I was reading this PDF.
On Page 2, the author mentions Vieta's Formula. Now I am familiar with it (the fact that a quadratic, for example, can be written as $(x-a)(x-b)$ and that the sum of the roots is equal to -(coefficient) of one of the terms and so on, but I didn't quite understand the form the author was proposing. I am also familiar with symmetric polynomials and how they are related to Vieta's formula. But what does the below equation mean?
The author says, and I quote: "Vieta’s Formulas state that $$\sigma_k={ (-1) }^k \cdot \frac { { a }_{ n-k } }{ { a }_{ k } }$$ for 1 ≤ k ≤ n."
How is this the statement of Vieta's Formula?
Could anybody elucidate?
Thanks in advance.
I think it is a mistake - the denominator should be $a_n$, which I assume is the coefficient of $x^n$ (it would be good if you put the polynomial in your question rather than a link).
Dividing through by $a_n$ is necessary to convert the polynomial to monic form.
For the benefit of the down voters the formula should be $$\sigma_k=(-1)^k\frac {a_{n-k}}{a_n}$$ and the formula given is $$\sigma_k=(-1)^k\frac {a_{n-k}}{a_k}$$
So if we have a polynomial $$p(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_0=0$$ we can factorise it as $$p(x)=a_n(x-t_1)(x-t_2)\dots (x-t_n)$$ where the $t_i$ are the roots. If $\sigma_k$ is the $k_{th}$ symmetric polynomial in the roots $t_i$ we can expand the factorisation to give $$p(x)=a_nx^n-a_n\sigma_1x^{n-1} + \dots + a_n(-1)^n\sigma _n$$
and the formula follows by equating coefficients.