Difference between revisions of "Newton's Sums"

Revision as of 12:24, 4 February 2023

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,

$a_nP_1 + a_{n-1} = 0$ $a_nP_2 + a_{n-1}P_1 + 2a_{n-2}=0$ $a_nP_3 + a_{n-1}P_2 + a_{n-2}P_1 + 3a_{n-3}=0$ $\vdots$ $\boxed{a_nP_k+a_{n-1}P_{k-1}+\cdots+a_{n-k+1}P_1+k\cdot a_{n-k}=0}$ (Define $a_j = 0$ for $j<0$ .)

We also can write:

$P_1 = S_1$ $P_2 = S_1P_1 - 2S_2$ $P_3 = S_1P_2 - S_2P_1 + 3S_3$ $P_4 = S_1P_3 - S_2P_2 + S_3P_1 - 4S_4$ $P_5 = S_1P_4 - S_2P_3 + S_3P_2 - S_4P_1 + 5S_5$ $\vdots$

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,

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

Note (Warning!): This technically only proves the statements for the cases where $k \geq n$ . For the cases where $k < n$ , an argument based on analyzing individual monomials in the expansion can be used (see http://web.stanford.edu/~marykw/classes/CS250_W19/Netwons_Identities.pdf, for example.)

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

2003 AIME II Problem 9

@@ Line 17: / Line 17: @@
 <cmath>a_nP_3 + a_{n-1}P_2 + a_{n-2}P_1 + 3a_{n-3}=0</cmath>
 <cmath>\vdots</cmath>
-<cmath>a_nP_k+a_{n-1}P_{k-1}+\cdots+a_{n-k+1}P_1+k\cdot a_{n-k}=0</cmath>
+<cmath>\boxed{a_nP_k+a_{n-1}P_{k-1}+\cdots+a_{n-k+1}P_1+k\cdot a_{n-k}=0}</cmath>
 (Define <math>a_j = 0</math> for <math>j<0</math>.)

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