Difference between revisions of "Vieta's formulas"
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== Statement == | == Statement == | ||
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_n</math> be the [[elementary symmetric polynomial]] of the roots with degree <math>n</math>. Vietas 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 written as if <math>j</math> is any integer such that <math>0<j<n</math>, then <math>s_j = (-1)^j \frac{a_{n-j}}{a_n}</math>. | 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_n</math> be the [[elementary symmetric polynomial]] of the roots with degree <math>n</math>. Vietas 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 written as if <math>j</math> is any integer such that <math>0<j<n</math>, then <math>s_j = (-1)^j \frac{a_{n-j}}{a_n}</math>. | ||
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== Proof == | == Proof == | ||
− | By the [[factor theorem]], <math>P(x) = a_n (x-r_1)(x-r_2) \cdots (x-r_n)</math>; we will then prove these formulas by expanding this polynomial. | + | 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 these formulas by expanding this polynomial. When expanding this polynomial, every term is generated by <math>n</math> choices whether to include <math>x</math> or <math>-r_{n-j}</math> from any factor <math>(x-r_{n-j})</math>. |
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− | If | + | Consider all the expanded terms of <math>P(x)</math> with degree <math>j</math>; they are formed by choosing <math>j</math> of the negative roots, then by making the remaining <math>n-j</math> choices <math>x</math>. Thus, every term is equal to a product of <math>j</math> of the negative roots multiplied by <math>x_{n-j}</math>. If one factors out <math>(-1^{j})x_{n-j}</math>, we are left with the <math>j</math>th elementary symmetric polynomial of the roots. Thus, when expanding this product, 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</math> = a_{n-j}<math>, or </math><math>s_j = (-1)^j \frac{a_{n-j}}{a_n}</math>, as required. <math>\square</math> |
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Provide links to problems that use vieta formulas: | Provide links to problems that use vieta formulas: |
Revision as of 14:59, 5 November 2021
In algebra, Vieta's formulas are a set of formulas that relate the coefficients of a polynomial to its roots.
(WIP)
Statement
Let be any polynomial with complex coefficients with roots
, and let
be the elementary symmetric polynomial of the roots with degree
. Vietas formulas then state that
This can be compactly written as if
is any integer such that
, then
.
Proof
Let all terms be defined as above. By the factor theorem, ; we will then prove these formulas by expanding this polynomial. When expanding this polynomial, every term is generated by
choices whether to include
or
from any factor
.
Consider all the expanded terms of with degree
; they are formed by choosing
of the negative roots, then by making the remaining
choices
. Thus, every term is equal to a product of
of the negative roots multiplied by
. If one factors out
, we are left with the
th elementary symmetric polynomial of the roots. Thus, when expanding this product, the coefficient of
is equal to
. However, we defined the coefficient of
to be
. Thus,
= a_{n-j}
, as required.
Provide links to problems that use vieta formulas: Examples: https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_23 https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_21
Proving Vieta's Formula
Basic proof:
This has already been proved earlier, but I will explain it more.
If we have
, the roots are
and
.
Now expanding the left side, we get:
.
Factor out an
on the right hand side and we get:
Looking at the two sides, we can quickly see that the coefficient
is equal to
.
is the actual sum of roots, however. Therefore, it makes sense that
. The same proof can be given for
.
Note: If you do not understand why we must divide by , try rewriting the original equation as