Difference between revisions of "Vieta's formulas"

Revision as of 19:07, 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 $P(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0$ be any polynomial with complex coefficients with roots $r_1, r_2, \ldots , r_n$ , and let $s_j$ be the $j$ th elementary symmetric polynomial of the roots.

Vieta’s formulas then state that $s_1 = r_1 + r_2 + \cdots + r_n = - \frac{a_{n-1}}{a_n}$ $s_2 = r_1r_2 + r_1r_3 + \cdots + r_{n-1}r_n = \frac{a_{n-2}}{a_n}$ $\vdots$ $s_n = r_1r_2r_3 \cdots r_n = (-1)^n \frac{a_0}{a_n}.$ This can be compactly written as $s_j = (-1)^j \frac{a_{n-j}}{a_n}$ for some $j$ such that $0<j \leq n$

Proof

Let all terms be defined as above. By the factor theorem, $P(x) = a_n (x-r_1)(x-r_2) \cdots (x-r_n)$ ;. When expanding this polynomial, every term is generated by $n$ choices of whether to include $x$ or $-r_{n-j}$ from any factor $(x-r_{n-j})$ . We will then prove Vieta’s formulas by expanding this polynomial and comparing the resulting coefficients with the original polynomial.

Consider all the expanded terms of $P(x)$ with degree $n-j$ ; they are formed by choosing $j$ of the negative roots, making the remaining $n-j$ choices $x$ , and finally multiplied by the constant $a_n$ . We note that when we multiply $j$ of the negative roots, we get $(-1)^j\cdot s_j$ .

So in mathematical terms, when we expand $P(x)$ , the coefficient of $x_{n-j}$ is equal to $(-1)^j a_n s_j$ .

However, we defined the coefficient of $x^{n-j}$ to be $a_{n-j}$ . Thus, $(-1)^j a_n s_j = a_{n-j}$ , or $s_j = (-1)^j a_{n-j}/a_n$ , which completes the proof. $\Box$

Problems

Here are some problems that test knowledge of Vieta's formulas.

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 == 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_j</math> be the <math>j</math>th elementary symmetric polynomial of the roots. 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 <math>s_j = (-1)^j \frac{a_{n-j}}{a_n}</math> for some <math>j</math> such that <math>0<j \leq 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_j</math> be the <math>j</math>th 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 written as <math>s_j = (-1)^j \frac{a_{n-j}}{a_n}</math> for some <math>j</math> such that <math>0<j \leq n</math>
 == 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>; 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>.
+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 expanding this polynomial, every term is generated by <math>n</math> choices of whether to include <math>x</math> or <math>-r_{n-j}</math> from any factor <math>(x-r_{n-j})</math>. We will then prove Vieta’s formulas by expanding this polynomial and comparing the resulting coefficients with the original polynomial.
+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>.
+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>.
-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 <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>\square</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</math>
 == Problems ==

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Difference between revisions of "Vieta's formulas"

Revision as of 19:07, 6 November 2021

Contents

Statement

Proof

Problems

Introductory

Intermediate

See also