Difference between revisions of "Vieta's formulas"
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== Proof == | == Proof == | ||
| − | Let all terms be defined as above. By the [[factor theorem]], <math>P(x) = a_n (x-r_1)(x-r_2) \cdots (x-r_n | + | Let all terms be defined as above. By the [[factor theorem]], <math>P(x) = a_n (x-r_1)(x-r_2) \cdots (x-r_n)</math>. We will then prove Vieta’s formulas by expanding this polynomial and comparing the resulting coefficients with the original polynomial’s coefficients. |
| − | + | When expanding this factorization of <math>P(x)</math>, each term is generated by a series of <math>n</math> choices of whether to include <math>x</math> or the negative root <math>-r_{i}</math> from every factor <math>(x-r_{i})</math>. Consider all the expanded terms of the polynomial with degree <math>n-j</math>; they are formed by multiplying a choice of <math>j</math> negative roots, making the remaining <math>n-j</math> choices in the product <math>x</math>, and finally multiplying by the constant <math>a_n</math>. | |
| − | + | Note that adding together every multiplied choice of <math>j</math> negative roots yields <math>-1)^ </math>(-1)^j s_j<math>. Thus, when we expand </math>P(x)<math>, the coefficient of </math>x_{n-j}<math> is equal to </math>(-1)^j a_n s_j<math>. However, we defined the coefficient of </math>x^{n-j}<math> to be </math>a_{n-j}<math>. Thus, </math>(-1)^j a_n s_j = a_{n-j}<math>, or </math>s_j = (-1)^j a_{n-j}/a_n<math>, which completes the proof. </math>\Box$ | |
| − | |||
| − | However, we defined the coefficient of <math>x^{n-j}< | ||
| − | |||
| − | Thus, <math>(-1)^j a_n s_j = a_{n-j}< | ||
== Problems == | == Problems == | ||
Revision as of 10:05, 7 November 2021
In algebra, Vieta's formulas are a set of results that relate the coefficients of a polynomial to its roots. In particular, it states that the elementary symmetric polynomials of its roots can be easily expressed as a ratio between two of the polynomial's coefficients.
It is among the most ubiquitous results to circumvent finding a polynomial's roots in competition math and sees widespread usage in all mathematics contests.
Statement
Let 
be any polynomial with complex coefficients with roots 
, and let 
be the 
elementary symmetric polynomial of the roots.
Vieta’s formulas then state that ![\[s_1 = r_1 + r_2 + \cdots + r_n = - \frac{a_{n-1}}{a_n}\]](http://latex.artofproblemsolving.com/4/8/f/48f4bdf0997f87131541283a8db9b4b3f9b9be09.png)
![\[s_2 = r_1r_2 + r_1r_3 + \cdots + r_{n-1}r_n = \frac{a_{n-2}}{a_n}\]](http://latex.artofproblemsolving.com/b/4/6/b463058fe37152140aeb6a5f5b291b04020aec11.png)
![\[\vdots\]](http://latex.artofproblemsolving.com/7/6/4/76403adfdd35fb846675aaeaac92daec680a4fba.png)
![\[s_n = r_1r_2r_3 \cdots r_n = (-1)^n \frac{a_0}{a_n}.\]](http://latex.artofproblemsolving.com/8/c/3/8c309b1382001729c054020a9b65792d6cc5800c.png)
This can be compactly summarized as 
for some 
such that 
.
Proof
Let all terms be defined as above. By the factor theorem, 
. We will then prove Vieta’s formulas by expanding this polynomial and comparing the resulting coefficients with the original polynomial’s coefficients.
When expanding this factorization of 
, each term is generated by a series of 
choices of whether to include 
or the negative root 
from every factor 
. Consider all the expanded terms of the polynomial with degree 
; they are formed by multiplying a choice of negative roots, making the remaining choices in the product , and finally multiplying by the constant 
.
Note that adding together every multiplied choice of negative roots yields $-1)^$ (Error compiling LaTeX. Unknown error_msg)(-1)^j s_j
P(x)
x_{n-j}
(-1)^j a_n s_j
x^{n-j}
a_{n-j}
(-1)^j a_n s_j = a_{n-j}
s_j = (-1)^j a_{n-j}/a_n
\Box$
Problems
Here are some problems with solutions that utilize Vieta's formulas.
