“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

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

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David Eugene Smith 33

American mathematician 1860–1944

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“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 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)

“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)

“[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

“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)

“Consider an event, for example the outburst if a nova… Suppose this event is observed from two stars in line with the nova, and suppose further that the two stars are moving uniformly with respect to each other in this line. Let the epoch at which these stars passed by each other be taken as the zero of time measurement, and let an observer A on one of the stars estimate the distance and epoch of the nova outburst to be x units of length and t units of time, respectively. Suppose the other star is moving toward the nova with velocity v relative to A. Let an observer B on the star estimate the distance and epoch of the nova outburst to be x' units of length and t' units of time, respectively. Then the Lorentz formulae, relating x' to t', arex' = \frac {x-vt}{\sqrt{1-\frac{v^2}{c^2}}}; \qquad t' = \frac {t-\frac{vx}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}}
These formulae are… quite general, applying to any event in line with two uniformly moving observers. If we let c become infinite then the ratio of v to c tends to zero and the formulae becomex' = x - vt; \qquad t' = t.”

Gerald James Whitrow (1912–2000) British mathematician

p, 125
The Structure of the Universe: An Introduction to Cosmology (1949)

“Measurements which may be made on the surface of the earth… is an example of a 2-dimensional congruence space of positive curvature K = \frac{1}{R^2}… [C]onsider… a "small circle" of radius r (measured on the surface!)… its perimeter L and area A… are clearly less than the corresponding measures 2\pi r and \pi r^2… in the Euclidean plane. …for sufficiently small r (i. e., small compared with R) these quantities on the sphere are given by 1):L = 2 \pi r (1 - \frac{Kr^2}{6} + …),
A = \pi r^2”

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

1 - \frac{Kr^2}{12} + …
Geometry as a Branch of Physics (1949)

“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 (1860–1944) American mathematician

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

“E = mc2 really applies only to isolated bodies at rest. In general, when you have moving bodies, or interacting bodies, energy and mass aren't proportional. E = mc2 simply doesn't apply. …For moving bodies, the correct mass-energy equation is
E=\frac {mc^2} {\sqrt{1-\frac{v^2} {c^2}}}
where v is the velocity. For a body at rest (v=0), this becomes E = mc2. …we must consider the special case of particles with zero mass… examples include photons, color gluons, and gravitons. If we attempt to put m = 0 and v = c in our general mass-energy equation, both the numerator and denominator on the right-hand-side vanish, and we get the nonsensical relation E = 0/0. The correct result is that the energy of a photon can take any value. …The energy E of a photon is proportional to the frequency f of the light it represents. …they are related by the Planck-Einstein-Schrödinger equation E = hf, where h is Plank's constant.”

Frank Wilczek (1951) physicist

when the velocity <math>v</math> approaches the speed of light c, the denominator approaches 0 thus E approaches infinity, unless m = 0.
Source: The Lightness of Being – Mass, Ether and the Unification of Forces (2008), Ch. 3, p. 19 & Appendix A

“Fermat died with the belief that he had found a long-sought-for law of prime numbers in the formula 2^{2^n} + 1 = a prime, but he admitted that he was unable to prove it rigorously. The law is not true, as was pointed out by Euler in the example 2^{2^5} + 1 = 4,294,967,297 = 6,700,417 times 641. The American lightning calculator Zerah Colburn, when a boy, readily found the factors but was unable to explain the method by which he made his marvellous mental computation.”

Florian Cajori book A History of Mathematics

Source: A History of Mathematics (1893), p. 180; also cited in Moritz, Memorabilia Mathematica; Or, The Philomath's Quotation-book (1914) pp. 156-157. https://books.google.com/books?id=G0wtAAAAYAAJ&pg=PA156

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