Difference between revisions of "Newton's Sums"
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Consider a polynomial <math>\displaystyle P(x)</math> of degree <math>n</math>, | Consider a polynomial <math>\displaystyle P(x)</math> of degree <math>n</math>, | ||
− | <center><math>\displaystyle | + | <center><math>\displaystyle P(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0</math></center> |
− | Let <math>P(x)=0</math> have roots <math>x_1,x_2,\ldots,x_n</math>. Define the following sums: | + | Let <math>\displaystyle P(x)=0</math> have roots <math>x_1,x_2,\ldots,x_n</math>. Define the following sums: |
<math>\displaystyle S_1 = x_1 + x_2 + \cdots + x_n</math> | <math>\displaystyle S_1 = x_1 + x_2 + \cdots + x_n</math> |
Revision as of 20:56, 2 July 2007
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.
Basic Usage
Consider a polynomial of degree
,
![$\displaystyle P(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$](http://latex.artofproblemsolving.com/0/4/8/048f3d9f71b9de5006ba1da753b7631c32eff33b.png)
Let have roots
. Define the following sums:
Newton sums tell us that,
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
.