Why $\pi r$ is not equal to $2r$?

Question

If there is infinity number of small arcs on top of diameter (can assume it is a simple line which has a length of $2r$) of a half circle (radius is “$r$”) why $\pi r$ is not equal to $2r$?

A diameter as a limit of semicircles

Searching this site for "arclength paradox" finds this commonly asked related question: http://math.stackexchange.com/questions/12906/is-value-of-pi-4/12919#12919 — Ethan Bolker, Jun 29 '16 at 18:27
To show that this phenomenon has nothing to do with the number $\pi$, you can produce a similar example as follows: Consider a sequence of sawtooth curves ${f_n(x)}$ defined over $x \in [0,1]$, where $f_n$ has $n$ teeth, each of height $1/n$. The length of each of these curves $f_n$ is $\sqrt{2}$, even though the curves converge to the line of length 1. Specifically, $$f_1(x) = 1-x, \quad f_2(x) = \left{ \begin{array}{ll} 1/2-x &\mbox{ if $x \in[0,1/2)$} \ 1-x & \mbox{ if $x \in [1/2, 1]$} \end{array} \right.$$ and so on. — Michael, Jun 29 '16 at 18:32
You don't get infitesimal regions with infitesimal bumps, you get regions of length 0 with no bumps. — Jacob Wakem, Jun 30 '16 at 13:52

score 18 · Accepted Answer · answered Jun 29 '16 at 18:25

18

You're making an assumption here: if the limit (in some sense) of a set of curves is some limit curve (the diameter in your case), then the limit of the LENGTHs of the curves must exist and equal the length of the limit curve.

That assumption is wrong* --- not everything in the world commutes with taking limits, even though many people who are not very familiar with limits assume this mistaken idea --- and for arclengths, the limit of the the curves and their tangents must be the limit curve and its tangents for this property to hold**. Since your curves all have vertical tangents, while the limit curve -- the diameter -- has only horizontal tangents, you don't get the arclength-limit property.

*Your example constitutes a proof that it's wrong!

** Proving this claim is comparatively difficult; you certainly need to know something about limits and calculus and perhaps even a bit of real analysis.

answered Jun 29 '16 at 18:25

John Hughes

93,729

2

A beautifully clear answer - can be understood by anyone - and you draw an important moral from it. – almagest Jun 30 '16 at 07:52
In short, arclength is not continuous in the pointwise convergence topology. – Jun 30 '16 at 14:26
Yes, @Rahul. But anyone who understands that sentence also knows the answer to this problem. That's not a complaint -- I like brief and pithy statements! -- but it means that its educational value is limited to those who know the answer, but not the condensed way to express it. – John Hughes Jun 30 '16 at 16:07
I agree, I did not intend that as a criticism or suggested change to your answer! – Jun 30 '16 at 16:37

score 1 · Answer 2 · answered Jun 29 '16 at 18:51

1

Simply because $c_n(r)=\pi r$ for all $n$, and thus, it's impossible to have $$\lim_{n\to \infty }c_n(r)=2r.$$

answered Jun 29 '16 at 18:51

Surb

55,662

1

I think that is too sophisticated an answer for the OP. I suspect he can see that and is bewildered by it. :) – almagest Jun 30 '16 at 07:50

Ng Chung Tak · Answer 3 · 2016-06-30T13:28:06.187

0

Note that the curvature $\kappa$ of each circular arcs grows with the number of division $n$, that is $$\kappa=\frac{n}{r}$$

The curve never approaches to a straight line in terms of smoothness, the curve is of class $C^0$ not $C^1$.

edited Jun 30 '16 at 13:28

answered Jun 30 '16 at 13:06

Ng Chung Tak

18,990

What exactly is an example is a continuous but everywhere non-differentiable curve? At each step $n$ the curve is differentiable almost everywhere, and the limit is the flat curve which is differentiable everywhere. What is true is that the sequence of curves does not converge for the $C^1$ metric to the flat curve. – Najib Idrissi Jun 30 '16 at 13:20
I mean the limiting curve. Note that the curvature blows up. – Ng Chung Tak Jun 30 '16 at 13:24
What do you think is the limiting curve? The only reasonable "limiting curve" I see is the limit for the $C^0$ metric, AKA the flat curve. – Najib Idrissi Jun 30 '16 at 13:25
It's rather an abstract structure to discuss here, so I delete my last sentence. – Ng Chung Tak Jun 30 '16 at 13:26
I don't want to be rude, but it's not abstract, it's inexistant...! – Najib Idrissi Jun 30 '16 at 13:27
I've deleted my last sentence. To be humble, I'd like to wait for your better answer. – Ng Chung Tak Jun 30 '16 at 13:31
1

There are already good answers, why should I need to post one? – Najib Idrissi Jun 30 '16 at 13:32
So, upvote them and/or downvote my own. – Ng Chung Tak Jun 30 '16 at 13:33
2

You would rather I had downvoted your answer with no comment instead of asking you about what I think is wrong with it? Okay... – Najib Idrissi Jun 30 '16 at 13:34
Is it OK after my deletion right now? – Ng Chung Tak Jun 30 '16 at 13:38
Yes, I think it's OK now. – Najib Idrissi Jun 30 '16 at 13:38
Many thanks to your feedback. – Ng Chung Tak Jun 30 '16 at 13:39

score 0 · Answer 4 · answered Jun 30 '16 at 13:48

Your function gets more and more bumpy as n approaches infinity; however when you take the limit all of the bumps get taken out (the bumps contributing to the perimeter).

Put another way the regions where the derivative is nonzero contribute to the extent that the perimeter is not 2r but when the limit is taken with respect to the function it is not taken with respect to the relevant differential structure.

Why $\pi r$ is not equal to $2r$?

4 Answers4