# Koch snowflake versus $\pi=4$

The only proof I could find of the Koch snowflake having infinite perimeter was by calculating the perimeter $P_n$ after the $n$th iteration $$P_n = 3s\left(\frac{4}{3}\right)^n,$$ where $s$ is the length of each side of the original equilateral triangle, and then taking the limit as $n$ approaches infinity (which is obviously infinity).

I was satisfied with this proof until I remembered the infamous "proof" that $\pi$ is equal to $4$ (see the link below). The length of the limit curve (a circle) is $\pi$ which is not the limit of the perimeter of the zig-zaged curve after the $n$th iteration because that's always $4$. This explanation http://qntm.org/trollpi of the false proof states that "the limit of a sequence need not necessarily share any properties with the members of that sequence" and that "we've seen a sequence of curves of length 4, whose limit does not have length 4".

So how come I can use this argument for proving that the Koch snowflake has an infinite perimeter? Is one of these derivations wrong or am I just missing something?

• If you can also argue that the perimeter at each stage is less than that of that of the final snowflake, i.e. letting $P$ be the actual perimeter, that $P\ge P_n$ for all $n$, then this would prove $P=\infty$. Can you show this is true? – Mike Earnest Oct 17 '14 at 6:31
• which source claims the perimeter of the Kock snowflake is $\infty$? – Ittay Weiss Oct 17 '14 at 6:32
• Length is a pretty weird concept. The problem with the $\pi=4$ proof is that the limit of a sequence of arclength integrals don't converge to the arclength of its limit. I haven't thought about it yet, but see if you can turn the perimeter of the snowflake into an integral-limit interchange question. The result may fall out pretty easily. – enthdegree Oct 17 '14 at 6:36
• It's also worth considering that the sequence of traces of pre-koch curves doesn't converge to something smooth as in the fake $\pi=4$ proof. – enthdegree Oct 17 '14 at 6:38
• @IttayWeiss: Well, I haven't thought deeply about it, but I don't see right away what's wrong. I'd be happy to learn, if you don't mind explaining a little more. We have a continuous function $\gamma:[0,1]\to X$, where $X$ is the perimeter, don't we? And by sampling it at the corners appearing at the $n$th stage, we find a partition $\pi$ such that $s(\pi,\gamma)=\text{constant} \times (4/3)^n$. (Using the notation from the article I linked to.) And therefore $\sup_\pi s(\pi,\gamma)=\infty$, so $\gamma$ doesn't parametrize a rectifiable curve. – Hans Lundmark Oct 17 '14 at 7:47

A bit easier concept is the arc length. What is the arc length of a circle? One could pick points on a circle. Then the length of the circle should be at least as big as the length of the resulting polygon, right? So you could ask: By doing this, how "long" polygon could you get? Everything works out nicely and the upper bound for those arc lengths of polygons will be $2\pi r$. Why? Well, one could prove that this definition leads to the normal integral definition for "nice" curves. Or do some approximation, essentially, it's all about $$\lim_{x \to 0}\frac{\sin(x)}{x} = 1.$$
Now the arc length of the Koch snowflake kinda makes sense. How long polygonal curve could you get by joining points on the curve? Well, if you consider the corner points of the iterates of snowflake, they will stay on the curve till the end. So joining those points will be joining points on the curve. Since choosing points on $n$:th iterate we can get arbitrarily long polygons, there is no upper bound on the arc length of Koch snowflake so it has, in some sense, infinite arc length.
What goes wrong with the $\pi = 4$ proof? The points won't be fixed like in the snowflake, except for the points on the circle. We could join them and note that the arc length of the resulting curve, if it exists, is at least $2\pi r$. Sure.