Difference between revisions of "Vieta's Formulas"

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=== Background ===
+
== Background ==
 +
 
 
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  
+
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>.
<center><math>P(x)=a_n(x-r_1)(x-r_2)\cdots(x-r_n)</math>,</center>
 
where <math>{r}_i</math> are the roots of <math>P(x)</math>.
 
  
Let <math>{\sigma}_k</math> be the <math>{}{k}</math>th [[symmetric sum]].
+
Also, let <math>{\sigma}_k</math> be the <math>{}{k}</math>th [[symmetric sum]].
  
=== Statement ===
+
== Statement ==
  
 
<math>\sigma_k = (-1)^k\cdot \frac{a_{n-k}}{a_n{}}</math>, for <math>{}1\le k\le {n}</math>.
 
<math>\sigma_k = (-1)^k\cdot \frac{a_{n-k}}{a_n{}}</math>, for <math>{}1\le k\le {n}</math>.
  
=== Proof ===
+
== Proof ==
  
 
[needs to be added]
 
[needs to be added]

Revision as of 15:50, 18 June 2006

Background

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

Also, let ${\sigma}_k$ be the ${}{k}$th symmetric sum.

Statement

$\sigma_k = (-1)^k\cdot \frac{a_{n-k}}{a_n{}}$, for ${}1\le k\le {n}$.

Proof

[needs to be added]