Lesson 00

How to use this course

John Wallis came to mathematics late, at about thirty, and became the most important English mathematician before Isaac Newton. His Arithmetica infinitorum of 1656 replaced the geometric quadratures of the Italians with arithmetic patterns, extended exponents to fractions and negatives, and found the first infinite product in mathematics. Newton read it as a student and discovered the binomial series by pushing Wallis’s interpolation one step further. Wallis was also a cryptographer for Parliament and the Crown, a founder of the Royal Society, a theologian, a grammarian and a teacher of the deaf.

Prerequisites for Part II

First-year calculus: integrals of powers, integration by parts.
Conic sections and their equations.
Elementary number theory for Lesson 06 (continued fractions are developed there).
Momentum and kinetic energy for Lesson 09; Euclid’s axioms for Lesson 10.

Dates follow the Julian (Old Style) calendar used in England in Wallis’s lifetime, with the year beginning on 1 January; Gregorian equivalents are given for birth and death. Each technical lesson opens with a paragraph for the general reader, then gives precise statements in modern notation, derivations and exercises with folded solutions. Where Wallis argued by analogy or incomplete induction, the lessons supply the modern proof.

Reference

Timeline

1616	Born 23 November at Ashford, Kent, third of five children of the Revd John Wallis, minister of Ashford, and Joanna Chapman.
1622	His father dies.
1625	Sent to James Movat’s school at Tenterden after plague reaches Ashford.
1631–32	Martin Holbeach’s school at Felsted, Essex. Learns arithmetic from his brother at Christmas 1631.
1632	Enters Emmanuel College, Cambridge.
1637	BA, Cambridge.
1640	MA; ordained. Chaplain in Yorkshire, Essex and London.
1642	Deciphers an intercepted Royalist letter within hours; begins work as Parliament’s cryptographer.
1643	Minister of St Gabriel Fenchurch, London. Scribe to the Westminster Assembly of Divines (to 1649).
1644	Fellow of Queens’ College, Cambridge.
1645	Marries Susanna Glyde on 14 March, vacating his fellowship. Joins the London meetings that later become the Royal Society.
1647	Reads William Oughtred’s Clavis mathematicae.
1649	Savilian Professor of Geometry at Oxford, a chair he holds until his death.
1653	Grammatica linguae Anglicanae.
1655	De sectionibus conicis, with the symbol \(\infty\). Elenchus geometriae Hobbianae opens the quarrel with Thomas Hobbes.
1656	Arithmetica infinitorum: quadrature by induction and interpolation; the Wallis product.
1657	Keeper of the University Archives. Fermat’s challenge on \(x^2-Ny^2=1\) reaches England.
1658	Commercium epistolicum, the Fermat correspondence.
1659	Rectification of Neile’s semicubical parabola; treatise on Pascal’s cycloid problems.
1660	At the Restoration, confirmed in his chair and made a royal chaplain.
1663	Fellow of the Royal Society. Oxford lecture on Euclid’s parallel postulate.
1668	Paper to the Royal Society on the laws of collision.
1669–71	Mechanica, sive de motu.
1685	A Treatise of Algebra, both Historical and Practical.
1689	Decipherer to William III after the Glorious Revolution.
1693–99	Opera mathematica, three volumes.
1703	Dies at Oxford on 28 October, aged 86.

Part I · Formation

Lesson 01

Ashford to Cambridge, 1616–1640

John Wallis was born at Ashford in Kent on 23 November 1616, the third of five children of the Revd John Wallis, minister of the town since 1602, and Joanna Chapman. His father died when the boy was about six. When plague came to Ashford in 1625 he was sent to a grammar school at Tenterden, and in 1631 to Martin Holbeach’s school at Felsted in Essex, where he learned Latin, Greek and Hebrew and the logic of the schools.

Mathematics was not taught at either school. At Christmas 1631 his brother showed him the rules of arithmetic, which he took up eagerly. It was almost the only formal mathematics he received. At the time mathematics in England was regarded as a practical skill for merchants, surveyors and navigators rather than an academic subject.

Emmanuel College

Around Christmas 1632 he entered Emmanuel College, Cambridge, a Puritan foundation. He read divinity and also anatomy and physic, and in a public disputation under Francis Glisson he defended the circulation of the blood, recently announced by William Harvey. He took the BA in 1637 and the MA in 1640. Cambridge offered him no mathematics; he read what little he could on his own. In 1640 he was ordained and left the university to become chaplain to Sir Richard Darley at Butterworth in Yorkshire, then at Hedingham in Essex and in London.

Lesson 02

Civil war and ciphers, 1640–1649

The English Civil War broke out in 1642. That year Wallis, a young chaplain of Parliamentary sympathies, was shown an intercepted Royalist letter in cipher concerning the fall of Chichester. He took it as a puzzle and solved it within two hours. It was the start of a career as Parliament’s decipherer and, eventually, the leading cryptanalyst in England. He guarded his methods closely.

His Puritan and Presbyterian sympathies brought further posts. In 1643 he became minister of St Gabriel Fenchurch in London, and from 1643 to 1649 he served as a scribe to the Westminster Assembly of Divines, which drafted the Westminster Confession and catechisms for a reformed Church of England. In 1644 he was elected a fellow of Queens’ College, Cambridge, but his marriage to Susanna Glyde on 14 March 1645 obliged him to resign. Their children included John, Anne and Elizabeth. His mother’s death in 1643 left him an estate in Kent and a modest independence.

Gresham and the London meetings

In 1645 Wallis joined a group that met weekly in London, often at Gresham College, to discuss the new experimental philosophy: anatomy, astronomy, magnetism, statics, and the work of Galileo and Francis Bacon. Its members included John Wilkins and Jonathan Goddard. After 1649 several of them moved to Oxford and continued meeting there. These meetings are one of the origins of the Royal Society.

Like many Presbyterians, Wallis opposed the trial of Charles I, and he reportedly signed a 1648 remonstrance against the king’s execution. This would help him at the Restoration.

Lesson 03

The Savilian chair, 1647–1656

In 1647 Wallis read William Oughtred’s Clavis mathematicae, a compact textbook of algebra in symbolic notation. He mastered it in a few weeks and began solving problems of his own, including the solution of equations of the fourth degree. His algebraic habits of mind come from Oughtred and from the Géométrie of René Descartes.

In 1648 the Parliamentary visitors of Oxford removed the Royalist Savilian professors. In 1649 Wallis was appointed in his place. The choice was political as much as mathematical, since his published mathematics at that date amounted to nothing. He proved worthy of it. He set himself to read the classical geometers and the moderns, among them François Viète, Johannes Kepler, Bonaventura Cavalieri and Evangelista Torricelli. In 1654 he took the degree of Doctor of Divinity.

Method

Wallis’s style was arithmetical. Where the Greeks and Cavalieri compared geometric figures, he compared sums of numbers, guessed general laws from a few cases, and accepted them by what he called induction. Fermat and Huygens thought this careless. Within seven years of his appointment, he produced two works that changed English mathematics: De sectionibus conicis (1655) and Arithmetica infinitorum (1656). Part II examines them.

Part II · The mathematics

Lesson 04

The arithmetic of infinites

For the general reader. To find the area under the curve \(y=x^k\), Cavalieri compared the figure to a stack of infinitely many thin lines. Wallis turned the stack into a sum of numbers, computed the sum for small cases, spotted the pattern, and extended it boldly to fractional and negative powers. The result was the rule \(\int_0^1 x^k\,dx=\frac1{k+1}\) for all rational \(k>-1\), a decade before calculus.

Goals

Follow Wallis’s ratio method for the quadrature of \(x^k\).
See how he extended exponents to fractions and negatives.
Understand what his induction proved and what it did not.

The ratio of sums

Divide the base of the region under \(y=x^k\), \(0\le x\le n\), into unit steps. Wallis compared the sum of the ordinates with the sum of as many ordinates equal to the greatest one:

\[ R_k(n)=\frac{0^k+1^k+2^k+\cdots+n^k}{n^k+n^k+\cdots+n^k}=\frac{\sum_{j=0}^{n}j^k}{(n+1)\,n^k}. \]

For \(k=1\) the ratio is exactly \(\tfrac12\). For \(k=2\) he computed \(\tfrac13+\tfrac16,\ \tfrac13+\tfrac1{12},\ \tfrac13+\tfrac1{18},\dots\), saw that the excess is \(\tfrac1{6n}\), and concluded that the ratio tends to \(\tfrac13\) as \(n\) becomes infinite. In the same way \(k=3\) gives \(\tfrac14\).

Theorem (Wallis, 1656; modern form)

For every real \(k\ge0\), \(\displaystyle\lim_{n\to\infty}R_k(n)=\frac1{k+1}\). Equivalently, \(\displaystyle\int_0^1x^k\,dx=\frac1{k+1}\). The same formula holds for \(-1<k<0\) as an improper integral.

Modern proof. \(R_k(n)=\frac{n}{n+1}\cdot\frac1n\sum_{j=0}^n(j/n)^k\), and the second factor is a Riemann sum for \(\int_0^1x^k\,dx\). Wallis verified finitely many integer cases and asserted the pattern. Fermat objected that this was not a proof; Wallis replied that the pattern was evident.

Fractional exponents

The curve \(y=x^{1/2}\) is the parabola \(x=y^2\) seen sideways. The region under it on \([0,1]\) is the complement in the unit square of the region under \(y=x^2\), with area \(1-\tfrac13=\tfrac23=\frac{1}{1/2+1}\). Wallis noticed that the rule \(\frac1{k+1}\) still works, and that the same complement argument gives \(\frac{q}{p+q}\) for \(y=x^{p/q}\) when \(p/q\) or \(q/p\) is an integer. He then extended it to all positive rational exponents by interpolation, and introduced the notation of fractional and negative indices, writing \(\frac1{x}\) as \(x^{-1}\) and \(\sqrt x\) as \(x^{1/2}\).

Exercise 4.1Prove \(R_2(n)=\frac13+\frac1{6n}\) exactly.

Show solution

\(\sum_{j=0}^nj^2=\frac{n(n+1)(2n+1)}{6}\), so \(R_2(n)=\frac{n(n+1)(2n+1)}{6(n+1)n^2}=\frac{2n+1}{6n}=\frac13+\frac1{6n}\).

Exercise 4.2Wallis computed \(\int_0^1(x-x^2)^k\,dx\) for \(k=0,1,2\) as \(1,\tfrac16,\tfrac1{30}\). Find the general formula and the value for \(k=3\).

Show solution

\(\int_0^1x^k(1-x)^k\,dx=B(k+1,k+1)=\frac{(k!)^2}{(2k+1)!}\). For \(k=3\): \(\frac{36}{5040}=\frac1{140}\).

Exercise 4.3Wallis applied \(\frac1{k+1}\) to \(k=-1\) and \(k=-2\) and concluded that the area under \(1/x^2\) on \((0,1]\) is “more than infinite”, being \(\frac1{-1}\). What is actually true about these integrals, and where does the formula legitimately apply for negative \(k\)?

Show solution

\(\int_0^1x^k\,dx\) diverges for every \(k\le-1\). The formula \(\frac1{k+1}\) is valid on \((0,1]\) only for \(k>-1\). For \(k<-1\) the correct statement concerns the tail: \(\int_1^\infty x^k\,dx=-\frac1{k+1}\), finite and positive. Wallis’s negative ratio mixes up these two regions.

Lesson 05

Interpolation and the product for π

For the general reader. Wallis wanted the area of a quarter circle, \(\int_0^1\sqrt{1-x^2}\,dx\), by his arithmetic method. The square root defeated direct computation. So he built a table of related areas he could compute, saw that the circle sat in a gap between the entries, and filled the gap by insisting that the table vary smoothly. The result, which he called interpolation, was an infinite product for \(\pi\).

Goals

Build Wallis’s table and locate the circle in it.
Derive the Wallis product by his squeeze and by Wallis integrals.
See how Newton’s binomial series grew from this table.

The table

For \(p,q\ge0\) set

\[ f(p,q)=\left(\int_0^1\big(1-x^{1/p}\big)^q\,dx\right)^{-1}=\frac{\Gamma(p+q+1)}{\Gamma(p+1)\,\Gamma(q+1)}. \]

For integers this is the binomial coefficient \(\binom{p+q}{p}\), and Wallis tabulated these as figurate numbers. The quarter circle corresponds to \(p=q=\tfrac12\), since \(x^{1/p}=x^2\). Its entry, which Wallis denoted by a small square \(\square\), is \(f(\tfrac12,\tfrac12)=4/\pi\).

Along the row \(p=\tfrac12\) the recurrence \(f(\tfrac12,q)=\frac{q+\frac12}{q}\,f(\tfrac12,q-1)\) holds. From \(f(\tfrac12,0)=1\) it gives \(1,\ \tfrac32,\ \tfrac{15}8,\ \tfrac{35}{16},\dots\) at integers, and \(\square,\ \tfrac43\square,\ \tfrac{8}{5}\square,\dots\) at half-integers. Wallis assumed the row increases throughout. That forces

\[ f(\tfrac12,n)<f(\tfrac12,n+\tfrac12)<f(\tfrac12,n+1), \]

bounds on \(\square\) which close in as \(n\) grows.

Theorem (Wallis product, 1656) \[ \frac{\pi}{2}=\prod_{n=1}^{\infty}\frac{2n}{2n-1}\cdot\frac{2n}{2n+1}=\frac21\cdot\frac23\cdot\frac43\cdot\frac45\cdot\frac65\cdot\frac67\cdots \]

Wallis wrote it as \(\square=\dfrac{3\times3\times5\times5\times7\times7\times\cdots}{2\times4\times4\times6\times6\times8\times\cdots}\).

Modern proof

Let \(I_m=\int_0^{\pi/2}\sin^m t\,dt\). Integration by parts gives \(I_m=\frac{m-1}{m}I_{m-2}\), with \(I_0=\frac\pi2\), \(I_1=1\). Hence

\[ I_{2n}=\frac\pi2\prod_{k=1}^n\frac{2k-1}{2k},\qquad I_{2n+1}=\prod_{k=1}^n\frac{2k}{2k+1}. \]

Since \(0\le\sin t\le1\), \(I_{2n+1}\le I_{2n}\le I_{2n-1}\), so \(1\le I_{2n}/I_{2n+1}\le I_{2n-1}/I_{2n+1}=\frac{2n+1}{2n}\to1\). Therefore \(\frac{\pi}{2}\prod_{k=1}^{n}\frac{(2k-1)(2k+1)}{(2k)^2}\to1\), which is the product. The proof is the rigorous form of Wallis’s monotonicity assumption: \(\log I_m\) is convex in \(m\).

Newton’s step

In 1664–65 Newton, reading the Arithmetica, interpolated instead the area functions \(\int_0^x(1-t^2)^n\,dt\) as polynomials in \(x\). The coefficients for \(n=\tfrac12\) followed the pattern of binomial coefficients with a fractional top entry, and so he found the general binomial series \((1+x)^\alpha=\sum_k\binom{\alpha}{k}x^k\).

Exercise 5.1Using only that the row \(f(\tfrac12,q)\) is increasing, show \(1<\square<\tfrac32\) and \(\tfrac98<\square<\tfrac{45}{32}\). What bounds on \(\pi\) follow?

Show solution

\(f(\tfrac12,0)<f(\tfrac12,\tfrac12)<f(\tfrac12,1)\) gives \(1<\square<\tfrac32\). Next, \(\tfrac32<\tfrac43\square<\tfrac{15}8\) gives \(\tfrac98<\square<\tfrac{45}{32}\). With \(\pi=4/\square\): \(2.84<\pi<3.56\). The bounds tighten slowly, like \(1/n\).

Exercise 5.2Show that the partial product \(P_n=\prod_{k=1}^n\frac{4k^2}{4k^2-1}\) satisfies \(P_n<\frac\pi2<P_n\cdot\frac{2n+1}{2n}\).

Show solution

From the proof, \(\frac{I_{2n}}{I_{2n+1}}=\frac{\pi}{2P_n}\). The inequalities \(1\le\frac{I_{2n}}{I_{2n+1}}\le\frac{2n+1}{2n}\) are strict for \(n\ge1\) because \(\sin^{2n+1}t<\sin^{2n}t\) on a set of positive measure. So \(P_n<\frac\pi2<\frac{2n+1}{2n}P_n\).

Exercise 5.3Verify \(f(\tfrac12,\tfrac12)=4/\pi\) from the Gamma-function formula, and from the integral directly.

Show solution

\(\frac{\Gamma(2)}{\Gamma(3/2)^2}=\frac{1}{(\sqrt\pi/2)^2}=\frac4\pi\). Directly, \(x^{1/p}=x^2\), so the integral is \(\int_0^1\sqrt{1-x^2}\,dx=\frac\pi4\), the quarter disc.

Lesson 06

Continued fractions and Fermat’s challenge

For the general reader. Wallis’s friend William Brouncker turned the product for \(\pi\) into a fraction whose denominators go on for ever. A year later, Brouncker and Wallis used similar endless fractions to answer a challenge from Pierre de Fermat: find whole numbers \(x,y\) with \(x^2-Ny^2=1\). Wallis gave the name continued fraction to these objects.

Goals

State Brouncker’s continued fraction for \(4/\pi\).
Solve \(x^2-Ny^2=1\) with the continued fraction of \(\sqrt N\).
Know the history of the 1657 challenge.

Brouncker’s formula (published by Wallis, 1656) \[ \frac4\pi=1+\cfrac{1^2}{2+\cfrac{3^2}{2+\cfrac{5^2}{2+\cfrac{7^2}{2+\cdots}}}} \]

Brouncker never published a derivation, and Wallis’s account in the Arithmetica reconstructs it from the interpolation of Lesson 05. The convergents, \(4/\pi\approx1,\ 1.5,\ 1.1538,\dots\), converge very slowly. Euler later showed how such formulas follow from series: the continued fraction is equivalent to Leibniz’s series \(\frac\pi4=1-\frac13+\frac15-\cdots\).

The challenge of 1657

In February 1657 Fermat challenged mathematicians, the English in particular, to show that for every non-square positive integer \(N\) the equation \(x^2-Ny^2=1\) has infinitely many integer solutions, and to find the smallest. His correspondent Bernard Frénicle de Bessy then challenged Wallis with \(N=151\) and \(N=313\). Brouncker found a general method, and Wallis wrote it up, with proofs, in the Commercium epistolicum of 1658. Leonhard Euler later attributed the method to John Pell by mistake, and the equation has carried Pell’s name ever since.

Theorem (Brouncker–Wallis method; proved in full by Lagrange, 1768)

Let \(N>0\) be a non-square integer. The continued fraction of \(\sqrt N\) is periodic, \(\sqrt N=[a_0;\overline{a_1,\dots,a_\ell}]\). If \(p/q\) is the convergent \([a_0;a_1,\dots,a_{\ell-1}]\), then \(p^2-Nq^2=(-1)^\ell\). The smallest solution of \(x^2-Ny^2=1\) is \((p,q)\) if \(\ell\) is even, and \((p^2+Nq^2,\,2pq)\) if \(\ell\) is odd. All positive solutions are given by \(x+y\sqrt N=(x_1+y_1\sqrt N)^k\).

For \(N=313\) the smallest solution is \(x=32\,188\,120\,829\,134\,849\), \(y=1\,819\,380\,158\,564\,160\).

Exercise 6.1Compute the continued fraction of \(\sqrt7\) and find the smallest solution of \(x^2-7y^2=1\).

Show solution

\(\sqrt7=[2;\overline{1,1,1,4}]\), period \(\ell=4\). Convergents: \(2,\ 3,\ \tfrac52,\ \tfrac83\). Then \(8^2-7\cdot3^2=64-63=1\). So \((x,y)=(8,3)\).

Exercise 6.2For \(N=13\), \(\sqrt{13}=[3;\overline{1,1,1,1,6}]\) has odd period. Find \(p,q\) with \(p^2-13q^2=-1\) and hence the smallest solution of \(x^2-13y^2=1\).

Show solution

Convergents: \(3,\ 4,\ \tfrac72,\ \tfrac{11}3,\ \tfrac{18}5\). \(18^2-13\cdot5^2=324-325=-1\). Squaring \(18+5\sqrt{13}\): \(x=18^2+13\cdot25=649\), \(y=2\cdot18\cdot5=180\). Check: \(649^2-13\cdot180^2=421201-421200=1\).

Exercise 6.3Show that if \((x_1,y_1)\) solves \(x^2-Ny^2=1\), so does \((x_1^2+Ny_1^2,\,2x_1y_1)\). Why does this give infinitely many solutions?

Show solution

The norm \(x^2-Ny^2=(x+y\sqrt N)(x-y\sqrt N)\) is multiplicative. Squaring \(x_1+y_1\sqrt N\) gives \((x_1^2+Ny_1^2)+2x_1y_1\sqrt N\), of norm \(1^2=1\). Since \(x_1+y_1\sqrt N>1\), its powers are distinct, giving infinitely many solutions.

Lesson 07

Conics as equations; ∞; arc length

For the general reader. For two thousand years conics had been studied as slices of a cone. In 1655 Wallis defined them instead as plane curves given by equations of the second degree, and derived their properties by algebra. In the same book he introduced the symbol \(\infty\). Four years later he published the first exact length of an algebraic curve.

Goals

Derive the plane equations of the conics from the cone.
See how Wallis used \(\infty\) and \(\frac1\infty\).
Rectify the semicubical parabola.

De sectionibus conicis (1655)

Wallis began from the cone only long enough to obtain, for each section, a relation between the abscissa \(x\) along the axis from the vertex and the ordinate \(y\). Apollonius had expressed these as relations between areas. Wallis wrote them as equations in the style of Descartes:

\[ \text{parabola: } y^2=px,\qquad \text{ellipse: } y^2=px-\frac{p}{d}x^2,\qquad \text{hyperbola: } y^2=px+\frac{p}{d}x^2, \]

with \(p\) the latus rectum and \(d\) the transverse diameter. From then on he worked only in the plane. The treatise did much to establish that a conic is simply a curve of the second degree.

The symbol

In the same treatise Wallis wrote \(\infty\) for an infinite number and treated a plane figure as composed of \(\infty\) parallelograms, each of height \(\frac1\infty\) of the whole. He also used this in the Arithmetica. The origin of the shape is uncertain; one suggestion is a variant of the Roman numeral for 1000.

Rectification

Descartes had written in the Géométrie that the ratio between straight and curved lines could not be known exactly. In 1657 William Neile, a young pupil of Wallis, found the length of an arc of the semicubical parabola \(y^2=x^3\). Wallis published the result in 1659; Hendrik van Heuraet found it independently in the Netherlands the same year.

Theorem (Neile, 1657)

The arc of \(y=x^{3/2}\) from \(x=0\) to \(x=a\) has length \(\displaystyle s(a)=\frac{8}{27}\Big[\big(1+\tfrac94a\big)^{3/2}-1\Big]\).

Exercise 7.1Prove Neile’s theorem.

Show solution

\(y'=\tfrac32x^{1/2}\), so \(s=\int_0^a\sqrt{1+\tfrac94x}\,dx=\tfrac49\cdot\tfrac23\big[(1+\tfrac94x)^{3/2}\big]_0^a=\tfrac8{27}\big[(1+\tfrac94a)^{3/2}-1\big]\). The integrand is a power of a linear function, which is why this curve, unlike the ordinary parabola, has an algebraic arc length.

Exercise 7.2Show that the ellipse equation \(y^2=px-\frac pdx^2\) describes an ellipse with semi-axes \(a=\frac d2\) and \(b\) satisfying \(b^2=\frac{pd}4\). Interpret \(p\) as the latus rectum.

Show solution

Complete the square: \(y^2=-\frac pd\big(x-\frac d2\big)^2+\frac{pd}4\), i.e. \(\frac{(x-d/2)^2}{(d/2)^2}+\frac{y^2}{pd/4}=1\). So \(a=d/2\), \(b^2=pd/4\), and \(p=\frac{4b^2}{d}=\frac{2b^2}{a}\), the length of the chord through a focus perpendicular to the major axis.

Lesson 08

Negative and imaginary quantities

For the general reader. In the seventeenth century many mathematicians still distrusted negative numbers and rejected square roots of negatives as impossible. Wallis argued in his Treatise of Algebra of 1685 that both have meaning. A negative is a step backwards along a line, and the square root of a negative can be pictured as a step off the line. These are early versions of the number line and of the complex plane.

Goals

See Wallis’s line model of negative numbers.
Work through his example of a cubic with three real roots.
Understand why imaginary numbers were unavoidable for cubics.

The Treatise of Algebra (1685)

The Treatise was both a history of algebra and a textbook. Its history is partisan: Wallis claimed that Descartes’s algebra derived from Thomas Harriot, a claim historians do not accept as shown. Its mathematics is lasting. Wallis explained negatives by motion along a line: a man who walks 5 yards forward and 8 back has advanced \(-3\) yards. He also wrote physical laws as formulas in letters, such as distance equals speed times time.

For the square root of a negative he proposed geometric constructions. Just as \(\sqrt{bc}\) is the mean proportional between segments \(b\) and \(c\) on the same side of a point, \(\sqrt{-bc}\) is the mean proportional when they lie on opposite sides, and he represented it by a segment perpendicular to the line. His constructions were not consistent, and the complex plane was set out properly only by Caspar Wessel (1799) and Jean-Robert Argand (1806).

Casus irreducibilis

If a real cubic \(x^3+px+q=0\) has three distinct real roots, that is \(\frac{q^2}4+\frac{p^3}{27}<0\), then Cardano’s formula expresses each root through cube roots of non-real complex numbers, and no formula using only real radicals exists.

Wallis’s example was \(a^3-7a=6\), whose roots are \(3,-1,-2\), while Cardano’s formula gives \(a=\sqrt[3]{3+\sqrt{-\frac{100}{27}}}+\sqrt[3]{3-\sqrt{-\frac{100}{27}}}\). Since the answer is real, the imaginary quantities must be meaningful intermediate steps.

Exercise 8.1Verify that \(3+\frac{10}{3\sqrt3}i\) has a cube root of the form \(\frac32+\frac{\beta}{2\sqrt3}i\) for a suitable real \(\beta\), and recover the root \(a=3\).

Show solution

Try \(u=\tfrac32+ti\). Then \(u^3=\tfrac{27}8-\tfrac92t^2+\big(\tfrac{27}4t-t^3\big)i\). Real part \(=3\) requires \(t^2=\tfrac1{12}\), \(t=\tfrac1{2\sqrt3}\). Imaginary part: \(\tfrac{27}{4}\cdot\tfrac1{2\sqrt3}-\tfrac1{24\sqrt3}=\tfrac{81-1}{24\sqrt3}=\tfrac{10}{3\sqrt3}\). So \(\beta=1\). Adding \(u\) and its conjugate gives \(a=3\).

Exercise 8.2Wallis’s geometric \(\sqrt{-bc}\). Place \(B\) at \(-b\) and \(C\) at \(c\) on the real axis with \(b,c>0\). Show that the point \(P\) on the imaginary axis with \(PB\perp PC\) is at height \(\sqrt{bc}\), so that \(P\) corresponds to \(i\sqrt{bc}=\sqrt{-bc}\).

Show solution

Let \(P=(0,h)\). \(\overrightarrow{PB}\cdot\overrightarrow{PC}=(-b)(c)+h^2=0\), so \(h=\sqrt{bc}\). This is Thales’ construction of the mean proportional; Wallis read the vertical segment as the square root of the negative product \((-b)c\).

Lesson 09

Collisions and the laws of motion

For the general reader. In 1668 the Royal Society asked for the true laws of impact, which Descartes had stated wrongly. Three answers arrived within weeks: Christopher Wren and Christiaan Huygens treated perfectly elastic bodies, and Wallis treated bodies that stick together. Wallis based his on what is now called conservation of momentum.

Goals

Derive Wallis’s rule for perfectly inelastic collisions.
Compare with the elastic rules of Wren and Huygens.
Compute energy loss.

Wallis sent his paper to the Royal Society late in 1668; Wren’s was read on 17 December 1668 and printed in the Philosophical Transactions in January 1669, alongside Wallis’s. Wallis expanded the theory in Mechanica, sive de motu (1669–71), a systematic treatise on statics and dynamics that treated motion, force and impetus quantitatively. He defined the momentum of a body as weight times speed, with direction taken into account by sign.

Perfectly inelastic collision (Wallis, 1668)

If bodies of masses \(m_1,m_2\) moving on a line with velocities \(u_1,u_2\) collide and move on together, their common velocity is

\[ v=\frac{m_1u_1+m_2u_2}{m_1+m_2}. \]

The kinetic energy lost is \(\displaystyle\Delta K=\frac12\,\frac{m_1m_2}{m_1+m_2}\,(u_1-u_2)^2\).

Wallis’s rule needs only conservation of momentum. Elastic rebound needs a second principle; Huygens used the relativity of motion and the conservation of \(\sum mv^2\). Together the three papers settled the laws of impact and fed directly into Newton’s third law.

Exercise 9.1Derive the energy-loss formula.

Show solution

\(K_{\text{before}}-K_{\text{after}}=\tfrac12m_1u_1^2+\tfrac12m_2u_2^2-\tfrac12\frac{(m_1u_1+m_2u_2)^2}{m_1+m_2}\). Over the common denominator, the numerator is \(m_1m_2(u_1^2+u_2^2-2u_1u_2)=m_1m_2(u_1-u_2)^2\). So \(\Delta K=\frac12\frac{m_1m_2}{m_1+m_2}(u_1-u_2)^2\), which depends only on the relative velocity and is the same in every inertial frame.

Exercise 9.2A ballistic pendulum: a bullet of mass \(m\) at speed \(u\) lodges in a block of mass \(M\) hanging at rest, which then rises a height \(h\). Find \(u\) in terms of \(h\).

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Wallis’s rule: \(v=\frac{mu}{m+M}\). Afterwards energy is conserved: \(\tfrac12v^2=gh\), so \(v=\sqrt{2gh}\) and \(u=\frac{m+M}{m}\sqrt{2gh}\). Using energy conservation through the collision itself would be wrong, since most of the bullet’s energy is lost there.

Lesson 10

The parallel postulate

For the general reader. Euclid’s fifth postulate, about parallel lines, had seemed less obvious than the others for two thousand years, and many tried to prove it from them. In an Oxford lecture of 1663 Wallis tried, by assuming instead that every figure has similar copies of every size. His proof was correct given that assumption. The assumption turns out to be equivalent to the postulate itself.

Goals

State Wallis’s postulate.
Prove the parallel postulate from it.
See why it fails in hyperbolic geometry.

Wallis’s postulate (1663)

Given any triangle and any segment, there is a triangle similar to the given one with a side equal to that segment.

Theorem

In neutral geometry (Euclid’s axioms without the fifth postulate), Wallis’s postulate is equivalent to the parallel postulate.

Sketch. Let a transversal \(t\) cross lines \(\ell\) and \(m\) at \(A\) and \(B\), with interior angles \(\alpha\) at \(A\) and \(\beta\) at \(B\) on one side and \(\alpha+\beta<\pi\). Slide \(m\) along \(t\) toward \(A\), keeping its angle \(\beta\) with \(t\). When its foot \(B'\) is close enough to \(A\), it meets \(\ell\) at a point \(C'\); Wallis argued this by continuity, and it can be proved in neutral geometry. Wallis’s postulate now gives a triangle \(ABC\) on the base \(AB\) similar to \(AB'C'\). Its angles at \(A\) and \(B\) are \(\alpha\) and \(\beta\), so its sides from \(A\) and \(B\) lie along \(\ell\) and \(m\), and these lines meet at \(C\). This is Euclid’s fifth postulate. The converse is the usual Euclidean construction of similar triangles with parallels. Later writers, including Girolamo Saccheri (1733), Johann Heinrich Lambert and Adrien-Marie Legendre, explored other equivalents. Gauss, János Bolyai and Nikolai Lobachevsky showed that the alternative geometry is consistent.

Exercise 10.1In the hyperbolic plane of curvature \(-1\), a triangle with angles \(\alpha,\beta,\gamma\) has area \(\pi-\alpha-\beta-\gamma\). Use this to show that Wallis’s postulate fails there.

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Suppose \(T'\) is similar to \(T\) with every side larger. Since the angles agree, place \(T'\) with one vertex on the corresponding vertex \(A\) of \(T\) and its two sides along the same rays. Then \(T\) lies inside \(T'\) and the difference has positive area. But the area formula gives both triangles the same area \(\pi-\alpha-\beta-\gamma\). This contradiction shows that similar hyperbolic triangles are congruent, so no larger similar copy exists.

Exercise 10.2Show that Wallis’s postulate implies that the angle sum of a triangle is two right angles. (Assume the parallel postulate has been derived as above.)

Show solution

With the parallel postulate, draw through one vertex the unique parallel to the opposite side. Alternate angles are equal, and the three angles at that vertex form a straight angle, so \(\alpha+\beta+\gamma=\pi\).

Part III · Life and legacy

Lesson 11

The Royal Society and the war with Hobbes

In Oxford in the 1650s Wallis met regularly with Wilkins, Seth Ward, Robert Boyle, Christopher Wren and others in the experimental club that, merged with the London group, became the Royal Society after 1660. He was elected a Fellow in 1663 and contributed for forty years to the Philosophical Transactions, on mathematics, the tides, music and language. In 1657 he was elected Keeper of the University Archives after a contested election.

Hobbes

In 1655 Thomas Hobbes, already famous for Leviathan, published De corpore, which included a claimed squaring of the circle. Wallis answered the same year with Elenchus geometriae Hobbianae, demolishing the geometry. Hobbes replied with Six Lessons to the Professors of the Mathematics (1656), and the exchange ran, through Hobbes’s attacks on the symbolic algebra he called a scab of symbols and Wallis’s replies, until 1678. Behind the mathematics lay deeper disputes: over religion and the universities, over whether algebra was real mathematics or merely shorthand, and, through Hobbes’s attack on Boyle’s air pump, over experimental science. Wallis was right on every geometric point, though often uncharitable.

Restoration

When Charles II returned in 1660, Wallis’s opposition to the regicide counted in his favour. He was confirmed in his chair, made a royal chaplain, and in 1661 appointed to the commission that revised the Book of Common Prayer. He conformed to the restored Church of England. His cipher work for the Commonwealth ended at the Restoration.

DiscussionHobbes held that geometry must rest on constructions and that algebraic symbols added nothing. Wallis held that symbols carry reasoning the eye cannot. Using Lessons 04 and 05, argue for one side.

Lesson 12

Late years and legacy, 1685–1703

Wallis’s later work extended in many directions. His Grammatica linguae Anglicanae (1653) included a careful account of how English sounds are produced, and from this phonetics he developed a method for teaching deaf people to speak, which he used at Littlecote House with a deaf boy; William Holder accused him of taking credit for Holder’s earlier work with another pupil. His Institutio logicae (1687) was a successful textbook. He edited Greek mathematical texts, including Ptolemy’s Harmonics, Aristarchus on the sizes and distances of the Sun and Moon, and Archimedes’ Sand-Reckoner. His mental arithmetic was famous: he reportedly extracted the square root of a 53-digit number in his head and dictated 27 digits of it the next morning.

After the Glorious Revolution of 1688 he again served as decipherer, now to William III. In 1697 he declined a request from Gottfried Wilhelm Leibniz to teach his methods to students from Hanover, holding that they belonged to his country. His grandson William Blencowe later became decipherer to Queen Anne. His Opera mathematica appeared in three volumes from 1693 to 1699. He was a vigorous defender of English priority, and gave Newton’s work on series a prominent place in his Algebra and Opera. He died at Oxford on 28 October 1703 (8 November by the Gregorian calendar), aged 86.

Legacy

Analysis: the arithmetical quadrature of powers, rational and negative exponents, interpolation, and the first infinite product, which led directly to Newton’s binomial series.
Notation and concepts: the symbol \(\infty\), \(\frac1\infty\) for an infinitesimal, the terms interpolation and continued fraction, and an early number line.
Number theory: with Brouncker, the continued-fraction method for \(x^2-Ny^2=1\).
Mechanics: momentum conservation in inelastic impact, and the Mechanica.
Institutions: fifty-four years in the Savilian chair, a founding role in the Royal Society, and the first systematic state cryptography in England.

Final projectChoose one result from Part II. Write a two-page account for a strong undergraduate that states it in Wallis’s form and in modern form, works one example in full, and explains exactly where his reasoning by induction or interpolation needed a later proof, and who supplied it.

Reference

Sources and further reading

Biography

J. J. O’Connor, E. F. Robertson, John Wallis, MacTutor History of Mathematics.
John Wallis, Wikipedia (with extensive references).
Hobbes–Wallis controversy, Wikipedia.
J. A. Stedall, A Discourse Concerning Algebra: English Algebra to 1685, Oxford University Press, 2002.
P. Beeley, C. J. Scriba (eds.), The Correspondence of John Wallis, Oxford University Press, 2003–.
Royal Society, Wren, Lex naturae de collisione corporum (1668), Making Science archive.

Mathematics

J. Wallis, Arithmetica infinitorum, Oxford, 1656. English translation: J. A. Stedall, The Arithmetic of Infinitesimals, Springer, 2004.
J. Wallis, De sectionibus conicis, Oxford, 1655.
J. Wallis, Commercium epistolicum, Oxford, 1658.
J. Wallis, Mechanica, sive de motu, tractatus geometricus, London, 1669–71.
J. Wallis, A Treatise of Algebra, both Historical and Practical, London, 1685.
Wallis product, Arithmetica Infinitorum, Pell’s equation (history), Wikipedia.
D. T. Whiteside, Patterns of mathematical thought in the later seventeenth century, Archive for History of Exact Sciences 1 (1961).
J. A. Stedall, Catching Proteus: the collaborations of Wallis and Brouncker, Notes and Records of the Royal Society 54 (2000).