“Of the contemporaries of Newton one of the most prominent was John Wallis. …Wallis was a voluminous writer, and not only are his writings erudite, but they show a genius in mathematics… He was one of the first to recognize the significance of the generalization of exponents to include negative and fractional as well as positive and integral numbers. He recognized also the importance of Cavalieri's method of indivisibles, and employed it in the quadrature of such curves as y=xn, y=x1/n, and y=x0 + x1 + x2 +… He failed in his attempts at the approximate quadrature of the circle by means of series because he was not in possession of the general form of the binomial theorem. He reached the result, however, by another method. He also obtained the equivalent of ds = \! dx \sqrt{1+(\frac{dy}{dx})^2} for the length of an element of a curve, thus connecting the problem of rectification with that of quadrature.”

History of Mathematics (1923) Vol.1

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American mathematician 1860–1944

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“Mice: What is the best early training for a writer?

Y. C.: An unhappy childhood.”

Ernest Hemingway (1899–1961) American author and journalist

Source: Ernest Hemingway on Writing

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“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)

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“This method of mine takes its beginnings where Cavalieri ends his Method of indivisibles.”

John Wallis (1616–1703) English mathematician

...for as his was the Geometry of indivisibles, so I have chosen to call my method the Arithmetic of infinitesimals.
Arithmetica Infinitorum (1656)

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