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authorNao Pross <np@0hm.ch>2022-04-26 00:08:07 +0200
committerNao Pross <np@0hm.ch>2022-04-26 00:08:07 +0200
commita8d3a3d63877539392da41028df8339fc49db27b (patch)
tree6fe861f7e52e5f6bd08340c51774bd3f75bca057 /notes
parentAdd draft of handout for presentation (diff)
downloadFourierOnS2-a8d3a3d63877539392da41028df8339fc49db27b.tar.gz
FourierOnS2-a8d3a3d63877539392da41028df8339fc49db27b.zip
Fix mistake in notes on Fourer on a flat surface
Diffstat (limited to 'notes')
-rw-r--r--notes/FourierOnS2.tex48
-rw-r--r--notes/build/FourierOnS2.pdfbin44725 -> 45497 bytes
2 files changed, 28 insertions, 20 deletions
diff --git a/notes/FourierOnS2.tex b/notes/FourierOnS2.tex
index 213a9d6..aeae7ef 100644
--- a/notes/FourierOnS2.tex
+++ b/notes/FourierOnS2.tex
@@ -72,7 +72,7 @@ where \(m, n \in \mathbb{Z}\), be our basis functions in the space of ``nice'' f
And finally by the Fourier theorem we can reconstruct the original function using an unoriginally named Fourier series:
\[
f(\mu, \nu) =
- \sum_{m = -\infty}^{\infty} \sum_{n = -\infty}^{\infty} c_{m, n} B_{m, n}(\mu, \nu).
+ \sum_{m\in\mathbb{Z}} \sum_{n\in\mathbb{Z}} c_{m, n} B_{m, n}(\mu, \nu).
\]
% \begin{definition}[\(L^2\) norm]
@@ -124,48 +124,56 @@ The solutions to this elementary ODE are of course complex exponentials, the sam
is easy to compute. This is shown in the next lemma.
\begin{lemma}
- Let \(f\) be a ``nice'' function. Then
+ Let \(f \in C(\mathbb{R}^2/\mathbb{Z}^2)\), then
\[
\langle \nabla^2 f, B_{m, n} \rangle
- = \left( \frac{1}{m^2} + \frac{1}{n^2} \right) \langle f, B_{m, n} \rangle.
+ = (2\pi i)^2 \left( m^2 + n^2 \right) \langle f, B_{m, n} \rangle.
\]
\end{lemma}
\begin{proof}
-To show this, we first expand the left side of the statement:
+To begin this proof, we first expand the left side of the statement:
\begin{gather}
\nonumber
\langle \nabla^2 f, B_{m, n} \rangle
- = \iint_{\mathbb{R}^2} \nabla^2 f B_{m,n} d\mu d\nu \\
- = \iint_{\mathbb{R}^2} \left(
+ = \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^{-im\mu} e^{-in\nu} d\mu d\nu.
+ \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_{\mathbb{R}} \frac{\partial^2 f}{\partial \xi^2} e^{-ix\xi} d\xi,
+ \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:
\[
- e^{-x\xi} \left(
- \frac{\partial f}{\partial \xi} + \frac{f}{ix}
- \right)\Bigg|_{-\infty}^{+\infty}
- - \frac{1}{(ix)^2} \int_\mathbb{R} f e^{-jx\xi} d\xi.
+ e^{-i2\pi x\xi} \left(
+ \frac{\partial f}{\partial \xi} - i2\pi f
+ \right)\Bigg|_0^1
+ + (i2\pi x)^2 \int_{[0, 1]} f e^{-i2\pi x\xi} d\xi.
\]
-However, because \(f\) is ``nice'', both \(f\) and its derivative vanish at infinity, leaving only the integral. By substituting back this result into \eqref{eqn:laplacian-coeffs-expanded} we get:
+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
- &= \frac{1}{m^2} \iint f e^{-im\mu} e^{-in\nu} d\mu d\nu \\
- &\qquad + \frac{1}{n^2} \iint f e^{-im\mu} e^{-in\nu} d\mu d\nu \\
- &= \frac{1}{m^2} \langle f, B_{m, n} \rangle
- + \frac{1}{n^2} \langle f, B_{m, n} \rangle
- \qedhere
+ \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{proof}
+% TODO: closing words: this is why fourier is useful
\section{Fourier on the Sphere}
diff --git a/notes/build/FourierOnS2.pdf b/notes/build/FourierOnS2.pdf
index e9a60dd..7316cc2 100644
--- a/notes/build/FourierOnS2.pdf
+++ b/notes/build/FourierOnS2.pdf
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