Applying Vieta's Formulas generally

Question

According to Wikipedia: I am having trouble understanding what is omitted by the $\vdots$ . How is the second statement reduced into the last?

Answer 1

The first line contains the sum $x_1+x_2+\cdots+x_n$ of all roots.

The second line contains the sum of all products $x_ix_j$ of two distinct roots.

The third (omitted line) contains the sum of all products $x_ix_jx_k$ of three distinct roots.

And so on.

The list ($n$^th) line contains the product of all $n$ roots.

Answer 2

The left-hand-side of the $k$-th line is the sum of all algebraically distinct products of $k$ distinct terms of the sequence $x_1,...,x_n$.

For each product in the sum, there are ${\large{\binom{n}{k}}}$ ways of choosing the $k$ factors (up to algebraic equivalence), so we get ${\large{\binom{n}{k}}}$ summands.

It's usually read as:

"The sum of the products of the roots, taken $k$ at a time."

In particular:

• For the first line ($k=1$), the left-hand-side $$x_1+\cdots + x_n$$ has $n$ summands (since ${\large{\binom{n}{1}}}=n$), and is just the sum of the roots.$\\[12pt]$
• For the last line ($k=n$), the left-hand-side $$x_1\cdots x_n$$ has only one summand (since ${\large{\binom{n}{n}}}=1$), and is just the product of the roots.

Note:

• When $k=1$, no actual multiplication is needed since each of the $n$ summands requires only one term.$\\[4pt]$
• When $k=n$, no actual addition is needed since there is only one summand.

As regards the right-hand side of the $k$-th line, it's just $(-1)^k\cdot{\large{\frac{a_{n-k}}{a_n}}}$. Note that the prefixed sign alternates, starting with a minus sign (i.e., for $k=1$, we have $(-1)^k=-1$).

As an example, letting $n=4$, we get $$ \begin{cases} x_1+x_2+x+x_3+x_4=-{\large{\frac{a_3}{a_4}}}\\[4pt] x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4={\large{\frac{a_2}{a_4}}}\\[4pt] x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4=-{\large{\frac{a_1}{a_4}}}\\[4pt] x_1x_2x_3x_4={\large{\frac{a_0}{a_4}}}\\[4pt] \end{cases} $$ To test the formula, and to get an idea as to why it holds, let's consider the second line of the above example . . .

By the Complete Factorization Theorem, we can express $P(x)$ as $$P(x)=a_4(x-x_1)(x-x_2)(x-x_3)(x-x_4)$$ In the expansion of $P$, you can verify that the $x^2$ coefficient, $a_2$, is equal to the sum of all products of the form $(a_4)(-x_i)(-x_j)$, where $1\le i < j\le 4$.

Thus, we get $$a_2=a_4(x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4)$$ which yields $$x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4={\small{\frac{a_2}{a_4}}}$$