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(I should say at the outset that this question is broad, and may need splitting up. Although I ask several questions, I present them as one because they are not independent of one another, and I am seeking a unified answer.)

My questions are:

  • How can we establish that the circumference $C$ and area $A$ of a circle of radius $r$ satisfy $C = 2\pi r$ and $A = \pi r^2$ for some constant, $\pi$?

  • How can we prove that $\pi$ is an element of the real field (e.g., a Dedekind cut)?

  • How can we prove (perhaps trivial, if the above are satisfied) that there are real functions $\sin(x)$ and $\cos(x)$, which have the usual analytic properties, and also satisfy the usual geometric intuition?


It seems like most calculus textbooks sort of weasel out on these questions. Usually, they ignore the first two questions pretty much completely, and their derivation on the third point is a filling-out of the following outline:

  • (1) Define $\sin(x)$ as height of triangle of central angle $x$ inscribed in the unit circle, where $x$ is in radians, and $\cos(x)$ as the length of its base. Assume the usual values for these functions at $k(\frac{\pi}{2}), k \in \mathbb{N}$.
  • (2) Notice, by assuming (a) that $A = \pi r^2$ and $C = 2\pi r$, and (b) that the area of a sector is proportionate to the length of arc subtended by the angle on the circumference, that the area $S$ of a sector of angle $x$ in a unit circle is $\frac{x}{2}$, since $$\frac{S}{\pi r^2} = \frac{x}{2 \pi r}, r = 1 \implies S = \frac{x}{2}$$ (Apparently, assumption (b) is Euclid VI 33. I haven't studied the proof, though.)
  • (3) Prove, using a geometric argument, that $\frac{\sin(x)}{2} < \frac{x}{2} < \frac{\tan(x)}{2}$ for $x \in [0,\frac{\pi}{2})$. Deal similarly (not identically) with $(\frac{-\pi}{2}, 0]$. Prove that we always have $1 > \frac{\sin(x)}{x} > \cos(x)$ on $(\frac{-\pi}{2}, \frac{\pi}{2})$.
  • (4) Derive (by a geometric argument, as done here) the usual angle addition formulas.
  • (5) Noting that $|\sin(x)| < |x|$ (geometrically), conclude that $\lim_{x \to 0} \sin(x) = 0$. Use the identity - derived from (4) - that $$\cos(x) = 1 - 2\sin^2(\frac{x}{2}) = (1 - \sqrt{2}\sin(\frac{x}{2}))(1 + \sqrt{2}\sin(\frac{x}{2}))$$ and the product theorem for limits to conclude that $\lim_{x \to 0} \cos(x) = 1$.
  • (6) Use the "Squeeze Theorem" and (3) to prove that $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$, $\lim_{x \to 0} \frac{\cos(x) - 1}{x} = 0$. Use (4) and (5) to establish continuity at all other values - e.g., $$\lim_{h \to 0} \sin(x_0 + h) = \lim_{h \to 0}\sin(x_0)\cos(h) + \sin(h)\cos(x_0) = \sin(x_0)$$
  • (7) Now prove that $\sin(x), \cos(x)$ are differentiable.

A notably different approach is that of Spivak's Calculus. Spivak tacitly assumes that the first question has been answered, and notices that in that case,

$$ \pi = 2\int_{-1}^{1} \sqrt{1-x^2} dx $$

which resolves the second question, albeit not too directly. Also assuming that the area of a sector of an angle of $x$ radians is $\frac{x}{2}$, he defines

$$ A(x) = \frac{x \sqrt{1- x^2}}{2} + \int_{x}^{1} \sqrt{1-t^2} dt$$

The area function is a function of the $x$-coordinate, not an angle $x$; it tends from $\frac{\pi}{2}$ to $0$ as $x$ goes from $-1$ to $1$. However, $\forall x \in [0, \pi]$, we have $\exists !y \in [-1, 1]: A(y) = \frac{x}{2}$; this $y$ we set as $\cos(x)$, and we define $\sin(x) = \sqrt{1 - \cos^2(x)}$. (The uniqueness of the value of $\cos(x)$ is guaranteed by the fact that $A(x)$ is decreasing and continuous.) The remainder of Spivak's derivation is about extending these functions (by symmetry) to the rest of $\mathbb{R}$.


Although I am familiar with these derivations, they are rather prominently silent about the first question; I've never seen any answer to that question which impressed me as rigorous. I am not at all sure that there is an actual cut which one can write down (i.e., in set-builder notation, as an explicit subset of $\mathbb{Q}$ in the usual way) for $\pi$.

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Regarding your last question (an explicit Dedekind cut for $\pi$), just write down the left and right Riemann sums approximating a suitable integral whose value is $\pi$. –  Qiaochu Yuan Oct 20 '12 at 3:21
    
@QiaochuYuan I do not follow you. How do we know that the cut of all rational approximating sums is in fact $\pi$? –  user1296727 Oct 20 '12 at 3:27
    
To write out details is lengthy. It is cleanest to start with properties of power series, in the reals or the complex numbers, and define $\cos x$ and $\sin x$ via these, or define $e^z$. Alternately but less popularly, start by establishing properties of second-order DE with constant coefficients. However, if we just want circumference and area, it is not hard to show that there is a constant $k$ such that the area of a circle of radius $k$ is $kr^2$, and then prove using the arclength formula, without introducing $\cos$ or $\sin$, that the circumference is $2kr$. –  André Nicolas Oct 20 '12 at 3:42
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@user1296727: It is all done in a very old book by Landau (Foundations of Analysis), and undoubtedly in quite a few other places. And it takes quite a few pages, which it would be unreasonable to type out in an answer. –  André Nicolas Oct 20 '12 at 3:59
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@EuYu: For the area of a circle, it is $4$ times $\int_0^r\sqrt{r^2-x^2}\,dx$. Make the change of variable $x=ru$. We get $r^2\int_0^1\sqrt{1-u^2}\,du$, so $r^2$ times a constant. For the circumference, write down the usual integral, and evaluate it not by substitution but by integration by parts. –  André Nicolas Oct 20 '12 at 4:04
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3 Answers

up vote 8 down vote accepted

The following answers your first question. We do not mention sine or cosine. We will use some basic integration techniques, but of course we will not use trigonometric substitution!

Define the circle of radius $r$ with centre the origin by the equation $x^2+y^2=r^2$, and the disk by $x^2+y^2\le r^2$.

We first show that the area of a disk of radius $r$ is a constant times $r^2$. By symmetry the area is $$4\int_0^r\sqrt{r^2-x^2}\,dx.$$ Make the change of variable $x=rt$. Our area is $$r^2\left(4\int_0^1 \sqrt{1-t^2}\,dt\right).$$ Thus $4\int_0^1 \sqrt{1-t^2}\,dt$ is the desired constant. We could call it $\pi$. But let us call it $4k$.

For the circumference, the usual arclength formula, after some simplification, gives that the circumference is $$4\int_0^r \frac{r\,dx}{\sqrt{r^2-x^2}}.$$ The change of variable $x=rt$ transforms this to $$r\left(4\int_0^1 \frac{dt}{\sqrt{1-t^2}}\right).$$ So we have $r$ times a constant. Which constant?

We will evaluate $\int_0^1\sqrt{1-t^2}\,dt$ (yes!) in a funny way, by parts. Let $u=\sqrt{1-t^2}$ and $dv=dt$. Then $du=-\frac{t}{\sqrt{1-t^2}}\,dt$ and $v=t$. After a little while we find that $$\int_0^1\sqrt{1-t^2}\,dt=\int_0^1 \frac{t^2\,dt}{\sqrt{1-t^2}}.$$ But the numerator on the right is $1-(1-t^2)$. And $\dfrac{1-t^2}{\sqrt{1-t^2}}=\sqrt{1-t^2}$. Thus $$\int_0^1\sqrt{1-t^2}\,dt=\int_0^1 \frac{dt}{\sqrt{1-t^2}}-\int_0^1\sqrt{1-t^2}\,dt.$$ We conclude that $$\int_0^1 \frac{dt}{\sqrt{1-t^2}}=2\int_0^1\sqrt{1-t^2}\,dt=2k.$$ It follows that the circumference of a circle of radius $r$ is $8kr$.

Remark: One can introduce the trigonometric functions via integrals, as mentioned in the OP. Then their basic properties, such as the addition laws, are not difficult to derive. It is, however, mildly tedious.

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+1 for writing these arguments out in full (hopefully they'll be as useful to others as they are to me :) ). –  user1296727 Oct 20 '12 at 6:13
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While I don't mean to deprive you of your triumph, there is actually something bothering me here. The integral you set up for arclength is an improper integral; the only theorem I know for dealing with arclength requires that the function be Riemann integrable on the interval in question. How can we account for that? –  user1296727 Oct 20 '12 at 21:22
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@user1296727: Sharp of you to notice! I had decided to elide over that detail. In principle I should have integrated from $0$ to $1-\epsilon$, and then let $\epsilon\to 0^+$. It is a standard purely technical process. And we can even bypass it completely, by working with the one-eighth of the circle that goes from $x=0$ to $x=r/\sqrt{2}$. –  André Nicolas Oct 20 '12 at 21:37
    
That's great! I'm wondering if this can be done more or less purely algebraically without integrals. For example if we look at the area of the n-gon inscribed in the circle, its area approaches that of the circle's as n approaces infinity. But the area is usually defined with trigonometric functions here. –  Aleks Vlasev Oct 21 '12 at 1:17
    
@AleksVlasev: That is a rather different approach, but not all that different, it is a form of integration in which one is repeatedly bisecting angles. It does not require explicit mention of trigonometric functions. The repeated bisection of the angles yields areas that can be computed via the Pythagorean Theorem. This puts us to the approach of Archimedes, and yields an expression for $\pi$ due to Vieta. –  André Nicolas Oct 21 '12 at 6:33
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I hope you won't mind if I answer in an outline. As André Nicolas says, the full details would take a long time to write out. My preferred strategy is to begin at the end; that is, to construct the trigonometric functions first.

  1. Prove that there exists a unique differentiable function $\text{cis}(t)$ from $\mathbb{R}$ to $\mathbb{C}$ satisfying the differential equation $\frac{d}{dt} \text{cis}(t) = i \text{cis}(t)$ with initial condition $\text{cis}(0) = 1$. (Existence can be proven using the matrix exponential. Uniqueness is straightforward.)
  2. Compute that $\frac{d}{dt} |\text{cis}(t)|^2 = 0$, hence by the mean value theorem, $|\text{cis}(t)| = 1$ identically.
  3. Show that $\text{cis}(s + t) = \text{cis}(s) \text{cis}(t)$. (Hint: both sides, as a function of $t$, satisfy the same differential equation with the same initial conditions.)
  4. Show that $\text{cis}(t)$ is periodic, and define $2 \pi$ to be its period. (I think I used to know an elementary proof of this but the current proof I have in mind requires a little topology.)
  5. By bullet 2, $\text{cis}(t)$ traces out the unit circle exactly once in a period and has constant speed $1$, so $2 \pi$ is the circumference of the unit circle.
  6. Conclude that the area of the unit circle is $\pi$. (There are several ways to do this; the most straightforward would just be to write the area as an integral over circumferences. A slick way is to use Green's theorem.)
  7. Conclude that $\pi = 2 \int_{-1}^1 \sqrt{1 - x^2} \, dx$. Approximating this integral by Riemann sums gives a Dedekind cut describing $\pi$.

If you believe that the integrals you were taught in calculus to compute areas and arc-lengths really do compute areas and arc-lengths, then the result for general circles follows by scaling. If not, how are you defining areas and arc-lengths?

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I'm sorry, but I haven't had any complex analysis, or topology. I don't know what to make of this derivation. –  user1296727 Oct 20 '12 at 4:01
    
There is no complex analysis here; a complex-valued function of a real variable is just a pair of real-valued functions of real variables. Have you taken either multivariable calculus or differential equations? –  Qiaochu Yuan Oct 20 '12 at 4:02
    
I had a weak course in the former and none in the latter. –  user1296727 Oct 20 '12 at 4:06
    
Okay. Come back to this answer after you've taken a course in differential equations; existence and uniqueness for ODEs is implicitly crucial in the first step (and is in my mind the main conceptual point of this outline). –  Qiaochu Yuan Oct 20 '12 at 4:07
    
+1 for the considered effort. :) –  user1296727 Oct 20 '12 at 4:10
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If you believe in limits, Archimedes' approach to the calculation of $\pi$ satisfies the first bullet. This is one definition of $\pi$.

For the second bullet, you just need the ability to compare any two rationals. Then follow through Archimedes' calculation and eventually any rational will be one side of $\pi$ or the other. Completeness says that $\pi$ exists.

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If you mean the method of exhaustion, well... no, actually, I wouldn't be thrilled to take that as a proof here. –  user1296727 Oct 20 '12 at 3:51
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@user: why not? –  Qiaochu Yuan Oct 20 '12 at 3:57
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@user1296727: I don't think this is exhaustion. $\pi$ is sandwiched between two calculations of rationals based on polygons of increasing numbers of sides. It certainly needs the completeness of the reals, as the proof fails in the rationals. –  Ross Millikan Oct 20 '12 at 4:01
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