Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Newton's Sums"

m (→‎Basic Usage)
m (→‎Basic Usage)
Line 5: Line 5:
 
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 <math>\displaystyle P(x)</math> = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0</math></center>
+
<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 19:56, 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 $\displaystyle P(x)$ of degree $n$,

$\displaystyle P(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$

Let $\displaystyle P(x)=0$ have roots $x_1,x_2,\ldots,x_n$. Define the following sums:

$\displaystyle S_1 = x_1 + x_2 + \cdots + x_n$

$\displaystyle S_2 = x_1^2 + x_2^2 + \cdots + x_n^2$

$\vdots$

$\displaystyle S_k = x_1^k + x_2^k + \cdots + x_n^k$

$\vdots$

Newton sums tell us that,

$\displaystyle a_nS_1 + a_{n-1} = 0$

$\displaystyle a_nS_2 + a_{n-1}S_1 + 2a_{n-2}=0$

$\displaystyle a_nS_3 + a_{n-1}S_2 + a_{n-2}S_1 + 3a_{n-3}=0$

$\vdots$


For a more concrete example, consider the polynomial $\displaystyle P(x) = x^3 + 3x^2 + 4x - 8$. Let the roots of $\displaystyle P(x)$ be $\displaystyle r, s$ and $\displaystyle t$. Find $\displaystyle r^2 + s^2 + t^2$ and $\displaystyle r^4 + s^4 + t^4$

Newton Sums tell us that:

$\displaystyle S_1 + 3 = 0$

$\displaystyle S_2 + 3S_1 + 8 = 0$

$\displaystyle S_3 + 3S_2 + 4S_1 - 24 = 0$

$\displaystyle S_4 + 3S_3 + 4S_2 - 8S_1 = 0$

Solving, first for $\displaystyle S_1$, and then for the other variables, yields,

$\displaystyle S_1 = r + s + t = -3$

$\displaystyle S_2 = r^2 + s^2 + t^2 = 1$

$\displaystyle S_3 = r^3 + s^3 + t^3 = 33$

$\displaystyle S_4 = r^4 + s^4 + t^4 = -127$

Which gives us our desired solutions, $\displaystyle 1$ and $\displaystyle -127$.

See Also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Newton%27s_Sums&oldid=15273"