# 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: $$$$\sum_{1\leq i_1 < i_2 \cdots i_k \leq n}\left(\prod_{j=1}^k \alpha_{i_j}\right)$$$$ However, I am struggling to disentangle this notation, in particular the expression beneath the summation and the notation subscripts for $$\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?

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

$$\sum_{S : k-\text{sequence which is strictly increasing between }1\text{ and }n} \left(\prod_{i\in S}\alpha_i\right)$$

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, $$\sum_{1\le i_1 which could also be written $$\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.