Simple geometry question

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Hawknc
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Simple geometry question

Postby Hawknc » Fri Aug 17, 2007 3:59 am UTC

I have to admit, I feel kinda stupid asking this, because it seems like it should be obvious. But for whatever reason, my brain won't give me the answer to it, so I'm hoping someone else can help me out here.

Say you want to move from point A to point B. You can go directly between the points, or you can move in the x direction, then the y direction, like so:

Code: Select all

A----------x
 \        |
  \       |
   \      |
    \     |
     \    |
      \   |
       \  |
        \ |
         B

Lousy drawing, sorry. Obviously the direct path is shortest here. You could also move halfway in the x direction, then halfway in y, then the rest of the way in x and rest of the way in y, so that the path looks like a step. Theoretically that is the same distance as going the whole way in x, then the whole way in y. If you keep reducing the interval between each direction change (say 1/3, then 1/4, then 1/10, etc), it should still remain the same distance as the first path, right? What happens when the interval goes to zero and the path is effectively the same as the direct path? Shouldn't it be shorter?

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Postby Nimz » Fri Aug 17, 2007 4:41 am UTC

If you move in alternating steps of dx and dy, you should still move the same distance as any other stairstep. When you approximate a curve (e.g. to find the arclength), you don't move dx then dy or vice versa. Rather, you take the diagonal, which has length (dx^2 + dy^2)^(1/2) != dx + dy.

*I tried =/= and \neq, but I decided to go with != for not equal.
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Postby Mathmagic » Fri Aug 17, 2007 5:18 am UTC

I've been pondering over this awhile, and I honestly can't answer your question.

I know it has something to do with limits, and something to do with infinities, but every time I try to form an expression for the distance travelled, things cancel and end up turning back into what you'd expect them to.

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Postby Xanthir » Fri Aug 17, 2007 5:29 am UTC

Ooh, this *is* interesting. It's similar in nature to things like the space-filling curve, in that it exhibits a certain property at every finite step and then breaks it in the limit.

Basically, yeah, the reason it works is because of the limit. At every finite step the traversed length is x+y. At the aleph-0th step, though, the length is (sqrt (+ (* x x) (* y y))) - the diagonal length. I don't have the experience to be able to prove this, but believe me, it happens because of the inherent weirdness of the infinite.

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Postby skeptical scientist » Fri Aug 17, 2007 6:27 am UTC

It just means that arclength is not preserved when you take limits in this way. That's because the arclength formula is based on derivatives, and the derivative of the limit is not the limit of the derivatives. The individual curves all have horizontal or vertical derivatives (or are not differentiable) at any point, but the limit curve has a diagonal derivative.
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Postby apeman5291 » Fri Aug 17, 2007 2:42 pm UTC

i think this is just a case of a function that lim (x->c) f(x) != f(c). In this case, the function is the distance from a to b, and while as the interval gets smaller, the function remains constant, but the actual value of the function at 0 is different from the limit.

(This is the first time I've actually tried to explain anything with limits, so I hope that makes sense to you)

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Postby jestingrabbit » Fri Aug 17, 2007 6:34 pm UTC

What they said.

If you are looking for values which bound the arc length above and below, and which converge to the arc length as the mesh of the partition goes to zero then let a=t(0)<t(1)<t(2)<...<t(k)=b and you have that

Sum( ((t(i)-t(i+1))^2 + (f(t(i))-f(t(i+1)))^2)^(1/2), i=0..k-1)
<= arc length of the curve {(t,f(t))} from t=a to t=b
<= 1/2( (t(1)-t(0))(1+f'(a)^2)^(1/2)+(t(k-1)-t(k))(1+f'(b)^2)^(1/2) +Sum( (t(i+1)-t(i-1))(1+f'(t(i))^2)^(1/2), i=1..(k-1)) )

and if you take the limits correctly you should end up with the formula for arc length.

edit: those values might not bound the arclenght for every curve, but I believe they will for convex (up or down) curves, and every differentiable curve is piecewise convex (or a straight line, which is a kind of convex anyway).

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Postby ihope127 » Fri Aug 17, 2007 8:38 pm UTC

When the interval goes to zero, you go nowhere. :-)
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Postby Hawknc » Sat Aug 18, 2007 3:23 am UTC

...Well at least that explains why my high school maths teacher wasn't able to tell me all those years ago.

Much thanks for the replies, I'll try and make some sense of it when I have a few minutes. ;) I knew it had something to do with limits, but I am just a lowly engineer and as such haven't delved too deeply into theoretical limits and the like.

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Postby orangeperson » Sat Aug 18, 2007 9:32 pm UTC

Ihope127 is a genius. It's much simpler than limits.
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Postby Xanthir » Mon Aug 20, 2007 9:47 pm UTC

Um, no. When dx = 0, you go nowhere. But in the limit case as dx -> 0, you start moving diagonally.

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Postby Quan » Mon Aug 20, 2007 9:58 pm UTC

This might interest you: http://en.wikipedia.org/wiki/Taxicab_geometry

I just had this open in a different tab as I was reading this post.

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Postby Nimz » Mon Aug 20, 2007 10:25 pm UTC

Do I hear my echo here? Hmmm. Must be just a case of déjà vu. If you only move horizontally and vertically, you will travel the same distance, even in the limiting case. The distance only decreases if you change your direction to somewhere between horizontal and vertical, which, by assumption, never happens.

Of course, this is physically unrealizeable. If you tried to follow an infinitessimal stairstep as closely as possible, you would perceive your distance travelled to be the diagonal distance. The assumption of only horizontal and vertical displacements would be violated.
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Postby Xanthir » Mon Aug 20, 2007 11:04 pm UTC

In every finite step, you are only moving horiz and vert. In the limit case, though, you are moving diagonally.

It's analogous to the Hilbert Curve. At every finite step it's isomorphic to the unit line. In the limit, though, it's isomorphic to the unit square (in fact, it *is* the unit square). Though it's still a line, it also fills an area in the limit.

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Postby 27b/6 » Tue Aug 21, 2007 1:05 am UTC

I think you're applying the notion of limit incorrectly.

In your setup, let's call "a" the total length in the x direction, "b" the total length in the y direction, and "t" be the number of times you "step" along the path.

For t=1, we move b up, then a over, so a+b=total length.
For t=2, we move b/2 up, then a/2 over, then b/2 up, then a/2 over.
For arbitrary t, we have t(b/t + a/t). Even if we let t->infinity, this always reduces to b+a.

What I think you're trying to prove is that lim t->inf (t(b/t+a/t) = sqrt(a^2+b^2), which obviously is not the case unless a=0 or b=0.
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Postby Vaniver » Tue Aug 21, 2007 2:02 am UTC

The way I thought about this was comparing a tightly wound, thin spring to a piece of wire. Since the spring is wound up (i.e. never pointing in the direction of the target), it takes more material (assuming the wire it is made of is the same). Er, this explanation is subpar. But hopefully you'll get it.
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Postby Alpha Omicron » Tue Aug 21, 2007 2:09 am UTC

So n(x/n + y/n) = n((x+y)/n) = x+y for finite n only? That's actually rather interesting.
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Postby 27b/6 » Tue Aug 21, 2007 2:39 am UTC

Alpha Omicron wrote:So n(x/n + y/n) = n((x+y)/n) = x+y for finite n only? That's actually rather interesting.


No. My point is that it's limit as n goes to infinity is x + y. The whole concept of limits is that you don't care what happens when you actually get there.

For it to be the case that the the distance would be the same as the diagonal case, the limit would have to be sqrt(x^2 + y^2). But it's not. It's x+y. Therefore, in the limit, the stepped case does NOT "collapse" to the diagonal case
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