editing theorem 2 proof

1 files changed, 7 insertions, 4 deletions

diff --git a/ClassJuly5.page b/ClassJuly5.page
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+++ b/ClassJuly5.page
@@ -20,11 +20,14 @@ So, if you only care about holomorphic functions you don't need to worry about t
 
 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 any confusing changes of 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
+An alternative proof of Theorem 2 (using contour integrals) goes as follows:  Since $f$ is holomorphic on a disk, it has a Laurent expansion.  Theorem 2 says precisely that the negative terms in this Laurent expansion are zero.  Intuitively, the reason for this is that terms like $\frac{1}{z}$ become infinite as $z \to 0$.  However, since we are dealing with an infinite sum it is theoretically possible that there could be massive oscillations that cancel out to give something which does not become infinite.  We will rule out this theoretical possibility by computing the coefficients $c_{-1},c_{-2},\dots$ individually and showing that we get zero each time.
+
+First let's prove that $c_{-1}$ is zero.  Since $c_{-1}$ is the residue of $f$ at zero, it is given by the integral
 $$c_{-1} = \frac{1}{2 \pi i}\int_{\gamma_r} f(z) dz$$
-where $\gamma$ is a small circle of radius $r$ that goes counterclockwise around the origin.  Taking the limit as $r \to 0$, we find ourselves integrating over a circle of radius $0$.  Therefore,  
-$$ c_{-1} = \frac{1}{2 \pi i} \lim_{r \to 0} \int_{\gamma_r} f(z) dz = \frac{1}{2 \pi i} \int_{\gamma_0} f(z) dz = 0  $$
-and we conclude that $c_{-1} = 0$.
+where $\gamma$ is a small circle of radius $r$ that takes one counterclockwise turn around the origin.  Taking the limit as $r \to 0$, we find ourselves integrating over a circle of radius $0$:
+$$ c_{-1} = \frac{1}{2 \pi i} \lim_{r \to 0} \int_{\gamma_r} f(z) dz = \frac{1}{2 \pi i} \int_{\gamma_0} f(z) dz = \frac{1}{2 \pi i} \int_0^{2\pi} f(0) \frac{d\gamma_0}{dt} dt$$
+But $\gamma_0(t)$ is constant, hence $\frac{d \gamma_0}{dt} = 0$. Therefore,
+$$c_{-1} = \frac{1}{2 \pi i} \int_0^{2\pi} 0 dt  = 0 $.
 
 Now let's prove that $c_{-2}$ has to be zero.  Consider the function 
 $$ g(z) = z f(z) = \cdots \frac{c_{-3}}{z^2} + \frac{c_{-2}}{z} + c_{-1} + c_0 z + c_1 z^2 + \cdots $$