Built on Facts

Sunday Function

I think we’ve developed a nice theme over the last few weeks, gradually working our way through a less well-behaved function – the triangle wave – and trying to find various series expansions for it. “Well-behaved” is kind of a term of art, which mathematicians use as shorthand for long strings of caveats about differentiability and continuity and which physicists use because who wants to memorize those long strings?

Well, to some extent we have to. While it’s not very common that badly-behaved functions arise in physics, there are functions which at least don’t always remember to say please and thank you. They have to be gently corrected, but they’re good at heart. The mathematicians are the ones who have to deal with the truly shady functions, the ones who form prison gangs and don’t play by the rules and obey the laws. Or theorems.

This one is in the latter category. It’s the ominous Blancmange function, a classic pathological case. It’s continuous everywhere and differentiable nowhere. There’s not a single point anywhere on its domain where it’s smooth, every single point is as prickly as a porcupine.

Via Wolfram, its graph:

i-ef5c9681359c41b848dc7b447c2c9a8b-BlancmangeFunction_1000.gif

Symbolically:

i-d4bb9fe09d6b78eb832985821d8a307f-1.png

Where

i-88516ca505d84f969b7ab15a965c3fb3-2.png

As ludicrous as it is to have continuity everywhere without differentiability, it’s not actually hopelessly bad. While it’s true that continuity doesn’t imply differentiability, it is true that continuity implies integrability. The integral of this function over its domain [0,1] is just 1/2. Since it’s integrable, expansion in terms of the various orthogonal sets of functions that we’ve been doing will still work. We could pretty easily do a Fourier series or a series in Legendre polynomials. Or we could if we could do the necessary integrals, which I don’t think I’d want to try but could do numerically if worse came to worst.

What can you do with the Blancmange function? Heck if I know. If I ever got to teach a Calc 1 course I could use it as a good example of the fact that you can’t always take a statement to be mathematically certain just because it’s intuitively reasonable. While most commonly encountered continuous functions are differentiable, not all of ‘em are. In fact by most set-theoretical standards such functions are actually the exception.

Fortunately they’re exceptions we don’t usually have to deal with. But like the outlaw mobsters of another era, they have their own rogue appeal.

Comments

  1. #1 Comrade PhysioProf
    August 30, 2009

    That’s fucking weird!

  2. #2 MPL
    August 30, 2009

    Is it just me, or does it look like a big mound of mashed potatoes?

    In fact by most set-theoretical standards such functions are actually the exception.

    But like the typical outlaw, badly-behaved functions rarely have permanent addresses, so they’re harder to find than the well-behaved ones.

  3. #3 CRM-114
    August 30, 2009

    Another fun one is the Dirac delta function. And, yes, some rule that it isn’t really a function, although it functions as one.

  4. #4 Matt Springer
    August 30, 2009

    That’s a really good point, somehow in all these posts we haven’t done the Dirac delta yet. That’ll have to be fixed soon. The fact that it’s not strictly a function isn’t a problem; it’s close enough and we’ve done a few pretty questionable functions already.

  5. #5 Uncle Al
    August 30, 2009

    It would be great for mountain climbing – nobody could legitimately complain about the slope.

  6. #6 Edy Levi
    August 30, 2009

    I love this example.

    Another interesting weird example is a bounded function that has infinite length. Calculus can get spooky.

  7. #7 MPL
    August 30, 2009

    Re: No. 5

    Somehow “undefined slope” sounds scarier than any defined slope at all.

  8. #8 agm
    August 30, 2009

    Regarding integrable functions, I take it we are talking about Riemann integration and not Lebesgue integration (which expands the scope of what can be integrated but is much less intuitive in some important ways)?

    We call the Dirac a function, but really, should we try to shoehorn the Dirac into the “function” category? I thought theses days the word function had basically been redefined to mean any mapping of the elements of a set onto the members of another set that followed certain rules. The Dirac’s sieving property does select out a member of a set, but it doesn’t seem useful to call it a function since you are mapping an entire domain onto a single member of another set.

  9. #9 CCPhysicist
    August 30, 2009

    @2: Sure. Now google “blancmange”, or watch some old Monty Python episode. (Yep, I checked and it’s on YouTube.) Far sillier than mashed potatoes.

    My favorite is a function built from Cantor’s ternary set on the interval [0,1]. It isn’t even continuous, so it can only be integrated with Lebesgue measure. What makes it particularly amusing is that this function has the value 1 at EXACTLY the same number of points in [0,1] as where your function is non-zero, so you might suspect its integral is 1 … but its integral is zero. This follows from the fact that this function is zero at the same number of points!

    Explaining how you can remove “everything” and still have the same number of points as you started with is as interesting as the contrast with Dirac’s delta function. The latter is only non-zero at one point (rather than at infinitely many), but its infinite value at that point gives it measure 1 rather than zero.

  10. #10 Douglas
    August 31, 2009

    I know calling the Dirac delta not a function is a defense mechanism, but it bothers me, too. The same thing with set functions. They *are* all functions – just with a different domain of definition than one might be used to. delta:Schwartz->R, mu:P(R)->R, etc.

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