Difference between revisions of "Newton's Sums"
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<cmath>a_nP_3 + a_{n-1}P_2 + a_{n-2}P_1 + 3a_{n-3}=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>\vdots</cmath> | ||
− | <cmath>a_nP_k+a_{n-1}P_{k-1}+\cdots+a_{n-k+1}P_1+k\cdot a_{n-k}=0</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>.) | (Define <math>a_j = 0</math> for <math>j<0</math>.) | ||
Revision as of 12:24, 4 February 2023
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 .