When Erwin Schrödinger devised his famous feline thought experiment, which he described as "quite ridiculous", he can't have imagined that it would give his surname a linguistic life of its own. His name popped into my head this morning as I struggled to describe a piece of maths that I both love and hate simultaneously. The pair of words
Schrödinger's Marmite sprang to mind. I was very pleased with my witty originality until I Googled the term and discovered that
Urban Dictionary had got there first. Of course they had, those
pony proud poseurs.
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| Schrödinger's Marmite - you either love it and you hate it |
Three weeks ago I started teaching at a new school in Wiltshire. Starting school is just as disorientating for new teachers as it is for new students. The environment is discombobulating; from one side of the maths block you can see a white chalk horse on a hill and, from the other, Europe's second-largest neolithic monument, also rumoured to be the resting place of Merlin (the wizard from Shrek 3).
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| Folklore provides indisputable proof that Merlin is buried yards from my classroom. |
While surrounded by a thousand unfamiliar faces, an unfamiliar lexicon, an extraordinary array of architecturally spectacular but geographically specious buildings, at least I could console myself with the fact that the maths remained the same.
Or so I thought.
I thought I'd lead my Hundreds (this means year 11 - no, I don't know why either) into the first Fixed Priv (a weekend when there are no lessons or activities and all the students have to leave) by solving some simultaneous equations where one is quadratic. A straightforward factorisation starter ended with me open-mouthed in bewildered awe at a technique that some members of the set unhesitatingly used to factorise a non-monic quadratic. They couldn't tell me how it works, just that it did, so I promised I'd look into it and blog about it.
Here's how the algorithm, if we might call it that, is used. Suppose we wished to factorise
Step 1: We multiply the constant term (-7) by the coefficient of the quadratic term (6).
Step 2: We consider how we'd factorise the resulting quadratic if it was monic. That is we factorise the expression
Determining that this factorises to
we, instead, write this down with the original quadratic coefficient multiplying each
x term.
Step 3: We see that each bracket contains a common factor (3 in the left and 2 in the right) and we divide the brackets by their respective common factors.
This, by some extraordinary miracle, is the correct factorisation of the original quadratic.
I hate it because none of the students could explain to me why on earth they were doing what they were doing. They computed the correct answer through a series of seemingly random mis-steps.
I love it because it works and there is a certain elegance to the reason why. It's a perfect example of
Schrödinger's Marmite.
Let's start an explanation from the rear. Suppose we were to expand the following brackets, which will represent the factorisation we are trying to reach:
We should note that the pairs A,B and C,D must be coprime.
Expanding the brackets give us the quadratic expression we are originally challenged by:
In
Step 1, we multiply the constant term,
BD, by
AC and then
Step 2 sees us consider the related monic quadratic
The factors of this are easy to spot. We want to find two numbers that add to give AD+BC and multiply to give ABCD. These are handed to us on a plate:
Assigning these to brackets as in Step 2, we reach:
Noting that the left-hand bracket contains a common factor of C and the right-hand bracket contains a common factor of A we perform Step 3 and reach our fully factorised destination via a lengthy detour of various pot-holed minor roads:
Perhaps it should be called the TomTom method because, in many ways, it is not a sensibly planned route but a hastily calculated compromise of the sort required when we get stuck in traffic.
But don't even get me started on SatNavs...
from matheminutes http://ift.tt/2hqyrPf