Difference between revisions of "Vieta's Formulas"

Revision as of 13:39, 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 $ax^2+bx+c=(x-p)(x-q)$

@@ Line 1: / Line 1: @@
-'''Vieta's Formulas''', otherwise called Viète's Laws, are a set of [[equation]]s relating the [[root]]s and the [[coefficient]]s of [[polynomial]]s.
+In [[algebra]], '''Vieta's formulas''' are a set of formulas that relate the coefficients of a [[polynomial]] to its roots.
-== Introduction ==
+(WIP)
-Vieta's Formulas were discovered by the French mathematician [[François Viète]].
+== Statement ==
-Vieta's Formulas can be used to relate the sum and product of the roots of a polynomial to its coefficients. The simplest application of this is with quadratics. If we have a quadratic <math>x^2+ax+b=0</math> with solutions <math>p</math> and <math>q</math>, then we know that we can factor it as:
+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>.
+== Proof ==
+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
-<center><math>ax^2+bx+c=(x-p)(x-q)</math></center>
-Using the distributive property to expand the right side we now have
-<center><math>x^2+ax+b=x^2-(p+q)x+pq</math></center>
-Vieta's Formulas are often used when finding the sum and products of the roots of a quadratic in the form <math>ax^2 + bx +c</math> with roots <math>r_1</math> and <math>r_2.</math> They state that:
-<cmath>r_1 + r_2 = -\frac{b}{a}</cmath> and <cmath>r_1 \cdot r_2 = \frac{c}{a}.</cmath>
-We know that two polynomials are equal if and only if their coefficients are equal, so <math>x^2+ax+b=x^2-(p+q)x+pq</math> means that <math>a=-(p+q)</math> and <math>b=pq</math>. In other words, the product of the roots is equal to the constant term, and the sum of the roots is the opposite of the coefficient of the <math>x</math> term.
-A similar set of relations for cubics can be found by expanding <math>x^3+ax^2+bx+c=(x-p)(x-q)(x-r)</math>.
-We can state Vieta's formulas more rigorously and generally. Let <math>P(x)</math> be a polynomial of degree <math>n</math>, so <math>P(x)={a_n}x^n+{a_{n-1}}x^{n-1}+\cdots+{a_1}x+a_0</math>,
-where the coefficient of <math>x^{i}</math> is <math>{a}_i</math> and <math>a_n \neq 0</math>. As a consequence of the [[Fundamental Theorem of Algebra]], we can also write <math>P(x)=a_n(x-r_1)(x-r_2)\cdots(x-r_n)</math>, where <math>{r}_i</math> are the roots of <math>P(x)</math>.  We thus have that
-<center><math> a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 = a_n(x-r_1)(x-r_2)\cdots(x-r_n).</math></center>
-Expanding out the right-hand side gives us
-<center><math> a_nx^n - a_n(r_1+r_2+\!\cdots\!+r_n)x^{n-1} + a_n(r_1r_2 + r_1r_3 +\! \cdots\! + r_{n-1}r_n)x^{n-2} +\! \cdots\! + (-1)^na_n r_1r_2\cdots r_n.</math></center>
-The coefficient of <math> x^k </math> in this expression will be the <math>(n-k)</math>-th [[elementary symmetric sum]] of the <math>r_i</math>.
-We now have two different expressions for <math>P(x)</math>.  These must be equal.  However, the only way for two polynomials to be equal for all values of <math>x</math> is for each of their corresponding coefficients to be equal.  So, starting with the coefficient of <math> x^n </math>, we see that
 <center><math>a_n = a_n</math></center>
@@ Line 59: / Line 35: @@
 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>
-SuperJJ
-==General Form==
-For a polynomial of the form <math>f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0</math> with roots <math>r_1,r_2,r_3,...r_n</math>, Vieta's formulas state that:
-<cmath>
-\begin{align*}
-s_1&= & r_1+r_2+r_3&+\cdots+r_n & &=-\frac{a_{n-1}}{a_n} \\
-s_2&= & r_1r_2+r_1r_3+r_1r_4&+\cdots+r_{n-2}r_{n-1} & &=\phantom{-}\frac{a_{n-2}}{a_n} \\
-s_3&= & r_1r_2r_3+r_1r_2r_4&+\cdots+r_{n-2}r_{n-1}r_n & &=-\frac{a_{n-3}}{a_n} \\
-& & &\vdots & & \\
-s_n&= & r_1r_2r_3&\cdots r_n & &=(-1)^n\frac{a_0}{a_n}.\\
-\end{align*}
-</cmath>
-These formulas are widely used in competitions, and it is best to remember that when the <math>n</math> roots are taken in groups of <math>k</math> (i.e. <math>r_1+r_2+r_3...+r_n</math> is taken in groups of 1 and <math>r_1r_2r_3...r_n</math> is taken in groups of <math>n</math>), this is equivalent to <math>(-1)^k\frac{a_{n-k}}{a_n}</math>.
 [[Category:Algebra]]
 [[Category:Polynomials]]
+[[Category:Theorems]]

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Difference between revisions of "Vieta's Formulas"

Revision as of 13:39, 5 November 2021

Statement

Proof

Proving Vieta's Formula