Difference between revisions of "Newton's Sums"
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− | <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> |
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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> | ||
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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] | ||
==See Also== | ==See Also== |
Revision as of 13:59, 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.
Contents
Statement
Consider a polynomial of degree ,
Let have roots . Define the following sums:
Newton sums tell us that,
(Define for .)
We also can write:
etc., 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,
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 .