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A method similar to Vieta's formula can be found in the work of the 12th century Arabic mathematician Sharaf al-Din al-Tusi. It is plausible that the algebraic advancements made by Arabic mathematicians such as al-Khayyam, al-Tusi, and al-Kashi influenced 16th-century algebraists, with Vieta being the most prominent among them. [2] [3]
Moreover from Vieta's formulas, yy ′ = x 2 - q, and y ′ = x 2 - q / y . Combining this equation with x < y , one can show that y ′ < x . The new constructed point Q = ( y ′ , x ) is then in the first quadrant, on the higher branch of H , and with smaller x , y -coordinates than the point P we started with.
Viète's formula may be rewritten and understood as a limit expression [3] = =, where = = +. For each choice of n {\displaystyle n} , the expression in the limit is a finite product, and as n {\displaystyle n} gets arbitrarily large, these finite products have values that approach the value of Viète's formula arbitrarily closely.
The characteristic polynomial of a square matrix is an example of application of Vieta's formulas. The roots of this polynomial are the eigenvalues of the matrix . When we substitute these eigenvalues into the elementary symmetric polynomials, we obtain – up to their sign – the coefficients of the characteristic polynomial, which are ...
Vieta's substitution is a method introduced by François Viète (Vieta is his Latin name) in a text published posthumously in 1615, which provides directly the second formula of § Cardano's method, and avoids the problem of computing two different cube roots.
Since α, β, and γ are the roots of , it is a consequence of Vieta's formulas that their product is equal to q 2 and therefore that √ α √ β √ γ = ±q. But a straightforward computation shows that √ α √ β √ γ = r 1 r 2 r 3 + r 1 r 2 r 4 + r 1 r 3 r 4 + r 2 r 3 r 4.
I think the last sentence summarizes it well. In English at least, the tradition has been to use Vieta, and Viete is an overcorrection (outside historical or biographic contexts), like saying "Hero's formula" with Greek pronunciation, instead than Heron's formula. 73.89.25.252 04:52, 13 June 2020 (UTC)
Since , the factors of 5 are addressed by noticing that since the residues of modulo 5 follow the cycle ,,, and those of follow the cycle ,,,, the residues of modulo 5 cycle through the sequence ,,,. Thus, 5 ∣ 149 n − 2 n {\displaystyle 5\mid 149^{n}-2^{n}} iff n = 4 k {\displaystyle n=4k} for some positive integer k {\displaystyle k} .