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Difference between revisions of "Vieta's formulas"
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In [[algebra]], '''Vieta's formulas''' are a set of formulas that relate the coefficients of a [[polynomial]] to its roots.
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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 polynomial | elementary symmetric polynomials]] of its roots can be easily expressed as a difference of the polynomial's coefficients.
  
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(WIP)
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It is among the most ubiquitous results to circumvent finding a polynomial's roots in competition math and sees widespread usage in all contests.
  
 
== Statement ==
 
== Statement ==
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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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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>
  
 
== 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>; 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>; 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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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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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 = a_{n-j}</math>, or <math>s_j = (-1)^j a_{n-j}/a_n</math>, as required. <math>\square</math>
  
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Provide links to problems that use vieta formulas:
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== Problems ==
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Examples:
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Here are some problems that test knowledge of Vieta's formulas.
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https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_23
 
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https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_21
 
  
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==Proving Vieta's Formula==
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=== Introductory ===
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Basic proof:
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* [[2010 AMC 10A Problems/Problem 21 | 2019 AMC 10A Problem 21]]
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This has already been proved earlier, but I will explain it more.
 
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If we have
 
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<math>x^2+ax+b=(x-p)(x-q)</math>, the roots are <math>p</math> and <math>q</math>.
 
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Now expanding the left side, we get: <math>x^2+ax+b=x^2-qx-px+pq</math>.
 
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Factor out an <math>x</math> on the right hand side and we get: <math>x^2+ax+b=x^2-x(p+q)+pq</math>
 
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Looking at the two sides, we can quickly see that the coefficient <math>a</math> is equal to <math>-(p+q)</math>. <math>p+q</math> is the actual sum of roots, however. Therefore, it makes sense that <math>p+q= \frac{-b}{a}</math>. The same proof can be given for <math>pq=\frac{c}{a}</math>.
 
  
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Note: If you do not understand why we must divide by <math>a</math>, try rewriting the original equation as <math>ax^2+bx+c=(x-p)(x-q)</math>
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=== Intermediate ===
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* [[2017 AMC 12A Problems/Problem 23 | 2017 AMC 12A Problem 23]]
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== See also ==
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* [[Polynomial]]
  
 
[[Category:Algebra]]
 
[[Category:Algebra]]
 
[[Category:Polynomials]]
 
[[Category:Polynomials]]
 
[[Category:Theorems]]
 
[[Category:Theorems]]

Revision as of 17:29, 5 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 difference 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 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. Vietas 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, $P(x) = a_n (x-r_1)(x-r_2) \cdots (x-r_n)$; we will then prove these formulas by expanding this polynomial. When expanding this polynomial, every term is generated by $n$ choices whether to include $x$ or $-r_{n-j}$ from any factor $(x-r_{n-j})$.

Consider all the expanded terms of $P(x)$ with degree $j$; they are formed by choosing $j$ of the negative roots, then by making the remaining $n-j$ choices $x$. Thus, every term is equal to a product of $j$ of the negative roots multiplied by $x_{n-j}$. If one factors out $(-1^{j})x_{n-j}$, we are left with the $j$th elementary symmetric polynomial of the roots. Thus, when expanding this product, 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 $s_j = (-1)^j a_{n-j}/a_n$, as required. $\square$

Problems

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

Introductory


Intermediate

See also

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