18 August 2022

The Differentiator

I have been very inspired by Leslie Dietiker's way of thinking about mathematics lessons as 'stories' (see Dietiker, 2015). In this post, I'm thinking more about how actual stories might be used within mathematics lessons (see https://www.mathsthroughstories.org/). I don't mean historical anecdotes about famous mathematicians (Gauss summing the positive integers, Galois fighting a duel, etc.). I am thinking of completely fictitious stories that get at some mathematical concept or idea. So, yes, I'm in 'holiday mode', but I'm hoping that this can be more than 'a bit of fun'. I'm going to give a calculus example suitable for ages 16-18, so as to counter the idea that stories are just for little children. But I'm hoping that you might be able to adapt the idea for any topic or age. I think something like this could be quite memorable but you would probably only want to do it occasionally...

In function-land, where all the elementary functions live, the polynomials were terrified. The Differentiator was stalking the land, striking fear into the hearts of all well-behaved functions. Poor $x^3$ had already been attacked, and, now as $3x^2$, had the scars to show it. Poor little $x$ was scared out of his mind – unlike $x^3$, he knew he had only two chances with The Differentiator before he would be reduced to nothing. But, it was really the constants who were most afraid - everyone knew that The Differentiator had had $\pi$ for lunch yesterday and not a crumb remained.

The Differentiator having $\pi$ for lunch

Everyone was looking to $x^{100}$ to help, and she was putting on as brave a face as she could. But even she knew that, sooner or later, her time would be up, and there was nothing anyone could do about it. The polynomials’ days were literally numbered.

The hopes of the entire community were pinned on their hero, $e^x$. They knew that $e^x$ could laugh in the face of The Differentiator: “Do your worst!” $e^x$ would say, and then stand back, completely unperturbed, as The Differentiator went into action. It was as though $e^x$ were inoculated against the terrors of The Differentiator.

Some of the polynomials, such as $x^2$, had decided to take refuge by hiding behind $e^x$, with some success:

$$\frac{d}{dx} \left( e^{x}x^2 \right) =e^x(x^2+2x).$$

or by climbing up onto $e$ for protection:

$$\frac{d}{dx} \left( e^{x^2} \right)=2xe^{x^2}.$$

But then, one fateful day, as The Differentiator was out prowling around, The Differentiator finally met their match, in the shape of $e^{2x}$. Unaware of what they were facing, The Differentiator attacked:

$$\frac{d}{dx} \left(e^{2x} \right) = 2e^{2x}.$$

Unbelievably, $e^{2x}$ immediately grew to twice the size! The Differentiator hit again:

$$\frac{d}{dx} \left(2e^{2x} \right) = 4e^{2x}.$$

And again, and again. But, the more times $e^{2x}$ was attacked, the stronger it became. Sixteen times its original height, $16e^{2x}$ stared down at The Differentiator, who – now defeated – turned on their heels and fled, and was never seen in function-land again.


Stories like this don't need to be lengthy - this one's a bit too long, I think. You could challenge students to make up their own - maybe a sequel, in which The Integrator comes to the rescue?

I have used this kind of story to lead in to thinking about 'differentiation-proof' functions, like $e^x$. At A-level, students meet the first-order differential equation:


and, with it, the idea of a differentiation-proof function.

“I’m thinking of a function, and when I differentiate it I get exactly the same function back. What might my function be?”

Starting with this prompt, students may suggest the zero function, $y=0$, and this is a trivial case of the general solution. They might think that $y=e^x$ is the only possibility for a general solution, but, of course, we need an arbitrary constant, and any constant multiple of this will also work, so the general solution is $y=Ae^x$, where $A$ is any constant, and the $A = 0$ case gives the zero function.

The general solution can be derived by separating the variables:


$$\int \frac{1}{y}\frac{dy}{dx}dx=\int 1dx$$

$$ \ln \lvert y \rvert=x+c$$


where $c$ and $A$ are constants.

We can differentiate $y=Ae^x$ as many times as we like, and it never changes, so it might seem that this is ‘the’ solution to the general equation $\frac{d^n y}{dx^n}=y$, where $n$ is any positive integer.

But, in fact it is only a solution, not the solution, because an $n$th-order differentiatial equation ought to have a general solution containing $n$ arbitrary constants, and $y=Ae^x$ contains only one arbitrary constant. Students of Further Mathematics may come across the second-order differential equation $\frac{d^2 y}{dx^2}=y$, which has general solution $y=Ae^x+Be^{-x}$, with two arbitrary constants, $A$ and $B$, this time. So, here is the general solution to finding a function which, when differentiated twice, returns to the original function. And note that this is not necessarily (unless $B= 0$) identical to the original function after just one differentiation.

It is natural for students to wonder about how this pattern might continue for the third-order differential equation $\frac{d^3 y}{dx^3}=y$. It is clear that $y=Ae^x$ will be one part of this, since we have seen that this satisfies any differential equation of the form $\frac{d^n y}{dx^n}=y$, where $n$ is a positive integer, since $Ae^x$ is differentiation-proof. But, now that we have a third-order differential equation, we should be expecting two more arbitrary constants, and how can we find them?

The term $Be^{-x}$ term has the wrong parity, since differentiating this three times gives $-Be^{-x}$, rather than $Be^{-x}$, and it seems like we have hit a brick wall. However, since terms involving $e^{kx}$ have served us very well so far, it may seem natural to use $y=e^{kx}$ as a trial solution in $\frac{d^n y}{dx^n}=y$. This gives us

$$k^n e^{kx}=e^{kx}$$ 

Since $e^{kx}$ is never zero, we require that $k^n=1$, so the $k$s that we need are the $n$th roots of unity.

When $n = 1$, $k = 1$, and we have $y=Ae^x$, as we found.

When $n = 2$, $k = \pm 1$, and we have $y=Ae^x+Be^{-x}$, as we also found.

Now that $n = 3$, we still have $k = 1$, but we also have two complex roots, $k = -\frac{1}{2} \pm \frac{\sqrt{3}}{2}i$, so our general solution should be

$$y=Ae^x+Be^{\left( -\frac{1}{2} + \frac{\sqrt{3}}{2}i \right) x} + Ce^{\left( -\frac{1}{2} - \frac{\sqrt{3}}{2}i \right) x},$$

which we can write as

$$y=Ae^x+B'e^{-\frac{x}{2}}\cos \left( \frac{\sqrt{3}x}{2} \right) + C'e^{-\frac{x}{2}}\sin \left( \frac{\sqrt{3}x}{2} \right),$$

or even, if we wish, as

$$y=Ae^x+B'\exp{\left( e^{ \frac{2 \pi i}{3} } x \right)}+C'\exp{\left( e^{- \frac{2 \pi i}{3} } x \right)}.$$

The $n=4$ case is much neater:


And we have explored the general terrain of 'differentiation-proof' functions!

Question to reflect on

1. What stories might you use in mathematics lessons to stimulate some worthwhile mathematical thinking?


Dietiker, L. (2015). Mathematical story: A metaphor for mathematics curriculum. Educational Studies in Mathematics, 90(3), 285-302. https://doi.org/10.1007/s10649-015-9627-x

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