Built on Facts

Sunday Function

This function is a two-dimensional one. It’s radially symmetric however, so we can specify it with only one coordinate – the distance from the origin r. It’s the two-dimensional Gaussian function, and it looks like this:

i-068617cf78ef07a0888beafc276772ae-gauss.png

i-d4bb9fe09d6b78eb832985821d8a307f-1.png

As r increases, -r^2 very quickly becomes a large negative number. The exponential function falls off rapidly toward zero at negative arguments, and so as you move away from the origin the value of the function is pretty much nil.

Now if you think of that graph as a clay sculpture, you’ll see that almost all of your clay will go to making that central bulge. Away from that bulge you’ll just need the barest smear of clay to cover the rest of the ground. In fact if the central bulge is taken to be a meter high by the time your radial distance is about r = 4.7 meters or so the clay will need to be only atomic thickness, and beyond that real clay simply couldn’t be spread thin enough to be an accurate representation.

So it’s intuitively plausible that if you model this whole thing with a magic clay that can be spread arbitrarily thin, you’ll still only need a finite amount of clay even if your model goes infinitely far beyond the 6×6 bounding box I’ve drawn here. How much clay? We just need to do a little calculus. It’s just the old area under the curve problem, or in this two dimensional case the volume V under the surface. Since we’ve already expressed the function in radial coordinates, the integral is easy to set up:

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Before we talk about the result, where did that extra r come from in the integral? We’re doing this integral in radial coordinates, and so we have to account for the fact that our coordinate system doesn’t preserve area for an arbitrary differential element dr*dθ. We have to take that into account, and long story short the way to fix it is to multiply by r. This is very convenient in this case because the integral is much, much harder to do without the r. In fact, when doing this problem for the one-dimensional Gaussian function the equivalent problem is quite problematic because we don’t have that extra r. We have to do a trick to put it in this 2-d form in order to get an answer. That’s something to be saved for another time though.

All that aside, the answer is pi. Pi cubic meters, or whatever units your graph happens to be drawn in. This particular pi is a little less mysterious than some of the other phantom pi we’ve encountered recently because at least the integral was in radial coordinates and radii have a connection to circles. Still, it’s a little weird to see it popping out of a real exponential function like that.

This has a very direct connection to physics. Aside from the Gaussian distribution in probability, we use these Gaussian functions to model quantum mechanical wave packets and other wave phenomena all the time. Like most math, not only is it cool but it’s also very useful.

Comments

  1. #1 dean
    November 23, 2008

    Technically this is ONE of the bivariate Guassians – the one in which the components are independent, with zero means and unit variances

    On the other hand – very cool discussion of a way to interpret the double integral.

  2. #2 bioephemera
    November 24, 2008

    Very cool indeed! pi is fascinating. If I’d gotten to see it “pop out” of equations like this in my math courses, maybe i’d have taken more of them. :)

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