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Difference between revisions of "Newton's Sums"

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==Practice==
 
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2019 AMC 12A #17
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[https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_17 2019 AMC 12A Problem 17]
  
 
==See Also==
 
==See Also==

Revision as of 12:59, 31 July 2020

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 $j<0$.)

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

See Also

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