Difference between revisions of "Newton's Sums"
(Added proof) |
m (Reverted edits by Marianasinta (talk) to last revision by Resources) (Tag: Rollback) |
||
(21 intermediate revisions by 14 users not shown) | |||
Line 7: | Line 7: | ||
<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 | + | Let <math>P(x)=0</math> have roots <math>x_1,x_2,\ldots,x_n</math>. Define the sum: |
− | < | + | <cmath>P_k = x_1^k + x_2^k + \cdots + x_n^k.</cmath> |
− | + | Newton's sums tell us that, | |
− | < | + | <cmath>a_nP_1 + a_{n-1} = 0</cmath> |
− | + | <cmath>a_nP_2 + a_{n-1}P_1 + 2a_{n-2}=0</cmath> | |
− | + | <cmath>a_nP_3 + a_{n-1}P_2 + a_{n-2}P_1 + 3a_{n-3}=0</cmath> | |
− | + | <cmath>\vdots</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>.) | |
− | |||
− | |||
− | |||
− | |||
− | < | ||
− | |||
− | < | ||
− | |||
− | <math> | ||
− | |||
We also can write: | We also can write: | ||
− | < | + | <cmath>P_1 = S_1</cmath> |
− | + | <cmath>P_2 = S_1P_1 - 2S_2</cmath> | |
− | < | + | <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> | ||
− | + | where <math>S_n</math> denotes the <math>n</math>-th [[elementary symmetric sum]]. | |
==Proof== | ==Proof== | ||
Line 49: | Line 42: | ||
− | <math>\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^ | + | <math>\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}</math> |
Line 64: | Line 57: | ||
− | <math>a_n\underbrace{(\alpha^k+\beta^k+...+\omega^k)}_{ | + | <math>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</math> |
Line 70: | Line 63: | ||
− | < | + | <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 77: | Line 71: | ||
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>. | 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: | + | Newton's Sums tell us that: |
<math>P_1 + 3 = 0</math> | <math>P_1 + 3 = 0</math> | ||
Line 99: | Line 93: | ||
Which gives us our desired solutions, <math>\boxed{1}</math> and <math>\boxed{-127}</math>. | Which gives us our desired solutions, <math>\boxed{1}</math> and <math>\boxed{-127}</math>. | ||
+ | |||
+ | ==Practice== | ||
+ | [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]] |
Latest revision as of 13:11, 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.
Contents
Statement
Consider a polynomial of degree
,
![$P(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$](http://latex.artofproblemsolving.com/d/a/4/da42fa5fa45147429584e2685f8dc2e22247c497.png)
Let have roots
. Define the sum:
Newton's sums tell us that,
(Define
for
.)
We also can write:
where denotes the
-th elementary symmetric sum.
Proof
Let be the roots of a given polynomial
. Then, we have that
Thus,
Multiplying each equation by , respectively,
Sum,
Therefore,
- Note (Warning!): This technically only proves the statements for the cases where
. For the cases where
, 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 . Let the roots of
be
and
. Find
and
.
Newton's Sums tell us that:
Solving, first for , and then for the other variables, yields,
Which gives us our desired solutions, and
.