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2011 May 15: Major updates on account of useful critical feedback in the comments.

Let’s see how well the storytelling framework holds up.

The Goods

Download the full archive [5.5 MB].

Act One

Play the question video.

[anyqs] Stacking Dolls – Question from Dan Meyer on Vimeo.

Ask your students what question interests them about it. Take some time here. This is the moment where we develop a shared understanding of the context. If a student has some miscellaneous question to ask or information to share about the dolls, encourage it. That isn’t off-task behavior. This task requires that behavior.

Then ask them to write down a guess at how many Russian dolls they think there are. Ask them to write down a number they think is too high and too low.

Act Two

Offer your students these resources:

  1. The first two dolls side-by-side.
  2. The second two dolls side-by-side.

After you show them the first set of two dolls, ask them how big they predict the third will be. As one of the commenters mentioned, they need to discover the fact that these guys aren’t decreasing by a fixed amount every time, that a new model is necessary.

Once they have this new model in mind, they’ll keep applying it until they reach a doll height they think is impossibly small.

Act Three

That task isn’t going to win anybody a Fields medal. As students finish, ratchet up the demand of the task with this sequel. Say:

I need you to design me a doll that’s as tall as the Empire State Building and is made up of 100 dolls total. Tell me everything you know about that doll.

Ask them to generalize. Ask them to graph.

Host a summary discussion of the activity. At this point you’ve identified different solution strategies around the room. Have those students explain and justify their work to their peers. Everyone is accountable for understanding everyone else’s strategy.

Then show them the answer video:

[anyqs] Stacking Dolls – Answer from Dan Meyer on Vimeo.

Find out whose guess was closest.

[h/t @baevmilena who gave me the idea when I met her in Doha.]

42 Responses to “[WCYDWT] Russian Stacking Dolls”

  1. on 13 May 2011 at 1:44 pmJohn Burk

    Dan,
    I really like this lesson, but I want to pick an important nit with you. Did you really measure the size of this doll to +/- 0.01 mm? If so, how did you do this? With a pair of 10 micron calipers and a microscope?

    I raise this because as a science teacher, I get infuriated when students tell me that an object that travels 20 meters in 30s is going at 0.666666666 m/s or even worse, 0.6 repeating, when all they did was measure the distance using the marks on a football field and the time using the timer on their iphone.

    I know teaching measurement theory isn’t the point of math class, but if we’re going to bring the real world into problems for our kids, the measurements we give them (or have them make) need to be appropriate, right?

    I think one element of pseudocontext must also be the made up measurements with fantastical precision that most textbooks hand to students without much explanation.

  2. on 13 May 2011 at 2:47 pmDan Meyer
    John: If so, how did you do this? With a pair of 10 micron calipers and a microscope?

    Digital calipers, actually, obviating the microscope.

  3. on 13 May 2011 at 4:28 pmThe Virtuosi

    I have to say Dan, I think this problem as you’ve set it up is a lot more closed than I;m used to from you. Once you give all three pieces of information, I don’t see how this is any different than a traditional textbook problem. A b^n = C

    I’d vote that you just have the starting video and let them have at it, have them estimate the size of the dolls just from the video, or put a dollar bill or something standing in the background.

  4. on 13 May 2011 at 4:38 pmDan Meyer
    TV: Once you give all three pieces of information, I don’t see how this is any different than a traditional textbook problem. A b^n = C

    I mean, let’s not go overboard here. Let’s A/B test this with another class that solves this problem out of the textbook:

    The first doll in a Russian stacking doll is 62.82 mm tall. The last is 1.26 mm tall and each new doll is 75.6% the height of the last. How many dolls are there?

    I’m guessing we’ll find a pronounced difference in engagement, if not achievement also.

    Which isn’t to discount your larger point that this is more closed than usual. (The extension is where the math gets rich.) I’m not exactly sure how to open up the exponential, though. Your suggestion adds estimation to the mix of skills required to solve the problem but doesn’t do anything to enrich the exponentiation. Maybe you can elaborate, though. I may be misreading you.

  5. on 13 May 2011 at 4:46 pmThe Virtuosi

    I agree that adding a dollar bill to the mix would only add to estimation (but let’s agree that estimation is important too), but I guess I’m worried that by offering those three particular clues, and only those three clues, you are already suggesting fairly strongly an approach to the problem, giving the average shrink percentage in particular hints strongly at exponentiation. I thought part of the value of ‘be less helpful’ was that it requires the students themselves to formulate the approaches. By giving the average shrink rate, you’ve already short circuited a whole dialog about what sort of governing number would be important for figuring the problem out. A thought process I think could reveal how well the student’s /think/ in terms of exponentiation.

  6. on 13 May 2011 at 4:50 pmScott

    Dan, I love this, but I have to say…

    Resource #2 is all wrong! The 12 second mark on the Act One video should be resource #2. Measurements can be taken or given on the slide. The percent can be calculated or given on the slide. But… it’s gotta be a comparison of two dolls. You can’t just throw that % out there on a essentially blank slide.

    Excuse my crude MS Paint, but I am on a photoshopless computer right now :) Here’s a mockup of what I mean: http://scottfarrar.com/algebra/dan%20doll%20slide%202.png

    Now we can SEE the geometric sequence. I would probably like to have another adjacent pair ready to go so that we can be more confident that each ratio is 75.6%.

    However, what do you mean “on average” ? That sets off some alarms in my head. Is it because of the base on the outermost one?

    On another note, I’d do without resource #3. The geometric sequence of heights will continue forever. But when do they stop becoming physically possible? When do they stop becoming *easily* possible? And when is the actual number that we’re going to stop at? (And how big should it be?)

  7. on 13 May 2011 at 4:52 pmDan Meyer
    TV: I thought part of the value of ‘be less helpful’ was that it requires the students themselves to formulate the approaches. By giving the average shrink rate, you’ve already short circuited a whole dialog about what sort of governing number would be important for figuring the problem out.

    You have me dead to rights there. Usually, in act two, I’ll ask students what information they need to know to solve the problem. That’s the “less helpful” aspect. Here, though, what are the students supposed to answer? The only measurable dimension is the height of the first doll, which isn’t sufficient. Any advice?

  8. on 13 May 2011 at 4:57 pmDan Meyer
    Scott: Resource #2 is all wrong! The 12 second mark on the Act One video should be resource #2. Measurements can be taken or given on the slide. The percent can be calculated or given on the slide. But… it’s gotta be a comparison of two dolls. You can’t just throw that % out there on a essentially blank slide.

    You people are relentless jackals and I wouldn’t have it any other way. This is right on the money, Scott. Before I fix it, though, what do you think about me just providing the two heights instead of the percent? Let them calculate the percent?

    Scott: However, what do you mean “on average” ? That sets off some alarms in my head. Is it because of the base on the outermost one?

    75% means (hypothetically) doll #2 is 80% the height of doll #1 while doll #3 is 70% the height of doll #2.

    Scott: On another note, I’d do without resource #3.

    This is interesting. You see the smallest doll and it’s the kind of “WTF?!” moment that usually belongs in the third act, not the second. You might be right. I need to think on it.

  9. on 13 May 2011 at 4:58 pmScott

    re: post [7]

    Measure directly off the screen. Yardstick on the ol’ projection screen, or fancy pixel measurements on a smartboard. Reshoot the video if necessary so that we are sure the height is not distorted by perspective (too much).

    Maybe that’s too far removed from real measurement, since we make assumptions about the camera, the angle, and the projection.

  10. on 13 May 2011 at 5:01 pmChristopher Danielson

    So John, how do you feel about 2/3 meters per second?

  11. on 13 May 2011 at 5:01 pmScott

    > what do you think about me just providing the two heights instead of the percent? Let them calculate the percent?

    Both slides could be prepared, and adapted to the class. My last post talked about the idea above about how to measure.

    > 75% means (hypothetically) doll #2 is 80% the height of doll #1 while doll #3 is 70% the height of doll #2.

    I guess I assumed they were similar under the same ratio. Are they not? If not, can we really answer anything about their progression and how many there are?

  12. on 13 May 2011 at 5:12 pmChristopher Danielson

    Here’s my storyboard, Dan…
    1. Question video as is.
    2. Largest, second-largest and smallest side-by-side
    3. Measurements of all displayed as in your Act 2 photo.

    Now we go to town.

    It’s messy data, since we don’t have an average scale factor to the nearest thousandth.

    As I discuss with Dan in comments elsewhere, I take issue with The Virtuosi’s implication that the information Dan offers needlessl narrows down the solution strategy. Here are my five solutions based on the given info. They all involve exponentiation, but that’s the mathematics of the situation. There is still plenty of room for a student to own their solution strategy.

  13. on 13 May 2011 at 5:40 pmScott

    Dan: I don’t think you’d want the arithmetic average of .7 and .8

    Suppose we have the first three doll heights, a, b, c. And lets say we have two ratios “r” and “s”.

    Such that r*a = b and s*b = c.
    Thus (rs)*a = c

    But you can’t average r and s using (r+s)/2.
    (r+s)/2 * (r+s)/2 * a is not equal to (rs)*a.

    We need the geometric mean (rs)^(1/2).

    Or, if we are finding the geometric mean of all rates, it would be the product of (r_1 * r_2 * … * r_n)^(1/n)

    I quickly ran the numbers and it makes such a small difference that it does not affect the outcome, but maybe that would be something to look for in another lesson. When would the arithmetic mean cause us trouble when we really need the geometric mean? And by trouble I mean real trouble.

  14. on 13 May 2011 at 5:42 pmJohn

    Dan,
    Awesome. I stand corrected. Would it be more powerful to include a shot of the digital calipers making the measurement?

  15. on 13 May 2011 at 6:11 pmDan Meyer

    @Scott, ugh. Thanks for the correction. I ran the numbers and the difference is a tenth of a percent — small. The damage to my pride is incalculable, though.

    Part of the trouble I’m having as I rework the visuals is that the height of the first, second, and third dolls don’t tell the whole story. Which is to say that if we use those to calculate the exponential base, their geometric mean is higher than the geometric mean of the entire set of fifteen. That’s why I gave their mean in the first place. The answer doesn’t work out otherwise.

    John: Would it be more powerful to include a shot of the digital calipers making the measurement?

    Almost certainly, yes. More visually distracting, though. It’s easier to get a sense of the relative size of the dolls when they’re positioned in exactly the same spot in front of a camera locked down on a tripod.

  16. on 13 May 2011 at 8:19 pmBowen Kerins

    I think the issue can be bailed out by giving students the heights of the first THREE dolls. This way, students could decide fairly well that there is a nonlinear rate of change in the dolls’ heights, and that there is a consistent proportional change in the heights.

    (This also could lead to one of the natural other questions, which is “How tall is the next doll?” Then a few more seconds of video verifies it.)

    I don’t think it’s obvious to all students that the heights will be proportional or exponential, especially after only seeing the second doll. Maybe students have a debate (an unresolved one) then offer some predictions about the height of Doll 3, then video reveal.

    I’d actually prefer students to decide how many dolls there might be without being told the height of the last doll, because it would then require them to decide what is a reasonable ending point for the whole business. I would have been surprised to learn there was a doll as small as 1.26 mm, but instead that information is handed to the kids.

    Still, as always this is some sweet stuff, and I continue to wonder how this sort of thing could translate into Textbook Land. The toughest part is the uncertainty of how much, if any, scaffolding students might need to reach the right questions or observations.

  17. on 13 May 2011 at 11:41 pmNumbat

    I’m thinking that if you gave them the height of the first and the second dolls then most students (depending upon year level of course) would assume a linear progression and work from there.

    Is giving them the first two heights in an almost deliberate attempt to send them off on a wild goose chase too nasty?

    I’m thinking that if you give them those two dimensions then ask if they need anything further. The student who asks for the third dimension is probably thinking along the right lines already?

  18. on 14 May 2011 at 5:56 amR. Wright

    I don’t think it’s obvious to all students that the heights will be proportional or exponential, especially after only seeing the second doll. Maybe students have a debate (an unresolved one) then offer some predictions about the height of Doll 3, then video reveal.

    This is exactly what I was thinking. When I teach college algebra, I like to cover exponential functions (and then, naturally, logarithms) as soon after linear functions as possible, because linear and exponential functions are so conceptually similar. Dan’s setup here could make a very nice bridge between the two (though to be honest, I feel that I already have better ones, which I’ll someday get around to blogging about.)

  19. on 14 May 2011 at 7:33 amThe Virtuosi

    Personally, I like the idea of not giving exact numbers and letting the students extract them from the video,

    But, I’d like to point out that I think that John’s original nit remains to be picked. I think you’ve fallen prey to your instruments Dan. Your calipers may know the distance between their ends to 0.01 mm, but that doesn’t mean you know the height of the doll to 0.01 mm. Grab the largest doll and your calipers and try to measure the height again. I’ll bet you a nickel you won’t get 62.82 mm.

    Set the doll down, open the calipers wide and set them down, stand up, walk around your chair, sit down and measure again. Do it ten times. I’ll bet another nickel that the standard deviation of your measurements is greater than 0.1 mm. In fact, I’ll throw another nickel on the measured error in your measurements being within a factor of 3 from 1.0 mm.

    I don’t think you should be bothered is the numbers don’t ‘work out’. Real data is messy and thats something we should embrace.

  20. on 14 May 2011 at 6:10 pmMylene

    Re: the unsimilar ratios… what if you gave Doll 1 and 2 side by side, along with an object of known size, then doll 7 and doll 8? Or, the two adjacent dolls with the largest ratio, and the two adjacent dolls with the smallest ratio.

    I liked the suggestion of not giving the dimensions of the last doll. ~1mm seems so implausible, and it looks great on the video (the WTF moment, as you call it).

    I realized, while watching the last video, that I wanted to see them side by side (I guess as a way of confirming the image in my head about what 75% looked like). I like the “hand reaches in and opens the doll” approach of the first video, and was looking forward to seeing it completed. Whether that makes videographic sense, I’m not sure — but it’s a question I wouldn’t have asked before reading your blog. Thanks for getting me thinking about visual design more carefully.

  21. on 15 May 2011 at 7:39 amDan Meyer

    Major updates.

  22. on 15 May 2011 at 10:48 amThe Virtuosi

    So, do I owe you 15 cents?

  23. on 15 May 2011 at 12:17 pmDan Meyer

    Not gonna lie: it’s a few items down my to-do list.

  24. on 15 May 2011 at 4:36 pmScott

    I’m a little unsatisfied that these ratios are different. I suppose after we have seen them all, we can create a best fit exponential with the exponent using the geometric mean, but what good does that best fit do us? We’ve seen all the dolls already!

    Also, how are students going to get any idea on where to stop? “Why is 15 the answer?” Will anybody guess 15 and have a good mathematical reason for it?

    I guess this means we have to show the last doll. Then the question becomes, “how many fit in-between?” And the lesson illustrates how the linear model doesn’t work (why?).

    But when I run the numbers using the data we *actually* have (no given Geometric Mean) I don’t get the right answer. http://scottfarrar.com/algebra/data%20stacking%20dolls%20dan%20meyer.png

    I’ll arrive at the last doll as #17. I think its cheating to have the geometric mean, since if we know that, we know how many dolls there are.

    And look at the % error we get using any of the three methods.

    Is it just this set of dolls?

  25. on 15 May 2011 at 6:17 pmChristopher Danielson

    If it helps, Scott, I can’t get any closer than 3 seconds on the escalator problem.

    This brings us back to “messy data”. If one of our goals is to help students see mathematics in the world, then messy data is unavoidable. Do we like the stacking dolls because they have an interesting but approximate mathematical pattern? Or do we like them only if we can shoehorn them into an exact pattern? I vote the former.

  26. on 15 May 2011 at 8:54 pmDan Meyer

    @Scott, now you see the constraints I was underneath, right?

    In the end, I don’t feel comfortable showing students the height of the last doll because a) that means the problem’s author knows how many dolls there are and b) the size of the last doll is one of the best parts about the problem. Those revelations belong in the third act.

    You all convinced me, for the same reason, not to include the geometric mean, to let students determine that for themselves.

  27. on 15 May 2011 at 10:26 pmScott

    Yes, this one is slippery. I’m ok with messy data — if its messy due to our collection. But this data is messy inherently.

    Compare with the basketball parabola: our predictions could be off because we may not fit the parabola onto the path of the ball precisely enough. But in the case of these dolls, our predictions could be off because the dolls do not automatically follow a predictable pattern. How are we to know if the next ratio will be 43% (14:15), 68% (12:13), or 85% (5:6)?

    I mean, the %error by the 4th Doll using geometric ratios is just as large as the %error using a linear model for the 3rd doll. So imagine this scenario:

    Call the dolls A, B, C, D, etc…

    You have an AB slide and a ABC slide. Especially if students are used to doing linear problems, then have them work based off that assumption from the AB slide. When you reveal the ABC slide we can see that the relationship is not linear. (due to the fact that we are so far off … 7mm (16% error))

    But how can we justify using a geometric relationship if we match that error with our geometric calculations? We obtain a measurement for doll D that is 5mm off (15% error) using the first ratio, or 7mm (21%) off again using the “cheater” geomean.

    If you consider the geomean of the first TWO ratios (B/A, and C/B) then you get pretty accurate data until the 12th doll when we’re back up to 13% error, and by the 15th doll we’re 200% off.

    I think a student that guesses 15 dolls would not have a good reason. Even if its the reason we are hoping her to give.

  28. on 16 May 2011 at 2:23 amBrian F

    I also find tremendous value in observing a different sort of three-acts taking place

    Act I: Sharing a compelling initial lesson idea
    Act II: Getting feedback and engaging with peers
    Act III: Using feedback to improve and refine the lesson

  29. on 16 May 2011 at 5:24 amR. Wright

    I’m ok with messy data — if its messy due to our collection. But this data is messy inherently.

    I have to strongly disagree with this sentiment, in general. Mathematical models are simplifications that capture only whatever behavior is considered “essential.” In fact, some useful mathematical models are only qualitative, not even attempting to fit the relevant data in a numerical way. One example that comes to mind is the use of Lotka–Volterra equations in population biology.

    Even in physics, mathematical models are often only very good approximations. For example, the path of a basketball is not really a parabola, because of air resistance and because gravity acts radially, not “constantly downward.” But again, as is usual in physics, parabolic motion is often a very, very good model.

  30. on 16 May 2011 at 7:52 am@thescamdog

    The updates have brought this along nicely. I like it a lot. I’m with Mylene, though. I would have found the answer video more satisfying if I could have seen all 15 emerging, and then placed side by side. Continue the question video until they are all out.

  31. on 16 May 2011 at 8:08 amScott Farrar

    @R. Wright

    I completely agree with you that mathematical models are only very good approximations.

    You’re right: the basketball is not *really* a parabola, because of all those physical and natural phenomena.

    Why are the dolls not *really* a geometric sequence? Can we study it mathematically? (like we study air resistance and radial gravity?) I’m just afraid these dolls are off due to human craftsmanship, *not* something inherent to the dolls structure.

  32. on 16 May 2011 at 8:54 amR. Wright

    I completely agree with you that mathematical models are only very good approximations.

    Actually, that only goes for physics and a very few other areas. In other areas, mathematical models aren’t really approximations at all, in the quantitative sense.

    Why are the dolls not *really* a geometric sequence? Can we study it mathematically?

    No more than we can study why small populations of organisms (including humans) only approximately exhibit exponential growth in rich environments. There is randomness (which can admittedly be somewhat quantified) and individual choice at work. Would you write off the use of simple mathematical models in biology, economics, epidemiology, sociology, etc.?

  33. on 17 May 2011 at 2:48 pmScott Farrar

    I guess my question is… if all I can see are the dolls A and B… why should I, as a student, use the ratio instead of the difference? Both of them lead to rather large errors off the actual measurements for Doll C.

    Is there anything physically stopping the dolls from following a linear pattern? Remember, we know only the heights A and B.

    I know our ideal set of stacking dolls would fit the geometric sequence, but how do we justify following that ideal? What is it about stacking dolls that makes them shrink geometrically?

  34. on 18 May 2011 at 6:29 amChristopher Danielson

    Scott:

    What is it about stacking dolls that makes them shrink geometrically?

    Similarity.

    But more importantly, don’t the dolls present us with a way of exploring the difference between these two models? If we see A and we see B, and if both linear and exponential models are in our repertoire, then the answer to How many dolls? is quite different if we choose a linear model (no more than 5 or 6) over an exponential one (theoretically infinite, but with some practical upper bound whose precise value is worth debating in class).

    So now we’re not so worried about the precise value of the scale factor as we are about which of these two mathematical models seems more reasonable. And we’re worried about what evidence will convince us which one better describes the situation at hand. Then we get to see that evidence.

  35. on 18 May 2011 at 11:32 amScott Farrar

    Are they really similar?

    Dan, could you provide measurements of the cross-sections? Does the wood’s thickness shrink? It should, if they were similar. If the heights of the dolls shrink by a ratio of B/A, so should the thickness.

    But I would hypothesize that they don’t. This would mean the ratios of of each successive doll would have to shrink. Which the data shows that they do (after a bit). Look at the “actual height ratios” http://scottfarrar.com/algebra/data%20stacking%20dolls%20dan%20meyer.png

    The first exception (the first ratio) is due to the pedestal on Doll A that is included in the height. No other dolls have this. **this could be something to have kids point out**

    Then it holds steady at about 84%. So perhaps the thickness is shrinking here. But all of a sudden the height ratios make a break, and start rapidly descending. I hypothesize that this is where the thickness stops obeying similarity to satisfy artistry (and physics). My point is… the artist CHOSE to make more dolls, even after it was impossible physically to construct the dolls with thin enough wood.

  36. on 18 May 2011 at 4:30 pmDan Meyer

    Have at it, Scott: guts.pdf.

  37. on 18 May 2011 at 4:46 pmChristopher Danielson

    Has no one else graphed this data? It provides awfully convincing evidence that Dan’s original task was phony (all due respect).

    Scott’s model is much, much better. Although it poorly predicts the 15th doll.

  38. on 18 May 2011 at 8:30 pmBowen Kerins

    First, that picture of the dolls’ innards is pretty f’in cool.

    Second, Daniel’s graph has an incorrect point for Doll 8, with the real point being much closer to the blue fitting curve.

    Third, actual Russian nesting dolls tend to have much more linear behavior than exponential behavior:

    http://www.russianlegacy.com/catalog/images/matryoshka_music/NDM017.jpg

    Back to the data, check out the heights on Dolls 7-15. The largest difference in height is just over 3mm — and it’s from 8 to 9, not 7 to 8. The smallest difference in height is just under 1.5mm — and it’s from 13 to 14, not 14 to 15. That’s fishy.

    Here’s what I think is happening — and compare it to that picture of the dolls’ innards: it’s the thickness of each doll’s material that controls how big the next one can be. If the dolls were truly similar, this thickness would be proportional to the same side ratio we see in the overall heights. But that’s not the case — the first one is thick, but then the next five (Dolls 2-6) look about the same thickness. Then there’s a substantial change in thickness for Doll 7, and that thickness seems relatively consistent (getting a little smaller, but not much) the rest of the way.

    And I think this matches (roughly) the pattern seen in the heights, which I would judge to be (roughly) piecewise linear as opposed to exponential.

    The last few dolls are the biggest giveaway: their size is dictated primarily by the thickness of the doll that comes before it, not by the shape of the original.

    It’s been an interesting problem for sure. I’ve learned I probably need to change the Russian nesting doll examples we use when learning about exponents in our Algebra 1 book!

  39. on 18 May 2011 at 9:58 pmThe Virtuosi

    Alright, this is getting interesting again. So, in light of the fact that the dolls do appear to run into a problem with their thickness near the end of the series, I propose a new model. Assume the dolls are each of width w, and attempt to get smaller by a factor of b at each step, until they run into the issue with their width.

    I.e. H_n = min( H_{n-1} * b , H_{n-1} – w )

    The height of the nth doll is the minimum of the height of the n-1th doll times b, and the height of the n-1th doll minus the thickness.

    Doing this, we have a nice two stage behavior. With H_1 = 61.4, b =0.8, w = 1.5, I obtain the following: Model graph

    This new model neatly incorporates an exponential decline, until the minimum becomes important, and then a linear decrease.

    In light of the issue with the pedestal, I took the initial height to be 61.4. Now we are in a position to predict the number of dolls in terms of our two parameters, b and w, wherein this model neatly terminates the number of dolls when H_n attempts to go negative. Doing so, I obtain this plot , showing nicely the models behavior as the parameters vary.

    Finally, to ease the eye, I’ve singled out the region of parameter space where 15 dolls are predicted here

  40. on 19 May 2011 at 8:01 amThe Virtuosi

    So, I also calculated the best fit for the piecewise model in my last comment above, you can see it here Notice that it doesn’t seem to line up perfectly with the data (note I used 61.4 for the height of the first dolls due to the pedestal and made it have a larger sigma, it has 10% error, the other points have 3% error.

    In light of this, I suggest yet another model, which isn’t so severe, Lets have both a scale factor and a width, but have them both act all the time, i.e.
    H_{n+1} = H_{n} * b – w

    So that this model will smoothly transition from the geometric scaling to the linear scaling at both ends. This model has a nicer best fit, as you can see here.

    Similarly, we can make a field of predictions with the new model for the number of dolls, shown here

    And show the region where this new model predicts 15 dolls here.

    So, the problem seems to be amenable to study afterall, and here especially there could be a lesson in model building and testing, though it has gone beyond the simple math puzzle it was originally intended to be.

    All of the files above, as well as the python code I used to generate the pictures is available here

  41. on 13 Jun 2011 at 6:31 amWhat Can You Do With This? | The Line

    [...] has made a successful recurring piece of posting multimedia materials  under the title “What Can You Do With This (WCYDWT)?” and soliciting comments on how to use them as the basis of math lessons. At the suggestion [...]

  42. [...] I have been reading some of  Dan Meyer‘s  series of “What can you do with this”   posts.  Cool Stuff, to be [...]