From 57c9fc6b05caa247a04a504d2d35b720e5a3c483 Mon Sep 17 00:00:00 2001
From: Erik Strand <erik.strand@cba.mit.edu>
Date: Fri, 15 Feb 2019 22:29:53 -0500
Subject: [PATCH] Equation formatting

---
 _posts/2019-02-15-ch2-notes.md | 10 ++++++----
 1 file changed, 6 insertions(+), 4 deletions(-)

diff --git a/_posts/2019-02-15-ch2-notes.md b/_posts/2019-02-15-ch2-notes.md
index 8578eb0..df038b4 100644
--- a/_posts/2019-02-15-ch2-notes.md
+++ b/_posts/2019-02-15-ch2-notes.md
@@ -114,11 +114,13 @@ real $$x$$.
 Let's see what the Fourier transform of the conjugate of a function looks like.
 
 $$
+\begin{align*}
 (\mathcal{F} \overline{f})(x)
-= \int_\mathbb{R} \overline{f(x')} e^{-2 \pi i x' x} \mathrm{d} x'
-= \int_\mathbb{R} \overline{f(x') e^{2 \pi i x' x}} \mathrm{d} x'
-= \overline{\int_\mathbb{R} f(x') e^{2 \pi i x' x} \mathrm{d} x'}
-= \overline{(\mathcal{F}^{-1} f)(x)}
+&= \int_\mathbb{R} \overline{f(x')} e^{-2 \pi i x' x} \mathrm{d} x' \\
+&= \int_\mathbb{R} \overline{f(x') e^{2 \pi i x' x}} \mathrm{d} x' \\
+&= \overline{\int_\mathbb{R} f(x') e^{2 \pi i x' x} \mathrm{d} x'} \\
+&= \overline{(\mathcal{F}^{-1} f)(x)}
+\end{align*}
 $$
 
 This implies that $$ \mathcal{F}^{-1} f$$ is the complex conjugate of $$\mathcal{F} \overline{f}$$.
-- 
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