Question about Fourier series

Factual Questions
21

And all of this works just as well in any number of dimensions, of course. Consider ANY equivalence relation whatsoever on x coordinates, and ANY separate equivalence relation whatsoever on y coordinates, and ANY separate equivalence relation whatsoever on z coordinates, etc. If each small block (in some block decomposition of a big block) runs between equivalent coordinates along at least one axial direction, then so does the big one.

Incidentally, instead of enumerating all the different types of points like “T-junction”, etc., perhaps the simpler way to have framed the argument would have been to consider only regular grids of blocks first; then there’s only very straightforward kinds of points. Specifically, around each point, there are then 2^n many blocks treating it with alternating sign in their g calculation, where the specific dimension n corresponds to whether it’s on an outside corner (n = 0), on an outside edge (n = 1), outside face (n = 2), etc. In all cases other than being an outside corner, the point ends up cancelling out in the sum g of over all blocks in the grid, so that this is just the same as the sum of f over the outermost corners (i.e., the sum of g over all the small blocks equals g for the overall big block they comprise).

Then, to handle non-regular-grid decompositions of a big block into small blocks, we note that such decompositions can always be embedded in regular grids, as though various cells of such grid happened to have been fused together, and this will make no difference to anything because, again, the calculation of g over each fused block is just the same as the sum of g over the individual grid cells making it up. So even with non-regular-grids, we maintain the result that the sum of g over individual blocks equals g for the overall big block they comprise.

All of this could actually still be framed in terms of sine integrals and/or specifically derived from our result on integer-sided rectangles, but, that would be bizarre and backwards and obfuscating.

Ok, I’m done rambling for now. [I know, I know, I serial-reply to myself. Whatever, the edit window is too short. Just think of the end result as one very long post; disorganized, but with some good stuff in there.]

22

I seem to have hit a chord with you, Indistinguishable. You can obviously use any integrable function(s) whose integral over an interval vanishes iff the interval is a whole number multiple of the period.

23

Right; I was just excited to see how far the result generalizes (“for any equivalence relation on x coordinates, for any equivalence relation on y coordinates, if each rectangle…”, etc.), and how the argument could be presented purely combinatorially, without depending on any calculus knowledge. It is, as you noted, a beautiful argument.

24

Naturally, I’ve been anticipated: the third proof at http://www.inference.phy.cam.ac.uk/mackay/rectangles/ is well-written and the sort of thing I was getting at, a simple purely combinatorial approach manifestly yielding the generalization for arbitrary equivalence relations (their edges connecting points whose respective coordinates are equivalent).

[Technically, the full generalization can be recovered from just knowing the integer length result, using suitable distortings of arbitrary block diagrams so that the finitely many equivalence classes used in each direction were made to correspond to equivalence classes under integer distance separation, but this would hardly be the most natural way to present it…]

25

Sometimes I think we should turn on unlimited editing just for Indistinguishable.

26

I fully endorse this plan.

27

Nope. If we did that he’d still be editing last weeks’ post. :smiley:

I like to watch the progression of his ideas. It’s the ultimate in “show your work” problem solving.

Since much of his work is beyond me I really need the breadcrumbs along the way to help me ascend as far up the mountain as I can before my brain runs out of oxygen.

28

Gosh, I never saw/heard of Khan Academy in OP cite.

Amazing stuff. Thanks.

29

Really? They’re pretty much the Big Name in online education.

Though I suppose you’d have less exposure to them if you’re past school age yourself and don’t have any school age children.

30

For what it’s worth, I like the concept of online education and thus the concept of Khan Academy, but I don’t really like their actual videos; I feel like they end up all in the same non-conceptual, rote memorization and execution of algorithms territory as every traditional math curriculum (e.g., “memorize this random arbitrary rule for how to multiply matrices”, with no concern given to why this definition was chosen or would be useful for anything; nothing conceptual). Basically, I don’t like Sal Khan’s explanations of anything, or find them to break from classroom convention in any way.

They did briefly hire Vi Hart, who is wonderful and one of the few people who does pop math stuff who I actually enjoy, but she doesn’t work there anymore. Oh well.

31

No. (x-1)/(x-1)=1 in all cases except when x=1, where it is undefined. There’s no “if you let me” involved. It is always undefined at 1.

OP has a similar problem, but backwards. sin(mt) = sin(0) = 0 for all t when m=0. No need to do the fancy integration at all. It’s just 0. Nothing undefined about that.

32

shrugs. Everything in math is “if you let me”. The point OldGuy was making was perfectly valid. This kind of response is pedantry without substance.

Indeed, we may ask whether there is any continuous function which, when multiplied by x - 1, yields (x - 1)[sup]2[/sup]. Yes. There is. There is a unique one. There is a unique continuous function which when multiplied by x - 1 yields (x - 1)[sup]2[/sup], and we might well call this (x - 1)[sup]2[/sup]/(x - 1) (interpreting the division not merely as a point-wise division, but as a division of continuous functions), and note that (x - 1)[sup]2[/sup]/(x - 1) = x - 1.

33

Well, you’re the expert, but the OP is studying basic calculus. I fail to see how (x-1)/(x-1) is defined at x=1 unless we’re using nonstandard notation or definitions or something (which defining (x - 1)^2/(x - 1) = x + 1 is).

You can’t just cancel out part of the definition of the function when it is relevant. f(x)=x/x=1 for all x except 0. x/x does not equal 1 when x is zero. I don’t know how else to explain it. This is stuff students of basic calculus need to know. It’s not pedantry without substance, unless every professor in college (not to mention high school math teachers around the world) are substanceless pedants.

Maybe it means something to talk about techniques for removing singular points in higher levels, but not at the Calc 2 level of “how to integrate sin(mt)”.

For example, removable singularities:

But notice you have to redefine the function, as sinc(z) is actually undefined at z=0, unless you go ahead and redefine the function. Which might be helpful in many cases, but not Calc 2. When you turn in your homework and say “I redefined the function so it was easier to work with”, you will get zero points.

34

No, you don’t have to redefine division. You have to define it in the first place. Are you constructing a function whose result is the quotient of the results of two other functions, or are you dividing the functions themselves, and then taking the result of that quotient of functions?

35

Alas, many professors in college, not to mention high school math teachers around the world, in fact DO repeat various substance-less pedantic canards. To my constant chagrin.

As I said, we can interpret the / of (x - 1)[sup]2[/sup]/(x - 1) = x - 1 not as a point-wise division but as a division of continuous functions, and then there is no doubt that the answer is indeed x - 1 with no need for special “except at 1” disclaimer, since x - 1 is the unique such thing which, when multiplied by x - 1, yields (x - 1)[sup]2[/sup].

It’s true that plugging in x = 1 on the left-hand side will yield the indeterminate form 0/0 while the right-hand side will yield the specific value 0, but that is the nature of indeterminate forms: they arise from a form which can be re-expressed differently to give more information.

For some purposes, one will want to distinguish “(x - 1)[sup]2[/sup]/(x - 1)” from “x - 1”, sure, reflecting these slightly different forms, but for many purposes (most purposes, frankly!), one will not and need not.

36

Oh, somehow I missed that Chronos had already replied. Well, I agree with Chronos in his agreeing with me.

37

Oh, and…

It’s interesting you should mention that, because just last week I gave full credit to a couple of students (the only ones who got full credit on that assignment) for redefining a problem to make it easier to work with.

And on the topic of Khan Academy, the goal isn’t really to present the material in a more fun or intuitive or conceptual way. The point is to enable a “flipped classroom”: The students get basically the same lecture material that they would traditionally have gotten from a classroom, but on their own time, freeing up classroom time (where there’s a live teacher to provide help) to do the practice work which has traditionally been assigned for homework. While videos can provide the sort of engaging conceptual instruction that we both favor (as evidenced by Vi Hart), that sort of instruction is really best suited for an interactive environment.

38

Yeah, I guess I just don’t find “Here’s an arbitrary algorithm; memorize it and practice running it by hand” a good use of time for either a classroom or online, and had hoped, with all the fuss I’d seen about what a great explainer Sal Khan was, that the opportunity was being grabbed to do better. But the world is as it is.

39

You want to have real fun. Define 0^0 = 1 (it really is the right definition), but admit that the function x^y is discontinuous at (0,0).

I’m going to quite while I’m ahead.

40

Well, it’s usually the right definition. But the limit of x^y as x and y both approach 0 can be anything, if you approach on the right path (this is why the function is discontinuous, after all), and sometimes you really are in a situation that corresponds to one of those other paths.

For real fun, try to find the Fourier series approximation for exp(-1/x^2). That discontinuity isn’t as removable as it looks.