From c2327bcc3571ead88ba2b0ce40656211a888f640 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Sun, 21 Feb 2021 16:46:21 +0100
Subject: Add "Convolution theorem" and "Parseval's theorem"

---
 latex/know/concept/fourier-transform/source.md | 8 +++-----
 1 file changed, 3 insertions(+), 5 deletions(-)

(limited to 'latex/know/concept/fourier-transform')

diff --git a/latex/know/concept/fourier-transform/source.md b/latex/know/concept/fourier-transform/source.md
index 58830df..3e25980 100644
--- a/latex/know/concept/fourier-transform/source.md
+++ b/latex/know/concept/fourier-transform/source.md
@@ -63,10 +63,9 @@ on whether the analysis is for forward ($s > 0$) or backward-propagating
 
 ## Derivatives
 
-The FT of a derivative has a very interesting property, let us take a
-look. Below, after integrating by parts, we remove the boundary term by
-assuming that $f(x)$ is localized, i.e. $f(x) \to 0$ for
-$x \to \pm \infty$:
+The FT of a derivative has a very interesting 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 $x \to \pm \infty$:
 
 $$\begin{aligned}
     \hat{\mathcal{F}}\{f'(x)\}
@@ -75,7 +74,6 @@ $$\begin{aligned}
     &= A \big[ f(x) \exp(i s k x) \big]_{-\infty}^\infty - i s k A \int_{-\infty}^\infty f(x) \exp(i s k x) \dd{x}
     \\
     &= (- i s k) \tilde{f}(k)
-    \qedhere
 \end{aligned}$$
 
 Therefore, as long as $f(x)$ is localized, the FT eliminates derivatives
-- 
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