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 $$\mathcal{F} \overline{f}$$. -- GitLab