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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]
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 ...
In Viète's formula, the numbers of terms and digits are proportional to each other: the product of the first n terms in the limit gives an expression for π that is accurate to approximately 0.6n digits. [4] [15] This convergence rate compares very favorably with the Wallis product, a later infinite product formula for π.
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.
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.
The total time is 1.1191 + 0.8672 = 1.9863 The conclusion, based on this particular model, is that equation 6 is slightly faster than equation 5, regardless of the fact that equation 6 has more terms. This result is typical of the general trend. The dominant factor is the ratio between and . In order to achieve a high ratio, it is necessary to ...
Gauss's circle problem asks how many points there are inside this circle of the form (,) where and are both integers. Since the equation of this circle is given in Cartesian coordinates by x 2 + y 2 = r 2 {\displaystyle x^{2}+y^{2}=r^{2}} , the question is equivalently asking how many pairs of integers m and n there are such that
But the terms of P which contain only the variables X 1, ..., X n − 1 are precisely the terms that survive the operation of setting X n to 0, so their sum equals P(X 1, ..., X n − 1, 0), which is a symmetric polynomial in the variables X 1, ..., X n − 1 that we shall denote by P̃(X 1, ..., X n − 1). By the inductive hypothesis, this ...