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Difference between revisions of "Newton's Sums"

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<center><math> P(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0</math></center>
 
<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>
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<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,
 
Newton's sums tell us that,
  
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<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>
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<math>a_nP_3 + a_{n-1}P_2 + a_{n-2}P_1 + 3a_{n-3}=0</math>
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<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>.)
−
<math>\vdots</math>
 
  
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(Define <math>a_j = 0</math> for <math>j<0</math>.)
 
  
 
We also can write:
 
We also can write:
  
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<math>P_1 = S_1</math>
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<cmath>P_1 = S_1</cmath>
−
 
+
<cmath>P_2 = S_1P_1 - 2S_2</cmath>
−
<math>P_2 = S_1P_1 - 2S_2</math>
+
<cmath>P_3 = S_1P_2 - S_2P_1 + 3S_3</cmath>
 +
<cmath>P_4 = S_1P_3 - S_2P_2 + S_3P_1 - 4S_4</cmath>
 +
<cmath>P_5 = S_1P_4 - S_2P_3 + S_3P_2 - S_4P_1 + 5S_5</cmath>
 +
<cmath>\vdots</cmath>
  
−
etc., where <math>S_n</math> denotes the <math>n</math>-th [[elementary symmetric sum]].
+
where <math>S_n</math> denotes the <math>n</math>-th [[elementary symmetric sum]].
  
 
==Proof==
 
==Proof==
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<math>\boxed{a_nP_k+a_{n-1}P_{k-1}+a_{n-2}P_{k-2}+...+a_0P_{k-n}=0}</math>
+
<cmath>a_nP_k+a_{n-1}P_{k-1}+a_{n-2}P_{k-2}+...+a_0P_{k-n}=0.</cmath>
 +
 
 +
*Note (Warning!): This technically only proves the statements for the cases where <math>k \geq n</math>. For the cases where <math>k < n</math>, 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==
 
==Example==
Line 101: Line 96:
 
==Practice==
 
==Practice==
 
[https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_17 2019 AMC 12A Problem 17]
 
[https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_17 2019 AMC 12A Problem 17]
 +
 +
[https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_9 2003 AIME II Problem 9]
 +
 +
[https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_7 2008 AIME II Problem 7]
  
 
==See Also==
 
==See Also==
  
 
*[[Vieta's formulas]]
 
*[[Vieta's formulas]]
−
 
 
*[[Newton's Inequality]]
 
*[[Newton's Inequality]]
  
 +
[[Category:Algebra]]
 
[[Category:Polynomials]]
 
[[Category:Polynomials]]
 
[[Category:Theorems]]
 
[[Category:Theorems]]

Revision as of 12:11, 20 February 2024

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,


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

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

2008 AIME II Problem 7

See Also

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