Notes on Deferred Computation – the Pythagorean Spiral

The following figure is quite interesting:


Imagine you start out home, in the center at the bottom, and move a distance of 1 to the right to A.  At A, you have traveled a total distance of 1, and you are also 1 away from home.  You then turn left, at exactly 90 degrees, at a right angle.  You travel another distance of 1.  At B, you’ve traveled a total distance of 2, but you’re not a distance of 2 away from home.  If you had to rush back home, you wouldn’t go back the way you came, you’d take the short cut, straight.  The straight path home would be along the diagonal (hypothenuse) of the right triangle, and the distance can be measured or it can be obtained from the Pythagorean theorem.  Clearly, it is more than 1 and less than 2.  But imagine you aren’t rushing back home – instead, from B, you continue on a new path, and the path is at a right angle to the straight way home.  On this new path, you also move a distance of 1, to C.  You’re now further away from home, but clearly you’re closer to home than the total distance you traveled, which is 3.  By always choosing a path at a right angle to the straight way home, and always traveling a distance of 1 on that path, you trace a spiral.  This spiral is often called the Pythagorean spiral.

If you ask middle schoolers what the distance home is after traveling a distance of 9 along the spiral path, most kids will realize that they can use the Pythagorean theorem to find the length of the path home.  For each triangle, the path home is the hypothenuse; and of the other two sides, one is always a length of 1, and the other one is the distance home at an earlier moment.  So by starting with the smallest triangle and then proceeding one triangle at a time, we are always looking at a right triangle for which we know two sides and for which we are supposed to find the hypothenuse.  Kids will settle in, grab their calculators, and get to work.  For the first triangle, they will write something like: a^2 + b^2 = c^2 \quad \quad 1^2 + 1^2 = c^2 \quad \quad c^2 = 2 \quad \quad c=\sqrt{2} \quad \quad c = 1.4

and then they will move on to the next triangle, and write

a^2 + b^2 = c^2 \quad 1.4^2 + 1^2 = c^2 \quad c^2 = 2.96 \quad c=\sqrt{2.96} \quad c = 1.7

If they continue without error all the way to the triangle representing the path of total distance 9, they would get some answer close to 3.

At this point, most kids have lost the forest for the trees, they are grinding through the motions. For them, the hardest part isn’t necessarily doing the computations on the calculator, it is keeping track of what they are doing by writing all the steps down.

Yet a sizable group of kids will have noticed something along the way.   They will have noticed that right after they get the square root of a number, they end up turning right around and squaring it again.  And since nobody asked for the distance home after traveling a distance of 2,3,4,… they stop doing the square root somewhere along the way and just write down the square of the hypothenuse, to be used in the next step as their a^2 .  They also notice that b^2 is always the same: 1.  So their work may look more like this: 1.4    1.7   3.89    4.89    5.89 … 8.89.  At this point, they have the square of the final hypothenuse, and they get the asked-for distance by taking the square root, obtaining 2.98, which they would round to 3.0.  Very few of these kids would go back and apply their insight to the earlier steps.  But their discovery of the benefits of deferred computation is an important one, both in terms of effort and in terms of their ability to see the forest over the trees.  The values of the square roots didn’t have any clear pattern to it, but the values of the squares of the hypothenuse were simply going up by one each step.

Some few kids defer computing square roots altogether, though they may not think of it in those terms.  They would write the hypothenuse of the smallest triangle as \sqrt{2} and leave it as that, not writing it as a decimal number at all.  Some kids do this because their teachers have “allowed them” to leave something written as \sqrt{2} , telling them something like “I will accept something with a square root symbol in it as an answer.”  Now, just because some kids write the first hypothenuse as \sqrt{2} doesn’t at all guarantee that they see that its square is 2.  If you ask a random middle schooler how much (\sqrt{17})^2 is, you won’t always hear 17.

The kids who do see they can simply keep track of the squares of the hypothenuses, and that these squares increase in a regular pattern, going up one each step, those kids will then see that the total distance traveled is the square of the distance home.  After a total distance traveled of 9, the distance home is 3.

The test if a particular kid is seeing the forest or is seeing the trees comes when you ask for the relationship between the distance traveled and the distance to home, e.g. after traveling a total distance of 100.  The number 100 is too big for kids to be tempted to compute each hypothenuse along the way, they will be looking for a short cut.  Many kids will be comfortable expressing the relationship as a formula, like \text{distance home } = \sqrt{\text{total distance traveled}} , a formula which is correct for any distance traveled, at least when the distance is a counting number.

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1 Response to Notes on Deferred Computation – the Pythagorean Spiral

  1. Bert Speelpenning says:

    I remember reading somewhere, decades ago, about a random walk and a square root. Best I can reconstruct it is that you imagine a drunk walking around, with no clue as to where he is going, and at any moment, the direction he proceeds in is completely random. That is, he is just as likely to go forward than to turn left, just as likely to turn right as to turn back the way he came – and so for each successive moment. The claim was that such a random walk would gradually move further away from the starting point, and the mean distance to the starting point would be proportional to the square root of the time walked.

    At the time I had no idea how such a thing could be analyzed, but having written the account of the Pythagorean spiral in terms of the distance from home versus the total distance traveled, it occurred to me there might be a direct link between the two. The Pythagorean spiral is done in discrete steps, but it isn’t terribly hard to imagine a continuous version of it. Conversely, the drunk’s random walk could be construed as done in discrete steps also, with the drunk always lurching a fixed distance in the new direction before lurching in a new (and independent) direction.

    Reasoning very crudely, you could say that “on average” the new direction is at right angles with the old direction: just as likely to be net forward as to be net backward. “On average”, then, the distance from home for the drunk would be just like the distance of the person traveling the Pythagorean spiral.

    It would be interesting to pursue if this line of reasoning holds any water.

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