Optimize last images, add proof template, improve CSS

1 files changed, 8 insertions, 15 deletions

diff --git a/source/know/concept/fourier-transform/index.md b/source/know/concept/fourier-transform/index.md
index 0bc849b..c86d997 100644
--- a/source/know/concept/fourier-transform/index.md
+++ b/source/know/concept/fourier-transform/index.md
@@ -67,6 +67,7 @@ on whether the analysis is for forward ($$s > 0$$) or backward-propagating
 ($$s < 0$$) waves.
 
 
+
 ## Derivatives
 
 The FT of a derivative has a very useful property.
@@ -113,6 +114,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 
+
 ## Multiple dimensions
 
 The Fourier transform is straightforward to generalize to $$N$$ dimensions.
@@ -150,11 +152,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-<div class="accordion">
-<input type="checkbox" id="proof-constants-ND"/>
-<label for="proof-constants-ND">Proof</label>
-<div class="hidden" markdown="1">
-<label for="proof-constants-ND">Proof.</label>
+
+{% include proof/start.html id="proof-constants-ndim" -%}
 The inverse FT of the forward FT of $$f(\vb{x})$$ must be equal to $$f(\vb{x})$$ again, so:
 
 $$\begin{aligned}
@@ -180,9 +179,8 @@ $$\begin{aligned}
     &= \frac{(2 \pi)^N A B}{|s|^N} \int f(\vb{x}') \: \delta(\vb{x}' - \vb{x}) \ddn{N}{\vb{x}'}
     = \frac{(2 \pi)^N A B}{|s|^N} f(\vb{x})
 \end{aligned}$$
+{% include proof/end.html id="proof-constants-ndim" %}
 
-</div>
-</div>
 
 Differentiation is more complicated for $$N > 1$$,
 but the FT is still useful,
@@ -197,11 +195,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-<div class="accordion">
-<input type="checkbox" id="proof-laplacian"/>
-<label for="proof-laplacian">Proof</label>
-<div class="hidden" markdown="1">
-<label for="proof-laplacian">Proof.</label>
+
+{% include proof/start.html id="proof-laplacian" -%}
 We insert $$\nabla^2 f$$ into the FT,
 decompose the exponential and the Laplacian,
 and then integrate by parts (limits $$\pm \infty$$ omitted):
@@ -236,9 +231,7 @@ $$\begin{aligned}
     &= - A s^2 \sum_{n = 1}^N k_n^2 \int f \exp(i s \vb{k} \cdot \vb{x}) \ddn{N}{\vb{x}}
     = - s^2 \sum_{n = 1}^N k_n^2 \tilde{f}
 \end{aligned}$$
-
-</div>
-</div>
+{% include proof/end.html id="proof-laplacian" %}