## Thursday, 2 October 2014

### I Refuse to Name This "Complex Analysis...Made Simple": 4

Hey guys! Having fun with your complex analysis assignment? =P

I've just got a few random comments that I wanted to make, and it's a bit random, but I think that you'll find it useful.

Comment 1: Taylor series/Laurent series expansions are AMAZEBALLS.

Most of the functions that you've encountered up until this point in life are super-super nice functions like $$\sin(x),\cos(x),\exp(x),\log(x),\frac{1}{1+x^n}$$ which all have nice Taylor series expansions and what have you. And so when you see this absolutely incredible fact that "complex differentiability of a function in a neighbourhood" implies "(local) Taylor expansions", you mightn't have an appreciation for how "phwoar" this is. In particular, you mightn't have realised that a smooth (infinitely differentiable) real function mightn't have Taylor expansions! In fact, "most" (*) real functions don't have Taylor expansions. The annoying thing is actually writing down one of these smooth real functions. Soo, here's one that I prepared earlier: \begin{align} f:\mathbb{R}\rightarrow\mathbb{R},\, f(x)=\left\{\right. \begin{array}{cc} \exp(\frac{-1}{x^2}) & \text{ if }x>0, \\ 0 & \text{ if }x\leq 0. \end{array}\end{align} Now, I'm going to leave it as an exercise for you guys to check that this function is actually smooth everywhere along the real line (**).

Now, assume for a second that this function had a Taylor expansion around $$x=0$$. Well, the fact that $$f(x)$$ is constant on the negative real line tells us that this Taylor expansion must just be $$0$$, but then the fact that $$f(x)$$ is non-zero on the positive real line contradicts this expansion. So, we see that this function has no Taylor expansion around $$x=0$$.

So, at this point you might think that well, that was an okay example, but $$f(x)$$ still has Taylor expansions at all the other points in its domain. Then, here's a function that's smooth but doesn't have a Taylor expansion ANYWHERE. (insert spooky music)

And so, you see, the fact that complex differentiability (and unlike the real case, you only need to check the first derivative) implies Taylor series expansions is TOTES AMAZEBALLS.

Comment 2: Question 20 of Practice Sheet 6 vs Question 24 of Practice sheet 8.

If you compare these two questions and the answers that you get for them (SPOILERS: YOU GET THE SAME ANSWER), you see that in fact: \begin{align} \int_0^{2\pi}\frac{\sin(t)}{(2+ \cos(t))^2}\mathrm{d}t=0.\end{align} This is annoyingly obvious in retrospect - you could have totally just solved this real integral explicitly to get $$0$$ instead of spending 30~40 minutes labouring over Question 24. But in fact, the fact that this integral is $$0$$ can be generalised somewhat to the following fact: \begin{align} \int_0^{2\pi} \sin(t) f(\cos(t))\mathrm{d}t=0.\end{align} Let's prove this: \begin{align} &\int_0^{2\pi} \sin(t) f(\cos(t))\mathrm{d}t\\ =&\int_0^{\pi} \sin(t) f(\cos(t))\mathrm{d}t+\int_\pi^{2\pi} \sin(t) f(\cos(t))\mathrm{d}t\\ =&\int_0^{\pi} \sin(t) f(\cos(t))\mathrm{d}t+\int_{\pi}^{0} \sin(2\pi-\tau) f(\cos(2\pi-\tau))\mathrm{d}(2\pi-\tau),\\ &\text{ with the substitution }\tau=2\pi-t,\\ =&\int_0^{\pi} \sin(t) f(\cos(t))\mathrm{d}t+\int_\pi^{0} \sin(\tau) f(\cos(\tau))\mathrm{d}\tau=0.\end{align} In fact, this really shouldn't be a surprise to anyone who's played around with Fourier series, but it's still a cute fact. =)

Also, here's something-else nice about Question 20 that Andre Ratiu (the other tutor for this subject) pointed out to me: \begin{align} \int_0^{2\pi} \frac{1}{(2+\cos(t))^2}\mathrm{d}t=\int_0^{2\pi} \frac{1}{(2+\sin(t))^2}\mathrm{d}t \end{align} And here're two ways to see this (without doing the actual computation):

Method 1 (real analysis): the function $$\frac{1}{(2+\cos(t))^2}$$ is periodic, and $$\frac{1}{(2+\sin(t))^2}$$ is just a $$\frac{\pi}{2}$$ shift of this. And you should be able to prove that for periodic function $$g(t)$$ with period $$p$$ that \begin{align}\int_0^p g(t)\mathrm{d}t=\int_b^{p+b} g(t) \mathrm{d}t\text{, for any }b\in\mathbb{R}.\end{align}

Method 2 (complex analysis): transform the two integrals into contour integrals around the unit disk on the complex plane and observe that the functions are rotations of each other by $$\frac{\pi}{2}$$. It's the same idea, as the periodicity thingy above, but phrased complex analytically. =)

Orrite, that's...been a fruitful morning of procrastination for me...time to get back to doing...other stuff. =P

*: the space of functions with Taylor expansions around a certain point $$x_0\in\mathbb{R}$$ is countably infinite dimensional, whereas the space of smooth real functions is uncountably infinite dimensional.

**: This is obvious everywhere except at $$x=0$$. One hint that I could give you is to first show that all of the derivatives of $$\exp(-x^{-2})$$ are in the form of $$P(x^{-1})\exp(-x^{-2})$$, where $$P$$ is a polynomial.