Archive for the ‘calculus’ Category

Irrationality of pi: the integral that wasn’t

Thursday, February 11th, 2010

Irrationality of pi

  1. Irrationality of pi
  2. Irrationality of pi: the unpossible function
  3. Irrationality of pi: derivatives of f
  4. Irrationality of pi: curiouser and curiouser
  5. Irrationality of pi: the impossible integral
  6. Irrationality of pi: the integral that wasn’t

And now for the punchline! Today we’ll show that, for large enough values of n,


0 < \int_0^\pi f(x) \sin (x)\, dx < 1,

completing the proof of the irrationality of \pi.

First, let’s show that f(x) \sin(x) is positive when 0 < x < \pi. We know that \sin(x) is positive for 0 < x < \pi. But I claim that f(x) is too. Remember that


\displaystyle f(x) = \frac{x^n(a - bx)^n}{n!}.

n! and x^n are clearly positive when x is positive; and a - bx is also positive when x < \pi = a/b. From here we simply note that if a function is positive over an entire interval, the integral of the function over that interval will be positive as well.

For the second part, note first that for 0 < x < \pi,


\displaystyle f(x) \sin(x) < \frac{\pi^n a^n}{n!}.

Why is this? Well, clearly x^n < \pi^n (since x < \pi), and also a - bx < a (since x > 0) and hence (a - bx)^n < a^n, so we conclude that


\displaystyle f(x) = \frac{x^n(a - bx)^n}{n!} < \frac{\pi^n a^n}{n!}.

This doesn’t yet include the \sin(x), but notice that multiplying by \sin(x) can only make things smaller, since \sin(x) is at most 1. Now, here’s the slightly sneaky part: I claim that we can make \frac{\pi^n a^n}{n!} as small as we want by making n big enough. Why is this? Notice that we can rewrite it as


\displaystyle \frac{\pi^n a^n}{n!} = \frac{\pi a}{1} \cdot \frac{\pi a}{2} \cdot \frac{\pi a}{3} \dots \frac{\pi a}{n}.

Now, a—the “denominator” of \pi—might be very large. It might have fourteen million zillion digits. But no matter how big a is, there will of course be an integer z which is bigger than \pi a, so \frac{\pi a}{z} < 1. And then \frac{\pi a}{z + 1} < 1, and \frac{\pi a}{z + 2} < 1, and so on... of course, multiplying by something less than one makes things smaller. And it might take a really really long time to cancel out the enormous product \frac{\pi a}{1} \cdot \frac{\pi a}{2} \dots \frac{\pi a}{z}, but if we just wait patiently it will get smaller and smaller... and eventually there will come some n for which


\displaystyle \frac{\pi a}{1} \cdot \frac{\pi a}{2} \dots \frac{\pi a}{z} \cdot \frac{\pi a}{z + 1} \dots \frac{\pi a}{n} < 1.

Actually, even this isn’t quite small enough: we want the integral from 0 to \pi of f(x) \sin(x) to be less than 1. But that’s not a problem; to ensure that we can just pick n big enough so that f(x) < 1/\pi (if the graph of f(x) fits inside a \pi by 1/\pi box, then its integral on this interval must be less than the area of the box).

Voila! An integral which is an integer absurdly between 0 and 1, all because we assumed \pi was rational.

The inescapable conclusion, which probably would have driven the ancient Greeks crazy, is that \pi is irrational!

« Previous in series |

Tags: , , , ,
Posted in algebra, calculus, convergence, famous numbers, proof, trig | 8 Comments »

Irrationality of pi: derivatives of f

Sunday, December 20th, 2009

Irrationality of pi

  1. Irrationality of pi
  2. Irrationality of pi: the unpossible function
  3. Irrationality of pi: derivatives of f
  4. Irrationality of pi: curiouser and curiouser
  5. Irrationality of pi: the impossible integral
  6. Irrationality of pi: the integral that wasn’t

In my previous post in this series, we defined the function


\displaystyle f(x) = \frac{x^n(a - bx)^n}{n!}

and showed that f(0) = f(\pi) = 0. Today we’ll show the surprising fact that, for every positive integer i, although f^{(i)}(0) and f^{(i)}(\pi) are not necessarily zero, they are always integers. (The notation f^{(i)} means the ith derivative of f; that is, take the derivative of f, then the derivative of that, then the derivative of that, … i times.) Put more succinctly: every derivative of f takes on integer values at x = 0 and x = \pi.

Why might this be surprising? It’s surprising because of the n! in the denominator of f. For example, consider the function (which I just made up):


g(x) = \frac{x^3 + 5x}{6}.

It’s easy to see that g(0) = 0. But let’s take the derivative: g'(x) = \frac{3x^2 + 5}{6}, so g'(0) = 5/6, which is clearly not an integer. For the derivatives of f to always give an integer at x = 0 (let alone at x = \pi) there must be some fancy canceling going on!

For now we will consider only f^{(i)}(0) (we’ll come back to f^{(i)}(\pi) later). Of course, substituting 0 for x causes every term containing x to disappear, so f^{(i)}(0) is just the constant term of f^{(i)}(x). Hence, we must show that the constant term of f^{(i)}(x) is always an integer.

Consider the numerator of f(x), that is,


n!f(x) = x^n (a - bx)^n

Note that (a - bx)^n, when expanded out, is a polynomial of the form a^n - \dots + (-b)^n x^n, where the ellipsis contains a bunch of terms with integer coefficients and powers of x between 1 and n-1. (In fact, we could use the Binomial Theorem to compute the precise coefficients—but it really doesn’t matter; all we will care about is that they are integers.) Multiplying by x^n, we see that


n!f(x) = a^n x^n - \dots + (-b)^n x^{2n}

so n!f(x) is a polynomial with terms of degree n through 2n, and hence so is f(x), since dividing by n! changes the coefficients but not the exponents. (Note that f(x) has no constant term, so f(0) = 0—but we already knew that.)

Recall that the derivative of x^k is k x^{k-1}, so taking the derivative of a polynomial reduces each of the exponents by one. So the first derivative of f(x) is a polynomial with terms of degree n-1 through 2n - 1 (and hence a constant term of zero); the second derivative has terms of degree n-2 through 2n - 2 (still no constant term); and so on. We can see that none of the first n-1 derivatives of f(x) will have a constant term, so f^{(i)}(0) = 0 (which is certainly an integer) for i < n. What about the nth derivative and higher? This is where the fancy canceling comes in!

As we noted above, when expanded out f(x) is a sum of a bunch of terms of the form


\displaystyle \frac{c_i x^{n+i}}{n!}

where 0 \leq i \leq n and c_i is some integer. When we take the derivative, this term will turn into (n+i) c_i x^{n+i-1}/n!; if we take the derivative again, it will become (n+i)(n+i-1) c_i x^{n+i-2}/n!; another derivative gives us (n+i)(n+i-1)(n+i-2) c_i x^{n+i-3}/n!, and so on. Do you see what is happening? After taking the derivative exactly n+i times, we will end up with the constant term


\displaystyle \frac{(n+i)! c_i}{n!}

and here’s our fancy canceling: (n+i)! is clearly divisible by n!, so this is some integer times c_i, which is also an integer. Voila! Said a different way, and more succinctly: since each term of f(x) has degree at least n, by the time we have taken the derivative enough times for it to yield a constant term, the n! will be canceled from the denominator, since we will have taken the derivative at least at each power of x from n down to 1.

Finally, if we take the derivative of f more than 2n times, we get 0, so no problems there.

Great, so f^{(i)}(0) is always an integer. But what about f^{(i)}(\pi)? Well, remember, last time we showed that f(\pi - x) = f(x). If we take the derivative of both sides with respect to x (being careful to use the chain rule on the left side, noting that the derivative of \pi - x with respect to x is -1), we get


$ \begin{align*}\frac{d}{dx}f(\pi - x) &= \frac{d}{dx} f(x) \\ -f'(\pi - x) &= f'(x) \end{align*} $

We can repeat this process to find that f^{(2)}(\pi - x) = f^{(2)}(x) (the two negatives cancel on the left side), -f^{(3)}(\pi - x) = f^{(3)}(x), and so on. But the extra negative sign for odd derivatives doesn’t really matter: in either case, f^{(i)}(\pi) = \pm f^{(i)}(\pi - \pi) = \pm f^{(i)}(0), which is an integer.

Getting closer! Next time, we will define another special function F(x) in terms of f(x) and its derivatives; this function F(x) will help us compute \int_0^\pi f(x) \sin (x) dx—which (if you recall the punchline) will turn out to be an integer strictly between 0 and 1 (which is impossible).

« Previous in series | Next in series »

Tags: , , , ,
Posted in calculus, famous numbers, proof | 10 Comments »

Irrationality of pi: the unpossible function

Saturday, December 12th, 2009

Irrationality of pi

  1. Irrationality of pi
  2. Irrationality of pi: the unpossible function
  3. Irrationality of pi: derivatives of f
  4. Irrationality of pi: curiouser and curiouser
  5. Irrationality of pi: the impossible integral
  6. Irrationality of pi: the integral that wasn’t

Recall from my last post what we are trying to accomplish: by assuming that \pi is a rational number, we are going to define an unpossible function! So, without further ado:

Suppose \pi = \frac{a}{b}, where a and b are positive integers. Define the function f like this:


\displaystyle f(x) = \frac{x^n(a - bx)^n}{n!}.

(In case you’ve forgotten, n!, pronounced “n factorial,” is the product of all the numbers from 1 to n.) “OK… but… what is n?” I hear you ask. Good question. The short answer is, it doesn’t matter: n can be any positive integer. We will show a bunch of things that are true about f no matter what n is. Later, we will see that we get a contradiction only for values of n which are “big enough.” But that’s OK; since everything we prove up to that point will be true no matter what n is, we can pick a value of n which is as big as we like.

Let’s explore some properties of f(x). First, it’s easy to see that f(0) = \frac{0^n a^n}{n!} = 0. It’s not too hard to see that f(\pi) = 0 as well (remembering that \pi = a/b, of course, which means that a-b\pi = a - a = 0):

$ \begin{align*}f(\pi) &= \frac{\pi^n (a - b\pi)^n}{n!} \\ &= \frac{\pi^n 0^n}{n!} = 0.\end{align*} $

So f(x) has zeros at x = 0 and x = \pi. But more is true: in fact, f(x) is symmetric (a mirror reflection of itself) around the line x = \pi/2. That is,


f(x) = f(\pi - x) = f(a/b - x).

Let’s prove this:

$ \begin{align*}f(a/b - x) &= \frac{(a/b - x)^n(a - b(a/b - x))^n}{n!} \\ &= \frac{(a/b - x)^n(a - a + bx)^n}{n!} \\ &= \frac{(a/b - x)^n b^n x^n}{n!} \\ &= \frac{(a - bx)^n x^n}{n!} = f(x). \end{align*} $

“I don’t see what’s so unpossible about f so far,” you say? Patience! (Of course, it isn’t really f itself which is the problem; the problem is our insistence that f is actually defined in terms of the “numerator” and “denominator” of \pi…)

Next time, we’ll see that the derivatives of f also have some special properties.

« Previous in series | Next in series »

Tags: , , , ,
Posted in calculus, famous numbers, proof | 6 Comments »

Irrationality of pi

Monday, December 7th, 2009

Irrationality of pi

  1. Irrationality of pi
  2. Irrationality of pi: the unpossible function
  3. Irrationality of pi: derivatives of f
  4. Irrationality of pi: curiouser and curiouser
  5. Irrationality of pi: the impossible integral
  6. Irrationality of pi: the integral that wasn’t

Everyone knows that \pi—the ratio of any circle’s diameter to its circumference—is irrational, that is, cannot be written as a fraction a/b. This also means that \pi’s decimal expansion goes on forever and never repeats …but have you ever seen a proof of this fact, or did you just take it on faith?

The irrationality of \pi was first proved (according to modern standards of rigor) in 1768 by Lambert, but his proof was rather complicated. A more elementary proof, using only basic calculus, was given in 1947 by Ivan Niven. You can read his original paper here, but it’s rather terse! Just as I did for Calkin and Wilf’s paper, Recounting the Rationals, I plan to write a series of posts explaining Niven’s proof in a bit more detail, with some accompanying intuition. I’ll assume a basic knowledge of calculus; if you don’t know calculus, just hang tight for a few posts!

Here’s the basic outline of the proof. We begin by supposing that \pi is rational: in particular, suppose \pi = a/b for some integers a and b. We’ll then use these values of a and b to define a special function f(x), about which we will show the following:


\int_0^\pi f(x) \sin(x) dx is an integer, AND

0 < \int_0^\pi f(x) \sin(x) dx < 1.

But this is absurd! There are no integers greater than zero and less than one. The inescapable conclusion will be that our initial assumption—that \pi = a/b—was false.

In my next post, we’ll define the special function f(x) and begin exploring some of its properties.

| Next in series »

Tags: , , ,
Posted in calculus, famous numbers, proof | 13 Comments »

Carnival of Mathematics #23: Haiku Edition

Friday, December 28th, 2007

Welcome to the 23rd Carnival of Mathematics: Haiku Edition! First, I must apologize for the delay: I usually have very little trouble with my hosting provider, but of course it went down just when the CoM was supposed to be posted. But it’s free, so I can’t complain! It’s back up now, and will hopefully stay that way.

For this edition of the CoM, I decided to write a short seventeen-syllable haiku about each of the excellent seventeen submissions I received (along with additional commentary of the more prosaic variety). I’ve arranged the posts more or less in order of required mathematical background, but don’t stop halfway through because then you’ll miss the pretty pictures at the end. Enjoy!

  1. English pols want to
    make math more interesting.
    It’s not already?

    From Naomi Stevens’s Diary From England: a government bid to make maths more interesting.

  2. Neat, use perfect spheres
    to define the kilogram!
    Off by just atoms…

    Heather Lewis, of 360, writes about Australian scientists who are trying to make a perfect sphere. Pretty incredible stuff!

  3. Freshmen work in groups,
    and answer their own questions.
    Effective? Discuss.

    JackieB of Continuities explains the pedagogical approach she takes with her freshman. Be sure to read (or contribute to!) the fascinating discussion that ensues in the comments section.

  4. Multiple choice, now
    with bonus choice enhancement!
    Hard tests, nice to grade.

    Maria Andersen, at the Teaching College Math Technology Blog, shows off a new sort of multiple-choice test that’s easy to grade, but avoids many of the well-known problems with traditional multiple-choice tests. I wish I’d thought of this when I was teaching high school!

  5. Are you learning two
    languages—math AND English?
    Great sites for you here.

    Larry Ferlazzo presents a list of the best math sites for english language learners.

  6. Mathematics blogs
    are many; which are the best?
    Here’s one opinion.

    Denise of Let’s play math! writes about her favorite math blogs.

  7. I have not yet read
    “Letters To A Young Mathster”.
    I’m not missing much.

    Andrée has written a (not-too-favorable) review of Ian Stewart’s book “Letters to a Young Mathematician”, over at her blog meeyauw.

  8. Albatrosses fly
    in fractal patterns! Oh wait–
    experiment sucked.

    Julie Rehmeyer discusses how scientists are revisiting some research on fractal patterns in the flight patterns of albatross at MathTrek. Apparently, just because an albatross’s feet are dry doesn’t necessarily mean it’s flying. Who knew?

  9. Eight ninety-eight, eight
    ninety-nine, nine hundred… sigh…
    infinity yet?

    Thad Guy has a funny comic about infinity. Check out some of his other comics, too—I’m a (new) fan!

  10. Need socks in the dark?
    The pigeonhole principle
    comes to your rescue!

    Mary Pat Campbell (aka meep) presents a cute video explaining the pigeonhole principle. Did you know that at least two people in the US have the exact same number of hairs on their body? You can’t argue with math!

  11. A counting problem:
    how many bracelets are there?
    Harder than it looks…

    MathMom came across an interesting MathCounts problem involving beaded bracelets, which generated some great discussion. How would you solve it?

  12. List of rationals,
    both elegant and complete?
    Is it possible?

    Yours truly has posted the first in a planned multi-part series explaining a particularly elegant way to enumerate the positive rational numbers.

  13. Koch snowflake fractal:
    Area? Perimeter?
    Fractals are so strange…

    Over at Reasonable Deviations, rod uses geometric series to calculate the area and perimeter of the Koch snowflake. The result is rather surprising!

  14. Twelve Days of Christmas?
    How many presents is that?
    Let’s figure it out!

    Over at Wild About Math!, Sol Lederman presents a seasonally-appropriate exploration in counting presents. Fun!

  15. A tricky puzzle:
    rectangles and angle sums.
    I solved it, can you?

    JD2718 shares a gem of a puzzle involving the sum of some angles. It’s tricky—are you up to the challenge? I would especially encourage would-be solvers to come up with a nice geometric solution (I couldn’t)!

  16. Pascal’s Triangle:
    writing it out is a chore.
    How fast does it grow?

    Foxy, of FoxMaths! fame, presents an interesting two-part analysis of the asymptotic growth of the rows of Pascal’s triangle—not the growth of the actual values in the rows, but of the space needed to write them!—making use of some clever algebraic gymnastics and asymptotic analysis.

  17. In how many ways
    can the Nauru graph be drawn?
    The answer: a lot!

    David Eppstein of 0xDE presents The many faces of the Nauru graph: a collection of diverse ways to visualize a particular graph which he dubs the “Nauru graph”, due to the similarity of one of its drawings to the flag of Nauru. Planar tesselation, hyperbolic tesselation, embedding on the surface of a torus… all that and much more, with, yes, pretty pictures for everything! Even those who don’t understand the article itself should still go take a look, solely for the sake of the pictures. =)

Thanks to everyone for the great submissions, I had a fun time reading them and putting this together. The next CoM will be hosted at Ars Mathematica. As always, email Alon Levy (including “Carnival of Mathematics” in the subject line) if you’d like to host an edition.

Wait! Before you go, in honor of the new year, here’s one last link from Mike Croucher at Walking Randomly, who wants to know: what is interesting about the number 2008?

Posted in algebra, books, calculus, challenges, counting, fractals, geometry, infinity, links, meta, number theory, pascal's triangle, pattern, people, sequences, trig, video | 17 Comments »

The Most Beautiful Equation in the World

Friday, April 13th, 2007

This post is a special shout-out to my former students who are now taking calculus (if you don’t know any calculus, just hang tight… there will be more calculus-less math goodness coming your way soon). This post is 100% money-back guaranteed to not really help you at all on the AP exam! But don’t worry, it’s still awesome.

First, consider this equation:

e^{i\theta} = \cos \theta + i \sin \theta

Perhaps you’ve seen it before… among other things, it gives us a very neat connection between complex numbers expressed in polar coordinates (re^{i\theta} is the complex number with polar coordinates (r,\theta)) and expressed in rectangular coordinates. But to see where this equation comes from, you need some calculus!

Let’s set z = \cos \theta + i \sin \theta. The key observation is that differentiating z is the same as multiplying it by i (remember, i^2 = -1):

$ \begin{equation*} \begin{split} \displaystyle\frac{dz}{d\theta} &= -\sin \theta + i \cos \theta \\ iz &= i \cos \theta + i^2 \sin \theta = -\sin \theta + i \cos \theta \end{split} \end{equation*} $

So we write this observation as a simple differential equation; after swapping d\theta and z we can integrate:

$ \begin{equation*}\begin{split} \displaystyle \frac{dz}{d\theta} &= iz \\ \int \frac{dz}{z} &= \int i \, d\theta \\ \ln z &= i\theta + C \\ z &= Ce^{i\theta} \end{split} \end{equation*} $

And since z = \cos \theta + i \sin \theta is equal to 1 when \theta is zero, C must be equal to 1. Therefore we have z = \cos \theta + i \sin \theta = e^{i\theta}, as promised! Note that this makes sense with our key observation from before: differentiating e^{i\theta} gives ie^{i\theta}, the same as multiplying by i.

Now for a really good time, set \theta = \pi. Noting that \cos \pi + i \sin \pi = -1, we end up with the Most Beautiful Equation in the World:


e^{i \pi} + 1 = 0.

Almost makes you cry, doesn’t it? It relates five of the Most Important Numbers in the World (0, 1, e, i, and \pi) using three of the Most Important Operations in the World (addition, multiplication, and exponentiation) and nothing else. Beauty, simplicity, elegance–it’s all right here.

And for a short time only, this equation can be yours for the low, low price of the number of grains of rice on the last square of a chessboard if you put just one grain on the first square, two grains on the second square, four grains on the third, and merely double the number of grains each time. Order now, supplies are limited!

Posted in calculus, famous numbers, proof | 42 Comments »