Difference between revisions of "Newton's Sums"
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− | <math>\begin{cases}a_n\alpha^n+a_{n-1}\alpha^{n-1}+...+a_0=0\\a_n\beta^n+a_{n-1}\beta^{n-1}+...+a_0=0\\~~~~~~~~~~~~~~~~~~\vdots\\a_n\omega^ | + | <math>\begin{cases}a_n\alpha^n+a_{n-1}\alpha^{n-1}+...+a_0=0\\a_n\beta^n+a_{n-1}\beta^{n-1}+...+a_0=0\\~~~~~~~~~~~~~~~~~~\vdots\\a_n\omega^n+a_{n-1}\omega^{n-1}+...+a_0=0\end{cases}</math> |
Revision as of 23:09, 8 January 2020
Newton sums give us a clever and efficient way of finding the sums of roots of a polynomial raised to a power. They can also be used to derive several factoring identities.
Contents
Statement
Consider a polynomial of degree
,
![$P(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$](http://latex.artofproblemsolving.com/d/a/4/da42fa5fa45147429584e2685f8dc2e22247c497.png)
Let have roots
. Define the following sums:
Newton sums tell us that,
(Define for
.)
We also can write:
etc., where denotes the
-th elementary symmetric sum.
Proof
Let be the roots of a given polynomial
. Then, we have that
Thus,
Multiplying each equation by , respectively,
Sum,
Therefore,
Example
For a more concrete example, consider the polynomial . Let the roots of
be
and
. Find
and
.
Newton Sums tell us that:
Solving, first for , and then for the other variables, yields,
Which gives us our desired solutions, and
.
Practice
2019 AMC 12A #17