From d3b96730bd01263098bbb96c15148878e5633a04 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Sat, 8 May 2021 16:08:41 +0200
Subject: Expand knowledge base, change text alignment

---
 content/know/concept/parsevals-theorem/index.pdc | 15 +++++++--------
 1 file changed, 7 insertions(+), 8 deletions(-)

(limited to 'content/know/concept/parsevals-theorem')

diff --git a/content/know/concept/parsevals-theorem/index.pdc b/content/know/concept/parsevals-theorem/index.pdc
index 8f653f8..ae34bda 100644
--- a/content/know/concept/parsevals-theorem/index.pdc
+++ b/content/know/concept/parsevals-theorem/index.pdc
@@ -13,8 +13,8 @@ markup: pandoc
 
 # Parseval's theorem
 
-**Parseval's theorem** relates the inner product of two functions $f(x)$ and $g(x)$ to the
-inner product of their [Fourier transforms](/know/concept/fourier-transform/)
+**Parseval's theorem** is a relation between the inner product of two functions $f(x)$ and $g(x)$,
+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:
@@ -29,12 +29,11 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-For this reason, physicists like to define their Fourier transform
-with $A = B = 1 / \sqrt{2\pi}$ and $|s| = 1$, because then the FT nicely
-conserves the total probability (quantum mechanics) or the total energy
-(optics).
+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 this, we insert the inverse FT into the inner product
+To prove the theorem, we insert the inverse FT into the inner product
 definition:
 
 $$\begin{aligned}
@@ -55,7 +54,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Where $\delta(k)$ is the [Dirac delta function](/know/concept/dirac-delta-function/).
-Note that we can just as well do it in the opposite direction,
+Note that we can equally well do the proof in the opposite direction,
 which yields an equivalent result:
 
 $$\begin{aligned}
-- 
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