Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've had this idea in my head for a while, but I've never told anybody because... well, I really don't know. I just never thought that it might even be remotely correct, but here goes. Here is just an example that, I think, makes a good job of explaining this principle, but there are other scenarios, this is just the simplest one I can think of: We can prove that the derivative of a function $f(x)=ax^n$ is equal to $f'(x)=anx^{n-1}$, for all rational numbers $\frac{n}{m}$ where $n$ and $m$ are integers, but that doesn't prove that this property holds for irrational/transcendental numbers, right? But I say it does, just from the fact that it works for all rational numbers, and here is why: say you have the expression $\frac{d}{dx}x^\pi$. Now we all know that $\pi$ is transcendental, but why can't we just "pretend" that we have a two immensely big numbers $n$ and $m$, such that $\frac{n}{m}$ is such a close approximation to $\pi$ that there is no way to distinguish them. This number is so close to $\pi$ that it differs on the googolplexth digit. But it is still a repeating decimal. So Why can't we just do that, and say that the power rule applies to irrational/transcendental numbers too?

I apologize if this questions is completely ludicrous.

share|improve this question
add comment

2 Answers

up vote 3 down vote accepted

The type of argument you describe is a big motivation for the definition of a continuous function. A basic result which captures the spirit of what you want is this: a continuous function $f : X \to Y$ between topological spaces is determined by its values on a dense subset of $X$. In particular, a continuous function $f : \mathbb{R} \to Y$ is determined by its values on the rationals. This gives rise to the following intuition:

Any "sufficiently continuous" statement you can prove for the rationals should be true of all real numbers.

The above result doesn't directly apply to this situation, at least not in a straightforward way: for one thing, knowing that the function $x^q$ exists for $q$ rational is not enough to immediately conclude that a function called $x^r$ even exists, let alone that it is continuous or differentiable. But it is still an important motivation for suspecting that such a thing ought to be true anyway.

For $x \ge 0$ it is a standard exercise in real analysis to prove that we can define

$$x^r = \lim_{n \to \infty} x^{q_n}$$

where $q_n$ is any sequence of rational numbers tending to $r$. Furthermore this function satisfies all of the expected algebraic identities, and so forth. But it is still not completely trivial to see that $x^r$ is differentiable for arbitrary real $r$ until you prove the basic properties of logarithms and exponentials, after which you then know that

$$x^r = e^{r \log x}.$$

Once you know this, you're in the money: the composite of differentiable functions is differentiable, and everything works out exactly the way you expect it to by the chain rule.


One might be tempted to argue as follows: surely if the function $x^r$ is a limit of a sequence of functions $x^{q_n}$, then the derivative of the function $x^r$ is a limit of the sequence of derivatives $q_n x^{q_n-1}$. But this argument is not so straightforward to formalize. First, there are various notions of convergence of a sequence of functions: there is pointwise convergence but also convergence with respect to various norms, and one must decide what notion is sensible here. Second, even if one takes a fairly restrictive definition of convergence (say uniform convergence), it is still false in general that a uniform limit of differentiable functions is differentiable (see Weierstrass function).

One needs the additional constraint that the derivatives themselves also converge uniformly. Now, the derivatives $q_n x^{q_n-1}$ do not converge uniformly to $rx^{r-1}$ on $\mathbb{R}_{\ge 0}$ (let's provisionally assume we only care about $r > 1$), but they do converge to $rx^{r-1}$ uniformly on the intervals $\left[ \frac{1}{k}, k \right]$ for all $k \in \mathbb{N}$ (say). I think that is enough to get this result without going through the stuff above about logarithms and exponentials, but I haven't checked it closely.

share|improve this answer
1  
"tending to zero" should be "tending to $r$", right? –  Zev Chonoles Jul 14 '11 at 6:10
    
@Zev: ha. Yes, it should. –  Qiaochu Yuan Jul 14 '11 at 6:29
add comment

The rationals are dense in the reals, and differentiation is, in some sense, continuous, and this can used to turn your argument into a rigorous proof, with some well-timed application of limits. One has to be careful - there are other properties of rationals that don't apply to reals in general, and that's why we have the apparatus of limits.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.