I've been struggling to understand the procedure shown below. Somehow, by setting x=0 I managed to get to the equation (1.12). However, I think didn't get it really as I do not know how to compare the coefficients (which is apparently important to get to (1.13) ).
Kind regards
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When you have a product of binomials like these $$ \alpha_n(x-\alpha_1)(x-\alpha_2)\dotsm(x-\alpha_n)=0 $$ Which for simplicity we are going to convert it to a monic polynomial: $$(x-\alpha_1)(x-\alpha_2)\dotsm(x-\alpha_n)=0$$ You can express it also like this: $$ (x-\alpha_j)\prod_{i\neq j}^n{}(x-\alpha_i)\iff 1\le j \le n $$ Now lets expand $\prod_{i\neq j}^n{}(x-\alpha_i)$: $$ (x-\alpha_j)(x^{n-1} + b_{n-2}x^{n-2} + b_{n-3}x^{n-3} + \dots + b_{1}x + b_{0}) = x(x^{n-1} + b_{n-2}x^{n-2} + b_{n-3}x^{n-3} + \dots + b_{1}x + b_{0}) - \alpha_j(x^{n-1} + b_{n-2}x^{n-2} + b_{n-3}x^{n-3} + \dots + b_{1}x + b_{0}) $$ After expanding the equation from above we focus on the term $ (b_{n-2}-\alpha_j)x^{n-1} $.
Now we need to figure out what $b_{n-2}$ is. Lets start with the first two cases.
I am going to omit non-relevant terms just for simplicity.
For n = 2: $$ (x-\alpha_1)(x-\alpha_2) = x^2+(-\alpha_1-\alpha_2)x+\alpha_1\alpha_2 $$
For n = 3: $$ (x-\alpha_1)(x^2+(-\alpha_2-\alpha_3)x+\alpha_2\alpha_3) = x^3+(-\alpha_1-\alpha_2-\alpha_3)x^2 +...-\alpha_1\alpha_2\alpha_3 $$
From this part it is easy to see that $b_{n-2} = b_{n-3}-\alpha_{n}$ where $b_{0} = -\alpha_2$.
That implies $a_{n-1}=b_{n-2}-\alpha_1=-(\alpha_1+\alpha_2+\alpha_3+..+\alpha_n)=-\sum_{k=1}^{n}{\alpha_k}$
$$a_{n-1}=-\sum_{k=1}^{n}{\alpha_k}\to -a_{n-1}=\sum_{k=1}^{n}{\alpha_k}$$
Now remember that the polynomial equation given in the picture is not necessarily monic so we fix that by dividing the left side of the equation by $a_n$. $$-\frac{a_{n-1}}{a_{n}}=\sum_{k=1}^{n}{\alpha_k}$$
All it means is that $\sum_{i=0}^n a_ix^i=\sum_{i=0}^n b_ix^i$ if and only if $a_i=b_i$ for all $i$. "Setting $x=0$" is one way of looking at the constant term of a polynomial, but it's not really necessary if you follow the fact I just mentioned.
You're presented with three ways of writing polynomials, but if you multiply them out, their coefficients have to match. The first equation gives you labels for the coefficients ($a_i$) and the latter two, if multiplied out, will give you coefficients in terms of $\alpha_i$, the roots.
The author's point is that the $a_0$ and $a_{n-1}$, are easy to compute in terms of the roots, but it is not very useful to compute the rest of the coefficients this way.