Lim Sup: Life Lessons.

August 22, 2011

I’ve always felt a little uneasy with Lim Sup and Lim Inf’s.  Today while reading Rudin’s Complex book, I finally "got" it.


To find the lim sup of your sequence: take every convergent subsequence’s limit and find the sup of those limits.


This is, of course, exactly what the definition is, but it wasn’t "visual" to me until I thought about it this way.  Is there something you learned early on that you weren’t quite clear with until much, much later?


I’ve been up for a while doing practice qualifying exam questions, and sometimes I hit a point where I just do whatever it is that comes to my mind first, no matter how tedious or silly it seems.  This is a bad habit.  I’ll show why with an example.


Here’s the question.  Let C be the unit circle oriented counterclockwise.  Find the integral

\displaystyle\int_{C}\frac{\exp(1 + z^{2})}{z}\, dz.


The sophisticated reader will immediately see the solution, but humor me for a moment.  I attempted to do this by Taylor expansion.  The following calculations were done:


= \displaystyle\int_{C} \frac{1 + (1+z^{2}) + (1 + z^{2})^{2} + \cdots}{z}\, dz


To which the binomial theorem was applied to the numerator terms to obtain:


= \displaystyle\int_{C} \frac{1 + (1+z^{2}) + \frac{\sum_{n=0}^{2}\binom{2}{n}z^{n}}{2!} + \frac{\sum_{n=0}^{3}\binom{3}{n}z^{n}}{3!} + \cdots}{z}\, dz


And at this point we note that everything is going to die off when we take the integral except the coefficient of the \frac{1}{z} term.  Our residue (the coefficient) will be:

1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots

which can also be written (slightly more suggestively) as:

\frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots

which we should recognize as the Taylor expansion of e^{z} at the point z = 1.  Nice!  Now we note that plugging in e^{i\theta} to take the contour integral (ignoring all those terms which don’t matter) will force us to integrate

\displaystyle e \int_{0}^{2\pi}\frac{1}{e^{i\theta}}ie^{i\theta}\, d\theta = ie\int_{0}^{2\pi}\, d\theta = 2\pi i e.

Cutely, if we think of the Greek letter \pi as being a "p", this solution spells out "2pie".


But now, readers, let’s slow down.  This is, indeed, the correct answer.  But if I had just looked at the form of the integrand, I would have seen an everywhere analytic function divided by a form of z - z_{0}.  This screams Cauchy Integral Formula.  Indeed, according to the CIF, we should get the solution as

2\pi i \exp(1 + 0^2) = 2\pi i e

which is exactly what we got before, but only took about 4 seconds to do.  It’s nice to be able to check yourself by doing something two different ways, but when time isn’t on your side (like in a qualifying exam situation, for example!) then remember:

Think before you Taylor Expand.