Difference between revisions of "Vieta's Formulas"

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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>,
 
Let <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>. 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
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where the coefficient of <math>\displaystyle x^{i}</math> is <math>\displaystyle {a}_i</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
  
<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-1}x^{n-1}+\cdots+a_1x+a_0 = a_n(x-r_1)(x-r_2)\cdots(x-r_n).</math>
  
 
Expanding out the right hand side gives us
 
Expanding out the right hand side gives us
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<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>
 
<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>
  
We can see that the coefficient of <math> x^k </math> will be the <math> k </math>th [[symmetric sum]].  The <math>k</math>th symmetric sum is just the sum of the roots taken <math>k</math> at a time.  For example, the 4th symmetric sum is <math>\displaystyle r_1r_2r_3r_4 + r_1r_2r_3r_5+\cdots+r_{n-3}r_{n-2}r_{n-1}r_n.</math>  Notice that every possible [[combination]] of four roots shows up in this sum.
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We can see that the coefficient of <math>\displaystyle  x^k </math> will be the <math> \displaystyle k </math>th [[symmetric sum]].  The <math>k</math>th symmetric sum is just the sum of the roots taken <math>k</math> at a time.  For example, the 4th symmetric sum is <math>\displaystyle r_1r_2r_3r_4 + r_1r_2r_3r_5+\cdots+r_{n-3}r_{n-2}r_{n-1}r_n.</math>  Notice that every possible [[combination]] of four roots shows up in this sum.
  
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
+
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
  
 
<center><math>\displaystyle a_n = a_n</math></center>
 
<center><math>\displaystyle a_n = a_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>
  
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>).
+
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>).
  
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>.
+
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>.
  
 
== See also ==
 
== See also ==

Revision as of 13:56, 24 July 2006

Introduction

Let $P(x)={a_n}x^n+{a_{n-1}}x^{n-1}+\cdots+{a_1}x+a_0$, where the coefficient of $\displaystyle x^{i}$ is $\displaystyle {a}_i$. 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 $\displaystyle {r}_i$ are the roots of $\displaystyle P(x)$. 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

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

We can see that the coefficient of $\displaystyle  x^k$ will be the $\displaystyle k$th symmetric sum. The $k$th symmetric sum is just the sum of the roots taken $k$ at a time. For example, the 4th symmetric sum is $\displaystyle r_1r_2r_3r_4 + r_1r_2r_3r_5+\cdots+r_{n-3}r_{n-2}r_{n-1}r_n.$ Notice that every possible combination of four roots shows up in this sum.

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

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

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

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

See also

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