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I'm trying to informally explain why $\pi$ holds for all circles. I would like to know if there is anything pertinent that I can add, or that is wrong with this explanation. It's an explanation, not an attempt at a rigorous proof, so go easy.

$\pi = \frac{C}{d}$ is the ratio of $C$, the circumference of the circle, to $d$, the diameter of the circle. The ratio holds for all circles because all circles are "similar". That is to say they are all the same essential shape and that they only differ in size, thus for "similarity" to hold this ratio must extend to all circles.

It is easier to see why $\pi$ holds for all circles if we rewrite it in terms of the radius $r$ as $\pi=\frac{2\pi r }{2r}$. According to this, any change in the radius $r$ (and thus circumference and ultimately the size of the circle) will affect an equivalent change in the numerator and denominator of the equation. So in essence any changes in the size of the radius $r$ and subsequently the circle are negated, so the ratio $\pi=\frac{C}{d}$ holds for all circle.

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seeker
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    What does it mean to say all circles are similar? Why is it that if you double the diamater you double the circumference? Why would it not be reasonable to expect some other kind of relationship? You haven't explained anything, you just used new words. – Pedro Jan 11 '15 at 19:52
  • @Pedro, thanks, I'll try and update my question with a new 'explanation'. – seeker Jan 11 '15 at 20:32
  • @Pedro, i've fleshed it out a little more, please tell me if the same shortcomings still hold – seeker Jan 11 '15 at 20:49
  • Is this a proof on why $\pi=\frac Cd$ and not something like $10=\frac Cd$ ? – Anonymous Computer Jan 11 '15 at 22:43
  • Oh... I thought you wanted to prove why only $\pi$ is equal to $\frac Cd$, and not another number like $10$. That's what I meant – Anonymous Computer Jan 11 '15 at 22:45
  • No, I want to explain how the ratio $\pi=\frac{C}{d}$ holds for all circles. – seeker Jan 11 '15 at 22:47
  • seeker, what JChau is describing is the same as what you are asking for. Showing $\pi = \frac{C}{d}$ for all circles is the same as showing that no circle satisfies $x = \frac{C}{d}$ for $x \not = \pi$. – dalastboss Jan 11 '15 at 23:31
  • @JChau, I've misread what you wrote, you're right, sorry – seeker Jan 11 '15 at 23:33
  • @JChau how would you show that only $\pi=\frac{C}{d}$? – seeker Jan 12 '15 at 13:00
  • When you rewrite the expression in terms of $r$ as you did, you can't write $C=2\pi r$ and conclude that $\pi = \frac{C}{d}$ for all circles from that, because $C = 2\pi r$ comes from the fact that $\pi = \frac{C}{d}$. – Integral Jan 12 '15 at 13:04
  • So it's circular logic.. I see.. – seeker Jan 12 '15 at 13:11

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A slightly better explanation (I believe): consider a regular $n$-gon inscribed in a circle, and decompose it into $n$ isoceles triangles with the two equal sides equal to the radius $r$ of the circle. The polygon's perimeter $P_n$ and the circle's radius are in a certain ratio, say $r_n:1$. (So $P_n/r=r_n)$. Then, if you vary the circle's radius, the polygon's perimeter changes proportionally, with proportionality factor $r_n$, because of similar triangles. So, if the $r_n$ approach a limit as $n$ increases (which you can make plausible numerically), then this limit must be the ratio of any circle's circumference to its radius. (Assuming that a circle may be viewed as an "$\infty$-gon".)

Kim Fierens
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  • Could you just clear something up please, you say "So, if the $r_n$ approach a limit as n increases (which you can make plausible numerically), then this limit must be the ratio of any circle's circumference to its radius". Shouldn't the limit be the circumference of the circle NOT the ratio of the circles circumference to its radius? I think you missed out the word 'factor' towards the beginning of the sentence? – seeker Jan 12 '15 at 16:33
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    I define $r_n$ as the ratio $P_n/r$, where $P_n$ is the perimeter (the sum of the lengths of the outer edges) of the $n$-gon. The idea is that, as $n$ increases, $P_n$ will converge to $2\pi r$, whereas $P_n/r$ will converge to $2\pi r/r=2\pi$, which is the ratio of the circle's circumference to its radius. – Kim Fierens Jan 12 '15 at 16:43
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    Thanks, that cleared it up, I think there's a little typo in your answer however. After "So, if the..." – seeker Jan 12 '15 at 16:44
  • I've rewritten the answer to make it even clearer. What is the typo? I use "the $r_n$" as a plural. This is relatively common. – Kim Fierens Jan 12 '15 at 16:48
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    It's all good now, thank you for your answer! – seeker Jan 12 '15 at 16:48
  • Perhaps you may be able to help with my other question: http://math.stackexchange.com/questions/1101838/how-can-i-improve-my-explanation-that-the-ratio-theta-fracsr-that-define – seeker Jan 12 '15 at 21:47
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Your explanation does little more than point out that he fact that $\frac Cd=\pi$ for all circles follows from the fact that all circles are similar.

Indeed, there's little else that you can do without diving into some pretty complicated analysis, so if that's all you're interested in saying, then that's fine.

A more succinct explanation is given by:

Choose some circle $V$ with circumference $C$ and diameter $d$, and define $\pi=\frac Cd$. Now choose some other circle $V'$ with circumference $C'$ and diameter $d'$. Any two circles are similar; in particular, $V$ and $V'$ are similar, and we have:

$$ \frac C{C'}=\frac d{d'} $$

Therefore, we can conclude that

$$ \frac{C'}{d'}=\frac Cd=\pi $$

Since the circle $V'$ was arbitrary, we conclude that the ratio of the circumference to the diameter of a circle is $\pi$ for all circles.

To get a rigorous proof, you ha to define what you mean by the length of a curve. Once you have done that, there is a good formula for finding the length of a curve parametrized by smooth functions:

If $\gamma=(\gamma_1,\gamma_2):[a,b]\to\mathbb R^2$ is a smooth curve, then the length of $\gamma$ is given by $$ L(\gamma)=\int_a^b \|\dot\gamma(t)\|dt=\int_a^b\sqrt{\gamma_1(t)^2+\gamma_2(t)^2}dt $$

In particular, if $\gamma$ is a circle of radius $r$ about a point $(X,Y)$, we can write it as

$$ \gamma:[0,2\pi]\to\mathbb R^2:t\mapsto(X+r\cos t,Y+r\sin t) $$

Then

$$ \dot\gamma(t)=(-r\sin t,r\cos t) $$

and

$$ C=L(\gamma)=\int_0^{2\pi} \sqrt{r^2\sin^2(t)+r^2\cos^2(t)}dt=2\pi r $$

Then we have the diameter $d=2r$, and so $\frac Cd=\pi$. Throughout, $\pi$ is defined to be the first positive zero of the (analytically defined) $\sin$ function. Of course, there's lots of analysis to do to check that $\cos$ and $\sin$ really do parametrize a circle, and that they behave the way they do with respect to differentiation, but this is the easiest way to get to a rigorous proof.

Update: having read Kim's answer, I've now changed my mind: inscribing a regular $n$-gon is the best way of proving this rigorously. It uses the definition of the length of a curve directly; that is:

$$ L(\gamma)=\sup\left\{\sum_{i=1}^n d\left(\gamma(t_i),\gamma(t_{i-1})\right)\;\middle|\;n\in\mathbb N,0=t_0<t_1<\dots<t_n=1\right\} $$

where $d(p,q)$ is the Euclidean distance. In addition, no calculus is required for that approach (just some care regarding limits).

John Gowers
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  • Thank you for your thorough explanation. The second part is beyond my current level, but I appreciate it nonetheless. – seeker Jan 12 '15 at 15:07
  • Great answer! I can follow how you do it rigorous but with your mentioned downside: Of course, there's lots of analysis to do to check that cos and sin really do parametrize a circle, and that they behave the way they do with respect to differentiation, but this is the easiest way to get to a rigorous proof. In your update you mentioned that it would be a lot easier to use the definition of a curve directly. How does that help? Could you clarify the steps? Do you not need to use $L(\gamma)=\int_a^b |\dot\gamma(t)|dt$? If so, then you have the same downsides right? – Adam Mar 11 '19 at 09:38
  • I figuered out how to prove it, but I needed some calculus: https://math.stackexchange.com/a/3149309/82101 – Adam Mar 15 '19 at 13:40