What pedagogical motivation is there for the existence/approximation of pi?

[Dr. Seafood]

This is being discussed in micromass's "Math stuff that hasn't been proven" thread, but I want to be particular about this topic.

Essentially, I think I'm looking for a "proof" or derivation of pi from few first principles. Honestly I have no idea which is the "purest", most motivating question to ask that reveals the number [itex]\pi[/itex] as a constant.

After defining some (relatively) intuitive axioms, I think it's okay to forsake some rigor (proofs of existence/uniqueness) so that we can really understand the meat of the origins of [itex]\pi[/itex]. Probably, a few first questions are: What is a circle? Why is the ratio of a circle's circumference to its diameter a constant; moreover, why is it a number between 3 and 4? Which "first principles" should we accept for the derivation of this number? How can we use a limiting expression to evaluate a decimal approximation for pi?
The goal is to carefully illustrate how we can begin with definitions, and arrive at a useful definition and approximation for pi.

To make analogy: teaching elementary calculus a few weeks ago, I tried to develop the use of e. I'm pretty much looking for a similar derivation of pi. Obviously it would rely more on a geometrically motivated problem, instead of a beginner's calculus one. Here is a quick recap of how I did this. Sorry for tl;dr.

Begin by defining exponentiation and exponential function, and making continuity arguments to extend "repeated multiplication" to entire real domain. The "motivating question" is "What is [itex]{d \over dx} {b^x}, \quad b > 1[/itex]"?

Using Newton's quotient (first principles) to differentiate, after some algebra which should be familiar to you all, we get
[tex]{{d \over dx} {b^x}} = {b^x} {\lim_{\Delta{x} \rightarrow 0} {{{b^{\Delta{x}} - 1} \over {\Delta{x}}}}} \qquad \forall x \in ℝ[/tex]
Now we make an assumption, for the sake of teaching, that the limit involved here exists. Let [itex]M = {\lim_{\Delta{x} \rightarrow 0} {{{b^{\Delta{x}} - 1} \over {\Delta{x}}}}}[/itex] so that we can write [itex]{{d \over dx} {b^x}} = M{b^x}[/itex]. Note that since M is just the definition of the derivative (Newton's quotient) of b^x at x = 0, geometrically this limit is the slope of the exponential function at x = 0. Another assumption, again for pedagogy instead of rigor: we assert that a unique number [itex]e[/itex] exists so that [itex]M = {\lim_{\Delta{x} \rightarrow 0} {{{e^{\Delta{x}} - 1} \over {\Delta{x}}}}} = 1[/itex]. The derivative of e^x at x = 0 is exactly 1. We persuade by claiming k ∈ ℝ can be chosen so that y = b^kx = (b^k)^x has slope 1 at x = 0, i.e. b^k = e: choose k = 1/M; we can always "stretch" an exponential function so that its slope at x = 0 is 1, or so that the base of the function is e. Obviously the coolest result here is that [itex]{d \over dx} e^x = e^x[/itex]!

Now define ln x to be the function such that [itex]\ln(e^x) = x[/itex] and [itex]e^{\ln x} = x[/itex]. Note then by this definition, ln 1 = 0 and ln e = 1. To find [itex]{d \over dx} \ln x[/itex], let y = ln x so x = e^y, and differentiate implicitly to yield [itex]{{d \over dx} \ln x} = {1 \over x}[/itex].

Observe that b = e^{ln b} holds, so (using the chain rule):
[tex]{{d \over dx} b^x} = {{d \over dx} e^{{x}{\ln b}}} = ({\ln b})e^{{x}{\ln b}} = ({\ln b})b^{x}[/tex]
We can see that the limit M = ln b, solving the problem of finding the derivative of a general exponential function. Now we want to achieve a decimal approximation for the "magic exponential number" e. Using the continuity of ln (limit of log is log of limit), we evaluate the following limit:
[tex]{\lim_{x \rightarrow \infty} (1 + {1 \over x})^x}[/tex]
Let [itex]{\Delta{x}} = {1 \over x}[/itex]. Take the log of the limiting expression to get
[tex]{\ln({(1 + {1 \over x})^x})} = x\ln(1 + {1 \over x})
= {1 \over \Delta{x}}{\ln(1 + \Delta{x})} = {1 \over \Delta{x}}({\ln(1 + \Delta{x}) - \ln 1})[/tex]
The far right-side is Newton's quotient for ln x at 0, but we know this is just 1. The result follows from continuity of log:
[tex]\ln {\lim_{x \rightarrow \infty} (1 + {1 \over x})^x} = {\lim_{x \rightarrow \infty} \ln(1 + {1 \over x})^x} = {\lim_{\Delta{x} \rightarrow 0} \ln(1 + \Delta{x})^{1 \over \Delta{x}}} = 1 \implies {\lim_{x \rightarrow \infty} (1 + {1 \over x})^x} = e^{\lim_{\Delta{x} \rightarrow 0} \ln(1 + \Delta{x})^{1 \over \Delta{x}}} = e^1 = e[/tex]
We can evaluate the limiting expression for large values of x to yield approximations to e, namely [itex]e \approx 2.71828[/itex].

 

[micromass]

I think that a pure derivation from first principles would inevitably involve measure theory. Why?? Because arc length and area are measure theoretic principles.

I'll think about it, but did you see measure theory already?? It would be rather difficult if you didn't...

 

[chiro]

Couldn't you use the idea of Archimedes himself?

Proscribe a polygon in a circle and find an expression of the length of all the sides as a function of the radius and the number of sides used to approximate the perimeter.

 

[micromass]

OK, here is something that should work.
Firstly, I'll assume that the exponential function [itex]e^x[/itex] is well defined on [itex]\mathbb{R}[/itex] (you did this already). We can extend the exponential function to complex numbers by

[tex]e^x=\sum{\frac{x^n}{n!}}[/tex]

We now define

[tex]\cos(x)=\frac{e^{ix}+e^{-ix}}{2}~\text{and}~\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}[/tex]

From the properties of the exponential, we can show that

• [itex]\overline{e^x}=e^{\overline{x}}[/itex], thus cos(x) and sin(x) are real if x is real.
• [itex]e^{ix}=\cos(x)+i\sin(x)[/itex]
• cos(0)=1, sin(0)=0
• [itex]|e^{ix}|=1[/itex] if x is real.
• [itex](\sin(x))^\prime=\cos(x),~(\cos(x))^\prime=\sin(x)[/itex]

Lemma: There exists a positive number x such that cos(x)=0
Proof: Suppose that this is not the case. Since cos(0)=1, it follows that cos(x)>0 for all positive x. Thus sin(x) is strictly increasing. And since sin(0)=0, it follows that sin(x)>0 for all positive x. Thus if 0<x<y, we have
[tex]\sin(x)(x-y)\leq \int_x^y{\sin(t)dt}=\cos(x)-\cos(y)\leq 2[/tex]
But since sin(x)>0, we have that the above cannot be true for large y. This is a contradiction.

Recall the the zero's of continuous functions form a closed set. Thus the following makes sense:

Definition: We define pi as the least number x such that cos(x/2)=0.

It can now be shown that

• the exponential function is periodic with period 2pi i
• sine and cosine are periodic with period 2pi
• Polar Decomposition: If z is a nonzero complex number, then there exists a unique r and a unique [itex]t\in [0,2\pi[[/itex] such that z=r(cos(t)+isin(t)).

The proofs of all of this can be found in Rudin, page 182.

In the following post, I'll make clear how to relate our pi to the circle.

 

[micromass]

The idea here is from Billingsley's "probability and measure". Exercise 12.6

Let [itex]\pi[/itex] be defined in my last post. And let [itex]\pi_0[/itex] be the surface area of the closed disk. The idea is to show [itex]\pi=\pi_0[/tex].

(a) Let [itex]T_\phi[/itex] be the linear map with matrix

[tex]\left(\begin{array}{cc} \cos(\phi) & -\sin(\phi)\\ \sin(\phi) & \cos(phi) \end{array} \right)[/tex]

This matrix rotates the plane through the angle [itex]\phi[/itex] (by the addition formula's). It also preserves area.

(b) Let [itex]S_{\alpha,\beta}[/itex] be the sector consisting of all the points in the unit disk with polar coordinates between alpha and beta. It can be shown that
[tex]area(S_{\alpha,\beta})=\frac{\pi_0(\theta_2-\theta_1)}{2\pi}[/tex]

(c) by Fubini's theorem, it follows for [itex]0<\theta<\pi/2[/itex] that

[tex]area(S_{0,\theta})= \frac{1}{2}\sin(\theta)\cos(\theta) + \int_{\cos(theta)}^1{\sqrt{1-x^2}dx}[/tex]

(d) equate (b) and (c) and differentiate both sides. This concludes that [itex]\pi=\pi_0[/itex].

 

[HallsofIvy]

Euclid showed, in his "Elements", that all "circles are similar". That is, if you take two circles, of radius r, circumference c, and radius R, circumference C, then r/R= c/C. From that, it follows that c/r= C/R.

He did it, if I recall correctly, by inscribing regular polygons with 3, 4, ..., n sides in the circle, showing that the ratio of the perimeter of a polygon to circumradius (distance from center to a vertex) is independent of the radius, then arguing that, since, as n gets larger and larger, the polygon approaches the circle, the same must be true for the circle. That argument can be made rigorous using limits.

 

[Dr. Seafood]

micromass, as much sense as that all makes, you delved into calculus (series, def integrals), arithmetic with complex numbers, and even some linear algebra to give a geometric result. Unfortunately not exactly what I'm looking for. I guess it doesn't have the 'beauty" I'm expecting, also that to get an approximation for [itex]\pi[/itex] in that manner you'd probably have to prove one of those crazy approximating series that converge to [itex]\pi[/itex]. But defining [itex]\pi[/itex] to be the least number [itex]\theta[/itex] such that [itex]\cos({\theta \over 2}) = 0[/itex] seems like a good idea that doesn't stray from plane geometry (we define [itex]\cos(\theta)[/itex] to be the x-coordinate of the point (1, 0) after rotation along the unit circle by [itex]\theta[/itex] radians). Still, I suppose I don't really have a good idea of what exactly I'm looking for here.

HallsofIvy, I think I'm going to read Elements. I spend a lot of time reading about math but I've never read it. MUST READ

 

[Bohrok]

··· = ¹/_Δx(ln(1 + Δx) - ln1)

The far right-side is Newton's quotient for ln x at 0, but we know this is just 1.

Could you explain what you mean by Newton's quotient for ln x at 0? Shouldn't it be ln(1 + x)?

I really like your derivation of e; I've been wanting to see something like that for a while.

 

[Dr. Seafood]

^ Woops, sorry, that is Newton's quotient for ln x at 1. My mistake, I'll fix it now.

EDIT: ... Except I can't edit the OP post anymore ... Well, this stands corrected.

I think that a pure derivation from first principles would inevitably involve measure theory. Why?? Because arc length and area are measure theoretic principles.

I'll think about it, but did you see measure theory already?? It would be rather difficult if you didn't...

Well, measure theory is a bit abstract for a specific problem like this one, I think. When I say "motivation" here, I mean for a person who wants to learn geometry with a full understanding of all the concepts while taking as little for granted as possible, but also without straying far from geometry.

Couldn't you use the idea of Archimedes himself?

Proscribe a polygon in a circle and find an expression of the length of all the sides as a function of the radius and the number of sides used to approximate the perimeter.

I've definitely been reading about this a lot in my pi quest here. I think the approximation to pi will be hiding in a limiting expression, similar to my "derivation" of e above. I'll read more about Archimedes and Euclid's work on this topic.

 

[Dr. Seafood]

I think we ought to start out as follows, with these definitions. Define a circle geometrically as the collection of points a fixed distance r (radius) from a fixed center (origin). By Pythagoras's theorem (which we will take for granted), it follows that all points (x, y) on the circle are such that x² + y² = r² (sum of squares of coordinates is equal to square of radius). Suppose an angle is subtended by an arc of a circle of radius r, and that this arc has length L. Define this angle to have radian measure L/r. Then one radian is the angle subtended by an arc of length r. It follows that a "half-rotation", or the angle subtended by a semi-circle, is (C/2)/r = C/2r where C is the circumference of the circle and r is its radius. Define [itex]\pi[/itex] as the half-rotation in radians (it would be so much nicer to define [itex]\tau[/itex] as the full rotation, but that's a story for a later day). Geometrically speaking, clearly [itex]\pi[/itex] both exists and is unique, and is a real number. We get [itex]C = 2\pi r[/itex]. Our job is to approximate [itex]\pi[/itex]. We can prove the lemma that all circles are similar to one another, i.e. ratios between dimensions are constant -- so we can proceed without loss of generality using the unit circle as our primary object of focus. Also, similarity of circles ensures that radians are well-defined. Note [itex]\pi \over 2[/itex] is a quarter rotation, [itex]2\pi[/itex] is a complete rotation, etc.

Define the ray (x, y) as the line segment from the origin to (x, y). Draw the ray from the origin to a point (x, y) on the circle, and measure the angle [itex]\theta[/itex] between the rays (1, 0) and (x, y) in radians. Note that [itex]\theta[/itex] is also the ratio of the arc length from (1, 0) to (x, y) to the radius of the circle. If the circle is the unit circle, then [itex]\theta[/itex] is simply the arc length. Define the functions "sine" and "cosine" to be such that, in this context, [itex]x = \cos\theta[/itex] and [itex]y = \sin\theta[/itex]. Essentially, if we rotate the point (1, 0) along the circumference of a circle by an angle [itex]\theta[/itex], its new coordinates are [itex](\cos\theta, \sin\theta)[/itex]. The Pythagorean identity immediately follows, as do the symmetrical, periodic, and bounded natures of sine and cosine (coordinates change sign roughly after each quarter rotation (plane quadrant); rotated coordinates "repeat" after a full rotation, i.e. [itex]2\pi[/itex] radians, etc). Perhaps it is nontrivial that sine and cosine are both bounded by [-1, 1], in which case a brief (geometric, since this is all we really have to work with here) proof should be provided.

It also follows that [itex]\cos(\pi) = -1, \sin(\pi) = 0[/itex]. Sine and cosine achieve these respective values (and any other values) for infinitely many arguments, by their periodicity; but note that [itex]\pi[/itex] is the least positive value for [itex]\theta[/itex] such that both [itex]\cos(\theta) = -1, \sin(\theta) = 0[/itex] hold.

Brb, reading Elements.

EDIT: Also observe that we have defined [itex]\pi[/itex] to be the number such that a circle with radius r has circumference [itex]C = 2\pi r[/itex] (it followed immediately from definition). Then the task here is not to prove this circumference formula -- it is to approximate our magic "half-rotation number".

Do we need sines and cosines? Do we make Archimedes's argument using limits of inscribed regular polygons?