Irrationality of pi: curiouser and curiouser

I’ve been remiss in posting here lately, which I will attribute to Christmas and New Year travelling and general craziness, and then starting a new semester craziness… but things have settled down a bit, so here we go again!

Since it’s been a while since my last post in this series, here’s a quick recap: I’m presenting a proof by Ivan Niven that $\pi$ is irrational, that is, that it cannot be represented as the ratio of two integers (and hence its decimal expansion goes on forever without repeating). My first post just gave some background and an outline of the general argument. In my second post, we began by assuming that $\pi$ is rational, and defined the function

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

(really, a family of functions, one for each value of ) where and are the “numerator” and “denominator” of $\pi$ . We then showed that $f(0) = f(\pi) = 0$ , and in fact that f(x) is symmetric, with $f(\pi - x) = f(x)$ . In my third post, we showed that all the derivatives of f(x) take on integer values when evaluated at both 0 and $\pi$ . We’re about halfway there! Today we’ll continue by defining a new function F(x) in terms of f(x) , and show some of its properties. Recall too our overall plan: we’re going to wind up with an integral which is strictly greater than 0, strictly less than 1, and also an integer! Since this is clearly nonsense (there are no integers between 0 and 1) we will conclude that our initial assumption—that $\pi$ is rational—was bogus, and that $\pi$ must be irrational after all.

So without further ado, here’s our new function F(x) . Actually, this too is technically a family of functions F_n(x) , one for each ; but again, everything we prove about it will be true no matter what is.

$\displaystyle F(x) = f(x) - f^{(2)}(x) + f^{(4)}(x) - \dots + (-1)^n f^{(2n)}(x).$

In words, F(x) is the alternating sum of all the even derivatives of f(x) . (I say “all” because, as noted in my last post, any derivative of f(x) higher than is zero.) Using Sigma notation, we can also write this more concisely as

$\displaystyle F(x) = \sum_{i = 0}^n (-1)^i f^{(2i)}(x).$

There are a few things to note. First, think what happens when we evaluate F(0) : since all the derivatives of f(x) take on integer values at 0, and F(x) is just a sum of a bunch of derivatives of f(x) , F(0) must be an integer too. Of course, the same thing goes for $F(\pi)$ .

Next, consider

$F^{\prime\prime}(x) + F(x).$

Since the derivative of a sum is the sum of the derivatives, we can compute $F^{\prime\prime}(x)$ as

$F^{\prime\prime}(x) = f^{(2)}(x) - f^{(4)}(x) + \dots + (-1)^{n-1}f^{(2n)}(x).$

That is, f(x) turns into $f^{(2)}(x)$ , $-f^{(2)}(x)$ turns into $-f^{(4)}(x)$ , and so on. “But wait a minute,” you say. “Shouldn’t the $(-1)^n f^{(2n)}(x)$ at the end of F(x) turn into $(-1)^n f^{(2n+2)}(x)$ in $F^{\prime\prime}(x)$ ?” In fact, it does—but as noted before, $f^{(2n+2)}(x)$ is zero, so that term just goes away. Now we note that every term of F(x) has a corresponding term in $F^{\prime\prime}(x)$ of the opposite sign, except f(x) , which has no corresponding term. So when we add F(x) and $F^{\prime\prime}(x)$ , everything cancels except f(x) :

$F^{\prime\prime}(x) + F(x) = f(x).$

Astute readers will note a funny resemblance between the definition of F(x) and the Taylor series for $\cos(x)$ … and indeed, next time we’ll start making some connections with our old trigonometric friends, $\sin$ and $\cos$ .

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