“Vieta was the first algebraist after Ferrari to make any noteworthy advance in the solution of the biquadratic. He began with the type x^4 + 2gx^2 + bx = c, wrote it as x^4 + 2gx^2 = c - bx, added gx^2 + \frac{1}{4}y^2 + yx^2 + gy to both sides, and then made the right side a square after the manner of Ferrari. This method… requires the solution of a cubic resolvent.
Descartes (1637) next took up the question and succeeded in effecting a simple solution… a method considerably improved (1649) by his commentator Van Schooten. The method was brought to its final form by Simpson”

— David Eugene Smith

1745
Source: History of Mathematics (1925) Vol.2, p.469

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David Eugene Smith 33

American mathematician 1860–1944

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“[Zuanne de Tonini] da Coi… impuned Tartaglia to publish his method, but the latter declined to do so. In 1539 Cardan wrote to Tartaglia, and a meeting was arranged at which, Tartaglia says, having pledged Cardan to secrecy, he revealed the method in cryptic verse and later with a full explanation. Cardan admits that he received the solution from Tartaglia, but… without any explanation. At any rate, the two cubics x^3 + ax^2 = c and x^3 + bx = c could now be solved. The reduction of the general cubic x^3 + ax^2 + bx = c to the second of these forms does not seem to have been considered by Tartaglia at the time of the controversy. When Cardan published his Ars Magna however, he transformed the types x^3 = ax^2 + c and x^3 + ax^2 = c by substituting x = y + \frac{1}{3}a and x = y - \frac{1}{3}a respectively, and transformed the type x^3 + c = ax^2 by the substitution x = \sqrt[3]{c^2/y}, thus freeing the equations of the term x^2. This completed the general solution, and he applied the method to the complete cubic in his later problems.”

David Eugene Smith (1860–1944) American mathematician

Source: History of Mathematics (1925) Vol.2, p.461

“In the work of Vieta the analytic methods replaced the geometric, and his solutions of the quadratic equation were therefore a distinct advance upon those of his predecessors. For example, to solve the equation x^2 + ax + b = 0 he placed u + z for x. He then hadu^2 + (2z + a)u +(z^2 + az + b) = 0.He now let 2z + a = 0, whence z = -\frac{1}{2}a,and this gaveu^2 - \frac{1}{4}(a^2 - 4b) = 0.
u = \pm \frac{1}{2} \sqrt{a^2 - 4b}.andx = u + z = -\frac{1}{2}a \pm \sqrt{a^2 - 4b}.</center”

David Eugene Smith (1860–1944) American mathematician

Source: History of Mathematics (1925) Vol.2, p.449

“The discovery of Hippocrates amounted to the discovery of the fact that from the relation
(1)\frac{a}{x} = \frac{x}{y} = \frac{y}{b}it follows that(\frac{a}{x})^3 = [\frac{a}{x} \cdot \frac{x}{y} \cdot \frac{y}{b} =] \frac{a}{b}and if a = 2b, [then (\frac{a}{x})^3 = 2, and]a^3 = 2x^3.The equations (1) are equivalent [by reducing to common denominators or cross multiplication] to the three equations
(2)x^2 = ay, y^2 = bx, xy = ab[or equivalently…y = \frac{x^2}{a}, x = \frac{y^2}{b}, y = \frac{ab}{x} ]Doubling the Cube
the 2 solutions of Menaechmusand the solutions of Menaechmus described by Eutocius amount to the determination of a point as the intersection of the curves represented in a rectangular system of Cartesian coordinates by any two of the equations (2).
Let AO, BO be straight lines placed so as to form a right angle at O, and of length a, b respectively. Produce BO to x and AO to y.
The first solution now consists in drawing a parabola, with vertex O and axis Ox, such that its parameter is equal to BO or b, and a hyperbola with Ox, Oy as asymptotes such that the rectangle under the distances of any point on the curve from Ox, Oy respectively is equal to the rectangle under AO, BO i. e. to ab. If P be the point of intersection of the parabola and hyperbola, and PN, PM be drawn perpendicular to Ox, Oy, i. e. if PN, PM be denoted by y, x, the coordinates of the point P, we shall have

\begin{cases}y^2 = b. ON = b. PM = bx\\ and\\ xy = PM. PN = ab\end{cases}whence\frac{a}{x} = \frac{x}{y} = \frac{y}{b}.
In the second solution of Menaechmus we are to draw the parabola described in the first solution and also the parabola whose vertex is O, axis Oy and parameter equal to a.”

Thomas Little Heath (1861–1940) British civil servant and academic

The point P where the two parabolas intersect is given by<center><math>\begin{cases}y^2 = bx\\x^2 = ay\end{cases}</math></center>whence, as before,<center><math>\frac{a}{x} = \frac{x}{y} = \frac{y}{b}.</math></center>
Apollonius of Perga (1896)

“Although Cardan reduced his particular equations to forms lacking a term in x^2, it was Vieta who began with the general formx^3 + px^2 + qx + r = 0and made the substitution x = y -\frac{1}{3}p, thus reducing the equation to the formy^3 + 3by = 2c.He then made the substitutionz^3 + yz = b, or y = \frac{b - z^2}{z},which led to the formz^6 + 2cz^2 = b^2,a sextic which he solved as a quadratic.”

David Eugene Smith (1860–1944) American mathematician

Source: History of Mathematics (1925) Vol.2, p.465

“The Rule of Three, or Golden Rule of Arithmeticall whole Numbers. Be the three termes given 2 3 4. …To finde their fourth proporcionall Terme: that is to say, in such Reason to the third terme 4, as the second terme 3, is to the first terme 2 [Modern notation: \frac{x}{4} = \frac{3}{2}]. …Multiply the second terme 3, by the third terme 4, & giveth the product 12: which dividing by the first terme 2, giveth the Quotient 6: I say that 6 is the fourth proportional terme required.”

Simon Stevin (1548–1620) Flemish scientist, mathematician and military engineer

Disme: the Art of Tenths, Or, Decimall Arithmetike (1608)

“All the light which is radiated… will, after it has traveled a distance r, lie on the surface of a sphere whose area S is given by the first of the formulae (3). And since the practical procedure… in determining d is equivalent to assuming that all this light lies on the surface of a Euclidean sphere of radius d, it follows…4 \pi d^2 = S = 4 \pi r^2 (1 - \frac{K r^2}{3} + …);whence, to our approximation 4)d = r (1- \frac{K r^2}{6} + …), or
r = d (1 + \frac{K d^2}{6} + …).</center”

Howard P. Robertson (1903–1961) American mathematician and physicist

Geometry as a Branch of Physics (1949)

“The solution of (1), which represents a homogeneous manifold, may be written in the form:ds^2 = \frac{d\rho^2}{1 - \kappa^2\rho^2} - \rho^2 (d\theta^2 + sin^2 \theta \; d\phi^2) + (1 - \kappa^2 \rho^2)\; c^2 d\tau^2, \qquad (2)where \kappa = \sqrt \frac{\lambda}{3}. If we consider \rho as determining distance from the origin… and \tau as measuring the proper-time of a clock at the origin, we are led to the de Sitter spherical world…”

Howard P. Robertson (1903–1961) American mathematician and physicist

"On Relativistic Cosmology" (1928)

“It is picked out from numbers progressing in continuous proportion. Of continuous progressions, an arithmetical is one which proceeds by equal intervals; a geometrical one which advances by unequal and proportionally increasing or decreasing intervals. Arithmetical progressions: 1, 2, 3, 4, 5, 6, 7, &c.; or 2, 4, 6, 8, 10, 12, 14, 16, &c, Geometrical progressions: 1, 2, 4, 8, 16, 32, 64, &c.; or 243, 81, 27, 9, 3, 1.”

John Napier (1550–1617) Scottish mathematician

The Construction of the Wonderful Canon of Logarithms (1889)

“Suppose we a certain Number of things exposed, different each from other, as a, b, c, d, e, &c.; The question is, how many ways the order of these may be varied? as, for instance, how many changes may be Rung upon a certain Number of Bells; or, how many ways (by way of Anagram) a certain Number of (different) Letters may be differently ordered?
Alt.1,21) If the thing exposed be but One, as a, it is certain, that the order can be but one. That is 1.
2) If Two be exposed, as a, b, it is also manifest, that they may be taken in a double order, as ab, ba, and no more. That is 1 x 2 = 2. Alt.3
3) If Three be exposed; as a, b, c: Then, beginning with a, the other two b, c, may (by art. 2,) be disposed according to Two different orders, as bc, cb; whence arise Two Changes (or varieties of order) beginning with a as abc, acb: And, in like manner it may be shewed, that there be as many beginning with b; because the other two, a, c, may be so varied, as bac, bca. And again as many beginning with c as cab, cba. And therefore, in all, Three times Two. That is 1 x 2, x 3 = 6.
Alt.34) If Four be exposed as a, b, c, d; Then, beginning with a, the other Three may (by art. preceeding) be disposed six several ways. And (by the same reason) as many beginning with b, and as many beginning with c, and as many beginning with d. And therefore, in all, Four times six, or 24. That is, the Number answering to the case next foregoing, so many times taken as is the Number of things here exposed. That is 1 x 2 x 3, x 4 = 6 x 4 = 24.
5) And in like manner it may be shewed, that this Number 24 Multiplied by 5, that is 120 = 24 x 5 = 1 x 2 x 3 x 4 x 5, is the number of alternations (or changes of order) of Five things exposed. (Or, the Number of Changes on Five Bells.) For each of these five being put in the first place, the other four will (by art. preceeding) admit of 24 varieties, that is, in all, five times 24. And in like manner, this Number 120 Multiplied by 6, shews the Number of Alternations of 6 things exposed; and so onward, by continual Multiplication by the conse quent Numbers 7, 8, 9, &c.;
6) That is, how many so ever of Numbers, in their natural Consecution, beginning from 1, being continually Multiplied, give us the Number of Alternations (or Change of order) of which so many things are capable as is the last of the Numbers so Multiplied. As for instance, the Number of Changes in Ringing Five Bells, is 1 x 2 x 3 x 4 x 5 = 120. In Six Bells, 1 x 2 x 3 x 4 x 5 x 6 = 120 x 6 = 720. In Seven Bells, 720 x 7 = 5040. In Eight Bells, 5040 x 8 = 40320, And so onward, as far as we please.”

John Wallis (1616–1703) English mathematician

Source: A Discourse of Combinations, Alterations, and Aliquot Parts (1685), Ch.II Of Alternations, or the different Change of Order, in any Number of Things proposed.

“The number 2 thought of by one man cannot be added to the number 2 thought of by another man so as to make up the number 4.”

Simone Weil (1909–1943) French philosopher, Christian mystic, and social activist

Oppression and Liberty (1958), p. 82

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David Eugene Smith 33

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