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komponisto comments on Inverse Speed - Less Wrong
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Comments (56)
I can't quite tell if your comment is analogous to "move your arms and legs!" or if instead you're just agreeing with what I said here (point 2), and mistaking the nature of my difficulty.
In any case, "wanting people to explain things to me intuitively" is not at all how I would put it. In my experience, asking for an "intuitive" explanation will often produce the very opposite kind of explanation from the kind I want: concrete and ad-hoc instead of abstract and general.
If I can't figure something out, it's because there's a concept that I'm missing, and what I want is to be told about that concept. The concept doesn't have to be "intuitive", it just has to be generalizable and (preferably) concisely expressible.
However, most people's approach to explanation is antithetical to this. They try their best to avoid introducing new concepts, tortuously beating around the bush as a result. It's not that I can't follow their explanations, but rather that I can't properly generalize from them, because they're not telling me the "real story".
In terms of your metaphor: my difficulties arise from lacking a hammer, not from being unwilling to smash.
My understanding has been that if you know how to set up a problem as a formula then you know how to make it out of toothpicks and rubber bands so you understand it.
This requires a certain ability to manipulate formulas:
2xy/(x+y)=2/(1/x+1/y)=1/(((1/x)+(1/y))/2)
as I'm sure you know.
(but I am 1-in-a-million atypical on this)
So like, the formula for concentrations goes like this:
If we add 1volume of xconcentration to 2volume of yconcentration then:
we are adding 1volume of stuff to 2volume of stuff and getting 1volume+2volume of stuff.
We are adding xconcentration x 1volume to yconcentration x 2volume of special stuff and getting xconcentration x 1volume+yconcentration x 2volume of stuff
so our final concentration is the amount of special stuff divided by the amount of stuff, or
(xconcetration x 1volume+ yconcentration x 2volume)/(1volume+2volume)=zconcentration
So, I mean, that's all there is. That's the formula, that's how the problem works, that's all that's at issue.
It's equivalent to the problem:
You spend 2volume time at yconcentration speed. How much time must you spend at xconcentration speed to attain an average speed of zconcentration?
Sorry, "move your arms and legs" was certainly not my intent. 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”.
I can think of a couple of reasons that I find more plausible: you didn’t realize that “average speed” means “total distance/total time”, or you don’t have enough experience with word problems involving rates, or in general, to know where to start setting this up algebraically. Could be other reasons too. You’re clearly pretty bright and capable of thinking in the abstract, though, so “bad at math” doesn’t ring true, and anyway it isn’t specific enough to do anything about it.
I actually do identify with your desire for highly abstract and general explanations, and I used to think that way to a fault, but for whatever reason I’ve gotten more concrete and practical over time. It probably isn’t only the tutoring, but that’s certainly contributed: 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, and half the time you realize you can just solve by substitution or something.
So, for the sake of argument, consider this explanation:
Speed is distance/time. Average speed is total distance/total time. We have numbers for speeds, but no numbers for distances or times. So call the distance traveled in the “first half” d, and the time it took to travel that distance t, and set this up:
20 = d/t
from which we get
20t = d
And then we write another equation, with the average speed of 40, we’ll want to set 40 equal to total distance/total time. We know that the total distance is twice the first half, so that’s 2d, but we don’t know the total time, except it must be some number added to the time for the first half. So we’ll call however long the second half took x, and the total time t + x.
40 = 2d/(t+x)
sub in 20t for d
40 = 2(20t)/(t+x)
Multiply both sides by (t+x) and distribute
40t+40x = 40t
x = 0
No time left for the second half of the trip. Speed = d/0 = undefined.
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?
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:
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.
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.
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.
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.
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.
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.)
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:
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.
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
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
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?").
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”.
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.
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.
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.)
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.
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).
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.
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.
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 you actually mean this:
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.
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.