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For any (a, b) satisfying the given condition, let k = a 2 + b 2 + 1 / ab and rearrange and substitute to get x 2 − (kb) x + (b 2 + 1) = 0. One root to this quadratic is a, so by Vieta's formulas the other root may be written as follows: x 2 = kb − a = b 2 + 1 / a . The first equation shows that x 2 is an integer and the ...
Viète's formula may be rewritten and understood as a limit expression [3] = =, where = = +.. For each choice of , the expression in the limit is a finite product, and as gets arbitrarily large, these finite products have values that approach the value of Viète's formula arbitrarily closely.
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]
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 ...
François Viète (French: [fʁɑ̃swa vjɛt]; 1540 – 23 February 1603), known in Latin as Franciscus Vieta, was a French mathematician whose work on new algebra was an important step towards modern algebra, due to his innovative use of letters as parameters in equations.
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)
In the case of two nested square roots, the following theorem completely solves the problem of denesting. [2]If a and c are rational numbers and c is not the square of a rational number, there are two rational numbers x and y such that + = if and only if is the square of a rational number d.
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} .