Vector path length of a hypotenuse

Question

Consider the red path from A that zigzags to B, which takes $n$ even steps of length $w$. The path length of the route $P_n$ will be equal to:

$ P_n = P_x + P_y = \frac{n}{2}\times w + \frac{n}{2}\times w = n \times w $

But $\frac{n}{2}\times w = 1$ beacuse it is the length of one of the sides of the triangle so:

$P_n = 2$

Which will be true no matter how many steps you take. However in the limit $n \to \infty, w \to 0$ the parth length $P_\infty$ suddenly becomes:

$P_\infty = \sqrt{1^2 + 1^2} = \sqrt{2}$

Due to Pythagoras. Why is this the case? It seems the path length suddenly decreases by 0.59!

Answer 1

The length of the $n$ path is define by : $ \int_{(a,b)} \gamma_n'(t)dt $ where $\gamma_n$ is your path. But you can't pass at the infinity because "you have too much point of discontinuity". For instance you have no simple convergence for $\gamma'_n$...

Answer 2

There are some misconceptions here regarding magnitude of a vector versus its direction+magnitude.

First, label the bottom left corner of the triangle as $O$ and declare the length of $OB=OA=1$ (you seem to imply this given your statement). Each one of these $n$ mini-paths (assuming we go up and over by equal amounts of $1/n$) tracing up hypotenuse $AB$ of the triangle has position vector given by $(-1/n,1/n)$. The magnitude of this position vector is given by the Pythagorean theorem, namely: $||(-1/n,1/n)||=\sqrt{2}/n$. Recall that there are $n$ such paths and so the total path $AB=\underbrace{\sqrt{2}/n+\cdots+\sqrt{2}/n}_n=\sqrt{2}$.