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Difference between revisions of "Newton's Sums"

(Added Newtons' Inequality to the "see also")
(Changed S to P, since I think it is more widely used, and S conflicts with the notation used in Newton's Inequality page.)
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Let <math>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>S_1 = x_1 + x_2 + \cdots + x_n</math>
+
<math>P_1 = x_1 + x_2 + \cdots + x_n</math>
  
−
<math>S_2 = x_1^2 + x_2^2 + \cdots + x_n^2</math>
+
<math>P_2 = x_1^2 + x_2^2 + \cdots + x_n^2</math>
  
 
<math>\vdots</math>
 
<math>\vdots</math>
  
−
<math>S_k = x_1^k + x_2^k + \cdots + x_n^k</math>
+
<math>P_k = x_1^k + x_2^k + \cdots + x_n^k</math>
  
 
<math>\vdots</math>
 
<math>\vdots</math>
Line 21: Line 21:
 
Newton sums tell us that,
 
Newton sums tell us that,
  
−
<math>a_nS_1 + a_{n-1} = 0</math>
+
<math>a_nP_1 + a_{n-1} = 0</math>
  
−
<math>a_nS_2 + a_{n-1}S_1 + 2a_{n-2}=0</math>
+
<math>a_nP_2 + a_{n-1}P_1 + 2a_{n-2}=0</math>
  
−
<math>a_nS_3 + a_{n-1}S_2 + a_{n-2}S_1 + 3a_{n-3}=0</math>
+
<math>a_nP_3 + a_{n-1}P_2 + a_{n-2}P_1 + 3a_{n-3}=0</math>
  
 
<math>\vdots</math>
 
<math>\vdots</math>

Revision as of 21:56, 27 June 2013

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.

Statement

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:

$P_1 = x_1 + x_2 + \cdots + x_n$

$P_2 = x_1^2 + x_2^2 + \cdots + x_n^2$

$\vdots$

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

$\vdots$

Newton sums tell us that,

$a_nP_1 + a_{n-1} = 0$

$a_nP_2 + a_{n-1}P_1 + 2a_{n-2}=0$

$a_nP_3 + a_{n-1}P_2 + a_{n-2}P_1 + 3a_{n-3}=0$

$\vdots$

(Define $a_j = 0$ for $j<0$.)

Example

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, $\boxed{1}$ and $\boxed{-127}$.

See Also

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