Path lengths on a unit square

Question

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?

Answer 1

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!

Answer 2

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.