z')
>>> Nf = sympy.fps((2-5*z)/(1-5*z+6*z*z))
>>> Nf.truncate(10)
2 + 5*z + 13*z**2 + 35*z**3 + 97*z**4 + 275*z**5
+ 793*z**6 + 2315*z**7 + 6817*z**8 + 20195*z**9 + O(z**10)