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 $n$ ) where $a$ and $b$ 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 $n$ ; but again, everything we prove about it will be true no matter what $n$ 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 $2n$ 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 $\cos$ .

About Brent

Associate Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.

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10 Responses to Irrationality of pi: curiouser and curiouser

Dave says:

January 30, 2010 at 7:17 pm

The long awaited 4th post! I’m still with you. However, I’ll have to do some research on the Taylor series for cos(x) between now and the next post. It’s been a long time since I’ve seen any Taylor series (though I know it’s back there in the cobwebs somewhere).
Brent says:

January 30, 2010 at 8:17 pm

Well, the similarity is that the Taylor series for cosine is also an alternating sum of “even things” (x^(2k)/(2k)!, to be precise).
Brent says:

January 30, 2010 at 8:18 pm

Also, knowing the Taylor series for cos(x) will NOT be a requirement for understanding the next post! I just threw that in there at the end as an interesting aside.
Dave says:

January 30, 2010 at 9:36 pm

I see. It still might be worth the time to go back and look at Talyor series (just for the sake of jogging my memory). That’s one of the things I like most about your site. It forces me to think about things that I haven’t studied rigorously in almost a decade. And many times I come out of the discussion with a better understanding than I had the first time around.

Also, it’s worth mentioning, AGAIN, that after seeing the F(x) function and the relationship, F”(x) + F(x) = f(x), I have the uncontrollable urge to climb inside Niven’s mind to find out how he came up with the F(x) function in the first place.
Brent says:

January 30, 2010 at 10:04 pm

Yes, you’re right, it’s probably still worth it. And I’m glad my blog spurs you to learn (or re-learn) things!

As for how Niven came up with F(x), I suspect it’s just f(x) that he came up with, and then used F(x) just as a convenient notational tool so he didn’t have to write out lots of long equations with sums of derivatives of f (I think this will make more sense once you see the next post—basically F(x) just helps to conveniently compute the integral of f(x) sin(x)). But I would also really like to know how he came up with f(x)! Hopefully in a wrap-up post I can do some research and try to get a bit of insight that I can share.
Sue VanHattum says:

January 31, 2010 at 9:13 am

I’m with you on this chunk, but when my brain is clearer, I’ll need to go over the first parts of the argument again to see the big picture more clearly.

Once we’re done with the details, I’d like to try to imagine/understand Niven’s thought process. We haven’t yet used the fact of what pi is, have we?
Brent says:

January 31, 2010 at 12:43 pm

Sue: you’re absolutely right, we haven’t actually used any properties of pi yet. That will come in the next post, when we bring some trig functions into the picture.
Jack says:

February 5, 2010 at 3:20 pm

I’m really enjoying this series =) I hope you do more like them.
Brent says:

February 6, 2010 at 11:41 am

Jack: thanks, glad you’re enjoying it! I enjoy doing this sort of expositional series so I imagine I will probably do more… I just have to come up with other topics that would work well. Suggestions welcome!
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