Explorations in mathematical beauty

## Irrationality of pi: the unpossible function

Posted on December 12, 2009 by

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 $a-b\pi = a - a = 0$):

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

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:

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

“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.

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### 7 Responses to Irrationality of pi: the unpossible function

1. Jack says:

I like this very-short style actually, it is quite nice just to read little bits. I hope you’ll be doing this with other interesting proofs too.

2. Brent says:

Jack: I’d love to do it with other interesting proofs too. If you find any interesting ones you’d like to see explained, feel free to send me a link!

3. Dave says:

I’m with you so far! I would like to have been a fly on the wall when Niven came up with the f(x) function. I took a look at the original paper and just as I suspected, the function seems to have arrived out of thin air. And so we can observe it and dissect it, but I can only imagine what his process was like in coming up with that function. Do you have any insight on that process?

4. Brent says:

Dave: I know exactly what you mean! This is a problem I have with the way math papers are often written in general: they present only the finished results and throw away everything used to get there in the first place—which is often the most interesting part! As for this paper in particular, I’m starting to get a bit of intuition as to the process Niven might have used to arrive at f(x); hopefully my intuition will continue to grow as I write more, and I’ll be able to give some insight later. But it will have to wait a few posts, since you have to see where we’re going with this function before you can appreciate why it is what it is.

5. Pingback: Irrationality of pi: derivatives of f « The Math Less Traveled

6. Pingback: Irrationality of pi: curiouser and curiouser « The Math Less Traveled

7. Pingback: Irrationality of pi: derivatives of f | The Math Less Traveled

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