Difference between revisions of "Newton's Sums"
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<cmath>P_2 = S_1P_1 - 2S_2</cmath> | <cmath>P_2 = S_1P_1 - 2S_2</cmath> | ||
<cmath>P_3 = S_1P_2 - S_2P_1 + 3S_3</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> | <cmath>\vdots</cmath> | ||
Revision as of 17:07, 21 July 2022
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: 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 .