courseshub.world · Lives in Mathematics and Physics

Ada Lovelace

Augusta Ada King, Countess of Lovelace, born Augusta Ada Byron

A course in two strands: the life of the poet’s daughter who learned university mathematics by letter and wrote the fullest contemporary account of Babbage’s Analytical Engine, and the mathematics of her Notes, from the calculus she studied to the Bernoulli-number program of Note G.

Born 10 December 1815, London Died 27 November 1852, Marylebone, London Notes A–G Taylor’s Scientific Memoirs, 1843 Level advanced undergraduate
The operation cards of Note G for \(n=4\). Operations 13–23 run twice, once for each earlier Bernoulli number; operation 24 leaves \(B_7=-\tfrac1{30}\) (her numbering) in column \(V_{24}\).
Lesson 00

How to use this course

Ada Lovelace published one scientific work in her lifetime: a translation of Luigi Menabrea’s 1842 article on Charles Babbage’s Analytical Engine, with seven notes of her own, lettered A to G, about three times the length of the article. Those notes explain what a general-purpose calculating machine is, how it is instructed, and what it might do beyond arithmetic. The last note contains a complete table of operations for computing Bernoulli numbers, often called the first published computer program.

Her reputation has swung between extremes, from prophet of the computer age to figurehead for another man’s work. This course tries to show the evidence: the mathematics she studied, the Notes themselves, and what historians have concluded from the surviving letters. It is a companion to the course on Babbage, whose Lessons 07 and 08 cover the engine’s architecture and the Bernoulli recurrence from his side.

Prerequisites for Part II
  • First-year calculus: limits, derivatives, Taylor series.
  • Basic combinatorics: binomial coefficients and generating functions.
  • For Lessons 06 and 08, any experience of tracing a simple program by hand helps.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements, derivations and exercises with folded solutions. Lessons 04 and 05 follow her calculus correspondence of 1840–41; Lessons 06 to 09 follow the Notes of 1843 in order, and Lesson 10 explains why the Bernoulli numbers were a natural showcase.

Reference

Timeline

1815Born 10 December in London, the only legitimate child of the poet Lord Byron and Anne Isabella Milbanke.
1816Her parents separate in January; Byron leaves England in April and never sees her again.
1824Byron dies in Greece.
1828At twelve, plans a flying machine and a book to be called Flyology.
1829Measles leaves her partly paralysed; she walks with crutches by 1831.
1833Presented at court. Meets Charles Babbage on 5 June and soon sees the demonstration piece of the Difference Engine.
1834Mary Somerville begins to guide her reading in mathematics.
18358 July: marries William King, 8th Baron King.
1836–39Children Byron (1836), Annabella (1837) and Ralph (1839).
1838Her husband is created Earl of Lovelace; she becomes Countess of Lovelace.
1840–41Correspondence course in calculus with Augustus De Morgan.
1842October: Menabrea’s account of the Analytical Engine appears in Geneva. She begins a translation.
1843The translation and Notes A–G appear in Richard Taylor’s Scientific Memoirs, signed A.A.L. Quarrel with Babbage over a preface.
1844Plans a mathematical model of the nervous system. De Morgan’s confidential assessment of her ability is sent to Lady Byron.
1850–51Heavy losses betting on horse races.
1852Dies of cancer on 27 November in Marylebone, aged 36. Buried beside her father at Hucknall, Nottinghamshire.
1950Alan Turing names “Lady Lovelace’s Objection” in Computing Machinery and Intelligence.
1980The US Department of Defense standardizes the Ada programming language as MIL-STD-1815.
2009First Ada Lovelace Day.
Part I · Formation
Lesson 01

Byron’s daughter, 1815–1832

Augusta Ada Byron was born in London on 10 December 1815. Her father was George Gordon, Lord Byron, the most famous poet in Europe. Her mother, Anne Isabella Milbanke, was a wealthy, devout and mathematically educated woman whom Byron called his princess of parallelograms. The marriage lasted barely a year. In January 1816 Lady Byron left with the baby, a deed of separation followed in April, and Byron left England within days. He never saw his daughter again and died in Greece in April 1824, when she was eight.

Lady Byron had sole custody and a clear plan. Mathematics and science would discipline the girl’s mind and keep her from her father’s temperament. Ada was taught at home by a succession of governesses and tutors, under a strict regime of lessons, rewards and punishments. Among them were William Frend, a Cambridge-trained mathematician who had once taught Lady Byron, and from 1829 Dr William King, a physician in Brighton (no relation of her future husband).

She was often ill. At about twelve she studied the anatomy of birds and planned a flying machine, with a book to be called Flyology. In 1829 measles left her partly paralysed, and she spent much of the next two years in bed; she was walking with crutches by 1831. Her mathematical education in these years was patchy. She learned arithmetic, geometry and some algebra, but her tutors were not research mathematicians, and nothing yet suggested unusual talent.

Lesson 02

Babbage, Somerville and marriage, 1833–1839

In 1833 Ada Byron was presented at court. On 5 June, at a party in London, she met Charles Babbage, then 41, Lucasian Professor at Cambridge and a famous host. About two weeks later she and her mother visited his house to see the demonstration section of the Difference Engine, about one seventh of the planned machine. Sophia Frend, William Frend’s daughter and later the wife of Augustus De Morgan, recalled that while other visitors stared, the young girl understood how it worked.

From 1834 the mathematician and science writer Mary Somerville became her friend and adviser. Somerville lent her books, set problems and encouraged her study. Through the Somerville family she met her future husband, William King, 8th Baron King. They married on 8 July 1835. Their children were Byron (born 1836), Anne Isabella, known as Annabella (1837), and Ralph Gordon (1839). In 1838 King was created Earl of Lovelace, and Ada became Countess of Lovelace.

She kept up her friendship with Babbage, attended his evening gatherings, and followed his work on the Analytical Engine, which he began to design from 1834. Domestic life left little time for study, and after her third child she resolved to take up mathematics seriously. Around 1839 she asked Babbage to find her a mathematics teacher in London.

Lesson 03

A correspondence course with De Morgan, 1840–1841

The teacher was Augustus De Morgan, professor of mathematics at University College London and a family friend through his wife Sophia. From the summer of 1840 to late 1841 he taught Lovelace by letter. About 63 letters survive, roughly two thirds of them hers, now in the Lovelace–Byron papers at the Bodleian Library in Oxford. The exchange paused between February and June 1841.

The syllabus was the one De Morgan gave his own students at University College: algebra and trigonometry, functional equations, limits, the differential and integral calculus, series and their convergence, and the foundations of algebra. She worked through his textbooks, including his Differential and Integral Calculus. Her letters are full of questions about the meaning of symbols and the legitimacy of steps. Why can one not simply multiply by \(dx\)? When does a divergent series mean anything? Lessons 04 and 05 reconstruct this material.

How good was she? In a confidential letter of 21 January 1844 to Lady Byron, De Morgan wrote that her power of thinking on these matters was far beyond that of any beginner he had known, and that in a young man it would have suggested an original mathematical investigator. He also warned, in the manner of his time, that the strain of study might harm a woman’s health. In 2017 Christopher Hollings, Ursula Martin and Adrian Rice published a full re-examination of the correspondence. They concluded that by 1842 she had a sound grasp of university mathematics, a critical attitude to foundations and a gift for perceptive questions, though she was probably not yet ready for original research.

DiscussionDe Morgan’s praise came in a private letter to the student’s mother, from a family friend. How much weight should a historian give it, and what other evidence from the letters would you look for?

Part II · The mathematics
Lesson 04

Limits and the derivative

For the general reader. In 1840 the foundations of calculus were still being rebuilt. British mathematics was moving from Newton’s fluxions to the continental notation of Leibniz, and teachers like De Morgan wanted to base everything on the idea of a limit. Lovelace learned calculus at exactly this moment, and her letters show her testing every step. This lesson works through the core of what she studied.

Goals
  • Define the derivative as a limit and compute it for powers.
  • See why some textbook proofs of the power rule are circular.
  • Understand why \(dx\) is not a number that can be multiplied through.

The derivative as a limit

De Morgan defined the differential coefficient of \(f\) at \(a\) as the limit of the ratio of increments,

\[ f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}. \]

The notation \(\frac{dy}{dx}\) names this limit as a single symbol. It is not a quotient of two quantities, and so the operation of multiplying by \(dx\), one of Lovelace’s early questions, needs a separate justification (in modern terms, via differentials or the chain rule for integrals).

Theorem (power rule)

For every rational \(r\) and every \(x>0\), \(\dfrac{d}{dx}x^{r}=r\,x^{r-1}\). For a positive integer \(r\) the formula holds for all real \(x\).

Proof

Positive integers. For \(n\ge1\) and \(x\ne a\), the identity \(x^n-a^n=(x-a)\sum_{k=0}^{n-1}x^{k}a^{n-1-k}\) gives

\[ \frac{x^n-a^n}{x-a}=\sum_{k=0}^{n-1}x^{k}a^{n-1-k}\ \longrightarrow\ n\,a^{n-1}\quad (x\to a). \]

Rationals. Let \(r=p/q\) with \(q\ge1\), and put \(u=x^{1/q}\), \(b=a^{1/q}\) for \(x,a>0\). Then

\[ \frac{x^{p/q}-a^{p/q}}{x-a}=\frac{u^p-b^p}{u^q-b^q}=\frac{(u^p-b^p)/(u-b)}{(u^q-b^q)/(u-b)}\ \longrightarrow\ \frac{p\,b^{p-1}}{q\,b^{q-1}}=\frac pq\,a^{p/q-1}, \]

using continuity of \(x\mapsto x^{1/q}\), so that \(u\to b\) as \(x\to a\). Negative \(p\) is handled by the same identity applied to \(u^{-p}\), or by the quotient rule.

A circular proof

A tempting shortcut writes \((a+h)^r=a^r+r\,a^{r-1}h+\cdots\) by the binomial theorem and reads off the coefficient of \(h\). For integer \(r\) this is sound. For general \(r\), the binomial series is usually proved by Taylor’s theorem, which already assumes the derivative of \(x^r\). Hollings, Martin and Rice record that Lovelace sent De Morgan an argument about this derivative and later recognised it herself as circular. Spotting circularity is a mark of exactly the critical reading her letters show.

Exercise 4.1Use the definition of the derivative to find \(\frac{d}{dx}\frac1x\) at \(a\ne0\), without the power rule.

Show solution
\(\dfrac{1/(a+h)-1/a}{h}=\dfrac{a-(a+h)}{h\,a(a+h)}=\dfrac{-1}{a(a+h)}\to-\dfrac1{a^2}\), in agreement with \(r=-1\).

Exercise 4.2Prove \(\frac{d}{dx}\sqrt x=\frac1{2\sqrt x}\) for \(x>0\) by rationalizing the difference quotient, and compare with the substitution \(u=x^{1/2}\) above.

Show solution
\(\dfrac{\sqrt{a+h}-\sqrt a}{h}=\dfrac{(a+h)-a}{h(\sqrt{a+h}+\sqrt a)}=\dfrac1{\sqrt{a+h}+\sqrt a}\to\dfrac1{2\sqrt a}\). This is the case \(p=1,q=2\): the ratio \((u-b)/(u^2-b^2)=1/(u+b)\).

Exercise 4.3Show that the chain rule makes “multiplying by \(dx\)” legitimate in the substitution rule: if \(F'=f\) and \(x=g(t)\), then \(\int f(g(t))\,g'(t)\,dt=F(g(t))+C\).

Show solution
By the chain rule \(\frac{d}{dt}F(g(t))=F'(g(t))\,g'(t)=f(g(t))\,g'(t)\). Hence \(F(g(t))\) is an antiderivative of the integrand. The formal step \(dx=g'(t)\,dt\) is shorthand for this theorem, not an algebraic multiplication.
Lesson 05

Functional equations and series

For the general reader. Some of De Morgan’s exercises asked which functions obey a given rule, such as the addition formula for cosines. Others concerned infinite sums, including sums that do not settle down to any value. British algebraists of the 1830s, led by George Peacock, held that algebraic formulas should keep their form whatever the symbols meant. Lovelace questioned De Morgan closely about where this principle breaks down.

Goals
  • Solve d’Alembert’s functional equation under a continuity hypothesis.
  • State Taylor’s theorem with remainder.
  • See how divergent series misled the “permanence of forms” and how Cesàro summation repairs part of it.

A functional equation

One exercise in the correspondence asked her to show that a given function satisfies

\[ \varphi(x+y)+\varphi(x-y)=2\,\varphi(x)\,\varphi(y). \]

This is d’Alembert’s functional equation, named after Jean le Rond d’Alembert. The cosine satisfies it because \(\cos(x+y)+\cos(x-y)=2\cos x\cos y\).

Theorem (continuous solutions)

If \(\varphi:\mathbb R\to\mathbb R\) is continuous and satisfies d’Alembert’s equation for all real \(x,y\), then \(\varphi\equiv0\), or \(\varphi(x)=\cos(ax)\) for some \(a\ge0\), or \(\varphi(x)=\cosh(ax)\) for some \(a>0\).

Proof sketch. Put \(x=y=0\): \(2\varphi(0)=2\varphi(0)^2\), so \(\varphi(0)\in\{0,1\}\). If \(\varphi(0)=0\), setting \(y=0\) gives \(2\varphi(x)=0\) for all \(x\). So assume \(\varphi(0)=1\). Setting \(x=0\) shows \(\varphi\) is even. Since \(\varphi\) is continuous with \(\varphi(0)=1\), choose \(\delta>0\) with \(c=\int_0^\delta\varphi(y)\,dy\ne0\). Integrating the equation in \(y\) over \([0,\delta]\),

\[ 2c\,\varphi(x)=\int_{x}^{x+\delta}\varphi(t)\,dt+\int_{x-\delta}^{x}\varphi(t)\,dt=\int_{x-\delta}^{x+\delta}\varphi(t)\,dt, \]

so \(\varphi\) is differentiable, and repeating the argument, infinitely differentiable. Differentiating the equation twice in \(y\) and setting \(y=0\) gives \(2\varphi''(x)=2\varphi(x)\varphi''(0)\). With \(k=\varphi''(0)\), the solution of \(\varphi''=k\varphi\), \(\varphi(0)=1\), \(\varphi'(0)=0\) is \(\cos(\sqrt{-k}\,x)\), \(1\) or \(\cosh(\sqrt k\,x)\) as \(k<0\), \(k=0\), \(k>0\).

Taylor’s theorem

Theorem (Taylor, with Lagrange’s remainder)

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

\[ f(x)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k+\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1} \]

for some \(\xi\) between \(a\) and \(x\).

The series \(\sum f^{(k)}(a)(x-a)^k/k!\) converges to \(f(x)\) exactly when the remainder tends to 0. For \(e^x\), \(\sin x\) and \(\cos x\) it does on the whole line; for \(1/(1+x)\) about \(a=0\) it does only for \(|x|<1\).

Divergent series and the permanence of forms

Peacock’s principle of the permanence of equivalent forms held, roughly, that an identity proved for one kind of quantity should hold for all. Applied to \(\frac1{1+x}=1-x+x^2-x^3+\cdots\) at \(x=1\), it yields \(\tfrac12=1-1+1-1+\cdots\), a series whose partial sums oscillate between 1 and 0. De Morgan took such series more seriously than most, and Lovelace pressed him on what they could mean. The modern answer is that a divergent series can be given a value by a stated summation method, and that the value depends on the method.

Theorem (Cesàro, 1890)

Let \(s_n=a_0+\cdots+a_n\) and \(\sigma_N=\frac1{N+1}\sum_{n=0}^{N}s_n\). If \(s_n\to s\), then \(\sigma_N\to s\). The converse fails.

Exercise 5.1Show that \(\cosh\) satisfies d’Alembert’s equation, and that \(\varphi(x)=\cos x+1\) does not.

Show solution
\(\cosh(x+y)+\cosh(x-y)=2\cosh x\cosh y\) by expanding in exponentials: \(\tfrac12(e^{x+y}+e^{-x-y}+e^{x-y}+e^{-x+y})=\tfrac12(e^x+e^{-x})(e^y+e^{-y})\). For \(\varphi=\cos x+1\), at \(x=y=0\) the left side is 4 and the right side is \(2\cdot2\cdot2=8\).

Exercise 5.2Compute the Cesàro sum of \(1-1+1-1+\cdots\), and show that it agrees with Peacock’s value \(\tfrac12\).

Show solution
The partial sums are \(1,0,1,0,\dots\). Among the first \(N+1\) of them, \(\lceil (N+1)/2\rceil\) equal 1, so \(\sigma_N\to\tfrac12\). Abel summation gives the same: \(\lim_{x\to1^-}\sum(-1)^nx^n=\lim\frac1{1+x}=\tfrac12\).

Exercise 5.3The series \(1-2+4-8+\cdots\) arises from \(\frac1{1+x}\) at \(x=2\). Show that its Cesàro means do not converge, so the “permanent form” value \(\tfrac13\) cannot be obtained this way.

Show solution
The partial sums are \(s_n=\frac{1-(-2)^{n+1}}{3}\), which grow without bound in absolute value and alternate in sign. Their averages \(\sigma_N=\frac13-\frac{1}{3(N+1)}\sum_{n=0}^{N}(-2)^{n+1}\) contain a term of size about \(2^{N+2}/(9(N+1))\) with alternating sign, which does not tend to a limit. The value \(\tfrac13\) needs a stronger method, such as analytic continuation.
Lesson 06

The engine in Notes A–D

For the general reader. Menabrea’s article described what the Analytical Engine would do. Lovelace’s notes explained what kind of thing it was. A difference engine computes one family of tables; the Analytical Engine could carry out any sequence of arithmetic operations it was given on punched cards, and repeat parts of that sequence. She counted operations, laid out their order, and showed how the same cards could act on different numbers.

Goals
  • Explain the contrast between the two engines drawn in Note A.
  • Describe operation cards, variable cards and the backing of cards (Notes B and C).
  • Reproduce the operation count of Note D for two linear equations.

Note A: two engines

Note A observes that the Difference Engine tabulates only functions whose seventh differences vanish, \(\Delta^7u_x=0\), using only addition, with one column per order of difference. The Analytical Engine performs all four operations directly and can therefore develop a function, not merely tabulate it. She called it the material embodiment of the science of operations, and compared its cards to those of the Jacquard loom, which weave flowers and leaves into silk.

Notes B and C: cards and store

Numbers live in columns of the store, labelled \(V_1,V_2,\dots\). Operation cards say which arithmetic operation the mill performs; variable cards say which columns supply the operands and which receive the result. Because the two card sets are separate, one operation sequence can be applied to different columns. Note C describes how a set of cards can be “backed”, returned so that it acts again, any number of times in the same problem. This is the loop.

Lovelace introduced an indexing convention that the Note G table uses throughout: \({}^{m}V_i\) denotes the \(m\)-th value held by column \(V_i\), with \({}^{0}V_i\) meaning the column is still empty (zero). The notation keeps track of state in a machine whose columns are overwritten.

Operation count (Note D)

The system \(mx+ny=d,\ m'x+n'y=d'\) is solved by

\[ x=\frac{dn'-d'n}{mn'-m'n},\qquad y=\frac{d'm-dm'}{mn'-m'n}, \]

using 11 operations: 6 multiplications, 3 subtractions and 2 divisions.

Note D sets out these 11 operations in a diagram listing, for each operation, the columns read, the column written and the algebraic meaning of the result. The point is the separation between the fixed pattern of operations and the variable data, which is what makes the cards reusable.

Exercise 6.1List the 11 operations of Note D in a valid order, giving each result a store column.

Show solution
Data in \(V_0\ldots V_5=m,n,d,m',n',d'\). (1) \(V_6=m\,n'\); (2) \(V_7=m'\,n\); (3) \(V_8=d\,n'\); (4) \(V_9=d'\,n\); (5) \(V_{10}=d'\,m\); (6) \(V_{11}=d\,m'\); (7) \(V_{12}=V_6-V_7\); (8) \(V_{13}=V_8-V_9\); (9) \(V_{14}=V_{10}-V_{11}\); (10) \(x=V_{13}/V_{12}\); (11) \(y=V_{14}/V_{12}\). Six products, three differences, two quotients.

Exercise 6.2The same coefficients \(m,n,m',n'\) are to be used with 100 different right-hand sides \((d,d')\). Which operations of Note D need be done only once, and what is the total count?

Show solution
The products \(mn'\), \(m'n\) and their difference (3 operations) do not depend on \(d,d'\). Each new right-hand side needs 4 products, 2 subtractions and 2 divisions: 8 operations. Total \(3+100\cdot8=803\), against \(1100\) if everything is recomputed. Keeping a result in a store column and reusing it is the economy Lovelace’s diagrams make visible.

Exercise 6.3A difference engine with \(k\) difference columns tabulates \(u_x\) with \(\Delta^ku_x=0\). Which functions are these, and why can it not tabulate \(\sin x\) exactly?

Show solution
\(\Delta^ku=0\) on integer steps exactly when \(u\) is a polynomial of degree less than \(k\). The sine is not a polynomial, so a difference engine can only approximate it piecewise, restarting from fresh initial differences. The Analytical Engine could instead compute each value from a series, which is the distinction Note A draws.
Lesson 07

Bernoulli numbers: the Note G formula

For the general reader. To show the engine at full stretch, Note G chose a hard example: the Bernoulli numbers, which appear in formulas for sums of powers and in the series for the tangent. They have no simple closed form, but each can be computed from the earlier ones. Lovelace derived the recurrence she needed from a power series, then arranged it for the machine.

Goals
  • Derive the Note G equation from the generating function.
  • Translate between her numbering and the modern one.
  • Compute the coefficients as products of simple factors, as the engine does.

Generating function and numbering

In modern notation, named after Jacob Bernoulli,

\[ \frac{x}{e^x-1}=\sum_{m\ge0}B_m\frac{x^m}{m!}=1-\frac x2+\frac16\frac{x^2}{2!}-\frac1{30}\frac{x^4}{4!}+\frac1{42}\frac{x^6}{6!}-\cdots, \]

with \(B_m=0\) for odd \(m\ge3\). Note G starts from the same expansion but names the nonzero even-index numbers with odd subscripts: her \(B_1,B_3,B_5,B_7\) are the modern \(B_2,B_4,B_6,B_8\). This course keeps her numbering in Lessons 07 and 08, as the Babbage course does.

Proposition (the equation programmed in Note G)

For \(n\ge1\), in Lovelace’s numbering,

\[ 0=A_0+A_1B_1+A_3B_3+\cdots+A_{2n-1}B_{2n-1},\qquad A_0=-\frac12\cdot\frac{2n-1}{2n+1},\quad A_{2k-1}=\frac{1}{2k}\binom{2n}{2k-1}. \]

In particular \(A_{2n-1}=1\), so \(B_{2n-1}=-\big(A_0+A_1B_1+\cdots+A_{2n-3}B_{2n-3}\big)\).

Derivation

Multiply the generating function by \(e^x-1\) and compare coefficients of \(x^{N}/N!\) with \(N\ge2\): \(\sum_{j=0}^{N-1}\binom{N}{j}B_j=0\) (modern numbering). Take \(N=2n+1\). With \(B_0=1\), \(B_1=-\tfrac12\) and the odd ones zero,

\[ 1-\frac{2n+1}{2}+\sum_{k=1}^{n}\binom{2n+1}{2k}B_{2k}=0. \]

Divide by \(2n+1\). The constant becomes \(\frac{1}{2n+1}\cdot\frac{1-2n}{2}=-\frac12\cdot\frac{2n-1}{2n+1}\), and \(\binom{2n+1}{2k}\big/(2n+1)=\frac{(2n)!}{(2k)!\,(2n+1-2k)!}=\frac1{2k}\binom{2n}{2k-1}\). Renaming \(B_{2k}\) as her \(B_{2k-1}\) gives the proposition.

Coefficients as running products

The engine never computes a binomial coefficient directly. It builds each \(A\) from the previous one:

\[ A_1=\frac{2n}{2},\qquad A_{2k+1}=A_{2k-1}\cdot\frac{2n-2k+1}{2k+1}\cdot\frac{2n-2k}{2k+2}. \]

Each step multiplies by two quotients, one numerator decreasing by 1 and one denominator increasing by 1 at a time. This is the inner loop of the program in Lesson 08.

Exercise 7.1Check the running-product formula against \(A_{2k-1}=\frac1{2k}\binom{2n}{2k-1}\).

Show solution
\(\dfrac{A_{2k+1}}{A_{2k-1}}=\dfrac{2k}{2k+2}\cdot\dfrac{\binom{2n}{2k+1}}{\binom{2n}{2k-1}}=\dfrac{2k}{2k+2}\cdot\dfrac{(2n-2k+1)(2n-2k)}{(2k+1)(2k)}=\dfrac{2n-2k+1}{2k+1}\cdot\dfrac{2n-2k}{2k+2}\).

Exercise 7.2For \(n=4\), list \(A_0,A_1,A_3,A_5,A_7\).

Show solution
\(A_0=-\tfrac12\cdot\tfrac79=-\tfrac7{18}\); \(A_1=4\); \(A_3=4\cdot\tfrac73\cdot\tfrac64=14\); \(A_5=14\cdot\tfrac55\cdot\tfrac46=\tfrac{28}3\); \(A_7=\tfrac{28}3\cdot\tfrac37\cdot\tfrac28=1\). Note that the last factor pair always produces 1, a useful check.

Exercise 7.3Use the generating function to show that \(B_m=0\) for odd \(m\ge3\) (modern numbering).

Show solution
\(\frac{x}{e^x-1}+\frac x2=\frac x2\cdot\frac{e^x+1}{e^x-1}=\frac x2\coth\frac x2\), which is an even function. So all odd-power coefficients other than the \(x^1\) term, which was removed by adding \(x/2\), vanish.
Lesson 08

The Note G program

For the general reader. The table at the end of Note G lists, line by line, every operation the engine would perform to compute the Bernoulli number \(B_7\), which store columns it reads and writes, and what each intermediate result means. It has a loop that is run more than once, counters that tell the machine when to stop, and one transcription error. No machine ever ran it; modern emulations confirm that, with the error corrected, it works.

Goals
  • Trace the program for \(n=4\) and obtain \(B_7=-\tfrac1{30}\).
  • Identify the loop and the counter.
  • Count the operations needed for general \(n\).
  • Locate and correct the error in operation 4.

Store layout

Columns \(V_1=1\), \(V_2=2\), \(V_3=n=4\) hold the data. The earlier results \(B_1,B_3,B_5\) sit in \(V_{21},V_{22},V_{23}\), and the answer goes to \(V_{24}\). Working columns \(V_4\)–\(V_{13}\) start empty; \(V_{13}\) accumulates the sum and \(V_{10}\) counts the remaining terms.

The program (operations 1–25)

1: \(V_4,V_5,V_6\leftarrow V_2\times V_3\) (\(=2n\)). 2: \(V_4\leftarrow V_4-V_1\) (\(2n-1\)). 3: \(V_5\leftarrow V_5+V_1\) (\(2n+1\)). 4: \(V_{11}\leftarrow V_4\div V_5\). 5: \(V_{11}\leftarrow V_{11}\div V_2\). 6: \(V_{13}\leftarrow V_{13}-V_{11}\) (\(=A_0\)). 7: \(V_{10}\leftarrow V_3-V_1\) (\(n-1\)).

8: \(V_7\leftarrow V_2+V_7\) (\(=2\)). 9: \(V_{11}\leftarrow V_6\div V_7\) (\(A_1\)). 10: \(V_{12}\leftarrow V_{21}\times V_{11}\) (\(B_1A_1\)). 11: \(V_{13}\leftarrow V_{12}+V_{13}\). 12: \(V_{10}\leftarrow V_{10}-V_1\).

13: \(V_6\leftarrow V_6-V_1\). 14: \(V_7\leftarrow V_1+V_7\). 15: \(V_8\leftarrow V_6\div V_7\). 16: \(V_{11}\leftarrow V_8\times V_{11}\). 17–20: the same four steps again, with the quotient in \(V_9\). 21: \(V_{12}\leftarrow V_{22}\times V_{11}\) (\(B_3A_3\)). 22: \(V_{13}\leftarrow V_{12}+V_{13}\). 23: \(V_{10}\leftarrow V_{10}-V_1\).

Operations 13–23 are then repeated with \(V_{23}\) in place of \(V_{22}\), for the term \(B_5A_5\). 24: \(V_{24}\) receives \(B_7\), the sum in \(V_{13}\) with its sign reversed (the printed table writes this step as an addition). 25: \(V_3\leftarrow V_1+V_3\), so that \(n\) becomes 5, ready for the next Bernoulli number.

The counter \(V_{10}\) starts at \(n-1\) and is decreased after each term; when it reaches zero the cycle stops. The inner pattern 13–16, 17–20 is the two-factor update of Lesson 07. In Lovelace’s table the operation cards for 13–23 are written once, with a note that they are to be repeated; the variable cards change on the second pass.

The error in operation 4

In the published table, the operation column for line 4 names the operands in the order \(V_5\div V_4\), while the statement of results gives \(\frac{2n-1}{2n+1}\), which is \(V_4\div V_5\). Run as printed, the program would give a wrong \(B_7\). Whether the slip was Lovelace’s, Babbage’s or the printer’s cannot be decided from the published text. Babbage recalled that she found a serious error in his own draft of the Bernoulli computation, and a surviving letter shows her correcting his assumption about the variable cards for the repeated cycle (Lesson 12).

Exercise 8.1Trace operations 1–12 for \(n=4\) and give \(V_{13}\) and \(V_{10}\) after operation 12.

Show solution
\(V_4=V_5=V_6=8\); \(V_4=7\); \(V_5=9\); \(V_{11}=\tfrac79\); \(V_{11}=\tfrac7{18}\); \(V_{13}=-\tfrac7{18}\); \(V_{10}=3\); \(V_7=2\); \(V_{11}=4\); \(V_{12}=4\cdot\tfrac16=\tfrac23\); \(V_{13}=-\tfrac7{18}+\tfrac23=\tfrac5{18}\); \(V_{10}=2\).

Exercise 8.2Continue the trace through both passes of 13–23 and operation 24.

Show solution
First pass: \(V_6=7,V_7=3,V_8=\tfrac73,V_{11}=\tfrac{28}3\); \(V_6=6,V_7=4,V_9=\tfrac32,V_{11}=14\); \(V_{12}=14\cdot(-\tfrac1{30})=-\tfrac7{15}\); \(V_{13}=\tfrac5{18}-\tfrac7{15}=-\tfrac{17}{90}\); \(V_{10}=1\). Second pass: \(V_6=5,V_7=5,V_8=1,V_{11}=14\); \(V_6=4,V_7=6,V_9=\tfrac23,V_{11}=\tfrac{28}3\); \(V_{12}=\tfrac{28}3\cdot\tfrac1{42}=\tfrac29\); \(V_{13}=-\tfrac{17}{90}+\tfrac{20}{90}=\tfrac1{30}\); \(V_{10}=0\), so the cycle ends. Operation 24: \(B_7=-\tfrac1{30}\), the modern \(B_8\).

Exercise 8.3How many operations does the program perform to compute \(B_{2n-1}\) for general \(n\ge2\)? How many distinct operation cards are needed?

Show solution
Operations 1–12 once, the 11-operation cycle 13–23 once for each of \(B_3,\dots,B_{2n-3}\), that is \(n-2\) times, then 24 and 25: \(12+11(n-2)+2=11n-8\). For \(n=4\) that is 36. Since the cycle is backed and reused, only 25 distinct operation cards are needed for every \(n\), and the store needs only one new column for each new Bernoulli number. Lovelace stressed both points.

Exercise 8.4Run operations 1–6 with operation 4 as printed (\(V_5\div V_4\)) and find the resulting \(B_7\).

Show solution
Operation 4 gives \(\tfrac97\), operation 5 gives \(\tfrac9{14}\), so \(V_{13}\) starts at \(-\tfrac9{14}\) instead of \(-\tfrac7{18}\), a difference of \(-\tfrac{9}{14}+\tfrac7{18}=-\tfrac{32}{126}=-\tfrac{16}{63}\). The rest is unchanged, so the final sum is \(\tfrac1{30}-\tfrac{16}{63}\) and \(B_7\) comes out as \(\tfrac{16}{63}-\tfrac1{30}=\tfrac{139}{630}\) instead of \(-\tfrac1{30}\).
Lesson 09

Symbols, music and the limits of machines

For the general reader. The most quoted passages of the Notes are not about arithmetic. Lovelace argued that the engine acts on symbols, and numbers are only one kind of symbol. If the rules of harmony could be expressed in its operations, it might compose music. She also set a limit: the machine does what we know how to order it to do, and originates nothing. A century later Alan Turing took that claim as an objection to be answered.

Goals
  • See how operating on coefficients amounts to operating on algebraic expressions (Note E).
  • Multiply trigonometric series by the product-to-sum rule.
  • State Lady Lovelace’s Objection and Turing’s replies.

Algebra as operations on coefficients

The engine stores only numbers. Lovelace pointed out that if a column is assigned to each coefficient of an expression, then arithmetic on columns is algebra on expressions. A polynomial \(\sum a_ix^i\) becomes the list \((a_0,a_1,\dots)\), and multiplication becomes the convolution \(c_k=\sum_{i+j=k}a_ib_j\). Note E carries this out for a product of trigonometric series, where the rule for cosines replaces the rule for powers.

Product of cosine series

If \(f(\theta)=\sum_{i\ge0}a_i\cos i\theta\) and \(g(\theta)=\sum_{j\ge0}b_j\cos j\theta\) are finite sums, then, since \(\cos i\theta\cos j\theta=\tfrac12[\cos(i+j)\theta+\cos(i-j)\theta]\),

\[ f g=\sum_{k\ge0}c_k\cos k\theta,\qquad c_k=\frac12\sum_{i+j=k}a_ib_j+\frac12\sum_{|i-j|=k}a_ib_j . \]

Each \(c_k\) is computed from the numbers \(a_i,b_j\) alone, so the engine produces the coefficients of the product series. Lovelace concluded that the machine’s processes are algebraical even when its outputs are figures, and that it could in principle arrange results so as to represent general symbols.

Lady Lovelace’s Objection

Note G warns against overrating the engine. It has, she wrote, “no pretensions whatever to originate anything”; it can do whatever we know how to order it to perform, and can follow analysis without anticipating any analytical truth. In Computing Machinery and Intelligence (1950), Turing gave the name Lady Lovelace’s Objection to the claim that a machine can never do anything new. He replied that machines often surprise their makers, that the objection assumes we can foresee all consequences of a program, and that a learning machine could change its own rules. He also suggested that the Analytical Engine was in principle a universal machine, a point developed in the Babbage course, Lesson 07.

Exercise 9.1Compute the cosine-series coefficients of \((1+2\cos\theta)(3+\cos\theta+\cos2\theta)\).

Show solution
\(3+\cos\theta+\cos2\theta+6\cos\theta+2\cos^2\theta+2\cos\theta\cos2\theta\). Use \(2\cos^2\theta=1+\cos2\theta\) and \(2\cos\theta\cos2\theta=\cos3\theta+\cos\theta\). Total: \(4+8\cos\theta+2\cos2\theta+\cos3\theta\). Check with the formula: \(c_0=a_0b_0+\tfrac12a_1b_1=3+1=4\); \(c_1=a_0b_1+a_1b_0+\tfrac12a_1b_2=1+6+1=8\); \(c_2=a_0b_2+\tfrac12a_1b_1=1+1=2\); \(c_3=\tfrac12a_1b_2=1\).

Exercise 9.2Write the multiplication of two polynomials of degree \(d\) as engine operations. How many multiplications and additions are needed?

Show solution
Store \(a_0,\dots,a_d\) and \(b_0,\dots,b_d\) in \(2(d+1)\) columns and reserve \(2d+1\) result columns \(c_0,\dots,c_{2d}\). For each pair \((i,j)\) multiply \(a_ib_j\) and add it into \(c_{i+j}\): \((d+1)^2\) multiplications and \((d+1)^2-(2d+1)=d^2\) additions. A double loop over \(i\) and \(j\) with backed cards keeps the card count independent of \(d\).

Exercise 9.3Turing wrote that a program can surprise its author. Give a short program for the Analytical Engine whose output its writer cannot predict without running it, and explain how this bears on Lovelace’s claim.

Show solution
Example: iterate \(n\mapsto n/2\) if even, \(3n+1\) if odd, from \(n=27\), counting steps until 1 (the answer is 111). The writer knows every rule but not the outcome. Lovelace’s claim survives in a narrow form, since every step was ordered, but fails in the broader sense that the results are new to the person who wrote the orders. Turing pressed exactly this distinction.
Lesson 10

Why Bernoulli numbers? Sums of powers

For the general reader. Jacob Bernoulli introduced these numbers in his Ars Conjectandi (published 1713) to add up powers: \(1^k+2^k+\cdots+n^k\). He boasted that with his formula he could sum the tenth powers of the first thousand numbers in a few minutes. The same numbers turn up in the series for the tangent and in the values of \(1+\frac1{4}+\frac1{9}+\cdots\). They were a fitting test for a machine meant to handle any formula, because computing them requires an unbounded sequence of steps, each depending on the previous results.

Goals
  • State and prove Bernoulli’s formula for sums of powers.
  • Relate Bernoulli numbers to \(\zeta(2k)\) and the tangent series.
  • Estimate how fast the numbers grow, and what that meant for a 40-digit store.
Theorem (Bernoulli’s summation formula)

For integers \(k\ge1\) and \(n\ge1\), with modern Bernoulli numbers and \(B_1=-\tfrac12\),

\[ \sum_{j=0}^{n-1}j^k=\frac1{k+1}\sum_{i=0}^{k}\binom{k+1}{i}B_i\,n^{k+1-i}. \]

Proof

Sum the geometric series \(\sum_{j=0}^{n-1}e^{jx}=\frac{e^{nx}-1}{e^x-1}=\frac{x}{e^x-1}\cdot\frac{e^{nx}-1}{x}\). The coefficient of \(x^k/k!\) on the left is \(\sum_j j^k\). On the right, \(\frac{e^{nx}-1}{x}=\sum_{r\ge0}\frac{n^{r+1}x^r}{(r+1)!}\). Multiplying by \(\sum_i B_ix^i/i!\) and reading off the coefficient of \(x^k\) gives \(\sum_{i}\frac{B_i}{i!}\cdot\frac{n^{k+1-i}}{(k+1-i)!}\); multiplying by \(k!\) gives the formula.

Theorem (Euler)

For \(k\ge1\),

\[ \zeta(2k)=\sum_{m\ge1}\frac1{m^{2k}}=\frac{(-1)^{k+1}B_{2k}\,(2\pi)^{2k}}{2\,(2k)!}. \]

Since \(\zeta(2k)\to1\), this shows \(|B_{2k}|\sim 2\,(2k)!/(2\pi)^{2k}\): the numbers alternate in sign and eventually grow faster than any exponential. Numerators and denominators grow too, so exact computation on a machine with fixed-length columns soon needs care with fractions. The Note G program works with the values as given, without addressing this.

Exercise 10.1Use the theorem with \(k=2\) to find \(1^2+2^2+\cdots+m^2\).

Show solution
With \(n=m+1\): \(\sum_{j=0}^{m}j^2=\tfrac13\big(n^3+3B_1n^2+3B_2n\big)=\tfrac13\big(n^3-\tfrac32n^2+\tfrac12n\big)=\frac{n(n-1)(2n-1)}{6}=\frac{m(m+1)(2m+1)}{6}\).

Exercise 10.2Use Euler’s formula with \(k=1\) and \(k=2\) to evaluate \(\zeta(2)\) and \(\zeta(4)\).

Show solution
\(k=1\): \(\zeta(2)=\frac{B_2(2\pi)^2}{2\cdot2}=\frac16\cdot\pi^2=\frac{\pi^2}6\). \(k=2\): \(\zeta(4)=\frac{-B_4(2\pi)^4}{2\cdot24}=\frac1{30}\cdot\frac{16\pi^4}{48}=\frac{\pi^4}{90}\).

Exercise 10.3Estimate \(|B_{50}|\) from the asymptotic formula. Could a store column of 40 decimal digits hold its integer part?

Show solution
\(|B_{50}|\approx2\cdot50!/(2\pi)^{50}\). With \(\log_{10}50!\approx64.48\) and \(50\log_{10}(2\pi)\approx39.91\), this is about \(2\times10^{24.57}\approx7.5\times10^{24}\) (the exact value is about \(7.5\times10^{24}\)). Twenty-five digits fit, but the exact fraction has denominator 66 and numerator of 27 digits, and \(B_{100}\approx-2.8\times10^{78}\) would not fit at all.
Part III · Life and legacy
Lesson 11

The last years, 1843–1852

Menabrea’s article appeared in the Bibliothèque universelle de Genève in October 1842. Charles Wheatstone suggested that Lovelace translate it for Richard Taylor’s Scientific Memoirs, a journal of foreign scientific papers in English. Babbage proposed that she add notes. She worked on them through the first half of 1843, sending drafts and questions to Babbage and receiving data and corrections in return. In the summer Babbage wanted to attach an unsigned preface attacking the government for abandoning his engine. Lovelace refused to have it bound into her work, and the dispute became bitter. The translation appeared without his preface, signed only A.A.L. On 14 August 1843 she wrote Babbage a long letter proposing that she manage the practical side of his project; on his copy he noted that he refused all the conditions. Their friendship survived. In September he addressed her as his Enchantress of Number.

She hoped to go on. In 1844 she wrote of building a calculus of the nervous system, a mathematical model of how the brain gives rise to thought, and visited the electrical experimenter Andrew Crosse to learn laboratory techniques. She drafted a review of Karl von Reichenbach’s work on magnetism and exchanged letters with Michael Faraday. None of these projects was finished. Her health, never strong, was treated with laudanum and other opiates.

In the later 1840s she began betting on horse races, and around 1851 a betting scheme run with a group of male friends left her heavily in debt. Lurid later stories of pawned family jewels and enormous losses rest on thin evidence. From early 1852 she suffered from what was then called uterine cancer, probably cervical. During long months of pain her mother took control of her household and her affairs. She died in Marylebone on 27 November 1852, aged 36, the same age as her father at his death. At her request she was buried beside him in the Byron vault of the Church of St Mary Magdalene, Hucknall.

Lesson 12

Who wrote the Notes? Legacy

No question about Lovelace is more disputed than her share in the Notes. The evidence is limited and comes mostly from three places: the published text, Babbage’s memoir, and the surviving letters between them. Historians read it differently.

The evidence

  • Babbage’s account (1864). In Passages from the Life of a Philosopher he wrote that he suggested she add notes, that she chose the illustrations and did the algebraic working of the problems, except the Bernoulli numbers, which he offered to do himself; she then found a serious error in his version and sent it back for correction.
  • Babbage’s earlier programs. Allan Bromley’s study of the manuscripts (1982) found that Babbage had written about two dozen programs for the engine between 1837 and 1840, never published. Bromley concluded that the programs in the Notes drew heavily on this earlier work and saw no evidence that Lovelace designed a program independently.
  • The letters of 1843. In July 1843 Lovelace asked Babbage for the data and formulae for the Bernoulli example, described herself as deep in difficulties with it, and later corrected his treatment of the variable cards in the repeated cycle. The letters show close collaboration, with Babbage supplying technical information and Lovelace writing, organizing and checking.

The interpretations

  • Sceptical. Bruce Collier’s thesis on the engines (1970) treated her as a publicist who added nothing to the design. Dorothy Stein’s biography (1985) argued that her mathematics was weak and that the Notes served Babbage’s purposes; Stein relied partly on a letter she dated to late 1842. Doron Swade, who led the Science Museum’s construction of Difference Engine No. 2, has said she published the first program rather than wrote it, while crediting her with seeing that the machine could manipulate symbols in general.
  • Positive. Betty Alexandra Toole’s edition of her letters (1992) presented her as a visionary. John Fuegi and Jo Francis (2003) argued from the correspondence that she was the principal author of the Notes and that contemporaries credited her. Stephen Wolfram (2015) read the letters as showing her driving the Bernoulli example and distilling the machine’s abstract operation in a way Babbage never did.
  • The mathematical re-appraisal. Hollings, Martin and Rice (2017, 2018) studied all her mathematical papers. They re-dated Stein’s key letter to 1840, near the start of her studies, which removes its force. Their verdict is measured: a capable, critical student of university mathematics, not yet a research mathematician in 1843, who might have become one.

A fair summary: the engine and its programming method were Babbage’s; the Bernoulli computation was a joint effort whose final table she prepared and corrected; the Notes as a text, with their account of the science of operations, symbolic computation and the limits of machines, are hers. Calling her the first programmer overstates the case; calling her a mere amanuensis ignores the letters.

Legacy

  • Rediscovery: the Notes were reprinted in B. V. Bowden’s Faster than Thought (1953), just as electronic computers appeared, and Turing’s 1950 paper had already brought her objection into debates about machine intelligence.
  • Names: the Ada programming language (standardized 1980; its military standard number, 1815, is her birth year); the Lovelace Medal of the British Computer Society; Ada Lovelace Day, founded in 2009 and now held on the second Tuesday of October.
  • Archive: her mathematical papers in the Bodleian Library, now digitized with the Clay Mathematics Institute, make it possible to judge her work directly.

Final projectUsing Lesson 08 and the published Note G table, write a two-page report that (a) explains the program to a programmer, (b) identifies which parts of the method rely on Babbage’s design and which on the organization of the Notes, and (c) states, with reasons, which of the historians’ positions in this lesson you find best supported.

Reference

Connections

People linked to Ada Lovelace across this collection. A name without a link has no page here yet. See Ada Lovelace in the network of all 337 lives →

Teachers and advisors

Influenced

Colleagues and collaborators

Family

  • Anne Isabella Milbanke, Lady Byron

Also mentioned in this life

Fields

Reference

Sources and further reading

Primary sources

  1. L. F. Menabrea, Sketch of the Analytical Engine invented by Charles Babbage, translated with notes by A. A. L., Scientific Memoirs 3 (1843), 666–731. Online text.
  2. C. Babbage, Passages from the Life of a Philosopher, Longman, 1864, chapter VIII.
  3. A. M. Turing, Computing machinery and intelligence, Mind 59 (1950), 433–460.
  4. The Lovelace–De Morgan correspondence, transcribed for the Clay Mathematics Institute from the Bodleian Library’s Lovelace–Byron papers.

Biography and history

  1. J. J. O’Connor and E. F. Robertson, Augusta Ada King, Countess of Lovelace, MacTutor.
  2. Ada Lovelace and Note G, Wikipedia.
  3. C. Hollings, U. Martin, A. Rice, The Lovelace–De Morgan mathematical correspondence: a critical re-appraisal, Historia Mathematica 44 (2017), 202–231.
  4. C. Hollings, U. Martin, A. Rice, The early mathematical education of Ada Lovelace, BSHM Bulletin 32 (2017), 221–234.
  5. C. Hollings, U. Martin, A. Rice, Ada Lovelace: The Making of a Computer Scientist, Bodleian Library, 2018.
  6. J. Fuegi and J. Francis, Lovelace & Babbage and the creation of the 1843 “Notes”, IEEE Annals of the History of Computing 25 (4) (2003), 16–26.
  7. A. G. Bromley, Charles Babbage’s Analytical Engine, 1838, Annals of the History of Computing 4 (1982), 196–217.
  8. D. Stein, Ada: A Life and a Legacy, MIT Press, 1985.
  9. B. A. Toole, Ada, the Enchantress of Numbers, Strawberry Press, 1992.
  10. S. Wolfram, Untangling the tale of Ada Lovelace, 2015.
  11. R. Rojas, How Charles Babbage invented the computer, arXiv:2311.04371.
  12. The mathematics of Ada Lovelace, Mathematical Institute, University of Oxford.