# Talk:Viète's Formulas

- Such stuff needs to be explained. Not every student has studied mathematics by the learn-by-rote-and-formula technique.

- In case someone else posts a gnomic proof using that technique, it may be worth while adding a page defining "Foil". It will need to be pondered. --prime mover (talk) 18:44, 11 November 2019 (EST)

## 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:

\(\displaystyle \paren {-1}^k \dfrac{ a_{n - k} } { a_n }\) | \(=\) | \(\displaystyle \sum_{1 \mathop \le i_1 \mathop < \dotsb \mathop < i_k \mathop \le n} z_{i_1} \dotsm z_{i_k}\) | for $k = 1, 2, \ldots, n$ in commutative ring with unity $R$. | ||||||||||

\(\displaystyle \paren {-1}^k \frac { a_{n - k} } { a_n }\) | \(=\) | \(\displaystyle e_{n-k} \paren { \set {z_1,\ldots,z_n} }\) | Elementary symmetric function for the ring $R$ of complex numbers. |

Listed explicitly:

\(\displaystyle \paren {-1} \dfrac { a_{n-1} } { a_n }\) | \(=\) | \(\displaystyle z_1 + z_2 + \cdots + z_n\) | |||||||||||

\(\displaystyle \paren {+1} \dfrac { a_{n - 2} } { a_n }\) | \(=\) | \(\displaystyle \paren {z_1 z_2 + \cdots + z_1 z_n} + \paren {z_2 z_3 + \cdots + z_2 z_n} + \cdots + \paren {z_{n - 1} z_n}\) | |||||||||||

\(\displaystyle \) | \(\vdots\) | \(\displaystyle \) | |||||||||||

\(\displaystyle \paren {-1}^n \dfrac { a_0 } { a_n }\) | \(=\) | \(\displaystyle z_1 z_2 \cdots z_n\) |

- 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)

## 2, 11 Nov 2019

Old statement and proof replaced. Problems inserting references. I tried to add text like "Theorem for fields" after the citation and CAPTCHA failed on a complaint of "new external links." I gave it up, no solution. --Gbgustafson (talk) 13:20, 2 November 2019 (EDT)

- Try using
`{{SpringerOnline}}`

-- info on Help:Editing/House Style/Sources. --prime mover (talk) 18:04, 2 November 2019 (EDT)

- That worked, eventually. No idea why appending some text caused a CAPTCHA failure.--Gbgustafson (talk) 12:00, 11 November 2019 (EST)

**11 Nov 2019**

Vieta's Formula for Pi was introduced (by Prime.mover) as a 'Source' but **it is not a source**. I removed the unrelated source.

- No I did not. I put it in the "Also see" section. --prime mover (talk) 12:32, 11 November 2019 (EST)

**Viete's Formula** previously appeared on the page because of name conflict: only the final letter **s** distinguishes **Viete's Formula** from **Viete's Formulas**. Editing has removed all mention of the historical name conflict.--Gbgustafson (talk) 12:00, 11 November 2019 (EST)