I have done most of the following problem, but I am stuck in the last part:
for some polynomial $x^n + a_{n-1}x^{n-1}+...+a_0$ I have to find $r_1^2+r_2^2+...+r_n^2$ in terms of the coefficients $a_i$, where $r_i$ is the ith root. I got the solution $\sum r_i^2 = a_{n-1}^2 -2a_{n-2}$. I am sure it is correct, but I cannot prove it. I am trying to use induction and I am stuck at the inductive case $n+1$. What I have is $r_1^2+...+r_{n+1}^2 = a_{n-1}^2-2a_{n-2} + r_{n+1}^2 \stackrel{?}{=} a_{n}^2-2a_{n-1}$
I don't know how to proceed from here. Any ideas? Thank you :)
Well, the easiest way is the using of Vieta's relations. But, your induction is not true. Let say the polynomial of degree $n+1$ with $n+1$ roots. We have
$$x^{n+1} + a_{n}x^{n} + \dots a_0 = \prod_{i=1}^{n+1}(x-r_i).$$
So, if you take n-roots $r_i$, where $i$ goes from 1 to n. Then, $$x^{n+1} + a_{n}x^{n} + \dots a_0 = P(x)(x-r_{n+1}),$$ where
$$P(x) = x^{n} + b_{n-1}x^{n} + \dots b_0$$ and by induction, you have $$\sum_{i=1}^n r_i^2 = b_{n-1}^2 - 2b_{n-2}.$$ So, you must check relation of $r_{n+1}$,$a_{n}$, $a_{n-1}$ and $b_{n-1}$, $b_{n-2}$. Hope it helps.
EDIT: Let me clarify. We assume that
$$r_1^2 + \dots r_n^2 = a_{n-1}^2 - a_{n-2},$$ for the polynomial of degree n.
We must prove that with the polynomial of degree $n+1$ which has $n+1$ roots, then $$r_1^2 + \dots r_{n+1}^2 = a_n^2 - 2a_{n-1}.$$ And you cannot apply like you wrote $r_1^2 + \dots r_{n+1}^2 = a_{n-1}^2 - 2a_{n-2} + r_{n+1}^2$.