Helping Ada Lovelace with her Homework: Classroom Exercises from a Victorian Calculus Course – Problem 6

Adrian Rice (Randolph-Macon College)

Another indispensable result in calculus is the Mean Value Theorem, which states that if a function \(f\) is continuous on a closed interval \([a,b]\) and differentiable on \((a,b)\), then there exists a value \(c\) in \((a,b)\) such that

\[\frac{f(b)-f(a)}{b-a}=f'(c).\] In his Differential and Integral Calculus, De Morgan stated it in the following form [De Morgan 1836–42, 67]:

\(\ldots\) between \(a\) and \(a+h\) \(\ldots\) it follows that \[\frac{\phi(a+h)-\phi(a)}{h}=\phi'(a+\theta h)\] is true for some positive value of \(\theta\) less than unity.

Recognizing that \(a+h=b\) and \(a+\theta h=c\in(a,b)\), since \(0<\theta<1\), it is clear that De Morgan’s statement is basically equivalent to the modern-day formulation. While Lovelace had no problem with his statement of the theorem itself, she was puzzled by his assumption that \(\theta\) was a function of \(a\) and \(h\), writing: ‘I see neither the truth of this assertion, nor do I perceive the importance of it (supposing it is true) to the rest of the argument’ [LB 170, 19 Feb. [1841], f. 100r].

De Morgan replied by asking her, rhetorically: ‘Why should \(\theta\) be independent of \(a\) and \(h\) [since] we have never proved it to be so’? [LB 170, [22 Feb. 1841], f. 42v] To demonstrate that \(\theta\) could be expressed as a function of the two values as he claimed, he let \(\psi\) be the inverse function of \(\phi'\) so that \(\psi(\phi'(x))=x\). Then, he wrote:

\frac{\phi(a+h)-\phi(a)}{h}=\phi'(a+\theta h)\\ \\
\psi \left (\frac{\phi(a+h)-\phi(a)}{h} \right )=\psi(\phi'(a+\theta h)))=a+\theta h\\ \\
\theta =\frac{ \psi \left (\frac{\phi(a+h)-\phi(a)}{h} \right ) - a}{h} \left \{\mbox{Say that this is not a function of \(a\) and \(h\), if you dare} \right .\end{array}\]

Convincing though this demonstration may seem, it turns out that Lovelace’s doubts about the validity of De Morgan’s assertion were well founded. In fact, the professor was actually incorrect: \(\theta\) is not necessarily a function of \(a\) and \(h\), and the above proof is wrong. The question is: why?

Return to Main Problems Page.
Continue to Problem 7.