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'''Newton sums''' give us a clever and efficient way of finding the sums of [[root]]s of a [[polynomial]] raised to a power.  They can also be used to derive several [[factoring]] [[identity|identities]].

==Basic Usage==

Consider a polynomial <math>P(x)</math> of degree <math>n</math>,

<center><math> 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:

<math>S_1 = x_1 + x_2 + \cdots + x_n</math>

<math>S_2 = x_1^2 + x_2^2 + \cdots + x_n^2</math>

<math>\vdots</math>

<math>S_k = x_1^k + x_2^k + \cdots + x_n^k</math>

<math>\vdots</math>

Newton sums tell us that,

<math>a_nS_1 + a_{n-1} = 0</math>

<math>a_nS_2 + a_{n-1}S_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>\vdots</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:

<math>S_1 + 3 = 0</math>

<math>S_2 + 3S_1 + 8 = 0</math>

<math>S_3 + 3S_2 + 4S_1 - 24 = 0</math>

<math>S_4 + 3S_3 + 4S_2 - 8S_1 = 0</math>

Solving, first for <math>S_1</math>, and then for the other variables, yields,

<math>S_1 = r + s + t = -3</math>

<math>S_2 = r^2 + s^2 + t^2 = 1</math>

<math>S_3 = r^3 + s^3 + t^3 = 33</math>

<math>S_4 = r^4 + s^4 + t^4 = -127</math>

Which gives us our desired solutions, <math>1</math> and <math>-127</math>.

==See Also==

*[[Vieta's formulas]]

[[Category:Polynomials]]
[[Category:Theorems]]