# Path lengths on a unit square

Suppose I'm at $(x=0,y=0)$ and I want to get to $(x=1,y=1)$. The shortest path is the diagonal and it has length $\sqrt{2}$. But what if I'm only allowed to make moves in coordinate directions---e.g., $1/2$ along $x$, $1/2$ along $y$, another $1/2$ along $x$, and a final $1/2$ along $y$. Then the length of my path is $2$. In fact, any coordinate-constrained path has length $2$. Let the path $p_n$ be $1/n$ along $x$, followed by $1/n$ along $y$, followed by $1/n$ along $x$, etc., until I get to $(1,1)$. Presumably, the limit of $p_n$ as $n\rightarrow\infty$ is the diagonal line. But the path length of each $p_n$ is $2$, while the path length of the limit is $\sqrt{2}$.

Weird, right? Is this just an example that shows that you can't exchange limit and path length?

• The square path $p_n$ that you have defined looks at first sight as a rectification (a polygonal approximation of a curve) but it is not. Hence the length will not converge to that of the diagonal. – Rogelio Molina Jun 18 '15 at 23:22
• This reminds me of the proof $\pi=4$ by circumscribing the circle with a square and cutting corners. – Teoc Jun 18 '15 at 23:31

It comes down to what kind of approximation you're working with. The sum of sides of a right triangle do NOT approximate it's hypotenuse. Even it the triangle is really small, the sides are still a bad approximation. To more specific, in an isocolese right triangle, with sides lengths $x$ and hypotenuse, $y$, we have $2x^2 = y^2$. So $y= \sqrt{2}x$. So $$\lim_{x\to 0} \frac{x}{y} = \lim_{x\to 0}\frac{x}{\sqrt{2}x} = 1/\sqrt{2}$$

This tells you your approximation will always be bad. We want the limit to be 1.

To contrast this, let's look at an example of good approximation. On the unit circle, consider two points $A$ and $B$. We want to approximate the arclength from $A$ to $B$ by the chord from $A$ to $B$. Call the arclength $\theta$ (we're mathematicians, we use radians) and call the length of the chord $x$. Doing some geometry, it's not hard to show that

$$x = 2\sin(\theta/2)$$

A standard result from calculus will tell us that $$\lim_{\theta\to 0} \frac{\theta}{2\sin{\theta/2}}=1$$

So the chord is a good approximation of the arc length for small arclength!

• I can measure the "distance" between the path $p_n$ and the diagonal by computing the area of the region bounded by the diagonal and $p_n$. That area is $1/(2n)$ which goes to $0$ as $n\rightarrow\infty$, right? – paulcon Jun 18 '15 at 23:22
• You'd have to prove that before you could use it. And you just showed me a counter example! The idea is that area is not a good proxy for perimeter. It's actually a really bad proxy, which is somewhat surprising and counter-intuitive. Measuring 2-D distance and expecting to know something about 1D distance doesn't usually work. – Zach Stone Jun 18 '15 at 23:24
• I can prove that the area of the region bounded by the diagonal and the path $p_n$ is $1/(2n)$. That's just counting triangles. And the $\lim 1/(2n)=0$. I think your point is that this area is not an appropriate measure of distance. – paulcon Jun 18 '15 at 23:29
• Yes, your calculation is fine. I meant that you'd need to prove that area is related to perimeter. I think you understand. By the way, this is a great question that gets at the heart of limits vs. geometry. If you came up with this on your own that's pretty good. – Zach Stone Jun 18 '15 at 23:32

This is indeed because you cannot exchange the limit that you have described and the path length. No matter how large your n gets, if we look at one change in x and one change in y, the total distance from start to finish will always be √2 * the change in x (or y, because they are both the same). The error does not disappear as n gets larger. So, yes this example just shows that you cannot exchange the limit and path length.