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I have sometimes wondered about a distance problem that involves travelling along the two triangular sides of distance between two points, then gradually shortcutting the distances into smaller and smaller chunks. I can't quite understand why, when the small distances approach zero, they are not the same as the hypotenuse distance.

I made a video that explains exactly what I mean. (Sorry, this service doesn't allow embeds)

https://drive.google.com/file/d/0BzU1txEbhPPUUFBnWlNteHJpNGM/edit?usp=sharing

thanks for your thinking

Chris

betchaboy
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2 Answers2

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This is related to something called the "taxi-cab metric". You're right when you say the distance doesn't approach the hypotenuse. In fact if you look closely you will realize that the total distance traveled is the same at every step. The "problem" is that the transformation you are applying to the path preserves its length.

Spencer
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The way I likw to explain it is this:

As the individual segments get smaller and smaller, consider the length (call it $h_n$) the segment on the hypotenuse and the length of the subdivided function from the segment (call it $f_n$).

If $f_n/h_n$ is bounded away from zero (as in the video), then the limit of the sums will be greater than the length of the hypotenuse.

If $f_n/h_n \to 1$ (as in the case of a circle being equally subdivided more and more), the limit of the sums will be equal to the length of the hypotenuse.

You can make this rigorous by estimating the lengths using the integral for arc length, but this is my intuitive understanding. If the function is concave ($f'' < 0$), you can probably work out bounds for the individual arc length integrals with the assumption that $\max |f'| \to 0$.

If I am wrong, please let me know so I can stop passing on incorrect information.

marty cohen
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