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)</math> | + | 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>. When we expand this polynomial, each term is generated by <math>n</math> choices of whether to include <math>x</math> or <math>-r_{n-i}</math> from any factor <math>(x-r_{n-i})</math>. 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 <math>P(x)</math> with degree <math>n-j</math>; they are formed by choosing <math>j</math> of the negative roots, making the remaining <math>n-j</math> choices <math>x</math>, and finally multiplied by the constant <math>a_n</math>. We note that when we multiply <math>j</math> of the negative roots, we get <math>(-1)^j\cdot s_j</math>. | Consider all the expanded terms of <math>P(x)</math> with degree <math>n-j</math>; they are formed by choosing <math>j</math> of the negative roots, making the remaining <math>n-j</math> choices <math>x</math>, and finally multiplied by the constant <math>a_n</math>. We note that when we multiply <math>j</math> of the negative roots, we get <math>(-1)^j\cdot s_j</math>. | ||
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So in mathematical terms, 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>. | So in mathematical terms, 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} | + | However, we defined the coefficient of <math>x^{n-j}</math> to be <math>a_{n-j}</math>. |
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+ | Thus, <math>(-1)^j a_n s_j = a_{n-j}</math>, or <math>s_j = (-1)^j \dfrac{a_{n-j}}{a_n}</math>, which completes the proof. <math>\Box</math> | ||
== Problems == | == Problems == |
Revision as of 20:44, 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
th elementary symmetric polynomial of the roots.
Vieta’s formulas then state that
This can be compactly written 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
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.