Irrationality of pi: derivatives of f

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{array}{rcl}  \frac{d}{dx}f(\pi - x) & = & \frac{d}{dx} f(x) \\   -f'(\pi - x) & = & f'(x)  \end{array}

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).

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10 Responses to Irrationality of pi: derivatives of f

  1. >We can take the derivative of both sides as many times as we want to see that f^{(i)}(\pi – x) = f^{(i)}(x) for any i, as well.

    This was the only step I couldn’t see right off. I tried to take the first derivative of f(pi-x), and I get -f'(x) instead of f'(x). I’m assuming I made a minor mistake. I’ll try it again later.

    Thanks again for doing this. I’m loving thinking about it.

  2. Dave says:

    Another beautifully laid out step to this proof. I’m happy (and somewhat surprised) to say that I am still with you! Looking forward to the next installment.

  3. Brent says:

    Sue: Ah, thanks, good catch! You’re absolutely right, it’s not quite as simple as what I wrote. The derivative of f(\pi - x) is -f'(\pi - x), you get a negative from the derivative of \pi - x (applying the chain rule). So for odd derivatives there is a negative sign and for even derivatives you get what I said (since the negatives cancel). But since we are talking about a case when one side is equal to zero, the extra negative sign doesn’t make a difference. I don’t know if that made sense, I’m just typing off the cuff here. I’ll think it through more carefully and correct it soon.

    Dave: Great! Actually, I think this was one of the most difficult parts, so if you’re still following, the rest should be gravy. ;-)

  4. Just getting back here… Hmm, I don’t remember what I did when I looked at this before. I’m sure it’s on a scrap of paper at my house in California, but I’m in Michigan for the holidays. I just thought it through again and …

    That line (We can take the derivative of both sides as many times as we want to see that f^{(i)}(\pi – x) = f^{(i)}(x) for any i, as well.) makes perfect sense to me now. If two functions are equal (for all values of the variable), I’d guess their derivatives must be equal. I have a few questions still:

    What does it mean to ‘take the derivative of f(pi-x)? Are we talking about df/dx or df/d(pi-x)? (Your last comment isn’t making sense to me, and I thought this might be where I’m not following you.)

    Calling f ‘a’ function, when it depends on values of n, is still bothering me. Does it work to think of this as a collection of functions, where f sub n is as you gave f?

    You mentioned on your previous post that you’d be interested in working through other interesting proofs. Sam Shah just mentioned the insolvability of the quintic on his blog. That’s something I’ve always been intrigued by, and never learned the proof of.

    This is something that never would have happened before the internet. I appreciate the opportunity to learn more mathematics.

  5. Brent says:

    Sue: We are indeed talking about df/dx. And of course you are right that if two functions are equal, their derivatives (taken with respect to the same variable) must also be equal. So it is true that

    \frac{d}{dx} f(\pi - x) = \frac{d}{dx} f(x)

    but this is NOT the same as

    f'(\pi - x) = f'(x).

    The former says “the derivative of f(\pi - x) is equal to the derivative of f(x)“, whereas the latter says “the derivative of f, when evaluated at \pi - x, is the same as the derivative of f when evaluated at x“. See the difference? In fact, the derivative of f(\pi - x) with respect to x is -f'(\pi - x) (by the chain rule; note the negative sign!).

    I still need to fix the post, which is currently wrong, but I’ve been hanging out with family and eating way too much turkey and pie. =)

    And yes, formally you should think of f as a family of functions, indexed by n. We are proving a bunch of properties which hold for the entire family, and at the end we will show that we can select a particular f_n which gives us a contradiction.

    Insolvability of the quintic would be fun! I’m sure I’ve seen the proof before, although I don’t remember how it goes off the top of my head. I’ll keep it in mind and maybe write about it in the future, thanks for the suggestion!

  6. Brent says:

    Sue: bad news, the proof of insolvability of the quintic requires an excursion through group theory and Galois theory, which is a bit too much background for my intended audience! I’ll still keep it in mind, though, perhaps it could work if I judiciously choose one or two places to say “trust me on this”…

  7. Brent says:

    OK, the post is now fixed. Let me know if it still isn’t clear (or if anyone finds any more mistakes)! Thanks for reading, another installment should be coming soon.

  8. Funny how being formal helps me think sometimes (f sub n instead of f), and other times gets in my way (nope, I’m not yet seeing how to think clearly about f'(pi-x)). I think I need to give myself some examples of the difference you describe.

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