Algorithmic Assertions - Craig Gidney's Computer Science Blog

My Misconceptions: The Unintegrable Function

17 Sep 2015

Sometimes, I have ideas. Dumb ideas. Usually I catch them, but sometimes I don't.

In "My Misconceptions" posts, I poke fun at myself while discussing wrong things I've thought.

From Undifferentiable to Unintegrable

Historically, there was a time when people thought that every continuous function was differentiable. People realized their error when pathological counter-examples, such as the Weierstrass_function, were discovered.

For example, the function $f(x) = \Sum{n=0}{\infty} 2^{-n} \sin(4^n x)$ is continuous (due to higher terms in the sum being bounded between $\pm 2^{-n}$). But, if you differentiate $f$, then the $2^{-n}$ term gets multiplied by the $4^n$ factor inside the $\sin$ and you end up with $f'(x) = \Sum{n=0}{\infty} 2^n \cos(4^n x)$. That sum fails to converge, because later terms are multiplied by $2^n$ instead of $2^{-n}$ and so can be arbitrarily large. The derivative does not exist.

When I learned about the above trick for breaking differentation, I had an idea: why not break integration with the same idea in reverse? Just replace $\sin(4^n x)$ with $\sin(4^{-n} x)$, since integrating will divide by the internal $\sin$ factor instead of multiplying by it:

$g(x) = \Sum{n=0}{\infty} 2^{-n} \sin(4^{-n} x)$

$(\int g)(x)$ $= \int \Sum{n=0}{\infty} 2^{-n} \sin(4^{-n} x) dx$

$= \Sum{n=0}{\infty} \int 2^{-n} \sin(4^{-n} x) dx$

$= -\Sum{n=0}{\infty} 2^n \cos(4^{-n} x)$

Notice that the infinite sum in the resulting expression for $\int g$ fails to converge (it always diverges towards infinity). And suppose for the moment that swapping the order of the infinite sum and the integral was justified. Clearly $\int g$ doesn't exist.

At least, that's what I thought for an embarassingly long time.

Fundamentally and Constantly Wrong

What's especially bone-headed about this mistake is how obvious it is, at least in hindsight.

How obvious? Well, my conclusion violates a somewhat well-known theorem. A theorem that says that every continuous function (e.g. $g$) has an integral. You may have heard of it; it's called "The Fundamental Theorem of Calculus".

Eventually someone took pitty on me (or rather, got angry at me for being wrong on the internet) and pointed something out:

$h(x) = -\Sum{n=0}{\infty} 2^n (\cos(4^{-n} x) - 1)$

$h'(x) = g(x)$

Oh.

Well.

That's definitely the worst forgetting-the-integration-constant mistake I've ever made.

Summary

If you're going to play fast and loose with the order of integrals and infinite sums, the consequences of forgetting your constant(s) of integation might get infinitely worse.

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