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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 ...
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 ...
By applying Vieta's Formulas, (x, qx − y) is a lattice point on the lower branch of H. Let y ′ = qx − y. From the equation for H, one sees that 1 + x y ′ > 0. Since x > 0, it follows that y ′ ≥ 0. Hence the point (x, y ′) is in the first quadrant. By reflection, the point (y ′, x) is also a point in the first quadrant on H.
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[6]: 202–207 If one is given a quadratic equation in the form x 2 + bx + c = 0, the sought factorization has the form (x + q)(x + s), and one has to find two numbers q and s that add up to b and whose product is c (this is sometimes called "Vieta's rule" [7] and is related to Vieta's formulas). As an example, x 2 + 5x + 6 factors as (x + 3)(x ...
By Vieta's formulas, s 0 is known to be zero in the case of a depressed cubic, and − b / a for the general cubic. So, only s 1 and s 2 need to be computed. They are not symmetric functions of the roots (exchanging x 1 and x 2 exchanges also s 1 and s 2 ), but some simple symmetric functions of s 1 and s 2 are also symmetric in the ...
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 ...