Lesson 00

How to use this course

Every student of calculus meets Taylor series, and few learn anything about Taylor. He was a wealthy Englishman trained in law, a fine amateur musician and painter, and for a few years one of the strongest mathematicians in Isaac Newton’s circle. Almost all his important work appeared between 1712 and 1719. It includes the first systematic calculus of finite differences, the series expansion that Joseph-Louis Lagrange later made the foundation of the calculus, the first mathematical treatment of a vibrating string, and a theory of perspective that influenced artists, geometers and surveyors.

Prerequisites for Part II

One-variable calculus: derivatives, the mean value theorem, integration by parts.
Some familiarity with series and convergence.
For Lesson 08, Newton’s second law and simple harmonic motion.
For Lesson 09, coordinates in three dimensions.

Each technical lesson opens with a paragraph for the general reader, then gives statements in modern notation, derivations and exercises with folded solutions. Taylor wrote in Newton’s notation of fluxions (dotted letters) and increments, and his style was famously compressed. The course translates freely and says where a modern statement goes beyond what he proved.

Reference

Timeline

1685	Born 18 August in Edmonton, Middlesex, to John Taylor of Bifrons, Kent, and Olivia Tempest.
1701	Enters St John’s College, Cambridge, as a fellow-commoner. Studies mathematics with John Machin and John Keill.
1708	Solves the problem of the centre of oscillation (published 1714).
1709	Bachelor of laws (LL.B.).
1712	Elected Fellow of the Royal Society. Sits on the committee on the Newton–Leibniz priority dispute. 26 July: letter to Machin stating his series.
1713	Paper on the motion of a stretched string in the Philosophical Transactions.
1714	Doctor of laws (LL.D.). January: elected Secretary of the Royal Society.
1715	Methodus incrementorum directa et inversa and Linear Perspective. Correspondence with Montmort on Malebranche.
1716	Visits Paris.
1717	Second edition of the Methodus. Paper on computing logarithms.
1718	21 October: resigns as Secretary, his health failing.
1719	New Principles of Linear Perspective. Takes the waters at Aix-la-Chapelle.
1721	Marries Miss Brydges of Wallington against his father’s wishes; estrangement.
1723	His wife dies in childbirth; the child dies too. Reconciliation with his father.
1725	Marries Sabetta Sawbridge of Olantigh, Kent.
1729	His father dies; he inherits Bifrons.
1730	Sabetta dies in childbirth; their daughter Elizabeth survives.
1731	Dies 29 December at Somerset House, London, aged 46. Buried at St Anne’s, Soho.
1772	Lagrange calls Taylor’s theorem the main foundation of the differential calculus.
1793	His grandson Sir William Young prints Contemplatio philosophica with a memoir.

Part I · Formation

Lesson 01

Edmonton and Bifrons, 1685–1701

Brook Taylor was born on 18 August 1685 at Edmonton, then a village north of London. His father, John Taylor, of Bifrons House at Patrixbourne near Canterbury, came from a prosperous family and sat in Parliament. His mother, Olivia, was a daughter of John Tempest. The family was wealthy and close to the gentry, and Brook grew up expecting an independent income rather than a profession.

His father was a stern disciplinarian with a love of painting and music, and the household received musicians and artists. Brook was taught at home by tutors in classics and mathematics, and he became an accomplished musician and a skilled draughtsman. His biographer and grandson later praised the precision of his drawings, most of which were watercolour landscapes. Both interests shaped his mathematics: his work on the vibrating string began from music, and his book on perspective began from drawing.

Lesson 02

Cambridge, Machin and Keill

Taylor entered St John’s College, Cambridge, as a fellow-commoner in 1701 (MacTutor gives his admission as 3 April 1703). He read law, graduating LL.B. in 1709 and receiving the LL.D. in 1714. His mathematics came from two of Newton’s most loyal followers. John Machin, later Gresham Professor of Astronomy, is remembered for his 1706 formula for \(\pi\). John Keill, Savilian Professor of Astronomy at Oxford from 1712, was Newton’s fiercest champion against Gottfried Wilhelm Leibniz. Taylor corresponded with both from his student years, and the letters show him working in Newton’s fluxional calculus.

By 1708 he had solved a problem in mechanics that had attracted the leading mathematicians of Europe: finding the centre of oscillation of a swinging body (Lesson 10). He did not publish the solution until 1714, and when he did, Johann Bernoulli disputed his priority. It was the first of many quarrels with Bernoulli.

Lesson 03

The Royal Society, 1712–1715

Taylor was elected a Fellow of the Royal Society on 3 April 1712, on the strength of what Machin and Keill knew of his ability; he had published almost nothing. In the same year he was named to the committee the Society set up to judge the dispute between Newton and Leibniz over the invention of the calculus. The committee reported within weeks, and its report, printed as the Commercium epistolicum, was largely shaped by Newton himself and found for him.

The year 1712 also gave Taylor his theorem. According to his own account, a remark Machin made at Child’s Coffee House in London, about using Newton’s series to solve Kepler’s problem and about Edmond Halley’s method for extracting roots of equations, set him thinking. On 26 July 1712 he wrote to Machin with a general expansion of a function in powers of the increment of its variable.

In January 1714 he was elected Secretary of the Royal Society. The next two years were the most productive of his life. Papers in the Philosophical Transactions treated the stretched string (1713), the centre of oscillation (1714), capillary rise and magnetic attraction. In 1715 he published two books: Methodus incrementorum directa et inversa (“The direct and inverse method of increments”), which contained the finite-difference calculus, his series, the string and much else; and Linear Perspective.

Part II · The mathematics

Lesson 04

The calculus of increments

For the general reader. Calculus studies how a quantity changes when its variable changes by an infinitely small amount. Taylor asked what happens when the change is finite: one step along a table, one year of interest, one row of a sum. The resulting “method of increments” is the calculus of finite differences. It has its own derivatives, its own integrals (sums), and its own integration by parts.

Goals

Define forward differences and compute difference tables.
Prove the Newton–Gregory interpolation formula.
Sum series with falling factorials and summation by parts.

Differences

For a function \(f\) and step \(h\), define \(\Delta f(x)=f(x+h)-f(x)\) and \(\Delta^{k}=\Delta(\Delta^{k-1})\). The shift \(E f(x)=f(x+h)\) satisfies \(E=1+\Delta\). The “direct method” finds differences; the “inverse method” recovers a function from its differences, which is summation. Taylor’s book is the first systematic treatment of both.

Theorem (Newton–Gregory forward-difference formula)

For every integer \(n\ge0\),

\[ f(a+nh)=\sum_{k=0}^{n}\binom{n}{k}\,\Delta^k f(a). \]

If \(f\) is a polynomial of degree \(d\), then \(\Delta^k f=0\) for \(k>d\), and the right side, read as a polynomial in \(n\), equals \(f(a+nh)\) for all real \(n\).

Proof. \(f(a+nh)=E^nf(a)=(1+\Delta)^nf(a)\), and the binomial theorem applies because \(1\) and \(\Delta\) commute. For a polynomial of degree \(d\), both sides are polynomials in \(n\) of degree at most \(d\) that agree at infinitely many integers, so they are identical. James Gregory found the formula in 1670, and Newton gave it in the Principia (Book III, Lemma 5). Taylor derived it by induction from a difference table.

Sums

With \(h=1\), the falling factorial \(n^{\underline k}=n(n-1)\cdots(n-k+1)\) satisfies \(\Delta n^{\underline k}=k\,n^{\underline{k-1}}\), the analogue of \(\frac{d}{dx}x^k=kx^{k-1}\). Hence

\[ \sum_{j=0}^{n-1}j^{\underline k}=\frac{n^{\underline{k+1}}}{k+1}. \]

Summation by parts \[ \sum_{j=0}^{n-1}u_j\,\Delta v_j=u_nv_n-u_0v_0-\sum_{j=0}^{n-1}v_{j+1}\,\Delta u_j. \]

It follows by summing the product rule \(\Delta(u_jv_j)=u_j\Delta v_j+v_{j+1}\Delta u_j\). Taylor gave both this and ordinary integration by parts in the Methodus.

Exercise 4.1Build the difference table of \(f(n)=n^3\) for \(n=0,\dots,4\) and use the forward-difference formula to write \(n^3\) in falling factorials.

Show solution

Values \(0,1,8,27,64\); first differences \(1,7,19,37\); second \(6,12,18\); third \(6,6\). So \(\Delta f(0)=1,\ \Delta^2f(0)=6,\ \Delta^3f(0)=6\), and \(n^3=n+6\binom n2+6\binom n3=n^{\underline1}+3n^{\underline2}+n^{\underline3}\).

Exercise 4.2Use Exercise 4.1 to find \(\sum_{j=0}^{n-1}j^3\).

Show solution

\(\sum j^3=\frac{n^{\underline2}}{2}+n^{\underline3}+\frac{n^{\underline4}}{4}\). Simplifying, \(\frac{n(n-1)}{4}\big[2+4(n-2)+(n-2)(n-3)\big]=\frac{n(n-1)}{4}\cdot n(n-1)=\Big(\frac{n(n-1)}{2}\Big)^2\).

Exercise 4.3Use summation by parts to evaluate \(\sum_{j=0}^{n-1}j\,2^j\).

Show solution

Take \(u_j=j\) and \(v_j=2^j\), so \(\Delta v_j=2^j\) and \(\Delta u_j=1\). Then \(\sum j2^j=n2^n-0-\sum_{j=0}^{n-1}2^{j+1}=n2^n-(2^{n+1}-2)=(n-2)2^n+2\). Check \(n=3\): \(0+2+8=10=1\cdot8+2\).

Lesson 05

Taylor’s theorem

For the general reader. If you know everything about a smooth function at one point, its value, its slope, how the slope is changing, and so on, can you predict its values nearby? Taylor’s series says yes, at least approximately, by a sum of simple powers. It turns the study of functions into the algebra of power series, and it is behind most of the approximations used in science and engineering.

Goals

Follow Taylor’s derivation from the interpolation formula.
State the theorem in modern form and compute standard expansions.
Know who had the series before Taylor and why it is still called his.

Taylor’s derivation

In the Methodus (Proposition VII, Theorem III, and its second corollary) Taylor started from the forward-difference formula of Lesson 04. Fix \(x\) and \(a\), set \(n=(x-a)/h\), and rewrite each term:

\[ \binom{n}{k}\Delta^kf(a)=\frac{(x-a)(x-a-h)\cdots(x-a-(k-1)h)}{k!}\cdot\frac{\Delta^kf(a)}{h^k}. \]

Now let the increment \(h\) vanish, keeping \(x\) fixed, so that \(n\) grows without bound. Each product tends to \((x-a)^k\), and each ratio \(\Delta^kf(a)/h^k\) tends to the \(k\)-th derivative (in Taylor’s language, the ratio of the \(k\)-th fluxions). Passing to the limit term by term gives

\[ f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+\cdots. \]

The interchange of limit and infinite sum is not justified, and Taylor said nothing about convergence. Those questions belong to Lesson 06.

Theorem (Taylor’s formula, modern form)

If \(f\) has \(n+1\) continuous derivatives on an interval containing \(a\) and \(x\), then

\[ f(x)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k+R_n(x),\qquad R_n(x)=\frac{1}{n!}\int_a^x f^{(n+1)}(t)\,(x-t)^n\,dt. \]

Proof. For \(n=0\) this is the fundamental theorem of calculus. Integrating the remainder by parts, with \(u=f^{(n+1)}(t)\) and \(dv=(x-t)^n\,dt/n!\), gives \(R_n=\frac{f^{(n+1)}(a)}{(n+1)!}(x-a)^{n+1}+R_{n+1}\), which is the induction step.

Priority

Taylor was not the first. James Gregory had expansions of this kind in a letter to John Collins of 1671. Newton had them in unpublished work. Leibniz, Johann Bernoulli (who published a related series in 1694) and Abraham de Moivre found versions independently. Taylor was the first to publish the general expansion as a theorem with a derivation and to use it systematically, for solving equations and differential equations. The special case \(a=0\) appears in the 1717 edition of the Methodus and was later popularized by Colin Maclaurin, whose name it now carries. The name “Taylor’s series” appears in print in the 1780s, in the work of Simon Lhuilier.

Exercise 5.1Derive the series for \(\ln(1+x)\) about \(0\) and use three terms to estimate \(\ln1.1\).

Show solution

\(f^{(k)}(x)=(-1)^{k-1}(k-1)!/(1+x)^k\), so \(f^{(k)}(0)/k!=(-1)^{k-1}/k\) and \(\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots\). With \(x=0.1\): \(0.1-0.005+0.000333=0.095333\). The true value is \(0.095310\).

Exercise 5.2Machin, Taylor’s teacher, computed \(\pi\) to 100 digits in 1706 using \(\frac{\pi}{4}=4\arctan\frac15-\arctan\frac1{239}\). Derive the series of \(\arctan x\) and use two terms of each to estimate \(\pi\).

Show solution

\(\arctan'x=1/(1+x^2)=1-x^2+x^4-\cdots\), so \(\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}-\cdots\). Then \(\arctan\frac15\approx0.2-0.0026667=0.1973333\) and \(\arctan\frac1{239}\approx0.0041841\). So \(\pi\approx4(4\times0.1973333-0.0041841)=3.14060\). Adding the next term \(x^5/5=0.000064\) for \(1/5\) gives \(3.14162\), against \(\pi=3.14159\).

Exercise 5.3Check Taylor’s limiting argument on \(f(x)=x^2\) with \(a=0\): write the forward-difference formula exactly, then let \(h\to0\).

Show solution

\(\Delta f(0)=h^2\) and \(\Delta^2f(0)=2h^2\). With \(n=x/h\): \(f(x)=0+n h^2+\frac{n(n-1)}{2}2h^2=xh+x(x-h)=x^2\), exactly, for every \(h\). The limit \(h\to0\) reproduces \(0+0\cdot x+\frac{2}{2}x^2\), since \(\Delta f(0)/h=h\to0=f'(0)\) and \(\Delta^2f(0)/h^2=2=f''(0)\).

Lesson 06

Remainders, convergence and Lagrange

For the general reader. Taylor’s series was a formal tool for half a century. Then Lagrange tried to rebuild all of calculus on it, defining derivatives as the coefficients of the series. The attempt failed, but in making it he discovered how to measure the error of a truncated series. Augustin-Louis Cauchy then showed that a function can have a perfectly good Taylor series that does not represent it.

Goals

State and use Lagrange’s form of the remainder.
Understand Cauchy’s counterexample.
Distinguish smooth from analytic functions.

In 1772 Lagrange called Taylor’s theorem the main foundation of the differential calculus. In his Théorie des fonctions analytiques (1797) he proposed defining \(f'(a)\) as the coefficient of \(h\) in the expansion of \(f(a+h)\), so avoiding infinitesimals.

Theorem (Lagrange remainder)

Under the hypotheses of Lesson 05, there is \(\xi\) strictly between \(a\) and \(x\) such that

\[ R_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}. \]

Proof. Suppose \(x>a\). Let \(m\) and \(M\) be the minimum and maximum of \(f^{(n+1)}\) on \([a,x]\). Since \((x-t)^n\ge0\), the integral form gives \(m\frac{(x-a)^{n+1}}{(n+1)!}\le R_n\le M\frac{(x-a)^{n+1}}{(n+1)!}\). The intermediate value theorem supplies \(\xi\).

Example (Cauchy, 1823)

Let \(f(x)=e^{-1/x^2}\) for \(x\ne0\) and \(f(0)=0\). Then \(f\) is infinitely differentiable and \(f^{(k)}(0)=0\) for every \(k\). Its Taylor series at \(0\) is identically zero and converges everywhere, yet it equals \(f(x)\) only at \(x=0\).

Functions equal to their Taylor series near every point are called analytic. Polynomials, \(e^x\), \(\sin x\) and \(\ln(1+x)\) are analytic; Cauchy’s function is smooth but not analytic. Lagrange’s programme fails because the derivative is a local property while the series encodes more. In complex analysis the two notions coincide: a complex-differentiable function is analytic, and its series converges in the largest disc avoiding singularities.

Exercise 6.1How many terms of the series for \(e^x\) about \(0\) guarantee \(e\) to within \(10^{-6}\)?

Show solution

With \(x=1\), \(|R_n|=e^\xi/(n+1)!<3/(n+1)!\). We need \((n+1)!>3\times10^6\). Since \(9!=362\,880\) and \(10!=3\,628\,800\), \(n=9\) suffices: ten terms, \(1+1+\frac12+\cdots+\frac1{9!}\).

Exercise 6.2For Cauchy’s function, show \(f'(0)=0\) from the definition.

Show solution

\(f'(0)=\lim_{h\to0}e^{-1/h^2}/h\). Put \(u=1/|h|\): \(|e^{-1/h^2}/h|=ue^{-u^2}\to0\) as \(u\to\infty\). The same argument with \(u^ke^{-u^2}\) gives all higher derivatives, since each derivative for \(x\ne0\) is a polynomial in \(1/x\) times \(e^{-1/x^2}\).

Exercise 6.3Why does the series \(1-x^2+x^4-\cdots\) for \(1/(1+x^2)\) diverge at \(x=2\), though the function is smooth on the whole real line?

Show solution

As a complex function, \(1/(1+z^2)\) has poles at \(z=\pm i\), at distance 1 from the origin. The series about 0 converges exactly in the disc \(|z|<1\). A real-variable view cannot see the obstruction.

Lesson 07

Inverse functions and singular solutions

For the general reader. The Methodus contains several results whose importance was recognized only later. Two are treated here: a rule for the derivatives of an inverse function, used for changing variables in differential equations, and the discovery that some differential equations have an extra solution that does not belong to the usual family.

Goals

Derive the first two derivatives of an inverse function.
Find general and singular solutions of first-order equations and interpret the singular one as an envelope.

Change of variables

Taylor expressed the fluxions of \(x\) with respect to \(y\) in terms of those of \(y\) with respect to \(x\), so that a differential equation could be rewritten with the roles of the variables exchanged.

Proposition (derivatives of the inverse)

If \(y=f(x)\) is twice differentiable with \(f'(x)\ne0\), and \(x=g(y)\) is the inverse, then

\[ \frac{dx}{dy}=\frac{1}{y'},\qquad \frac{d^2x}{dy^2}=-\frac{y''}{(y')^3}. \]

Differentiate \(g'(f(x))=1/f'(x)\) with respect to \(x\): \(g''(f(x))\,f'(x)=-f''(x)/f'(x)^2\).

Singular solutions

Taylor found and identified a solution of a first-order differential equation that cannot be obtained from the general solution by choosing the constant. The general theory came later, with Alexis Clairaut (1734), Leonhard Euler and Lagrange. The standard example is Clairaut’s equation.

Proposition (Clairaut’s equation)

The equation \(y=xp+\phi(p)\), \(p=y'\), with \(\phi\) twice differentiable, has the general solution \(y=cx+\phi(c)\), a family of lines. Where \(\phi''\ne0\), it also has the singular solution given parametrically by \(x=-\phi'(p),\ y=-p\,\phi'(p)+\phi(p)\), which is the envelope of the family.

Proof. Differentiating, \(p=p+xp'+\phi'(p)p'\), so \(p'\,(x+\phi'(p))=0\). Either \(p'=0\), giving the lines, or \(x=-\phi'(p)\), giving the singular curve. At each point the singular curve is tangent to one line of the family, so uniqueness of solutions fails along it.

Exercise 7.1Solve \(y=xp-p^2\). Sketch the general and singular solutions.

Show solution

General: \(y=cx-c^2\). Singular: \(x=2p\), \(y=2p^2-p^2=p^2\), so \(y=x^2/4\). Each line \(y=cx-c^2\) touches the parabola at \(x=2c\).

Exercise 7.2Show that \((y')^2=4y\) has the solutions \(y=(x-c)^2\) and \(y=0\), and explain why \(y=0\) is singular.

Show solution

For \(y=(x-c)^2\), \(y'=2(x-c)\) and \((y')^2=4(x-c)^2=4y\). The zero function also satisfies the equation. No value of \(c\) gives \(y\equiv0\), and through each point \((c,0)\) pass two solutions, \(y=0\) and \(y=(x-c)^2\), tangent there. The line \(y=0\) is the envelope of the parabolas.

Exercise 7.3Check the inverse formula for \(y=e^x\), \(x=\ln y\).

Show solution

\(y'=y''=e^x=y\). The formula gives \(d^2x/dy^2=-y/y^3=-1/y^2\), which is the second derivative of \(\ln y\).

Lesson 08

The vibrating string

For the general reader. Why does a shorter or tighter string sound higher? Marin Mersenne had found the rules by experiment in the 1630s. Taylor, a musician, derived them from Newton’s laws of motion in 1713. His analysis found the shape of the vibrating string and its pitch, and it started a line of work that led to the wave equation and to Fourier series.

Goals

Reconstruct Taylor’s argument for the fundamental mode.
Derive the frequency of a string and Mersenne’s laws.
See how d’Alembert and Daniel Bernoulli extended the problem.

Taylor’s assumptions

Taylor treated a uniform string of length \(L\), weight \(w\) and tension \(P\) (expressed as a stretching weight), fixed at both ends and vibrating with small amplitude. He assumed that all points pass through the straight position at the same moment, and that the restoring force on each small piece is proportional to the curvature of the string there. The second assumption is the physical content of the wave equation: the net vertical force on a piece of length \(dx\) is \(P\,y_{xx}\,dx\). The first restricts attention to a single mode. In his paper and in the Methodus (Propositions XXII–XXIII), he concluded that each point moves like a pendulum on a cycloid, that is, in simple harmonic motion with the same period.

Result (Taylor, 1713; modern form)

If every point of the string moves in simple harmonic motion with the same frequency and no interior point is at rest, the shape of the string is a sine arc, \(y=A\sin(\pi x/L)\), and the frequency is

\[ f=\frac{1}{2L}\sqrt{\frac{P}{\mu}}, \]

where \(\mu\) is the mass per unit length and \(P\) the tension in force units.

Derivation. Newton’s law for a piece of string gives \(\mu\,y_{tt}=P\,y_{xx}\). If \(y(x,t)=Y(x)\cos\omega t\), then \(-\mu\omega^2Y=PY''\). With \(Y(0)=Y(L)=0\) and \(Y\) positive inside, \(Y=A\sin(\pi x/L)\) and \(\omega^2=\pi^2P/(\mu L^2)\). Taylor expressed the period by comparison with a pendulum (Exercise 8.2), so the prediction could be checked against a clock and against Mersenne’s experiments.

After Taylor

In 1747 Jean le Rond d’Alembert wrote the wave equation \(y_{tt}=c^2y_{xx}\), \(c^2=P/\mu\), and solved it by \(y=F(x+ct)+G(x-ct)\). Euler argued that the initial shape could be any continuous curve. Daniel Bernoulli, starting from Taylor’s single mode, proposed that every motion is a superposition \(\sum_k A_k\sin(k\pi x/L)\cos(k\pi ct/L)\) of modes. The dispute over whether such sums can represent arbitrary functions was settled only after Joseph Fourier.

Exercise 8.1A steel string of length 0.65 m and mass per length \(\mu=7.0\times10^{-4}\) kg/m is under tension 70 N. Find its fundamental frequency. By what factor must the tension change to raise the pitch an octave?

Show solution

\(\sqrt{P/\mu}=\sqrt{10^5}\approx316.2\) m/s, so \(f=316.2/1.3\approx243\) Hz. An octave doubles \(f\), which requires four times the tension (Mersenne’s law \(f\propto\sqrt P\)).

Exercise 8.2Taylor compared the string with a pendulum. Show that the string vibrates in the same time as a simple pendulum of length \(\ell=\dfrac{wL}{\pi^2P}\), where \(w\) is the string’s weight and \(P\) the stretching weight.

Show solution

Treat \(w\) and \(P\) as forces. The mass per length is \(\mu=w/(gL)\), so \(\omega^2=\dfrac{\pi^2P}{\mu L^2}=\dfrac{\pi^2Pg}{wL}\). A pendulum of length \(\ell\) has \(\omega^2=g/\ell\). Equating, \(\ell=\dfrac{wL}{\pi^2P}\). A light string under a heavy stretching weight corresponds to a very short pendulum, hence a high pitch.

Exercise 8.3Verify that \(y=F(x+ct)+G(x-ct)\) satisfies \(y_{tt}=c^2y_{xx}\), and show that the mode \(\sin(\pi x/L)\cos(\pi ct/L)\) is of this form.

Show solution

\(y_{tt}=c^2F''+c^2G''=c^2y_{xx}\). And \(\sin\alpha\cos\beta=\tfrac12[\sin(\alpha+\beta)+\sin(\alpha-\beta)]\) gives \(\tfrac12\sin\frac{\pi(x+ct)}{L}+\tfrac12\sin\frac{\pi(x-ct)}{L}\): two travelling waves.

Lesson 09

Linear perspective

For the general reader. Painters since the Renaissance knew that parallel lines in a picture seem to meet at a point on the horizon. Taylor made perspective a branch of geometry. He defined the vanishing point of every line and the vanishing line of every plane, proved general theorems about them, and showed how to work backward from a picture to the scene it represents.

Goals

Compute the image of a point and the vanishing point of a line.
Prove that parallel lines share a vanishing point and parallel planes a vanishing line.
Use vanishing points to recover the viewing distance.

Setting

Put the eye at the origin and the picture plane at \(z=d\). A point \((X,Y,Z)\) with \(Z>0\) is seen at the picture point \((u,v)=\big(dX/Z,\ dY/Z\big)\). Taylor’s terms: the intersection of a line is where it meets the picture plane; its vanishing point is where the line through the eye parallel to it meets the picture plane.

Theorem (Taylor, 1715)

The image of a straight line not parallel to the picture is the straight line through its intersection and its vanishing point.
Parallel lines have the same vanishing point. Parallel planes have the same vanishing line, the intersection of the picture with the plane through the eye parallel to them.

Proof. A line \(\mathbf P+t\mathbf w\) with \(w_z\ne0\) has image \(\big(d(P_x+tw_x)/(P_z+tw_z),\ d(P_y+tw_y)/(P_z+tw_z)\big)\). As \(t\to\infty\) this tends to \(d(w_x/w_z,\ w_y/w_z)\), which depends only on the direction \(\mathbf w\). The image of the whole line is the intersection of the picture plane with the plane through the eye and the line, so it is straight; it contains the intersection point (\(t\) with \(P_z+tw_z=d\)) and the vanishing point. For a plane with normal \(\mathbf n\), every direction in it satisfies \(\mathbf n\cdot\mathbf w=0\), so its vanishing points fill the line \(n_xu+n_yv+n_zd=0\). For horizontal planes, \(\mathbf n=(0,1,0)\), this is the horizon \(v=0\).

Taylor also treated the inverse problem: given a correct perspective picture, recover the position of the eye and the shapes in the scene. This side of his work influenced Johann Heinrich Lambert and, much later, photogrammetry. His books were terse. Johann Bernoulli judged them abstruse to all and unintelligible to artists, and the painter Joshua Kirby wrote a popular exposition, Dr Brook Taylor’s Method of Perspective Made Easy (1754).

Exercise 9.1With \(d=1\), find the vanishing point of the lines with direction \((1,-1,2)\). Where does the line through \((0,1,1)\) in this direction cross the picture?

Show solution

Vanishing point \((1/2,-1/2)\). The line \((t,1-t,1+2t)\) meets \(z=1\) at \(t=0\), so its intersection is \((0,1)\). Its image is the segment of the line through \((0,1)\) and \((1/2,-1/2)\).

Exercise 9.2A square floor tile is seen in two-point perspective: its two horizontal edge directions \((\cos\alpha,0,\sin\alpha)\) and \((-\sin\alpha,0,\cos\alpha)\) are perpendicular. Show that their vanishing points \(u_1,u_2\) on the horizon satisfy \(u_1u_2=-d^2\). How can a viewer recover the correct viewing distance from a picture?

Show solution

\(u_1=d\cot\alpha\) and \(u_2=-d\tan\alpha\), so \(u_1u_2=-d^2\). In a picture, measure the two vanishing points of a pair of perpendicular horizontal directions from the centre of the picture; then \(d=\sqrt{-u_1u_2}\). This is a simple case of Taylor’s inverse problem.

Exercise 9.3Show that lines parallel to the picture plane have no vanishing point and that their images are parallel to them.

Show solution

If \(w_z=0\), then \(Z=P_z\) is constant along the line, and the image \(d(P_x+tw_x,\,P_y+tw_y)/P_z\) moves in direction \((w_x,w_y)\) at constant scale \(d/P_z\). Parallel lines of this kind stay parallel in the picture.

Lesson 10

Centres of oscillation and curvature

For the general reader. A swinging door or a swinging bar keeps time like a simple pendulum of some definite length. Finding that length was a classic problem, solved by Christiaan Huygens in 1673. Taylor’s first paper gave a solution by Newton’s calculus. In the Methodus he also defined the curvature of a curve as the curvature of the circle that best fits it.

Goals

Find the equivalent length of a compound pendulum.
Derive the radius of curvature as the limit of circles through three points.

Theorem (centre of oscillation)

A rigid body of mass \(m\) swinging about a horizontal axis, with moment of inertia \(I\) about the axis and centre of mass at distance \(h\) from it, oscillates for small amplitudes like a simple pendulum of length

\[ \ell=\frac{I}{mh}. \]

The point at distance \(\ell\) from the axis, on the line through the centre of mass, is the centre of oscillation.

Derivation. The torque equation is \(I\ddot\theta=-mgh\sin\theta\approx-mgh\,\theta\), so \(\omega^2=mgh/I\). A simple pendulum has \(\omega^2=g/\ell\).

Proposition (radius of curvature)

Let \(y=f(x)\) be twice differentiable with \(f''(x_0)\ne0\). The circles through three points of the curve near \(x_0\) tend, as the points merge, to the circle of radius

\[ \rho=\frac{\big(1+f'(x_0)^2\big)^{3/2}}{|f''(x_0)|}. \]

Sketch. A circle \((x-a)^2+(y-b)^2=r^2\) through three points of the curve meets it to second order in the limit, so \(x-a+(y-b)y'=0\) and \(1+y'^2+(y-b)y''=0\) at \(x_0\). Solving, \(y-b=-(1+y'^2)/y''\) and \(x-a=y'(1+y'^2)/y''\), so \(r^2=(1+y'^2)^3/y''^2\). Taylor’s definition by limiting circles connects curvature to the classical Greek problem of the angle of contact. Newton’s own account of curvature was published only in 1736.

Exercise 10.1A uniform rod of length \(L\) swings about one end. Find the length of the equivalent simple pendulum and the period for \(L=1\) m.

Show solution

\(I=mL^2/3\), \(h=L/2\), so \(\ell=\frac{mL^2/3}{mL/2}=\frac{2L}{3}\). For \(L=1\) m, \(T=2\pi\sqrt{(2/3)/9.81}\approx1.64\) s.

Exercise 10.2Show that if the rod is hung from its centre of oscillation instead, the period is unchanged (Huygens’s reciprocity), using \(I=I_{\text{cm}}+mh^2\).

Show solution

\(\ell=h+I_{\text{cm}}/(mh)\). Hanging from the centre of oscillation puts the centre of mass at \(h'=\ell-h=I_{\text{cm}}/(mh)\), so the new length is \(h'+I_{\text{cm}}/(mh')=I_{\text{cm}}/(mh)+h=\ell\). For the rod: \(h=L/2\), \(h'=L/6\), and \(\ell=2L/3\) both times.

Exercise 10.3Find the radius of curvature of \(y=x^2/(2p)\) at the vertex and of \(y=\sqrt{r^2-x^2}\) at \(x=0\).

Show solution

Parabola: \(y'(0)=0\), \(y''=1/p\), so \(\rho=p\). Circle: \(y'(0)=0\), \(y''(0)=-1/r\), so \(\rho=r\), as it must be.

Part III · Life and legacy

Lesson 11

Controversy, France and philosophy, 1715–1721

Taylor took Newton’s side in the priority war with Leibniz and carried it on against Johann Bernoulli in the journals. He issued a challenge to “non-English mathematicians” to integrate a certain differential; Bernoulli pointed out that Leibniz had already solved it in the Acta eruditorum, and Giulio Fagnano also replied. The exchanges grew bitter, at one point involving a wager of fifty guineas. When Bernoulli privately suggested a more gentlemanly tone, Taylor answered that he had meant “to show an indignation”. Bernoulli in turn disputed Taylor’s priority and attacked the obscurity of the Methodus.

The quarrels did not cut him off from the Continent. From 1715 he corresponded with Pierre Rémond de Montmort about infinite series, probability and the philosophy of Nicolas Malebranche, and sometimes passed messages between Montmort and Abraham de Moivre. He visited Paris in 1716, returning early in 1717. His experimental papers in these years measured the force between magnets at different distances, the rise of water between glass plates, and the expansion of liquid in a thermometer. He explained that atmospheric pressure falls exponentially with altitude, so that altitude is proportional to the logarithm of the pressure ratio, and in 1717 published a method for computing logarithms.

His health broke under the pace of work. He resigned the secretaryship on 21 October 1718 and in 1719 went to Aix-la-Chapelle to take the waters. After his return he turned largely to religion and philosophy, writing unfinished treatises on the Jewish sacrifices and on the lawfulness of eating blood, and visited Lord Bolingbroke in France in 1720–21.

Exercise 11.1For an isothermal atmosphere, pressure satisfies \(dp/dh=-\rho g\) with \(\rho=p/(gH)\). Show \(p=p_0e^{-h/H}\), and find the altitude where pressure has halved for \(H=8\) km.

Show solution

\(dp/dh=-p/H\) gives \(p=p_0e^{-h/H}\), so \(h=H\ln(p_0/p)\). Halving: \(h=8\ln2\approx5.5\) km.

Lesson 12

Losses, death and legacy

In 1721 Taylor married a Miss Brydges of Wallington in Surrey. His father objected, and father and son were estranged. In 1723 she died in childbirth, and the child died too. Taylor was reconciled with his father. In 1725 he married Sabetta Sawbridge of Olantigh, Kent, with his father’s approval. His father died on 4 April 1729, and Taylor inherited Bifrons. In 1730 Sabetta died in childbirth; their daughter Elizabeth survived. Taylor never recovered. He died at Somerset House in London on 29 December 1731, aged 46, and was buried in the churchyard of St Anne’s, Soho.

His grandson Sir William Young printed his philosophical essay Contemplatio philosophica privately in 1793, with a life of Taylor and an appendix of correspondence.

Legacy

Analysis: the expansion in the Methodus became, through Lagrange, the central tool of eighteenth-century analysis and the starting point for the theory of remainders, analytic functions and asymptotic expansions.
Finite differences: his calculus of increments was the first systematic treatment of the subject, developed later by James Stirling, Euler and George Boole into a branch of numerical analysis.
Mathematical physics: the 1713 string analysis is the first application of the calculus to a vibrating continuum and the root of the wave equation and of Fourier analysis.
Geometry and art: Linear Perspective went through many editions, translations and adaptations; its vanishing-point theory is the basis of descriptive and projective treatments of perspective.
Memory: the lunar crater Taylor is named after him.

Final projectChoose one result from Part II. Write a two-page account for a strong undergraduate that states it in modern form, sketches how Taylor arrived at it, works one example in full, and explains what later mathematicians (Lagrange, Cauchy, d’Alembert, Bernoulli or Lambert) added.

Reference

Sources and further reading

Biography

J. J. O’Connor and E. F. Robertson, Brook Taylor, MacTutor History of Mathematics.
P. S. Jones, Brook Taylor, Dictionary of Scientific Biography.
Brook Taylor, Wikipedia (with references to the Dictionary of National Biography).
A. Chalmers, Taylor, Brook, General Biographical Dictionary, 1812–17.
W. Young, “Life of Brook Taylor”, prefixed to B. Taylor, Contemplatio philosophica, London, 1793.

Mathematics

B. Taylor, Methodus incrementorum directa et inversa, London, 1715; 2nd ed. 1717.
B. Taylor, De motu nervi tensi, Philosophical Transactions 28 (1713).
B. Taylor, Linear Perspective, London, 1715; New Principles of Linear Perspective, London, 1719.
K. Andersen, Brook Taylor’s Work on Linear Perspective, Springer, 1992.
L. Feigenbaum, Brook Taylor and the method of increments, Archive for History of Exact Sciences 34 (1985).
R. C. Archibald, Mathematicians and music, MacTutor.
J.-L. Lagrange, Théorie des fonctions analytiques, Paris, 1797.
C. Truesdell, The rational mechanics of flexible or elastic bodies 1638–1788, in Euler, Opera omnia ser. II vol. 11, part 2, 1960.