Helping Ada Lovelace with her Homework: Classroom Exercises from a Victorian Calculus Course – Problem 1

We begin with a very simple problem. For any student intending to study higher mathematics, the ability to manipulate and re-arrange algebraic expressions is essential. However, this is a skill that does not necessarily come easily to all students, and at first Ada Lovelace also found it difficult, as we saw in her questions to Mary Somerville. Her problems with symbolic manipulation are further evident in her correspondence with De Morgan, particularly early on in her studies. In one letter, from November 1840, she noted how difficult she found trying to verify a relatively simple point in algebra.

Figure 7. Extract from page 226 of De Morgan’s Elements of Algebra (2nd ed., 1837).

The problem arose from a discussion concerning logarithmic expressions in De Morgan’s Elements of Algebra. As can be seen from Figure 7, at the bottom of page 226 De Morgan made a simple substitution, letting

\[\frac{1+b}{1-b}=\frac{1+x}{x}\]

which he said resulted in

\[b=\frac{1}{2x+1}.\]

Lovelace had two problems with De Morgan’s claim [LB 170, 10 Nov. [1840], ff. 64r-64v]:

In the first place I do not feel satisfied that the form \(\frac{1+b}{1-b}\) is capable of being changed into the form \(\frac{1+x}{x}\). There are three suppositions we may make upon it, (supposing that it is capable of this second form).  \(x\) may be less than \(b\), in which case the denominator must also be less than \(1-b\), and less in a certain given proportion, in order that the Fractional Expression may remain the same.  \(x\) may \(=b\), in which case the second form can only be true on the supposition that \(1-b=x=b\), or \(b=\frac{1}{2}\).  \(x\) may be greater than \(b\), in which case the denominator of the second form must also be greater than \(1-b\), in a certain given proportion, in order that the Fractional expression may remain the same.

I will leave it to the reader to determine whether her three suppositions are correct. However, it is Lovelace’s second issue that was of more concern [LB 170, 10 Nov. [1840], ff. 64v-65r]:

But secondly supposing \(\frac{1+b}{1-b}\) to be under all circumstances susceptible of the form \(\frac{1+x}{x}\), I cannot deduce from this equation \(b=\frac{1}{2x+1}\).

In other words, given the equation \[\frac{1+b}{1-b}=\frac{1+x}{x}\] Lovelace was unable to re-arrange it to solve for \(b\). This was clear evidence that there were significant gaps in her mathematical knowledge and De Morgan was quick to warn her of the periodic digressions into more elementary mathematics that would be required to fill them: ‘You understand of course that your Differential Calculus must be delayed from time to time while you make up those points of Algebra and Trigonometry which you have left behind’ [LB 170, 27 Sept. 1840, f. 16r]. And in a subsequent letter, he explained how to solve the problem. You can see his solution here.

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