Difference between revisions of "Vieta's Formulas"

 
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=== Background ===
 
=== Background ===
Let  
+
Let <math>P(x)={a_n}x^n+{a_{n-1}}x^{n-1}+\cdots+{a_1}x+a_0</math>,
<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>,
+
<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>.
 
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]].
+
Let <math>{\sigma}_k</math> be the <math>{}{k}</math>th [[symmetric sum]].
  
 
=== Statement ===
 
=== Statement ===
 +
Vieta's says that <math>\sigma_k = (-1)^k\cdot \frac{a_{n-k}}{a_n{}}</math>,
  
<math>\sigma_k = (-1)^k\cdot \frac{a_{n-k}}{a_{k}}</math>,
+
for <math>{}1\le k\le {n}</math>.
 
 
for <math>1\le {k}\le {n}</math>.
 
  
 
=== Proof ===
 
=== Proof ===
  
 
[needs to be added]
 
[needs to be added]

Revision as of 14:26, 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 $P(x)$.

Let ${\sigma}_k$ be the ${}{k}$th symmetric sum.

Statement

Vieta's says that $\sigma_k = (-1)^k\cdot \frac{a_{n-k}}{a_n{}}$,

for ${}1\le k\le {n}$.

Proof

[needs to be added]