Difference between revisions of "Vieta's Formulas"

Revision as of 16:49, 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

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

Proof

[needs to be added]

@@ Line 8: / Line 8: @@
 === Statement ===
-Vieta's says that <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 ===
 [needs to be added]

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Difference between revisions of "Vieta's Formulas"

Revision as of 16:49, 18 June 2006

Background

Statement

Proof