From ac8f1dd9177b5b4c38674d6f2cf944d245ca8af4 Mon Sep 17 00:00:00 2001
From: Nao Pross <np@0hm.ch>
Date: Thu, 28 Apr 2022 12:50:55 +0200
Subject: Fix indentation

---
 notes/FourierOnS2.tex | 78 +++++++++++++++++++++++++--------------------------
 1 file changed, 39 insertions(+), 39 deletions(-)

(limited to 'notes/FourierOnS2.tex')

diff --git a/notes/FourierOnS2.tex b/notes/FourierOnS2.tex
index de15f01..24b972f 100644
--- a/notes/FourierOnS2.tex
+++ b/notes/FourierOnS2.tex
@@ -40,7 +40,7 @@
 
 \section{Fourier on a flat surface}
 
-%% TODO: replace everywhere "nice" with this definition
+%% FIXME
 First, we need to discuss about the class of functions on which the Fourier works at all. We will denote the set of such functions with \(C(\Omega; \mathbb{C})\), that means: continuous bounded functions from \(\Omega\) to the complex numbers. Informally, we will refer to these as ``nice'' functions. 
 
 %% TODO: Check that R^2 / Z^2 = (R / Z) \times (R / Z)
@@ -134,46 +134,46 @@ is easy to compute. This is shown in the next lemma.
   \]
 \end{lemma}
 \begin{proof}
-To start, we first expand the left side of the statement:
-\begin{gather} 
-  \nonumber
-  \langle \nabla^2 f, B_{m, n} \rangle
+  To start, we first expand the left side of the statement:
+  \begin{gather} 
+    \nonumber
+    \langle \nabla^2 f, B_{m, n} \rangle
     = \iint_{[0, 1]^2} \nabla^2 f B_{m,n} d\mu d\nu \\
     = \iint_{[0, 1]^2} \left(
-        \frac{\partial^2 f}{\partial \mu^2} 
-        + \frac{\partial^2 f}{\partial \nu^2} 
-      \right) e^{-i2\pi m\mu} e^{-i2\pi n\nu} d\mu d\nu.
+    \frac{\partial^2 f}{\partial \mu^2} 
+    + \frac{\partial^2 f}{\partial \nu^2} 
+    \right) e^{-i2\pi m\mu} e^{-i2\pi n\nu} d\mu d\nu.
     \label{eqn:laplacian-coeffs-expanded}
-\end{gather}
-Since the integrand is a sum of partial derivatives, we now have 2 integrals. Notice that inside each integral we have an expressions of the form:
-\begin{equation} \label{eqn:inner-by-parts}
-  \int_{[0, 1]} \frac{\partial^2 f}{\partial \xi^2} e^{-i2\pi x\xi} d\xi,
-\end{equation}
-once with \(x = m, \xi = \mu\) and the second time with \(x = n, \xi = \nu\).
-The integral \eqref{eqn:inner-by-parts} can be integrated by parts twice resulting in this ugly expression:
-\[
-  e^{-i2\pi x\xi} \left(
+  \end{gather}
+  Since the integrand is a sum of partial derivatives, we now have 2 integrals. Notice that inside each integral we have an expressions of the form:
+  \begin{equation} \label{eqn:inner-by-parts}
+    \int_{[0, 1]} \frac{\partial^2 f}{\partial \xi^2} e^{-i2\pi x\xi} d\xi,
+  \end{equation}
+  once with \(x = m, \xi = \mu\) and the second time with \(x = n, \xi = \nu\).
+  The integral \eqref{eqn:inner-by-parts} can be integrated by parts twice resulting in this ugly expression:
+  \[
+    e^{-i2\pi x\xi} \left(
     \frac{\partial f}{\partial \xi} - i2\pi f
-  \right)\Bigg|_0^1
-  + (i2\pi x)^2 \int_{[0, 1]} f e^{-i2\pi x\xi} d\xi.
-\]
-However, actually this is not too bad. That is because once we substitute the bounds two things happen: the exponential in the front always equals 1 and what is inside of the parenthesis can be rewritten as
-\[
-  \frac{\partial f}{\partial\xi}(1) - \frac{\partial f}{\partial\xi}(0)
-  + i2\pi \left[ f(1) - f(0) \right],
-\]
-which equals zero, since \(f\) and its derivative are continuous and periodic. Hence, we are left with two integrals, that when substituted back into \eqref{eqn:laplacian-coeffs-expanded} give:
-\begin{align*}
-  \langle \nabla^2 & f, B_{m, n} \rangle = \\
+    \right)\Bigg|_0^1
+    + (i2\pi x)^2 \int_{[0, 1]} f e^{-i2\pi x\xi} d\xi.
+  \]
+  However, actually this is not too bad. That is because once we substitute the bounds two things happen: the exponential in the front always equals 1 and what is inside of the parenthesis can be rewritten as
+  \[
+    \frac{\partial f}{\partial\xi}(1) - \frac{\partial f}{\partial\xi}(0)
+    + i2\pi \left[ f(1) - f(0) \right],
+  \]
+  which equals zero, since \(f\) and its derivative are continuous and periodic. Hence, we are left with two integrals, that when substituted back into \eqref{eqn:laplacian-coeffs-expanded} give:
+  \begin{align*}
+    \langle \nabla^2 & f, B_{m, n} \rangle = \\
     &(i2\pi m)^2 \iint_{[0, 1]^2} f e^{-i2\pi m\mu} e^{-i2\pi n\nu} d\mu d\nu \\
     &\quad + (i2\pi n)^2 \iint_{[0, 1]^2} f e^{-i2\pi m\mu} e^{-i2\pi n\nu} d\mu d\nu.
-\end{align*}
-Finally, to complete the proof we rewrite the right side using the compact notation:
-\[
-  (i2\pi m)^2 \langle f, B_{m, n} \rangle
-  + (i2\pi n)^2 \langle f, B_{m, n} \rangle.
-  \qedhere
-\]
+  \end{align*}
+  Finally, to complete the proof we rewrite the right side using the compact notation:
+  \[
+    (i2\pi m)^2 \langle f, B_{m, n} \rangle
+    + (i2\pi n)^2 \langle f, B_{m, n} \rangle.
+    \qedhere
+  \]
 \end{proof}
 
 % TODO: closing words: this is why fourier is useful
@@ -244,10 +244,10 @@ Thus we first need examine the solutions to this equation before constructing th
 \end{proposition}
 
 \begin{lemma} The expression
-\begin{equation} \label{eqn:legendre-rodrigues}
-  P_n(x) = \frac{1}{n!2^n}\frac{d^n}{dx^n}(x^2-1)^n.
-\end{equation}
-is equivalent to \eqref{eqn:legendre-poly}.
+  \begin{equation} \label{eqn:legendre-rodrigues}
+    P_n(x) = \frac{1}{n!2^n}\frac{d^n}{dx^n}(x^2-1)^n.
+  \end{equation}
+  is equivalent to \eqref{eqn:legendre-poly}.
 \end{lemma}
 
 \begin{proof}
-- 
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