Seems at least partially related to cognitive apprenticeship, a type of teaching that aims at explicitly making the teacher’s “thinking visible”, so that the pupils can find out what it is that the teacher is tracking at any moment when solving the problem. For instance, they might carry out an assignment in front of pupils and try to explicitly speak out loud their thoughts while doing it.
For instance, when writing an essay:
Assignment
(Suggested by students)
Write an essay on the topic “Today’s Rock Stars Are More Talented than Musicians of Long Ago.”
THINKING-ALOUD EXCERPT
I don’t know a thing about modern rock stars. I can’t think of the name of even one rock star. How about, David Bowie or Mick Jagger… But many readers won’t agree that they are modern rock stars. I think they’re both as old as I am. Let’s see, my own feelings about this are… that I doubt if today’s rock stars are more talented than ever. Anyhow, how would I know? I can’t argue this… I need a new idea… An important point I haven’t considered yet is… ah… well… what do we mean by talent? Am I talking about musical talent or ability to entertain—to do acrobatics? Hey, I may have a way into this topic. I could develop this idea by…
or doing math:
A MATHEMATICIAN THINKS OUT LOUD
(from Schoenfeld, 1983)
Problem
Let P(x) and Q(x) be two polynomials with “reversed” coefficients:
P(x)=anxn+an−1xn−1+...+a2x2a1x+a0
Q(x)=a0xn+alxn−1+...+an−2x2an−1x+an,
where an ≠ 0 ≠ a0. What is the relationship between the roots of P(x) and those of Q(x)? Prove your answer.
Expert Model
What do you do when you face a problem like this? I have no general procedure for finding the roots of a polynomial, much less for comparing the roots of two of them. Probably the best thing to do for the time being is to look at some simple examples and hope I can develop some intuition from them. Instead of looking at a pair of arbitrary polynomials, maybe I should look at a pair of quadratics: at least I can solve those. So, what happens if
P(x)=ax2+bx+c
and
Q(x)=cx2+bx+a?
The roots are
(−b±√(b2−4ac))/2a
and
(−b±√(b2−4ac))/2c
respectively.
That’s certainly suggestive, because they have the same numerator, but I don’t really see anything that I can push or that’ll generalize. I’ll give this a minute or two, but I may have to try something else...
Well, just for the record, let me look at the linear case. If P(x) = ax + b and Q(x) = bx + a, the roots are –b/a and –a/b respectively.
They’re reciprocals, but that’s not too interesting in itself. Let me go back to quadratics. I still don’t have much of a feel for what’s going on. I’ll do a couple of easy examples, and look for some sort of a pattern. The clever thing to do may be to pick polynomials I can factor; that way it’ll be easy to keep track of the roots. All right, how about something easy like (x + 2)(x + 3)?
Then P(x) = x2+ 5x + 6, with roots −2 and −3. So, Q(x) = 6x2 + 5x + 1 = (2x + 1)(3x + 1), with roots − 1⁄2 and −1/3.
Those are reciprocals too. Now that’s interesting.
How about P(x) = (3x + 5)(2x − 7) = 6x2 − 11x − 35? Its roots are −5/3 and 7⁄2; Q(x) = −35x2 − 11x + 6 = -(35x2 + 11x − 6) = -(7x − 2)(5x + 3).
All right, the roots are 2⁄7 and −3/5. They’re reciprocals again, and this time it can’t be an accident. Better yet, look at the factors: they’re reversed! What about
Aha! It works again, and I think this will generalize…
At this point there are two ways to go. I hypothesize that the roots of P(x) are the reciprocals of the roots of Q(x), in general. (If I’m not yet sure, I should try a factorable cubic or two.) Now, I can try to generalize the argument above, but it’s not all that straightforward; not every polynomial can be factored, and keeping track of the coefficients may not be that easy. It may be worth stopping, re-phrasing my conjecture, and trying it from scratch:
Let P(x) and Q(x) be two polynomials with “reversed” coefficients. Prove that the roots of P(x) and Q(x) are reciprocals.
All right, let’s take a look at what the problem asks for. What does it mean for some number, say r, to be a root of P(x)? It means that P(r) = 0. Now the conjecture says that the reciprocal of r is supposed to be a root to Q(x). That says that Q(1/r) = 0. Strange. Let me go back to the quadratic case, and see what happens.
Let P(x) = ax2 + bx + c, and Q(x) = cx2 + bx + a. If r is a root of P(x), then P(r) = ar2 + br + c = 0. Now what does Q(1/r) look like?
So it works, and this argument will generalize. Now I can write up a proof.
Proof:
Let r be a root of P(x), so that P(r) = 0. Observe that r ≠ 0, since a0 ≠ 0. Further, Q(1/r) = a0(1/r)n + a1(1/4)n-1+… +an-2(1/r) + an = (1/rn)(a0 + a1r + a2r2 +… + an-2rn-2 + an-1rn-1 + anrn) – (1/rn)P(r) = 0, so that (1/r) is a root of Q(x).
Conversely, if S is a root of Q(x), we see that P(1/S) = O. Q.E.D.
All right, now it’s time for a postmortem. Observe that the proof, like a classical mathematical argument, is quite terse and presents the results of a thought process. But where did the inspiration for the proof come from? If you go back over the way that the argument evolved, you’ll see there were two major breakthroughs.
The first had to do with understanding the problem, with getting a feel for it. The problem statement, in its full generality, offered little in the way of assistance. What we did was to examine special cases in order to look for a pattern. More specifically, our first attempt at special cases—looking at the quadratic formula—didn’t provide much insight. We had to get even more specific, as follows: Look at a series of straightforward examples that are easy to calculate, in order to see if some sort of pattern emerges. With luck, you might be able to generalize the pattern. In this case, we were looking for roots of polynomials, so we chose easily factorable ones. Obviously, different circumstances will lead to different choices. But that strategy allowed us to make a conjecture.
The second breakthrough came after we made the conjecture. Although we had some idea of why it ought to be true, the argument looked messy, and we stopped to reconsider for a while. What we did at that point was important, and is often overlooked: We went back to the conditions of the problem, explored them, and looked for tangible connections between them and the results we wanted. Questions like “what does it mean for r to be a root of P(x)?”, “what does the reciprocal of r look like?” and “what does it mean for (1/r) to be a root of Q(x)?” may seem almost trivial in isolation, but they focused our attention on the very things that gave us a solution.
Seems at least partially related to cognitive apprenticeship, a type of teaching that aims at explicitly making the teacher’s “thinking visible”, so that the pupils can find out what it is that the teacher is tracking at any moment when solving the problem. For instance, they might carry out an assignment in front of pupils and try to explicitly speak out loud their thoughts while doing it.
For instance, when writing an essay:
or doing math:
I tried to follow my own thoughts on the polynomial example. They were pretty brief; the whole problem took only a few seconds. Basically:
Whelp, roots are a PITA
Can I transform Q(x) in a way which swaps the order of the coefficients?
Pattern match: yup! I’ve done that before. Divide by x^n.
Oh I see, roots of one will be reciprocals of roots of the other.
… so I guess +1 point for the “bag of tricks” model of expertise.
There are a lot of 1⁄4 instead of 1/r in the formulas (I guess you wrote some of them with 1⁄4 initially but then replaced but overlooked some).