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.

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