Difference between revisions of "Newton's Sums"
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(Define <math>a_j = 0</math> for <math>j<0</math>.) | (Define <math>a_j = 0</math> for <math>j<0</math>.) | ||
+ | |||
+ | We also can write: | ||
+ | |||
+ | <math>P_1 = S_1</math> | ||
+ | |||
+ | <math>P_2 = S_1P_1 - 2S_2</math> | ||
+ | |||
+ | etc., 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>\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> | ||
+ | |||
+ | |||
+ | Multiplying each equation by <math>\alpha^{k-n},\beta^{k-n},...,\omega^{k-n}</math>, respectively, | ||
+ | |||
+ | |||
+ | <math>\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}</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, | ||
+ | |||
+ | |||
+ | <math>\boxed{a_nP_k+a_{n-1}P_{k-1}+a_{n-2}P_{k-2}+...+a_0P_{k-n}=0}</math> | ||
==Example== | ==Example== | ||
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− | Solving, first for <math> | + | Solving, first for <math>P_1</math>, and then for the other variables, yields, |
<math>P_1 = r + s + t = -3</math> | <math>P_1 = r + s + t = -3</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== | ||
+ | 2019 AMC 12A #17 | ||
==See Also== | ==See Also== |
Revision as of 23:09, 8 January 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 Sums tell us that:
Solving, first for , and then for the other variables, yields,
Which gives us our desired solutions, and .
Practice
2019 AMC 12A #17