Vieta's formulae - looking for a plain English explanation of the general-case formulae I am working with a set of equations within which arise functions of the coefficients of a polynomial expressed in terms of the roots. The expressions get rather complicated as the order of the polynomial goes above 3, and so I want to write these coefficients using Vieta's formulae:
\begin{equation}
\sum_{1\leq i_1 < i_2 \cdots i_k \leq n}\left(\prod_{j=1}^k \alpha_{i_j}\right)
\end{equation}
However, I am struggling to disentangle this notation, in particular the expression beneath the summation and the notation subscripts for $\alpha$. I know what the result should be for a polynomial of order 1, 2, 3 4 and so on, but I would appreciate a plain-language description of how the notation describes this - and any suggestions as to how the expression could be made more straightforward to interpret?
 A: This particular formula is for a particular $k$.
The summation is for every strictly increasing sequence of $k$ elements, in which each element ranges from $1$ to $n$ inclusive. The inequality $1\le i_1 < i_2 < \ldots < i_k \le n$ denotes that the sequence $(i_1, i_2, \ldots, i_k)$ should have $k$ elements, the elements are strictly increasing, and the elements are between $1$ and $n$ inclusive.
There are $\binom n k$ such sequences, so there are $\binom nk$ summands. Each summand is a product, which may be written as
$$\prod_{j=1}^k \alpha_{i_j} = \prod_{i\in\{i_1, i_2,\ldots,i_k\}}\alpha_i.$$
The product means that for each unique sequence, the sequence would choose $k$ of the $n$ roots, and take their product.
For example, $k = 4$ and $n=6$, one of the sequences is $(1,2,3,5)$, and the sequence corresponds to the product $\prod_{i\in\{1,2,3,5\}} \alpha_i = \alpha_1\alpha_2\alpha_3\alpha_5$. There are $\binom64 = 15$ such products to sum.
So I may reword the summation and product sign as
$$\sum_{S : k-\text{sequence which is strictly increasing between }1\text{ and }n} \left(\prod_{i\in S}\alpha_i\right)$$
A: This summation merely expresses the sum of all possible products of $k$ roots.
The roots, $\alpha_1$, $\alpha_2$, $\ldots,$ $\alpha_n$ have indices $1$, $2$, $\ldots$, $n$, so call the set $\{1,2,\ldots,n\}$ the index set. The summation,
$$
\sum_{1\le i_1<i_2<\ldots<i_k\le n},
$$
which could also be written
$$
\sum_{i_1=1}^{n-k+1}\sum_{i_2=i_1+1}^{n-k+2}\ldots\sum_{i_k=i_{k-1}+1}^n
$$
effectively runs over all subsets of the index set that are composed of $k$ elements from that set. So the first subset will be $\{1,2,\ldots,k\}$, the second will be $\{1,2,\ldots,k-1,k+1\}$, and the last will be $\{n-k+1,n-k+2,\ldots,n\}$. In the particular case $n=5$, $k=3$, the complete list of subsets is
\begin{align}
&\{1,2,3\},\ \{1,2,4\},\ \{1,2,5\},\ \{1,3,4\},\ \{1,3,5\},\\
&\{1,4,5\},\ \{2,3,4\},\ \{2,3,5\},\ \{2,4,5\},\ \{3,4,5\}
\end{align}
In the summation, the fourth of these subsets, $\{1,3,4\}$, to take an example, corresponds to $i_1=1$, $i_2=3$, $i_3=4$ and the summand will be the product $\alpha_1\alpha_3\alpha_4$.
You could express the summation as
$$
\sum_{\text{all size-$k$ subsets $S$ of the index set}}(\text{product of roots indexed by elements of $S$}).
$$
If you multiply out
$$
(x-\alpha_1)(x-\alpha_2)(x-\alpha_3)(x-\alpha_4)(x-\alpha_5),
$$
you will get $2^5=32$ terms:
\begin{align}
&x x x x x+x x x x (-\alpha_5)+xxx(-\alpha_4)x+xxx(-\alpha_4)(-\alpha_5)\\
&+xx(-\alpha_3)xx+xx(-\alpha_3)x(-\alpha_5)+xx(-\alpha_3)(-\alpha_4)x+xx(-\alpha_3)(-\alpha_4)(-\alpha_5)\\
&+x(-\alpha_2)xxx+\ldots\\
&\vdots\\
&+\ldots+(-\alpha_1)(-\alpha_2)(-\alpha_3)(-\alpha_4)(-\alpha_5).
\end{align}
If you think about collecting the $x^2=x^{5-3}$ terms in this expansion, you'll realize that, apart from a minus-sign issue, the resulting coefficient of $x^2$ is precisely the sum of products indexed by the ten three-element subsets listed above. I hope this suffices to illustrate the general principle.
