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Difference between revisions of "Vieta's formulas"

 
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#REDIRECT [[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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(WIP)
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== 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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== 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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<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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Also, <math>-b/a = p + q, c/a = p \cdot q</math>.
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Provide links to problems that use vieta formulas:
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Examples:
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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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Basic proof:
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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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[[Category:Algebra]]
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[[Category:Polynomials]]
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[[Category:Theorems]]

Revision as of 13:41, 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

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. Let $j$ be any integer such that $0<j<n$. We wish to find a process that generates every term with degree $j$. If


$a_n = a_n$
$a_{n-1} = -a_n(r_1+r_2+\cdots+r_n)$
$a_{n-2} = a_n(r_1r_2+r_1r_3+\cdots+r_{n-1}r_n)$
$\vdots$
$a_0 = (-1)^n a_n r_1r_2\cdots r_n$

More commonly, these are written with the roots on one side and the $a_i$ on the other (this can be arrived at by dividing both sides of all the equations by $a_n$).

If we denote $\sigma_k$ as the $k$-th elementary symmetric sum, then we can write those formulas more compactly as $\sigma_k = (-1)^k\cdot \frac{a_{n-k}}{a_n{}}$, for $1\le k\le {n}$. Also, $-b/a = p + q, c/a = p \cdot q$.

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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