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

Author(s):

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. , 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?