komponisto comments on Inverse Speed - Less Wrong

14 Post author: komponisto 27 March 2011 05:57AM

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Comment author: komponisto 30 March 2011 03:44:47PM 2 points [-]

I wasn’t trying to make any awfully astute or insightful point at all, just disagreeing with your assessment that you cannot do this problem without an elaborate theoretical underpinning of what inverse speed means because you are (implied intrinsically) “bad at math”. ...You’re clearly pretty bright and capable of thinking in the abstract, though, so “bad at math” doesn’t ring true

Well, it shouldn't -- this is a case of deliberate irony, since the linked comments (as well as the post itself) imply that I have considerable background in mathematics.

This post has a "secret agenda" that goes far beyond figuring out how to solve this particular problem. You in fact put your finger on a significant part of it with this:

if, when faced with a problem you do not immediately know how to do, you stare off into space and go “hmmm,” the student quickly loses patience with you, but if you start writing and say “ok call this x and this y and here are some relationships we know must be true,” you look very competent

This is a failure mode. In fact, this is exactly what Yudkowsky is warning about in posts such as Grasping Slippery Things Yes, in the short term, it may get a student through a word problem, or at a higher level, get you a publication. But, in the long term, it's a bad habit, to the extent it prevents you from noticing your own confusion.

Now, maybe when you see it done this way the answer still feels wrong, or something, but there’s no reason why you couldn’t just set the thing up and solve it like I did. Is there? Am I still not getting where you’re coming from?

By way of clarifying where I'm coming from, I should emphasize that my interest is not in guessing teachers' passwords. Yes,of course I am capable of performing the algebra you demonstrated, but doing so would not leave me with the feeling that the answer is obvious. Contrast with the following problem:

If your speed for the first hour of a trip was 20 mph and you want your average speed for the whole two-hour trip to be 40 mph, what must your speed be for the second hour?

Now that problem I understand. It's not just that I can do the algebra involved; I can do it mentally. I probably wouldn't have been tricked by it if it came up in conversation. This is what it means, to me, to "be able to do" a math problem. This is what knowledge feels like, as opposed to confusion.

You're arguing that I can do without the "elaborate theoretical underpinning of what inverse speed means", but I don't want to do without that elaborate theoretical underpinning, because once I have that theoretical underpinning, the two problems become entirely symmetrical, and I can answer one just as easily as the other.

Comment author: [deleted] 30 March 2011 04:19:41PM 0 points [-]

I don't have a lot of time here because LeechBlock is about to cut me off, and I don't want to make a well-written reply anyway because karma is kind of a whore. But you did say you didn't have a hammer. And now you seem to be switching to "This hammer is bad! It breaks things!"

I agree btw that always brute-forcing problems could be a bad habit if you stopped there. Not infrequently, I pause afterwards and say, "Ok, that's the answer, but I think I just did that the stupid way. Gimme a sec and I'll see if I can tell you the smart way." But it's an adaptive habit in my line of work.

Comment author: komponisto 30 March 2011 04:47:25PM 0 points [-]

But you did say you didn't have a hammer. And now you seem to be switching to "This hammer is bad! It breaks things!"

No, that's not right. The problem is that it's not a hammer at all, it's a fake hammer. It lets you pretend you're driving in nails when you're really not. The hammer isn't too powerful ("breaking things"), it's not powerful enough. That might be okay if your only goal is to be seen "hammering", but if you actually want to hammer the nails in, you need a real hammer.

Comment author: wnoise 30 March 2011 05:27:28PM 2 points [-]

When you don't have the proper concepts, working out things with mere algebra lets you develop the concepts by focusing on constraints and other properties the concepts must have. Sure, it doesn't force you to develop the concepts, but if you're planning on doing so, it is extremely valuable for getting a grasp on this concept.

This is different than most slippery philosophical problems -- the math actually fights back in a revealing way.

Comment author: komponisto 31 March 2011 08:46:58PM *  0 points [-]

This is different than most slippery philosophical problems -- the math actually fights back in a revealing way.

I don't see any particular asymmetry, actually. (Which is no surprise when you realize that I consider mathematics to be rigorous philosophy.) Sometimes the way it fights back is (sufficiently) revealing, and sometimes it isn't.

There remain deeply mysterious unsolved problems in mathematics, for which merely fiddling with existing tools has not produced answers. The point of view I take (which is implicitly advocated by this post) is that whenever you have a problem that you can't solve, it's because your existing tools are inadequate, and you need to develop better tools. How does one develop better tools? Well, you can hope to discover them by accident in the course of analyzing the unsolved problem, or you can try to develop them systematically by figuring out how to better solve problems you are already able to solve. The latter is my preferred approach.

(I would also recommend this comment for context.)

Comment author: [deleted] 30 March 2011 07:13:28PM 0 points [-]

So wait, you’re saying algebra doesn’t work? Because it definitely does. It’s nice that it keeps my hands busy, but I also sincerely believe I’m helping my students by teaching them to approach problems this way. Algebra works equally well for the other problem you suggested:

If your speed for the first hour of a trip was 20 mph and you want your average speed for the whole two-hour trip to be 40 mph, what must your speed be for the second hour?

You just set it up with your unknown in the numerator instead of the denominator:

20 = d/1

40 = (d+x)/2

Or say another problem, one you might not recognize as similar if you’re stuck on speed and inverse speed: My dad is borrowing my mom’s hybrid car to drive to Miami. She’s very hung up on what the display says her mileage is, and she’ll get grumpy if it’s less than 50 mpg. My dad drives like a hyperactive child and gets 25 mpg on his way to Miami. What mileage will he need to get on his way back to make 50 mpg overall?

25 = d/g

50 = 2d/(g+x)

Solve the system for x, turns out he can’t use any gas on the way back. His only hope is to reset the mileage calculator and drive like a sane person on the way home. (Or drive around the neighborhood really efficiently several hundred times after he gets home and hope she doesn’t notice the extra miles.)

(Or convert the Insight into a plug-in in my cousin’s garage.)

What I’m saying is that algebra is a fully general tool for solving word problems, and you should embrace it. Thanks for reminding me to train my intuition too, though. And I’m sorry if I came off as patronizing.

Comment author: komponisto 31 March 2011 05:54:52PM *  1 point [-]

So wait, you’re saying algebra doesn’t work?

No -- I'm afraid you misunderstand. This is by no means a question of "algebra" versus "non-algebra": in order to solve such problems, algebra necessarily has to be performed in some manner. The point is that the calculations you present are too bare to serve the goal of dissolving confusion; they do not invoke the sort of higher-level concepts that are necessary for me to mentally store (and reproduce) them efficiently.

Let me explain. It is in fact ironic that you write

Or say another problem, one you might not recognize as similar if you’re stuck on speed and inverse speed

because "recognizing different things as similar" is exactly what higher-level concepts are for! As it turns out, I have no trouble recognizing the other problem as similar, because the difficulty is exactly the same: it asks about the inverse of the quantity that you need to think in terms of in order to solve it. That is, it asks about "mileage" when it should be asking about "gallonage".

The problem is not that "algebra" -- solving for an unknown variable -- is required. The problem is that the unknown variable is in the denominator, and that's confusing because it means that the quantity associated with the whole journey is not the sum (or average) of the quantities associated with the parts of the journey. That is, x miles per gallon on the first half does not combine with y miles per gallon on the second half to yield x+y (or even, say, (x+y)/2) miles per gallon for the whole trip. As a result, the correct equation to solve is not 25+x = 50, nor (25/2)+(x/2) = 50. Instead, it's something else entirely, namely (1/25)(1/2) + (1/x)(1/2) = 1/50, or equivalently 50x/(x+25) = 50.

Now, there are actually two ways of making this comprehensible to me. One would be the way I've been talking about, which is to switch from talking about miles per gallon to talking about its inverse, gallons per mile. That way, the quantities do combine properly: x gpm for the first half and y gpm for the second half yields (1/2)x+(1/2)y gpm for the whole trip. (This is how I was able to write down the equation above!) The other way would be to explicitly teach the rule for combining mileages on parts of a journey to obtain the mileage for the whole journey, which is that x mpg on the first half combines with y mpg on the second half to yield 2xy/(x+y) mpg for the total.

Now I think the first way is preferable, but the second way would also be tolerable, and it illustrates the distinction between an "intuitive" explanation and what I'm seeking. There's nothing particularly "intuitive" about the formula 2xy/(x+y); it's just something I would have to learn for the purposes of doing these problems. But it makes the algebra make sense. What I would do is regard this as a new arithmetic operation, and give it a name, say #, and I would learn x#y = 2xy/(x+y). Then, when you asked me

My dad drives like a hyperactive child and gets 25 mpg on his way to Miami. What mileage will he need to get on his way back to make 50 mpg overall?

I would set up the equation

25#x = 50

which automatically converts to

50x/(25+x) = 50

which I could then easily solve.

So do you see what I mean? The issue isn't algebra, it's having the right higher-level concepts to organize the algebra mentally. In this context, that means either thinking about the inverse of the quantity asked for (inverse speed instead of speed, or "gallonage" instead of mileage), or else learning a new operation of arithmetic (that is, asking "well, what's the general rule for combining speeds on segments of a journey?" or "what's the general rule for combining mileages on segments of a journey?").

Comment author: [deleted] 01 April 2011 12:29:15PM 0 points [-]

"recognizing different things as similar" is exactly what higher-level concepts are for!

I am fully on board with this. Higher-level concepts ftw. Of course, just setting up the algebra can help you recognize two problems as similar too. By “setting up the algebra” I mean “accepting the statements made in the word problem and translating them into math, introducing symbols to represent unknown quantities as necessary”.

As it turns out, I have no trouble recognizing the other problem as similar, because the difficulty is exactly the same: it asks about the inverse of the quantity that you need to think in terms of in order to solve it. That is, it asks about "mileage" when it should be asking about "gallonage".

Well yes, I expect you did recognize it, in context; I did everything I could to make the similarity explicit. And naturally, having recognized it as the same problem, you can then solve it by the same method you used for the first problem. But does having “inverse speed” in your arsenal really make it natural to approach this problem in the same way? I guess I’ll have to take your word for it.

Now, there are actually two ways of making this comprehensible to me. One would be the way I've been talking about, which is to switch from talking about miles per gallon to talking about its inverse, gallons per mile. That way, the quantities do combine properly: x gpm for the first half and y gpm for the second half yields (1/2)x+(1/2)y gpm for the whole trip. (This is how I was able to write down the equation above!)

Yeah, I get that you can do it that way. I agree that it’s kinda neat, even. It’s sort of like how you calculate the total resistance of resistors wired in parallel. The only thing I take issue with is that you cannot, or should not, do this problem without it.

The other way would be to explicitly teach the rule for combining mileages on parts of a journey to obtain the mileage for the whole journey, which is that x mpg on the first half combines with y mpg on the second half to yield 2xy/(x+y) mpg for the total…There's nothing particularly "intuitive" about the formula 2xy/(x+y); it's just something I would have to learn for the purposes of doing these problems. But it makes the algebra make sense. What I would do is regard this as a new arithmetic operation, and give it a name, say #, and I would learn x#y = 2xy/(x+y).

Ok, using formulas by rote totally is a failure mode, so I would be against that. (Btw on the subject of “intuitive explanations”, so we can avoid arguing about terms, I understand that to mean not “an explanation that appeals to the intuition you already have,” but “an explanation that fixes your intuition.” Explanations like that are fun and cool, but I don’t insist on having one before I do every new problem.)

So do you see what I mean? The issue isn't algebra, it's having the right higher-level concepts to organize the algebra mentally.

I guess I don't get why you can't just use "average rate of change = (total change in one variable)/(total change in other variable)" as your organizing concept, introduce symbols to represent the various quantities, and grind it out. That approach is certainly useful for much more than just finding average speeds or average mileages in this particular special case. You don’t need a new formula; you don’t even need a new concept. You can just use the ones you’ve already got.

Comment author: komponisto 01 April 2011 05:55:44PM 0 points [-]

But does having “inverse speed” in your arsenal really make it natural to approach this problem in the same way?

The idea in my arsenal is not just inverse speed, of course; it's inverses in general, and the fact that finding a rate may first require finding the inverse of that rate.

I recognized the two problems as similar not merely because the context implied that they were constructed to be similar (in point of fact the mileage one contained a large amount of distracting detail), but because they involve the same difficulty -- I "got stuck" at the same point, for the same reason: not knowing the relationship between the "partial" rates (those associated with the segments of the journey) and the "total" rate (that associated with the whole journey).

Ok, using formulas by rote totally is a failure mode

Not necessarily, but that isn't even the point here. I wouldn't actually need to memorize 2xy/(x+y), because I could easily derive it by taking inverses. (Indeed I wouldn't be satisfied until I understood the derivation.) The point is that the existence of a rule for combining mileages (or speeds) -- and indeed, the abstract concept of a binary operation other than the usual operations of arithmetic -- be acknowledged.

The only thing I take issue with is that you cannot, or should not, do this problem without it...I don't get why you can't just use "average rate of change = (total change in one variable)/(total change in other variable)" as your organizing concept...You don’t need a new formula; you don’t even need a new concept. You can just use the ones you’ve already got.

What you're missing is that the avoidance of new concepts is not a desideratum. If I can't figure out how to solve a problem, an explanation that uses only concepts I already had available will never be satisfactory, because there was some reason I couldn't figure it out, and such an explanation necessarily fails to address that reason.

In this case, a = b/c wasn't enough for me. That's just a fact. The speed problem crashed my brain until I came up with the concept of inverse speed. Now, in retrospect, now that I have a satisfactory understanding of these problems, I can go back and look at the solutions that just use a = b/c, translate them into my way of understanding, and come up with a way that I could have written down those same solutions that just use a = b/c while privately having a more sophisticated interpretation of what I was writing, so that I could appear to be doing the same thing as everyone else. But I wouldn't be doing the same thing as everyone else.

Now, when I say I couldn't have done it without this higher-level conceptual understanding, do I mean that literally, in the sense that no amount of mere fiddling with variables would have eventually allowed me to stumble upon the correct answer? Of course not. However, that wouldn't be satisfactory. For one thing, it would be difficult for me to do that quickly: if this had been some sort of two-minute test (or worse, a conversation among mathematically knowledgeable people, with status at stake, where you have only a few seconds), I would probably have been out of luck. But more importantly, I wouldn't "believe" the solution, or to put it differently, I wouldn't feel that I myself had solved it, but rather that "someone else" had and that I was taking their word for it. There would be a feeling of discomfort, of inadequacy. I would acutely sense that I was missing some insight. And, as it turns out, I would be entirely right! The insight I would be missing would be that described in this post and these comments, which is a legitimate and powerful insight that makes things clearer. I wouldn't want to do without it, even if I could manage to do so.

Comment author: [deleted] 01 April 2011 06:53:06PM 0 points [-]

Oh, I am perfectly happy for you to like your way better. It's a good way of doing the problem. But it seems like asking for trouble to say this:

If for any reason anyone is ever tempted to describe me as "good at math", I will invite them to reflect on the fact that an explicit understanding of the concept of "inverse speed" as described above (i.e. as a function that sends distances to times) was a necessary prerequisite for my being able to solve this problem, and then to consider that problems of this sort are customarily taught in middle- or high school, by middle- and high school teachers.

if you actually mean this:

But more importantly, I wouldn't "believe" the solution, or to put it differently, I wouldn't feel that I myself had solved it, but rather that "someone else" had and that I was taking their word for it. There would be a feeling of discomfort, of inadequacy. I would acutely sense that I was missing some insight. And, as it turns out, I would be entirely right! The insight I would be missing would be that described in this post and these comments, which is a legitimate and powerful insight that makes things clearer. I wouldn't want to do without it, even if I could manage to do so.

You know, they seem to be saying different things, to me.

And I don't agree that fiddling with variables is somehow cheating, or that it's a "fake hammer." It only gets easier to understand a problem once you know how to find the right answer.

If I can't figure out how to solve a problem, an explanation that uses only concepts I already had available will never be satisfactory, because there was some reason I couldn't figure it out, and such an explanation necessarily fails to address that reason.

Huh, is that actually true? Plenty of times I have failed to understand a problem just because I misinterpreted it-- you know, like holding the wrong variable constant? If I introduced a new concept every time, my picture of the world would get pretty baroque. It might be better to go in and knock down old, wrong concepts instead, or just clarify the ones I have, or remind myself that yes, they really do apply here.

Anyway. The idea of solving for a different variable (like you're doing here with your inverse rates, taking time as a function of distance instead of distance as a function of time) often seems to confuse my students. I wonder if it comes from introducing the concepts of dependent and independent variables too early, and teaching them how to do all kind of operations with functions, but only spending a couple of days talking about how to find the inverse of a function.

Comment author: komponisto 03 April 2011 08:28:53PM *  1 point [-]

it seems like asking for trouble to say this:...if you actually mean this:

I really don't think there's much of a difference. Yes, there was some rhetorical exaggeration in the first quote; to make it literally accurate I should have written about a necessary prerequisite for comfortably or reliably solving the problem. But since I take it for granted that a solid understanding is the goal, the difference between barely being able to solve it and not being able to solve it at all isn't particularly significant from my point of view.

There's a real psychological phenomenon having to do with difficulty that I'm pointing to here, something that goes beyond mere aesthetic preference (as important as that might also be, to me). I have a history of having trouble with problems of this type, at multiple points during my life. Believe me, I've been exposed to the standard explanations -- the three-row tables and so on. I've tried to learn them. I've even had sporadic success. But it just doesn't work reliably. There is some piece of knowledge or cognitive habit that these explanations presuppose that is always left implicit, but that I don't actually possess. To the extent that I can understand them, it is always by figuring out the solution in my own way and then "translating" -- a process which greatly increases the effort required. Failing that, I'm stumbling in the dark and guessing -- which sometimes works, and sometimes doesn't.

If I introduced a new concept every time, my picture of the world would get pretty baroque. It might be better to go in and knock down old, wrong concepts instead, or just clarify the ones I have, or remind myself that yes, they really do apply here

I think most people's picture of the world is probably not "baroque" enough. However, yes, it's always possible to overdo anything; the way you avoid having too many concepts is by constantly upgrading the ones you have -- making sure they're as general and powerful as possible.

This is my conception of mathematical research: "concept R&D." Whereas for most people, it's about solving problems (by any means, and then you're done once they're solved), for me, it's about developing concepts that make the solution easy.

The idea of solving for a different variable (like you're doing here with your inverse rates, taking time as a function of distance instead of distance as a function of time) often seems to confuse my students. I wonder if it comes from introducing the concepts of dependent and independent variables too early, and teaching them how to do all kind of operations with functions, but only spending a couple of days talking about how to find the inverse of a function.

My personal saying is bite the abstraction bullet early. My experience is that students have trouble with the very idea of a function, and will try to avoid actually learning it (and similarly abstract ideas) for as long as they can -- instead seeking lower-level "shortcuts" that allow them to get the right answer (sometimes). They need to be broken out of this as early as possible, lest they show up in a calculus class incapable of understanding the meaning of "f(x+h)", as all too often happens.

That's what I suspect is going on here: what's confusing them is that they're not used to thinking about independent and dependent variables at all: they think of d = rt as a rule for manipulating symbols, not as a representation of a relationship between (abstract) quantities. It's like confusing numbers with numerals. (You can ultimately think of mathematics as rules for maniplulating symbols, but in that case the rules are much more complicated than "d = rt"! That's their problem: they want the rules to be both concrete and simple, but they can't have it both ways!)

Comment author: [deleted] 04 April 2011 12:23:22AM 2 points [-]

There is some piece of knowledge or cognitive habit that these explanations presuppose that is always left implicit, but that I don't actually possess. To the extent that I can understand them, it is always by figuring out the solution in my own way and then "translating" -- a process which greatly increases the effort required. Failing that, I'm stumbling in the dark and guessing -- which sometimes works, and sometimes doesn't.

Hm, hell if I know. I can tell you that your talk about speed as a "mapping of times to distances" seemed downright weird to me-- I intuitively think of the relationship as being two-way, and to say that you're going faster means both that you'll go further in the same length of time and that you'll travel the same distance in a shorter period. If the price of chocolate bars goes down, I can buy the same number of chocolate bars for less, or I can buy more chocolate bars for the same amount of money.

Is it possible that you love functions too much? Not every situation has a natural choice of independent and dependent variables, after all. It's not any more meaningful to say that pressure depends on volume than that volume depends on pressure; pV just equals nRT.

It always fascinates me to find how different people are in the types of explanations that work for them. For example, all that "distracting detail" in my mileage story would be motivating detail for me, and for many of my students I've found that the more concrete I can make a problem, the more sense it makes to them.