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 | + | 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> |
− | <math> | + | 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>.) | ||
+ | |||
+ | |||
+ | 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== | ||
+ | |||
+ | Let <math>\alpha,\beta,\gamma,...,\omega</math> be the roots of a given polynomial <math>P(x)=a_nx^n+a_{n-1}x^{n-1}+..+a_1x+a_0</math>. Then, we have that | ||
+ | |||
+ | <math>P(\alpha)=P(\beta)=P(\gamma)=...=P(\omega)=0</math> | ||
+ | |||
+ | |||
+ | Thus, | ||
+ | |||
+ | |||
+ | <math> | ||
+ | |||
+ | |||
+ | Multiplying each equation by <math>\alpha^{k-n},\beta^{k-n},...,\omega^{k-n}</math>, respectively, | ||
+ | |||
+ | |||
+ | <math> | ||
+ | |||
+ | |||
+ | <math>\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}</math> | ||
− | |||
− | + | Sum, | |
− | |||
− | + | <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> | |
− | |||
− | + | Therefore, | |
− | |||
− | < | + | <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== | ||
− | 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's Sums tell us that: | ||
+ | |||
+ | <math>P_1 + 3 = 0</math> | ||
+ | |||
+ | <math>P_2 + 3P_1 + 8 = 0</math> | ||
+ | |||
+ | <math>P_3 + 3P_2 + 4P_1 - 24 = 0</math> | ||
− | + | <math>P_4 + 3P_3 + 4P_2 - 8P_1 = 0</math> | |
− | |||
− | <math> | + | Solving, first for <math>P_1</math>, and then for the other variables, yields, |
− | <math> | + | <math>P_1 = r + s + t = -3</math> |
− | <math> | + | <math>P_2 = r^2 + s^2 + t^2 = 1</math> |
+ | <math>P_3 = r^3 + s^3 + t^3 = 33</math> | ||
− | + | <math>P_4 = r^4 + s^4 + t^4 = -127</math> | |
− | <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] | |
− | + | [https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_13#Solution_2 2024 AIME II Problem 13] | |
==See Also== | ==See Also== | ||
*[[Vieta's formulas]] | *[[Vieta's formulas]] | ||
+ | *[[Newton's Inequality]] | ||
+ | [[Category:Algebra]] | ||
[[Category:Polynomials]] | [[Category:Polynomials]] | ||
[[Category:Theorems]] | [[Category:Theorems]] |
Latest revision as of 07:28, 24 November 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
[hide]Statement
Consider a polynomial of degree ,
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 .