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The big problem that people don't seem to understand is, what you were taught in maths class wasn't mathematics. We should rename the popular K12 class "calcula
by drostie 15y ago
The big problem that people don't seem to understand is, what you were taught in maths class wasn't mathematics. We should rename the popular K12 class "calculations", except for the one called "geometry", which I suppose should be called "formal geometry" or "Euclid class" or something like that.
The issue is the scope of the course. I know plenty of professional mathematicians. None of them can calculate a tip. The way to excite their mathematical minds is to drag out the Cosmic Wimpout dice, play it with them until they understand the rules, and then ask them, "so, it's obvious that strategy doesn't matter much, but what's the optimal strategy?" Bam, they're writing Markov diagrams and fussing over the fact that one of the dice has a wildcard which you can technically choose to be anything in certain rolls, so which one should it be, and so forth.
The most accessible and iconic piece of plain mathematics that I know is the Fundamental Theorem of Calculus. In Math class in high school, you may have been taught this by saying "it's very complicated and rests on some very difficult theorems and trust us, the mathematicians have proved it. But no. That's not mathematics.
I occasionally get the opportunity to talk to high school students and college freshmen about calculus, not in any formal setting, but just one-to-one. I tell them that the fundamental theorem of calculus is about + and −. The − is confusingly called a "derivative" and the + is confusingly called an "integral", but the fundamental theorem just says: + undoes −, and − undoes +, and that's it.
They look at me funny. Suppose we start defining sequences (F₀ F₁ F₂ ...), and suppose we start defining ΔF as a new sequence, Fᵢ − Fᵢ₋₁ . And let ΣF be a new sequence F₀ + F₁ + ... + Fᵢ .
Then it's not too hard to see that Δ(ΣF) = F, in other words, − undoes +. And it's not too hard to see also that Σ(ΔF) = F, because if you look at it:
[Σ(ΔF)]ᵢ = Fᵢ − Fᵢ₋₁ + Fᵢ₋₁ − Fᵢ₋₂ + ... + F₁ − F₀ + F₀.
And those cancel out by pairs into just Fᵢ.
Σ of course forms the basis for a Riemann sum and Δ forms the basis for a difference quotient, and we take limits because the product rule says:
δ(fg) = (f + δf)(g + δg) − fg = g δf + f δg + δf δg
...and we want to say δf δg → 0 to keep the algebra pretty, because if δf and δg are one in a million, g δf and f δg are a million times larger than δf δg, and so on until δf δg is negligible by comparison.
I kind of lost my point, so let me return to it and say that being able to see ∫dx as + and d/dx as −, being able to feel these patterns, that is actual mathematics. Being able to write out the matrix transform of complex multiplication and just seeing, "oh wow, that's a scaled rotation matrix!" and the implication -- "wow, complex-differentiable functions must preserve angles!" -- that sort of thing is mathematics.
I think it would be nice if we taught this sort of thing in a mathematics class, but in the meantime, I'd prefer if we just insisted that "math class" doesn't teach mathematics until you get toward college.
- jjtheblunt 15y agoI'd opine that when one first encounters mathematics which is not just rote calculation depends, at least, upon whom one has for a teacher. It can happen long before college that a teacher drives students towards the discoveries you allude to.
- mturmon 15y agoYes. A teacher with talent can steer even an elementary school student to "real math" insights. Something like Pascal's triangle is very interesting and contains a lot of patterns, relations, and theorems, which open out very wide and deep in the right hands. Without contortions, you can go all the way to the rudiments of information theory if you want to.
- JadeNB 15y ago> Then it's not too hard to see that Δ(ΣF) = F, in other words, − undoes +. And it's not too hard to see also that Σ(ΔF) = F, because if you look at it …. Note that your derivation assumes that F₀ = 0; but that, on the other hand, we can't assume this blanket condition, since it's not preserved by Δ (unless we restrict further to sequences with F₀ = F₁ = 0, etc., until eventually you wind up considering only the 0 sequence, which barely interests even mathematicians). One is thus forced into the definition of the indefinite integral (which I suppose we would call the indefinite summation in this discretised view? :-) ), which, in our context, says that Σ(ΔF) = F + F₀. (None of this is to take away from your very fine main point, or even your equally fine smaller point; but it's a shame to come so close to a rigorous-and-yet-intelligible picture, and then to fall just shy!) (Perhaps you were implicitly identifying sequences that differ by a constant.)
- drostie 15y agoSorry that it took me two days to get back to this; if you're not reading the comment threads, that's okay. I did indeed forget the one statement which makes this rigorous (and nontrivial): technically this definition for [ΔF]ᵢ = Fᵢ − Fᵢ₋₁ is undefined for [ΔF]₀ , because our sequences don't have a -1'th element. For these we take [ΔF]₀ = F₀, which in some sense means that F₋₁ = 0. I don't know why you want to require that "F₀ = F₁ = 0, etc." -- perhaps you were, in the back of your mind, iterating through the negative numbers, thinking that they were positive. ^_^;;