Newton sums (Newton identities for power sums over the roots) proof question Kindly take a look at the proof.
I understand the entire proof and I get the final result:
$$a_nP_k + a_{n-1}P_{k-1} + \cdots + a_0P_{k-n}=0$$
 But I don't see how the above result allows us to have below:
$$\begin{align}
&a_nP_1+a_{n-1}=0 \\
&a_nP_2+a_{n-1}P_1 + \color{red}{2}a_{n-2}=0 \\
&\vdots
\end{align}
$$
I don't see the connection.. there is no way we can ever get $\color{red}{2}$ using the proof. Any help?
 A: You have by FTA and with the roots $z_1,...,z_n$ of the given polynomial 
$$
p(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0=a_n(x-z_1)...(x-z_n).
$$
Define
$$
q(t)=a_n(1-tz_1)...(1-tz_n)=t^np(1/t).
$$
Then for $t$ sufficiently small expand the logarithmic derivatve using geometric series,
\begin{align}
\frac{q'(t)}{q(t)}&=\frac{d}{dt}\ln|q(t)|
\\
&=\frac{d}{dt}(\ln|a_n|+\ln|1-tz_1|+...+\ln|1-tz_n|)
\\
&=-\frac{z_1}{1-tz_1}-...-\frac{z_z}{1-tz_z}
\\
&=-\sum_{k=0}^\infty z_1^{k+1}t^k-...-\sum_{k=0}^\infty z_n^{k+1}t^k
\\
&=-(P_1+tP_2+t^2P_3+t^3P_4+...)
\end{align}
Now multiply out the denominator and compare coefficients in
\begin{multline}
0=\underbrace{(a_{n-1}+2a_{n-2}t...+(n-1)a_1t^{n-2}+na_0t^{n-1})}_{=q'(t)}
\\
+\underbrace{(a_n+a_{n-1}t+...+a_1t^{n-1}+a_0t^n)}_{=q(t)}(P_1+tP_2+t^2P_3+t^3P_4+...)
\end{multline}
to read off the Newton identities
\begin{align}
0&=~~a_{n-1}+a_nP_1\\
0&=2a_{n-2}+a_{n-1}P_1+a_nP_2\\
0&=3a_{n-2}+a_{n-2}P_1+a_{n-1}P_2+a_nP_3\\
&\vdots\\
0&=na_{0}+a_{1}P_1+...+a_{n-1}P_{n-1}+a_nP_n\\[1em]
0&=\qquad a_{0}P_1+a_{1}P_2+...+a_{n-1}P_{n}+a_nP_{n+1}\\
0&=\qquad a_{0}P_2+a_{1}P_3+...+a_{n-1}P_{n+1}+a_nP_{n+2}\\
\end{align}
etc.
A: To be honest, I don't understand Newton's sums well enough to explain them, but what I've figured out is that the equation $$a_nP_k+a_{n-1}P_{k-1}+\ldots +a_0P_{k-n}=0$$ only makes sense when $k\geq n$. By the way, the proof you posted doesn't mention this, but $$P_0=x_1^0+x_2^0+\dots+x_n^0=n$$
In the case that $k < n$ the equation you should actually use is $$a_nP_k+a_{n-1}P_{k-1}+\ldots+a_{n-k+1}P_1+k*a_{n-k}$$ If you want to see the proof for this second equation, then check out this article. It's a pretty rough read, but I couldn't find anything better.
