From 16555851b6514a736c5c9d8e73de7da7fc9b6288 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Thu, 20 Oct 2022 18:25:31 +0200
Subject: Migrate from 'jekyll-katex' to 'kramdown-math-sskatex'
---
source/know/concept/fourier-transform/index.md | 64 +++++++++++++-------------
1 file changed, 33 insertions(+), 31 deletions(-)
(limited to 'source/know/concept/fourier-transform')
diff --git a/source/know/concept/fourier-transform/index.md b/source/know/concept/fourier-transform/index.md
index f8b5b2a..0bc849b 100644
--- a/source/know/concept/fourier-transform/index.md
+++ b/source/know/concept/fourier-transform/index.md
@@ -10,11 +10,11 @@ layout: "concept"
---
The **Fourier transform** (FT) is an integral transform which converts a
-function $f(x)$ into its frequency representation $\tilde{f}(k)$.
+function $$f(x)$$ into its frequency representation $$\tilde{f}(k)$$.
Great volumes have already been written about this subject,
so let us focus on the aspects that are useful to physicists.
-The **forward** FT is defined as follows, where $A$, $B$, and $s$ are unspecified constants
+The **forward** FT is defined as follows, where $$A$$, $$B$$, and $$s$$ are unspecified constants
(for now):
$$\begin{aligned}
@@ -35,9 +35,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-Clearly, the inverse FT of the forward FT of $f(x)$ must equal $f(x)$
+Clearly, the inverse FT of the forward FT of $$f(x)$$ must equal $$f(x)$$
again. Let us verify this, by rearranging the integrals to get the
-[Dirac delta function](/know/concept/dirac-delta-function/) $\delta(x)$:
+[Dirac delta function](/know/concept/dirac-delta-function/) $$\delta(x)$$:
$$\begin{aligned}
\hat{\mathcal{F}}^{-1}\{\hat{\mathcal{F}}\{f(x)\}\}
@@ -49,29 +49,29 @@ $$\begin{aligned}
= \frac{2 \pi A B}{|s|} f(x)
\end{aligned}$$
-Therefore, the constants $A$, $B$, and $s$ are subject to the following
+Therefore, the constants $$A$$, $$B$$, and $$s$$ are subject to the following
constraint:
$$\begin{aligned}
\boxed{\frac{2\pi A B}{|s|} = 1}
\end{aligned}$$
-But that still gives a lot of freedom. The exact choices of $A$ and $B$
+But that still gives a lot of freedom. The exact choices of $$A$$ and $$B$$
are generally motivated by the [convolution theorem](/know/concept/convolution-theorem/)
and [Parseval's theorem](/know/concept/parsevals-theorem/).
-The choice of $|s|$ depends on whether the frequency variable $k$
-represents the angular ($|s| = 1$) or the physical ($|s| = 2\pi$)
-frequency. The sign of $s$ is not so important, but is generally based
-on whether the analysis is for forward ($s > 0$) or backward-propagating
-($s < 0$) waves.
+The choice of $$|s|$$ depends on whether the frequency variable $$k$$
+represents the angular ($$|s| = 1$$) or the physical ($$|s| = 2\pi$$)
+frequency. The sign of $$s$$ is not so important, but is generally based
+on whether the analysis is for forward ($$s > 0$$) or backward-propagating
+($$s < 0$$) waves.
## Derivatives
The FT of a derivative has a very useful property.
Below, after integrating by parts, we remove the boundary term by
-assuming that $f(x)$ is localized, i.e. $f(x) \to 0$ for $x \to \pm \infty$:
+assuming that $$f(x)$$ is localized, i.e. $$f(x) \to 0$$ for $$x \to \pm \infty$$:
$$\begin{aligned}
\hat{\mathcal{F}}\{f'(x)\}
@@ -82,7 +82,7 @@ $$\begin{aligned}
&= (- i s k) \tilde{f}(k)
\end{aligned}$$
-Therefore, as long as $f(x)$ is localized, the FT eliminates derivatives
+Therefore, as long as $$f(x)$$ is localized, the FT eliminates derivatives
of the transformed variable, which makes it useful against PDEs:
$$\begin{aligned}
@@ -92,8 +92,8 @@ $$\begin{aligned}
\end{aligned}$$
This generalizes to higher-order derivatives, as long as these
-derivatives are also localized in the $x$-domain, which is practically
-guaranteed if $f(x)$ itself is localized:
+derivatives are also localized in the $$x$$-domain, which is practically
+guaranteed if $$f(x)$$ itself is localized:
$$\begin{aligned}
\boxed{
@@ -115,9 +115,9 @@ $$\begin{aligned}
## Multiple dimensions
-The Fourier transform is straightforward to generalize to $N$ dimensions.
-Given a scalar field $f(\vb{x})$ with $\vb{x} = (x_1, ..., x_N)$,
-its FT $\tilde{f}(\vb{k})$ is defined as follows:
+The Fourier transform is straightforward to generalize to $$N$$ dimensions.
+Given a scalar field $$f(\vb{x})$$ with $$\vb{x} = (x_1, ..., x_N)$$,
+its FT $$\tilde{f}(\vb{k})$$ is defined as follows:
$$\begin{aligned}
\boxed{
@@ -127,7 +127,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where the wavevector $\vb{k} = (k_1, ..., k_N)$.
+Where the wavevector $$\vb{k} = (k_1, ..., k_N)$$.
Likewise, the inverse FT is given by:
$$\begin{aligned}
@@ -138,9 +138,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-In practice, in $N$D, there is not as much disagreement about
-the constants $A$, $B$ and $s$ as in 1D:
-typically $A = 1$ and $B = 1 / (2 \pi)^N$, with $s = \pm 1$.
+In practice, in $$N$$D, there is not as much disagreement about
+the constants $$A$$, $$B$$ and $$s$$ as in 1D:
+typically $$A = 1$$ and $$B = 1 / (2 \pi)^N$$, with $$s = \pm 1$$.
Any choice will do, as long as:
$$\begin{aligned}
@@ -155,7 +155,7 @@ $$\begin{aligned}
-The inverse FT of the forward FT of $f(\vb{x})$ must be equal to $f(\vb{x})$ again, so:
+The inverse FT of the forward FT of $$f(\vb{x})$$ must be equal to $$f(\vb{x})$$ again, so:
$$\begin{aligned}
\hat{\mathcal{F}}^{-1}\{\hat{\mathcal{F}}\{ f(\vb{x}) \}\}
@@ -180,14 +180,15 @@ $$\begin{aligned}
&= \frac{(2 \pi)^N A B}{|s|^N} \int f(\vb{x}') \: \delta(\vb{x}' - \vb{x}) \ddn{N}{\vb{x}'}
= \frac{(2 \pi)^N A B}{|s|^N} f(\vb{x})
\end{aligned}$$
+
-Differentiation is more complicated for $N > 1$,
+Differentiation is more complicated for $$N > 1$$,
but the FT is still useful,
-notably for the Laplacian $\nabla^2 \equiv \idv{ {}^2}{x_1^2} + ... + \idv{ {}^2}{x_N^2}$.
-Let $|\vb{k}|$ be the norm of $\vb{k}$,
-then for a localized $f$:
+notably for the Laplacian $$\nabla^2 \equiv \idv{ {}^2}{x_1^2} + ... + \idv{ {}^2}{x_N^2}$$.
+Let $$|\vb{k}|$$ be the norm of $$\vb{k}$$,
+then for a localized $$f$$:
$$\begin{aligned}
\boxed{
@@ -201,9 +202,9 @@ $$\begin{aligned}
-We insert $\nabla^2 f$ into the FT,
+We insert $$\nabla^2 f$$ into the FT,
decompose the exponential and the Laplacian,
-and then integrate by parts (limits $\pm \infty$ omitted):
+and then integrate by parts (limits $$\pm \infty$$ omitted):
$$\begin{aligned}
\hat{\mathcal{F}}\{\nabla^2 f\}
@@ -216,7 +217,7 @@ $$\begin{aligned}
\end{aligned}$$
Just like in 1D, we get rid of the boundary term
-by assuming that all derivatives $\idv{f}{x_n}$ are nicely localized.
+by assuming that all derivatives $$\idv{f}{x_n}$$ are nicely localized.
To proceed, we then integrate by parts again:
$$\begin{aligned}
@@ -228,13 +229,14 @@ $$\begin{aligned}
\end{aligned}$$
Once again, we remove the boundary term
-by assuming that $f$ is localized, yielding:
+by assuming that $$f$$ is localized, yielding:
$$\begin{aligned}
\hat{\mathcal{F}}\{\nabla^2 f\}
&= - A s^2 \sum_{n = 1}^N k_n^2 \int f \exp(i s \vb{k} \cdot \vb{x}) \ddn{N}{\vb{x}}
= - s^2 \sum_{n = 1}^N k_n^2 \tilde{f}
\end{aligned}$$
+