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authorNao Pross <np@0hm.ch>2022-04-28 12:50:55 +0200
committerNao Pross <np@0hm.ch>2022-04-28 12:55:41 +0200
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treeb5c17ea3c57762c1b81493056e353268e20735eb /notes
parentStart copying proofs (diff)
downloadFourierOnS2-ac8f1dd9177b5b4c38674d6f2cf944d245ca8af4.tar.gz
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Fix indentation
Diffstat (limited to 'notes')
-rw-r--r--notes/FourierOnS2.tex78
-rw-r--r--notes/build/FourierOnS2.pdfbin54623 -> 54623 bytes
2 files changed, 39 insertions, 39 deletions
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}
diff --git a/notes/build/FourierOnS2.pdf b/notes/build/FourierOnS2.pdf
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