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authorluccul <luccul@gmail.com>2010-07-13 14:02:45 +0000
committerbnewbold <bnewbold@adelie.robocracy.org>2010-07-13 14:02:45 +0000
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**[Lucas' Notes for Lecture 3](/SummerCourseHeatWave.pdf)**
+The goal of the lecture (and the only take-away point, as far as the rest of the course is concerned!) was to prove the following two theorems:
+
+Theorem 1: If $f:A \to \C$ is a holomorphic function on an annulus, then it has a Laurent series expansion
+$$ f(z) = \sum_{n = -\infty}^{\infty} c_n z^n = \cdots \frac{c_{-2}}{z^2} + \frac{c_{-1}}{z} + c_0 + c_1 z + c_2 z^2 + \cdots $$
+
+Theorem 2: If $f: D \to \C$ is a holomorphic function on a disk, then it has a power series expansion
+$$ f(z) = \sum_{n = 0}^{\infty} c_n z^n = c_0 + c_1 z + c_2 z^2 + \cdots $$
+
+The reason we discussed convergence of Fourier series was to give some taste for the type of mathematical analysis that goes in to proving things rigorously using Fourier series.
+
+The reason we discussed the Heat and Wave equations was to illustrate other examples of the methods we used to prove Theorems 1 and 2. So, if you only care about holomorphic functions you don't need to worry about those examples.
+
+You may find it helpful to think about other ways of deriving Theorems 1 and 2. For an alternate proof of Theorem 1 (which may be more comprehensible, since it doesn't involve changing coordinates), see Problems 7 and 8 of Problem Set 3.
+
+An alternate proof of Theorem 2 goes as follows: Since $f$ is holomorphic on a disk, it has a Laurent expansion. The statement of Theorem 2 says that the negative terms in this Laurent expansion are zero. First let's prove that $c_{-1}$ is zero. Since $c_{-1}$ is the residue of $f$ at zero, it is given by
+$$c_{-1} = \int_{\gamma_r} f(z) dz$$
+where $\gamma$ is a small circle of radius $r$ that goes counterclockwise around the origin. As we shrink the radius of this circle, its length goes to zero. On the other hand since $f(z)$ tends to $f(0)$. Taking the limit as $r \to 0$,
+$$ c_{-1} = \lim_{r \to 0} \int_{\gamma_r} f(z) dz = \lim_{r \to 0} 2 \pi r f(0) = 0 $$
+so we conclude that $c_{-1} = 0$.
+
+Now let's prove that $c_{-2}$ has to be zero. Consider the function
+$$ g(z) = z f(z) = \cdots \frac{c_{-3}}{z^2} + \frac{c_{-2}}{z} + c_{-1} + c_0 z + c_1 z^2 + \cdots $$
+Since $g$ is also holomorphic on $D$, and its residue is $c_{-2}$, we conclude by the above argument that $c_{-2} = 0$. More generally, the function $z^{k-1}f(z)$ is holomorphic and its residue at $0$ is $c_{-k}$, so we conclude that $c_{-k} = 0$.
+
# Unanswered Questions
Below are answers to some questions that came up during the lecture. If anyone remembers other unanswered questions please post them here as well, and hopefully they'll get answered.