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  2. Vieta's formulas - Wikipedia

    en.wikipedia.org/wiki/Vieta's_formulas

    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]

  3. Vieta jumping - Wikipedia

    en.wikipedia.org/wiki/Vieta_jumping

    Using Vieta's formulas, show that this implies the existence of a smaller solution, hence a contradiction. Example. Problem #6 at IMO 1988: Let a and b be positive integers such that ab + 1 divides a 2 + b 2. Prove that ⁠ a 2 + b 2 / ab + 1 ⁠ is a perfect square. [8] [9] Fix some value k that is a non-square positive integer.

  4. Viète's formula - Wikipedia

    en.wikipedia.org/wiki/Viète's_formula

    The formula can be derived as a telescoping product of either the areas or perimeters of nested polygons converging to a circle. Alternatively, repeated use of the half-angle formula from trigonometry leads to a generalized formula, discovered by Leonhard Euler, that has Viète's formula as a special case. Many similar formulas involving nested ...

  5. Elementary symmetric polynomial - Wikipedia

    en.wikipedia.org/wiki/Elementary_symmetric...

    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 ...

  6. Cubic equation - Wikipedia

    en.wikipedia.org/wiki/Cubic_equation

    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. [35]

  7. Quartic function - Wikipedia

    en.wikipedia.org/wiki/Quartic_function

    If this number is −q, then the choice of the square roots was a good one (again, by Vieta's formulas); otherwise, the roots of the polynomial will be −r 1, −r 2, −r 3, and −r 4, which are the numbers obtained if one of the square roots is replaced by the symmetric one (or, what amounts to the same thing, if each of the three square ...

  8. François Viète - Wikipedia

    en.wikipedia.org/wiki/François_Viète

    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.

  9. Template:Did you know nominations/Viète's formula - Wikipedia

    en.wikipedia.org/wiki/Template:Did_you_know...

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