Talk:Viète's Formulas

19 October 2019

 * The 'Proof' is an outline of a proof. If foils is changed to expands then it is fixed enough for an outline of a proof. Does anyone care that there is no proof? --Gbgustafson

29 Oct, 1 Nov 2019
Viète Theorem is stated for commutative rings with unity. The proof uses linear independence of the powers $x^k$. Currently, I found on proofWiki no ring theory support for such independence arguments.
 * Suggestion: replace commutative ring with unit by the set $R$ of complex numbers. ProofWiki section Corollary efficiently handles ring theory extensions. --Gbgustafson (talk) 06:06, 29 October 2019 (EDT)

History. Available is a translation of Viète's work The Analytic Art (1983 translation republished by Dover 2006), but the history of Viète's Theorem remains a mystery after scanning the book. The most reliable history is that Viète introduced algebraic notation with symbols and that Girard stated the theorem for at least the field of reals; Viète may have considered only positive real roots (Encyc. of Math.). The spot in the Dover publication was never located. I lack further sources. --Gbgustafson (talk) 06:06, 1 November 2019 (EDT)

Below is the state of the original page after edits for correct definitions and references. The proof of Viète's Formulas is broken except for complex numbers: --Gbgustafson (talk) 06:06, 29 October 2019 (EDT)

Viète Theorem
Let
 * $\map P x = a_n x^n + a_{n - 1} x^{n - 1} + \dotsb + a_1 x + a_0$

be a polynomial of degree $n$ over a commutative ring with unity $R$.

Suppose $a_n$ is invertible in $R$ and:
 * $\displaystyle \map P x = a_n \prod_{k \mathop = 1}^n \paren {x - z_k}$

where $z_1, \ldots, z_k \in R$ are roots of $P$, not assumed unique.

Then:

Listed explicitly:


 * Feel free to post this up and replace the letter salad that's in there at the moment.


 * Also, feel free to take it back to real polynomials -- that's how Viete had it. --prime mover (talk) 15:32, 29 October 2019 (EDT)