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Difference between revisions of "Vieta's Formulas"
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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>,
 
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>,
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where the coefficient of <math>\displaystyle x^{i}</math> is <math>\displaystyle {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>\displaystyle {r}_i</math> are the roots of <math>\displaystyle P(x)</math>.  We thus have that
+
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>
 
<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>
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<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>
 
<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>
  
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The coefficient of <math>\displaystyle  x^k </math> in this expression will be the <math> \displaystyle k </math>th [[symmetric sum]] of the <math>r_i</math>.   
+
The coefficient of <math> x^k </math> in this expression will be the <math>k </math>th [[symmetric sum]] of the <math>r_i</math>.   
  
−
We now have two different expressions for <math>\displaystyle P(x)</math>.  These must be equal.  However, the only way for two polynomials to be equal for all values of <math>\displaystyle x</math> is for each of their corresponding coefficients to be equal.  So, starting with the coefficient of <math>\displaystyle  x^n </math>, we see that
+
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
  
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<center><math>\displaystyle a_n = a_n</math></center>
+
<center><math>a_n = a_n</math></center>
 
<center><math> a_{n-1} = -a_n(r_1+r_2+\cdots+r_n)</math></center>
 
<center><math> a_{n-1} = -a_n(r_1+r_2+\cdots+r_n)</math></center>
 
<center><math> a_{n-2} = a_n(r_1r_2+r_1r_3+\cdots+r_{n-1}r_n)</math></center>
 
<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>a_0 = (-1)^n a_n r_1r_2\cdots r_n</math></center>
 
<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>\displaystyle a_i</math> on the other (this can be arrived at by dividing both sides of all the equations by <math>\displaystyle a_n</math>).
+
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>\displaystyle \sigma_k</math> as the <math>\displaystyle k</math>th symmetric sum, then we can write those formulas more compactly as <math>\displaystyle \sigma_k = (-1)^k\cdot \frac{a_{n-k}}{a_n{}}</math>, for <math>\displaystyle 1\le k\le {n}</math>.
+
If we denote <math>\sigma_k</math> as the <math>k</math>th 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>.
  
 
== See also ==
 
== See also ==
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* [[Newton sums]]
 
* [[Newton sums]]
  
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== Related Links ==
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== External Links ==
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[http://mathworld.wolfram.com/VietasFormulas.html Mathworld's Article]
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*[http://mathworld.wolfram.com/VietasFormulas.html Mathworld's Article]
 +
 
 +
[[Category:Elementary algebra]]

Revision as of 20:57, 10 December 2007

Vieta's formulas are a set of equations relating the roots and the coefficients of polynomials.

Introduction

Let $P(x)$ be a polynomial of degree $n$, so $P(x)={a_n}x^n+{a_{n-1}}x^{n-1}+\cdots+{a_1}x+a_0$, where the coefficient of $x^{i}$ is ${a}_i$ and $a_n \neq 0$. As a consequence of the Fundamental Theorem of Algebra, we can also write $P(x)=a_n(x-r_1)(x-r_2)\cdots(x-r_n)$, where ${r}_i$ are the roots of $P(x)$. We thus have that

$a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 = a_n(x-r_1)(x-r_2)\cdots(x-r_n).$

Expanding out the right hand side gives us

$a_nx^n - a_n(r_1+r_2+\!\cdots\!+r_n)x^{n-1} + a_n(r_1r_2 + r_1r_3 +\! \cdots\! + r_{n-1}r_n)x^{n-2} +\! \cdots\! + (-1)^na_n r_1r_2\cdots r_n.$

The coefficient of $x^k$ in this expression will be the $k$th symmetric sum of the $r_i$.

We now have two different expressions for $P(x)$. These must be equal. However, the only way for two polynomials to be equal for all values of $x$ is for each of their corresponding coefficients to be equal. So, starting with the coefficient of $x^n$, we see that

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

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 symmetric sum, then we can write those formulas more compactly as $\sigma_k = (-1)^k\cdot \frac{a_{n-k}}{a_n{}}$, for $1\le k\le {n}$.

See also

External Links

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