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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 ==
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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 <math>j</math> be any integer such that <math>0<j<n</math>. We wish to find a process that generates every term with degree <math>j</math>. If
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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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<center><math>a_n = a_n</math></center>
 
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<center><math> a_{n-1} = -a_n(r_1+r_2+\cdots+r_n)</math></center>
 
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<center><math> a_{n-2} = a_n(r_1r_2+r_1r_3+\cdots+r_{n-1}r_n)</math></center>
 
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<center><math>\vdots</math></center>
 
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<center><math>a_0 = (-1)^n a_n r_1r_2\cdots r_n</math></center>
 
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More commonly, these are written with the roots on one side and the <math>a_i</math> on the other (this can be arrived at by dividing both sides of all the equations by <math>a_n</math>).
 
  
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If we denote <math>\sigma_k</math> as the <math>k</math>-th elementary symmetric sum, then we can write those formulas more compactly as <math>\sigma_k = (-1)^k\cdot \frac{a_{n-k}}{a_n{}}</math>, for <math>1\le k\le {n}</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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Also, <math>-b/a = p + q, c/a = p \cdot q</math>.
 
  
 
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 $P(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0$ be any polynomial with complex coefficients with roots $r_1, r_2, \ldots , r_n$, and let $s_n$ be the elementary symmetric polynomial of the roots with degree $n$. 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 if $j$ is any integer such that $0<j<n$, then $s_j = (-1)^j \frac{a_{n-j}}{a_n}$.

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 \frac{a_{n-j}}{a_n}$, as required. $\square$

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 $x^2+ax+b=(x-p)(x-q)$, the roots are $p$ and $q$. Now expanding the left side, we get: $x^2+ax+b=x^2-qx-px+pq$. Factor out an $x$ on the right hand side and we get: $x^2+ax+b=x^2-x(p+q)+pq$ Looking at the two sides, we can quickly see that the coefficient $a$ is equal to $-(p+q)$. $p+q$ is the actual sum of roots, however. Therefore, it makes sense that $p+q= \frac{-b}{a}$. The same proof can be given for $pq=\frac{c}{a}$.

Note: If you do not understand why we must divide by $a$, try rewriting the original equation as $ax^2+bx+c=(x-p)(x-q)$

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