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Difference between revisions of "Newton's Sums"

m (Newton sums moved to Newton's sums: AoPS Volume II calls them Newton's Sums, and it makes sense because they are the Sums of Newton.)
(categories?)
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==Basic Usage==
 
==Basic Usage==
  
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Consider a polynomial <math>\displaystyle P(x)</math> of degree <math>n</math>,
+
Consider a polynomial <math>P(x)</math> of degree <math>n</math>,
  
−
<center><math>\displaystyle  P(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0</math></center>
+
<center><math> P(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0</math></center>
  
−
Let <math>\displaystyle P(x)=0</math> have roots <math>x_1,x_2,\ldots,x_n</math>.  Define the following sums:
+
Let <math>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>S_1 = x_1 + x_2 + \cdots + x_n</math>
  
−
<math>\displaystyle S_2 = x_1^2 + x_2^2 + \cdots + x_n^2</math>
+
<math>S_2 = x_1^2 + x_2^2 + \cdots + x_n^2</math>
  
 
<math>\vdots</math>
 
<math>\vdots</math>
  
−
<math>\displaystyle S_k = x_1^k + x_2^k + \cdots + x_n^k</math>
+
<math>S_k = x_1^k + x_2^k + \cdots + x_n^k</math>
  
 
<math>\vdots</math>
 
<math>\vdots</math>
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Newton sums tell us that,
 
Newton sums tell us that,
  
−
<math>\displaystyle a_nS_1 + a_{n-1} = 0</math>
+
<math>a_nS_1 + a_{n-1} = 0</math>
  
−
<math>\displaystyle a_nS_2 + a_{n-1}S_1 + 2a_{n-2}=0</math>
+
<math>a_nS_2 + a_{n-1}S_1 + 2a_{n-2}=0</math>
  
−
<math>\displaystyle a_nS_3 + a_{n-1}S_2 + a_{n-2}S_1 + 3a_{n-3}=0</math>
+
<math>a_nS_3 + a_{n-1}S_2 + a_{n-2}S_1 + 3a_{n-3}=0</math>
  
 
<math>\vdots</math>
 
<math>\vdots</math>
  
  
−
For a more concrete example, consider the polynomial <math>\displaystyle P(x) = x^3 + 3x^2 + 4x - 8</math>.  Let the roots of <math>\displaystyle P(x)</math> be <math>\displaystyle r, s</math> and <math>\displaystyle t</math>.  Find <math>\displaystyle r^2 + s^2 + t^2</math> and <math>\displaystyle r^4 + s^4 + t^4</math>
+
For a more concrete example, consider the polynomial <math>P(x) = x^3 + 3x^2 + 4x - 8</math>.  Let the roots of <math>P(x)</math> be <math>r, s</math> and <math>t</math>.  Find <math>r^2 + s^2 + t^2</math> and <math>r^4 + s^4 + t^4</math>
  
 
Newton Sums tell us that:
 
Newton Sums tell us that:
  
−
<math>\displaystyle S_1 + 3 = 0</math>
+
<math>S_1 + 3 = 0</math>
  
−
<math>\displaystyle S_2 + 3S_1 + 8 = 0</math>
+
<math>S_2 + 3S_1 + 8 = 0</math>
  
−
<math>\displaystyle S_3 + 3S_2 + 4S_1 - 24 = 0</math>
+
<math>S_3 + 3S_2 + 4S_1 - 24 = 0</math>
  
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<math>\displaystyle S_4 + 3S_3 + 4S_2 - 8S_1 = 0</math>
+
<math>S_4 + 3S_3 + 4S_2 - 8S_1 = 0</math>
  
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Solving, first for <math>\displaystyle S_1</math>, and then for the other variables, yields,
+
Solving, first for <math>S_1</math>, and then for the other variables, yields,
  
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<math>\displaystyle S_1 = r + s + t = -3</math>
+
<math>S_1 = r + s + t = -3</math>
  
−
<math>\displaystyle S_2 = r^2 + s^2 + t^2 = 1</math>
+
<math>S_2 = r^2 + s^2 + t^2 = 1</math>
  
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<math>\displaystyle S_3 = r^3 + s^3 + t^3 = 33</math>
+
<math>S_3 = r^3 + s^3 + t^3 = 33</math>
  
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<math>\displaystyle S_4 = r^4 + s^4 + t^4 = -127</math>
+
<math>S_4 = r^4 + s^4 + t^4 = -127</math>
  
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Which gives us our desired solutions, <math>\displaystyle 1</math> and <math>\displaystyle -127</math>.
+
Which gives us our desired solutions, <math>1</math> and <math>-127</math>.
  
 
==See Also==
 
==See Also==
  
 
*[[Vieta's formulas]]
 
*[[Vieta's formulas]]
 +
 +
[[Category:Polynomials]]
 +
[[Category:Theorems]]

Revision as of 22:46, 25 December 2008

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 $P(x)$ of degree $n$,

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

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

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

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

$\vdots$

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

$\vdots$

Newton sums tell us that,

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

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

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

$\vdots$


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

Newton Sums tell us that:

$S_1 + 3 = 0$

$S_2 + 3S_1 + 8 = 0$

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

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

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

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

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

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

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

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

See Also

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