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

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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:

$\begin{array}{l} \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?

Adrian Rice (Randolph-Macon College), "Helping Ada Lovelace with her Homework: Classroom Exercises from a Victorian Calculus Course – Problem 6," Convergence (September 2021)