Sunday Function

Here's a straightforward function of two variables, x and y:

i-d4bb9fe09d6b78eb832985821d8a307f-1.png

Its domain is all real x and y, with the single exception of x = y = 0, which would make the denominator 0. But we have experience with functions such as sin(x)/x, where we can find the limiting value as x approaches 0 and just make a fiat declaration that that limit isthe value of the function at the origin. Can we do it here?

Well, let's just tabulate the value of f(x,y) for different values of x close to 0 and see what happens:

f(0.1, 0.1) = 0.0990099
f(0.01, 0.01) = 0.009999
f(0.001, 0.001) = 0.000999999

This suggests that if we just define f(0, 0) = 0 then we'll have a continuous function. But there's a nagging question - we just approached the point (0, 0) from the line y = x. What if we approached from (say) y = 2x?

f(0.1, 0.2) = 0.344828
f(0.01, 0.02) = 0.0399361
f(0.01, 0.02) = 0.00399994

So far so good. But we're really paranoid people. What if we want to approach the origin from every single linear path and see if anything comes out weird?

i-88516ca505d84f969b7ab15a965c3fb3-2.png

Which, I assure you, will approach 0 as x -> 0 for any m you care to name.

Maybe we're sufficiently convinced by this that we're ok to just declare that the limit of f(x,y) as x and y approach 0 is 0. But maybe we're just not paranoid enough. Dale Gribble might come up to us and say "That's just what the feds want you to think. I bet if you approach the origin from the parabolic path x = y^2 then you won't get 0 as the limit."

We can try it out:

i-cba0cb33fdea362b22886cdd25435eca-fix.png

Oh geez, Dale is right. The path of approach to the origin does matter, and so the limit does not exist. Let's take a look at the graph:

i-fd7004caed9c3008a7945cd9a08ecfbd-sheetGraph.png

There's a parabolic ridge and a parabolic canyon along those contours. Approach via those veatures and you never start descending. Approach via a straight-line path to the origin and sure enough you'll start vertically descending toward (0,0,0). A contour map might help clarify:

i-1d0c5603edf1a42faa7150c4fc6d9b4c-contourPlot.png

This illustrates something that's actually a serious pain in real analysis, the formal name for the study of functions involving real numbers. Real functions can be very weird animals in general, and you have to worry about seemingly straightforward limits not existing and directional derivatives behaving very badly. Perhaps counterintuitively, in complex analysis (the study of complex numbers) this sort of thing is actually much easier. The real and imaginary parts of complex functions are tightly bound together in a rather profound way that makes working with these limits and derivatives much easier than in the case of purely real numbers.

Go figure.

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Hi,
Really enjoy you blogging.

You have a y2 instead of y4 in the second last term in the f(y2,y) function.

Cheers,
Brent.
(Feel free to delete this comment once when you fix the post)

The Cauchy-Riemann equations don't hold for all complex functions. "Complex functions" (by which you mean those for which CR hold) are nicer than those in real analysis solely because you're excluding the not-nice functions from consideration.

Ditto Miko @2.

We could try doing complex analysis with non-analytic functions. But we wouldn't like it.

Good post, though.

Just for grins, I entered the function into my iPad SpaceTime app, and got the plot you gave. I think this is a baby Mathematica. I'm not familiar with Mathematica, so is there a quick description of how to go from the 3D plot to the contour map? My math is 40+yrs rusty.

By Gray Gaffer (not verified) on 18 Oct 2010 #permalink

@Gray Gaffer

I'm not really familiar with SpaceTime, but I looked up their manual on the website. I'm guessing you were plotting using Plot3D(...). It seems you'll get what you want if you just change that to ContourPlot(...). http://www.spacetime.us/manual/ContourPlot

Comment on #2 & #4

All complex differentiable functions are complex analytic.

A study of say merely continuous complex-valued functions
of a single complex variable with domain the complex plane
would be completely identical to the study of continuous
functions from R2 to R2.

So there is no theory of complex functions that are not
complex analytic which is not already part of real analysis.

In real analysis there is a hierarchy of successively better
behaved functions starting with continuous, differentiable,
C1, ....., Coo, real analytic.

Each class in the hierarchy is strictly larger than any
preceeding class.

But if a complex function is complex differentiable than it
is actually analytic,

Thus real analysis is like a First World Economy. There are
riff-raff (continuous non-differentiable functions),
respectable poor (C1), upper middle class (Coo) and finally
the aristocrats (real analytic).

Complex analysis is like a Third World Economy. There are
just the riff-raff and the aristocrats. Nothing is in between.

By annonymous (not verified) on 19 Oct 2010 #permalink

Sorry, I should have said "strictly smaller" not "strictly larger"

By annonymous (not verified) on 19 Oct 2010 #permalink

How much easier it is to see the function now that we have programs to make plots of the function easily, and in 3d to boot, the 3d makes the behavior much more visible than a 2 d plot.