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by schoen
2942 days ago
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Your review is a very interesting take. It reminds me of someone's observation (I forgot whose) that mathematics is often presented as a set of successful proofs and derivations, without an explanation of the motivation behind them or how they were discovered. So proof tactics may seem somewhat magical, even though they might in fact be a result of a mathematician's tinkering and blundering around, including alternative approaches that didn't work. An interesting example that someone gave from elementary mathematics is the derivation of the quadratic formula. ax²+bx+c=0 (a≠0) 4a(ax²+bx+c)=4a(0)=0 4a²x²+4abx+4ac=0 4a²x²+4abx+4ac+b²=b² 4a²x²+4abx+b²=b²-4ac (2ax+b)(2ax+b)=b²-4ac 2ax+b=±√(b²-4ac) 2ax=-b±√(b²-4ac) x=(-b±√(b²-4ac))/2a Someone discussing this pointed out that this derivation is easy to follow, but extraordinarily mysterious in terms of how we knew to do various things at various steps, such as mysteriously multiplying both sides by 4a or mysteriously adding b² to each side at some point. How did we know to do that? Of course there are different explanations of the underlying motivations and the history of how people discovered this proof, but it's easy to be given the proof without any of that context, and that kind of thing is in fact the rule rather than the exception in many parts of math study. Again, I think this was someone else's observation but I don't remember where I came across it. |
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A tangential point: this is indeed a horrible way to derive the formula, for the reasons you mentioned.
If anyone is curious, here's a better way to think about it. The graph of ax² + bx + c is just the graph of ax² translated. Keeping that in mind, let's investigate.
First, consider a very easy problem: find roots of ax² = 0. The graph intersects the x-axis at x=0, done.
Now, let's shift the whole graph down by Q, and solve the problem again. The equation for that graph is ax² - Q, and it intersects the x-axis at ±√(Q/a). Still easy.
Now, let's shift the whole graph again to the right by R. The equation for that new graph is a(x-R)² - Q.
What of the roots? Oh, we don't need to do much work here! The places where the graph intersects the x axis simply shifted to the right by R. So the roots are R±√(Q/a).
So, to recap: the roots of a(x-R)² - Q = 0 are R±√(Q/a).
What if our equation is written in the form ax² + bx + c? Well, now is the time for algebra. Open up the parentheses:
a(x-R)² - Q
=a(x² - 2xR + R²) - Q
= ax² + (-2aR)x + (aR² - Q)
= ax² + bx + c
Solve the following system for Q and R:
-2aR = b
aR² - Q = c
Obtain:
R = -b/2a
Q = b²/4a - c
Now plug these Q and R into the formula we already have: R±√(Q/a) - to obtain the all-familiar result
x=(-b±√(b²-4ac))/2a
What the formula is hiding is the simple idea that the roots of a parabola are easily found if you know where the vertex is. So assume you do, and work backwards from there.
A deeper idea is solving an easier version of the problem, and then changing the problem back to the more general original question, refining the solution on each step.
And this is, in fact, how mathematics is often done.
>and that kind of thing is in fact the rule rather than the exception in many parts of math study.
There's work done to change it[1]. Note that in the argument above, I could have left out all the "work", leaving only the questions, and many people would still be able to do the work. And with the right preparation, the student would be led to ask the same questions.
[1]https://en.wikipedia.org/wiki/Inquiry-based_learning