From bd13537ee2fb704b02b961b5d06dd4f406f19a71 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Sat, 21 Oct 2023 14:21:59 +0200
Subject: Improve knowledge base

---
 source/know/concept/parsevals-theorem/index.md | 10 ++++++----
 1 file changed, 6 insertions(+), 4 deletions(-)

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

diff --git a/source/know/concept/parsevals-theorem/index.md b/source/know/concept/parsevals-theorem/index.md
index a7ce0bf..b2db490 100644
--- a/source/know/concept/parsevals-theorem/index.md
+++ b/source/know/concept/parsevals-theorem/index.md
@@ -17,9 +17,11 @@ where $$A$$, $$B$$, and $$s$$ are constants from the FT's definition:
 $$\begin{aligned}
     \boxed{
         \begin{aligned}
-            \inprod{f(x)}{g(x)} &= \frac{2 \pi B^2}{|s|} \inprod{\tilde{f}(k)}{\tilde{g}(k)}
+            \inprod{f(x)}{g(x)}
+            &= \frac{2 \pi B^2}{|s|} \inprod{\tilde{f}(k)}{\tilde{g}(k)}
             \\
-            \inprod{\tilde{f}(k)}{\tilde{g}(k)} &= \frac{2 \pi A^2}{|s|} \Inprod{f(x)}{g(x)}
+            \inprod{\tilde{f}(k)}{\tilde{g}(k)}
+            &= \frac{2 \pi A^2}{|s|} \inprod{f(x)}{g(x)}
         \end{aligned}
     }
 \end{aligned}$$
@@ -71,8 +73,8 @@ $$\begin{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.
+with $$A\!=\!B\!=\!1 / \sqrt{2\pi}$$ and $$|s|\!=\!1$$,
+because then it nicely preserves the functions' normalization.
 
 
 
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