# Colin Foster's Mathematics Education Blog

## 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:

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

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:

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

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

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

$$y=e^{x+c}=Ae^x,$$

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:

$$y=Ae^x+B''e^{-x}+C''e^{ix}+D''e^{-ix}$$

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?

### Reference

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

## 04 August 2022

### Misremembering Goldbach’s Conjecture

It's the holiday, so a shorter, lighter blogpost today, and only one reflection question. I hope you are having a good break!

When I went on Craig Barton’s podcast for the first time (Note 1), he asked me (as he asks all his guests) to recount a ‘favourite failure’ - a situation where things didn't go to plan. I had so many to choose from that I had plenty of ideas leftover after the episode, so I thought I’d relate another one here…

This is about the time when I misremembered Goldbach’s Conjecture, which states:

Every even integer greater than 2 is the sum of two primes.

Unfortunately, for some reason, I misremembered it as:

Every integer greater than 2 is the sum of two primes.

If I had taken a moment to reflect on this, I would have realised that this obviously couldn’t be right, but it was one of those situations where I was distracted or ill or something (I can’t remember the specifics of my excuses!). And so I noticed nothing and carried on...

I wanted my Year 8 class (age 12-13) to work on something a bit exploratory and to understand the notion of a ‘counterexample’ – and also get in a bit of incidental practice on recognising prime numbers, which we had just been working on. So, this seemed like a good way to address all of that.

So, I told the class that Goldbach’s Conjecture was one of the best-known unsolved problems in all of mathematics, and I explained what a counterexample was. No one knows how to prove that Goldbach’s Conjecture is true, but, if it is false, all it needs is one counterexample to demonstrate that. A single counterexample can do a great deal of work!

The students seemed interested in this:

“Would we be famous if we found a counterexample?”
“Sure!”

There was immediately a bit of confusion about the number 3, which should have alerted me to the fact that something was wrong. Some pupils had written $3 = 1 + 2$, but others were – correctly – saying that 1 is not a prime number, in which case 3 would be a counterexample. I knew that 1 did used to be considered as a prime number (see Foster, 2016), so I thought perhaps this was just a historical glitch, so I decided that we would say “every integer greater than 3”, rather than 2, to avoid that problem.

And so the students began work. Of course, I knew very well that they would not find a counterexample, since all numbers at least as far as $4 \times 10^{18}$ have been checked (Note 1). If ever a teacher knew ‘the right answer’, I knew that the right answer here was that there would be no counterexample today!

The students began work on their own or in pairs, writing (at least, those working more systematically!) things like this:

$4 = 2 + 2$
$5 = 2 + 3$
$6 = 3 + 3$
$7 = 2 + 5$
$8 = 3 + 5$
$9 = 2 + 7$
$10 = 5 + 5$
$11 = …$

I walked around casually observing what was going on, my mind drifting a little, perhaps. I engaged in some discussion with students about who Goldbach was, why prime numbers matter, and so on, in quite a relaxed way. This was basically routine practice with primes in a more interesting context (a kind of mathematical etude, see Foster, 2018).

I gradually became aware that I could hear the number 11 being muttered quite a bit.

Then a few people started to say that they had found a counterexample, and it was 11. I decided that this would be a good opportunity to stop everyone and highlight the importance of ‘being systematic’. There's 'being systematic' in the sense of choosing your numbers according to some pattern, rather than haphazardly, but there's also 'being systematic' when you check each number. It’s easy to think you have found a number which can’t be made by summing two primes, and it may just be because you haven’t thoroughly checked all the possibilities. You haven’t managed to find a pair of primes that sum to 11, but that doesn’t mean that there isn’t one. The only way to be sure is to be systematic and check all the possibilities in such a way that you can be sure that you haven’t missed any. “Go back and check – be systematic – make sure you haven’t missed a possibility!” All good advice, to be sure.

I vividly remember the moment that one student came to the board and said, essentially:

Look, the only possibilities for 11 are:

1 and 10, but neither is prime
2 and 9, but 9 isn’t prime
3 and 8, but 8 isn’t prime
4 and 7, but 4 isn’t prime
5 and 6, but 6 isn’t prime
So, 11 is a counterexample.

Ordinarily, I would have been very pleased with such a proof by exhaustion. But, I remember staring at the board thinking, “What am I missing?” Even if we included 1 as prime, it would have to go with 10, which had certainly never been prime in anyone’s book!

As I tried to figure out what was going on, the class became more excited at my puzzlement:
“We’ve done it! We’ve solved this big maths problem – and it wasn’t even that hard!”, “Are we going to be on the news, sir?”, “Maybe no one ever bothered to check 11 because they assumed someone else had already checked it? Sometimes it’s the easy things that get missed!”, etc.

Obviously, if there were a counterexample, it was going to be considerably higher than 11. So, feeling desperate, I now Googled “Goldbach’s Conjecture”: “Every even integer greater than 2 is the sum of two primes.” ‘Even, even, even!’ (Having computers connected to the internet in every classroom has to be one of the great pedagogical advances of recent decades.)

Of course, with hindsight it is very clear that the only way to make an odd number by summing two integers is if one of the integers is odd and the other one is even. And the only even prime is 2. So, the only way my version of Goldbach’s Conjecture could be true is if every odd number were 2 greater than a prime. This is equivalent to saying that every odd number is prime, and although it is true that (almost, with the exception of 2), every prime number is odd, the reverse is not the case. This is why we had the problem with 3, because 1, which is $3-2$, is not prime. But then 5, 7 and 9 all have primes 2 less than them, so everything seems fine for a while, but then 11 doesn’t, because 9 is not prime, and that’s why it had appeared to be a counterexample. So, at least there was something mildly interesting to understand in relation to my mistake. Obviously, a counterexample to one conjecture is not necessarily a counterexample to a different conjecture.

There was understandably limited enthusiasm now for going back and checking for counterexamples to the real Goldbach Conjecture. It felt like the moment had passed, and perhaps the objectives of understanding what a counterexample is and gaining facility with primes had been accomplished more or less anyway.

I reflected afterwards on the strange feeling of seeing the student's apparently flawless proof and yet not believing it – the feeling that ‘there must be something wrong even though I can’t see what it is’. However rational we might aspire to be about mathematics, we are influenced by more than merely logical arguments. I was quite sure that the students must have made a mistake and omitted a possibility, and I was reluctant to believe even the very simple mathematics of their proof until I had appreciated the wrong assumption that I had begun the whole lesson with.

### Question to reflect on

1. Do you have any 'armchair responses' (AssocTeachersMaths, 2020; Foster, 2019) to my ‘favourite failure’ or to any of your own?

### Notes

1. You can listen to the episode here: http://www.mrbartonmaths.com/blog/colin-foster-mathematical-etudes-confidence-and-questioning/

### References

AssocTeachersMaths. (2020, July 13). Armchair Responses to Classroom Events - with Colin Foster [Video]. YouTube. https://youtu.be/L0ovhillL0c

Foster, C. (2016). Questions pupils ask: Why isn’t 1 a prime number? Mathematics in School, 45(3), 12–13. https://www.foster77.co.uk/Foster,%20Mathematics%20in%20School,%20Why%20isn't%201%20a%20prime%20number.pdf

Foster, C. (2018). Developing mathematical fluency: Comparing exercises and rich tasks. Educational Studies in Mathematics, 97(2), 121–141. https://doi.org/10.1007/s10649-017-9788-x

Foster, C. (2019). Armchair responses. Mathematics in School, 48(3), 26–27. https://www.foster77.co.uk/Foster,%20Mathematics%20in%20School,%20Armchair%20responses.pdf

### Teaching specific tactics for problem solving

This is my final blogpost, as my year as President of the Mathematical Association draws to a close, so I've allowed myself to go on at ...