Difference between revisions of "Vieta's formulas"
m |
m |
||
| Line 6: | Line 6: | ||
Let <math>P(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0</math> be any polynomial with [[Complex number | complex]] coefficients with roots <math>r_1, r_2, \ldots , r_n</math>, and let <math>s_j</math> be the <math>j^{\text{th}}</math> elementary symmetric polynomial of the roots. | Let <math>P(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0</math> be any polynomial with [[Complex number | complex]] coefficients with roots <math>r_1, r_2, \ldots , r_n</math>, and let <math>s_j</math> be the <math>j^{\text{th}}</math> elementary symmetric polynomial of the roots. | ||
| − | Vieta’s formulas then state that <cmath>s_1 = r_1 + r_2 + \cdots + r_n = - \frac{a_{n-1}}{a_n}</cmath> <cmath>s_2 = r_1r_2 + r_1r_3 + \cdots + r_{n-1}r_n = \frac{a_{n-2}}{a_n}</cmath> <cmath>\vdots</cmath> <cmath>s_n = r_1r_2r_3 \cdots r_n = (-1)^n \frac{a_0}{a_n}.</cmath> This can be compactly summarized as <math>s_j = (-1)^j \frac{a_{n-j}}{a_n}</math> for some <math>j</math> such that <math>1 \leq j \leq n</math> | + | Vieta’s formulas then state that <cmath>s_1 = r_1 + r_2 + \cdots + r_n = - \frac{a_{n-1}}{a_n}</cmath> <cmath>s_2 = r_1r_2 + r_1r_3 + \cdots + r_{n-1}r_n = \frac{a_{n-2}}{a_n}</cmath> <cmath>\vdots</cmath> <cmath>s_n = r_1r_2r_3 \cdots r_n = (-1)^n \frac{a_0}{a_n}.</cmath> This can be compactly summarized as <math>s_j = (-1)^j \frac{a_{n-j}}{a_n}</math> for some <math>j</math> such that <math>1 \leq j \leq n</math>. |
== Proof == | == Proof == | ||
Revision as of 20:03, 6 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, 
. When we expand this polynomial, each term is generated by the 
choices of whether to include 
or 
from any factor 
. We will then prove Vieta’s formulas by expanding this polynomial and comparing the resulting coefficients with the original polynomial’s coefficients.
Consider all the expanded terms of 
with degree 
; they are formed by choosing of the negative roots, making the remaining choices , and finally multiplied by the constant 
. We note that when we multiply of the negative roots, we get 
.
So in mathematical terms, when we expand , the coefficient of 
is equal to 
.
However, we defined the coefficient of 
to be 
.
Thus, 
, or 
, which completes the proof. 
Problems
Here are some problems that test knowledge of Vieta's formulas.
