Difference between revisions of "Newton's Sums"
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− | <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> |
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− | <math>\boxed{ | + | <math>\boxed{a_nP_k+a_{n-1}P_{k-1}+a_{n-2}P_{k-2}+...+a_0P_{k-n}=0}</math> |
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==Example== | ==Example== |
Revision as of 07:17, 22 January 2018
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 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 .