Difference between revisions of "Newton's Sums"

Revision as of 09:22, 12 June 2021

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 sum:

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

Newton's 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 $j<0$ .)

We also can write:

$P_1 = S_1$ $P_2 = S_1P_1 - 2S_2$

etc., where $S_n$ denotes the $n$ -th elementary symmetric sum.

Proof

Let $\alpha,\beta,\gamma,...,\omega$ be the roots of a given polynomial $P(x)=a_nx^n+a_{n-1}x^{n-1}+..+a_1x+a_0$ . Then, we have that

$P(\alpha)=P(\beta)=P(\gamma)=...=P(\omega)=0$

Thus,

$\begin{cases}a_n\alpha^n+a_{n-1}\alpha^{n-1}+...+a_0=0\\a_n\beta^n+a_{n-1}\beta^{n-1}+...+a_0=0\\~~~~~~~~~~~~~~~~~~\vdots\\a_n\omega^n+a_{n-1}\omega^{n-1}+...+a_0=0\end{cases}$

Multiplying each equation by $\alpha^{k-n},\beta^{k-n},...,\omega^{k-n}$ , respectively,

$\begin{cases}a_n\alpha^{n+k-n}+a_{n-1}\alpha^{n-1+k-n}+...+a_0\alpha^{k-n}=0\\a_n\beta^{n+k-n}+a_{n-1}\beta^{n-1+k-n}+...+a_0\beta^{k-n}=0\\~~~~~~~~~~~~~~~~~~\vdots\\a_n\omega^{n+k-n}+a_{n-1}\omega^{n-1+k-n}+...+a_0\omega^{k-n}=0\end{cases}$

$\begin{cases}a_n\alpha^{k}+a_{n-1}\alpha^{k-1}+...+a_0\alpha^{k-n}=0\\a_n\beta^{k}+a_{n-1}\beta^{k-1}+...+a_0\beta^{k-n}=0\\~~~~~~~~~~~~~~~~~~\vdots\\a_n\omega^{k}+a_{n-1}\omega^{k-1}+...+a_0\omega^{k-n}=0\end{cases}$

Sum,

$a_n\underbrace{(\alpha^k+\beta^k+...+\omega^k)}_{P_k}+a_{n-1}\underbrace{(\alpha^{k-1}+\beta^{k-1}+...+\omega^{k-1})}_{P_{k-1}}+a_{n-2}\underbrace{(\alpha^{k-2}+\beta^{k-2}+...+\omega^{k-2})}_{P_{k-2}}+...+a_0\underbrace{(\alpha^{k-n}+\beta^{k-n}+...+\omega^{k-n})}_{P_{k-n}}=0$

Therefore,

$\boxed{a_nP_k+a_{n-1}P_{k-1}+a_{n-2}P_{k-2}+...+a_0P_{k-n}=0}.$

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's Sums tell us that:

$P_1 + 3 = 0$

$P_2 + 3P_1 + 8 = 0$

$P_3 + 3P_2 + 4P_1 - 24 = 0$

$P_4 + 3P_3 + 4P_2 - 8P_1 = 0$

Solving, first for $P_1$ , and then for the other variables, yields,

$P_1 = r + s + t = -3$

$P_2 = r^2 + s^2 + t^2 = 1$

$P_3 = r^3 + s^3 + t^3 = 33$

$P_4 = r^4 + s^4 + t^4 = -127$

Which gives us our desired solutions, $\boxed{1}$ and $\boxed{-127}$ .

Practice

2019 AMC 12A Problem 17

@@ Line 7: / Line 7: @@
 <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:
+Let <math>P(x)=0</math> have roots <math>x_1,x_2,\ldots,x_n</math>.  Define the sum:
-<math>P_1 = x_1 + x_2 + \cdots + x_n</math>
+<cmath>P_k = x_1^k + x_2^k + \cdots + x_n^k.</cmath>
-<math>P_2 = x_1^2 + x_2^2 + \cdots + x_n^2</math>
-<math>\vdots</math>
-<math>P_k = x_1^k + x_2^k + \cdots + x_n^k</math>
-<math>\vdots</math>
 Newton's sums tell us that,
-<math>a_nP_1 + a_{n-1} = 0</math>
+<cmath>a_nP_1 + a_{n-1} = 0</cmath>
+<cmath>a_nP_2 + a_{n-1}P_1 + 2a_{n-2}=0</cmath>
-<math>a_nP_2 + a_{n-1}P_1 + 2a_{n-2}=0</math>
+<cmath>a_nP_3 + a_{n-1}P_2 + a_{n-2}P_1 + 3a_{n-3}=0</cmath>
+<cmath>\vdots</cmath>
-<math>a_nP_3 + a_{n-1}P_2 + a_{n-2}P_1 + 3a_{n-3}=0</math>
-<math>\vdots</math>
 (Define <math>a_j = 0</math> for <math>j<0</math>.)
@@ Line 33: / Line 22: @@
 We also can write:
-<math>P_1 = S_1</math>
+<cmath>P_1 = S_1</cmath>
+<cmath>P_2 = S_1P_1 - 2S_2</cmath>
-<math>P_2 = S_1P_1 - 2S_2</math>
 etc., where <math>S_n</math> denotes the <math>n</math>-th [[elementary symmetric sum]].

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