In my previous post in this series, we defined the function
and showed that . Today we’ll show the surprising fact that, for every positive integer , although and are not necessarily zero, they are always integers. (The notation means the th derivative of ; that is, take the derivative of , then the derivative of that, then the derivative of that, … times.) Put more succinctly: every derivative of takes on integer values at and .
Why might this be surprising? It’s surprising because of the in the denominator of . For example, consider the function (which I just made up):
It’s easy to see that . But let’s take the derivative: so , which is clearly not an integer. For the derivatives of to always give an integer at (let alone at ) there must be some fancy canceling going on!
For now we will consider only (we’ll come back to later). Of course, substituting for causes every term containing to disappear, so is just the constant term of . Hence, we must show that the constant term of is always an integer.
Consider the numerator of , that is,
Note that , when expanded out, is a polynomial of the form , where the ellipsis contains a bunch of terms with integer coefficients and powers of between 1 and . (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 , we see that
so is a polynomial with terms of degree through , and hence so is , since dividing by changes the coefficients but not the exponents. (Note that has no constant term, so —but we already knew that.)
Recall that the derivative of is , so taking the derivative of a polynomial reduces each of the exponents by one. So the first derivative of is a polynomial with terms of degree through (and hence a constant term of zero); the second derivative has terms of degree through (still no constant term); and so on. We can see that none of the first derivatives of will have a constant term, so (which is certainly an integer) for . What about the th derivative and higher? This is where the fancy canceling comes in!
As we noted above, when expanded out is a sum of a bunch of terms of the form
where and is some integer. When we take the derivative, this term will turn into ; if we take the derivative again, it will become ; another derivative gives us , and so on. Do you see what is happening? After taking the derivative exactly times, we will end up with the constant term
and here’s our fancy canceling: is clearly divisible by , so this is some integer times , which is also an integer. Voila! Said a different way, and more succinctly: since each term of has degree at least , by the time we have taken the derivative enough times for it to yield a constant term, the will be canceled from the denominator, since we will have taken the derivative at least at each power of from down to .
Finally, if we take the derivative of more than times, we get , so no problems there.
Great, so is always an integer. But what about ? Well, remember, last time we showed that . If we take the derivative of both sides with respect to (being careful to use the chain rule on the left side, noting that the derivative of with respect to is ), we get
We can repeat this process to find that (the two negatives cancel on the left side), , and so on. But the extra negative sign for odd derivatives doesn’t really matter: in either case, , which is an integer.
Getting closer! Next time, we will define another special function in terms of and its derivatives; this function will help us compute —which (if you recall the punchline) will turn out to be an integer strictly between and (which is impossible).