From cc295b5da8e3db4417523a507caf106d5839d989 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Wed, 2 Jun 2021 13:28:53 +0200
Subject: Introduce collapsible proofs to some articles
---
content/know/concept/parsevals-theorem/index.pdc | 32 ++++++++++++++----------
1 file changed, 19 insertions(+), 13 deletions(-)
(limited to 'content/know/concept/parsevals-theorem/index.pdc')
diff --git a/content/know/concept/parsevals-theorem/index.pdc b/content/know/concept/parsevals-theorem/index.pdc
index 824afa6..9f440f2 100644
--- a/content/know/concept/parsevals-theorem/index.pdc
+++ b/content/know/concept/parsevals-theorem/index.pdc
@@ -17,24 +17,24 @@ markup: pandoc
and the inner product of their [Fourier transforms](/know/concept/fourier-transform/)
$\tilde{f}(k)$ and $\tilde{g}(k)$.
There are two equivalent ways of stating it,
-where $A$, $B$, and $s$ are constants from the Fourier transform's definition:
+where $A$, $B$, and $s$ are constants from the FT's definition:
$$\begin{aligned}
\boxed{
- \braket{f(x)}{g(x)} = \frac{2 \pi B^2}{|s|} \braket*{\tilde{f}(k)}{\tilde{g}(k)}
- }
- \\
- \boxed{
- \braket*{\tilde{f}(k)}{\tilde{g}(k)} = \frac{2 \pi A^2}{|s|} \braket{f(x)}{g(x)}
+ \begin{aligned}
+ \braket{f(x)}{g(x)} &= \frac{2 \pi B^2}{|s|} \braket*{\tilde{f}(k)}{\tilde{g}(k)}
+ \\
+ \braket*{\tilde{f}(k)}{\tilde{g}(k)} &= \frac{2 \pi A^2}{|s|} \braket{f(x)}{g(x)}
+ \end{aligned}
}
\end{aligned}$$
-For this reason, physicists like to define the Fourier transform
-with $A\!=\!B\!=\!1 / \sqrt{2\pi}$ and $|s|\!=\!1$, because then it nicely
-conserves the functions' normalization.
-
-To prove the theorem, we insert the inverse FT into the inner product
-definition:
+<div class="accordion">
+<input type="checkbox" id="proof-fourier"/>
+<label for="proof-fourier">Proof</label>
+<div class="hidden">
+<label for="proof-fourier">Proof.</label>
+We insert the inverse FT into the defintion of the inner product:
$$\begin{aligned}
\braket{f}{g}
@@ -54,7 +54,7 @@ $$\begin{aligned}
\end{aligned}$$
Where $\delta(k)$ is the [Dirac delta function](/know/concept/dirac-delta-function/).
-Note that we can equally well do the proof in the opposite direction,
+Note that we can equally well do this proof in the opposite direction,
which yields an equivalent result:
$$\begin{aligned}
@@ -73,6 +73,12 @@ $$\begin{aligned}
&= \frac{2 \pi A^2}{|s|} \int_{-\infty}^\infty f^*(x) \: g(x) \dd{x}
= \frac{2 \pi A^2}{|s|} \braket{f}{g}
\end{aligned}$$
+</div>
+</div>
+
+For this reason, physicists like to define the Fourier transform
+with $A\!=\!B\!=\!1 / \sqrt{2\pi}$ and $|s|\!=\!1$, because then it nicely
+conserves the functions' normalization.
--
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