Introduce collapsible proofs to some articles

1 files changed, 20 insertions, 6 deletions

diff --git a/content/know/concept/convolution-theorem/index.pdc b/content/know/concept/convolution-theorem/index.pdc
index 9d1a666..1454cc0 100644
--- a/content/know/concept/convolution-theorem/index.pdc
+++ b/content/know/concept/convolution-theorem/index.pdc
@@ -32,7 +32,12 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-To prove this, we expand the right-hand side of the theorem and
+<div class="accordion">
+<input type="checkbox" id="proof-fourier"/>
+<label for="proof-fourier">Proof</label>
+<div class="hidden">
+<label for="proof-fourier">Proof.</label>
+We expand the right-hand side of the theorem and
 rearrange the integrals:
 
 $$\begin{aligned}
@@ -45,8 +50,8 @@ $$\begin{aligned}
     = A \cdot (f * g)(x)
 \end{aligned}$$
 
-Then we do the same thing again, this time starting from a product in
-the $x$-domain:
+Then we do the same again,
+this time starting from a product in the $x$-domain:
 
 $$\begin{aligned}
     \hat{\mathcal{F}}\{f(x) \: g(x)\}
@@ -57,6 +62,8 @@ $$\begin{aligned}
     &= B \int_{-\infty}^\infty \tilde{g}(k') \tilde{f}(k - k') \dd{k'}
     = B \cdot (\tilde{f} * \tilde{g})(k)
 \end{aligned}$$
+</div>
+</div>
 
 
 ## Laplace transform
@@ -76,9 +83,14 @@ $$\begin{aligned}
     \boxed{\hat{\mathcal{L}}\{(f * g)(t)\} = \tilde{f}(s) \: \tilde{g}(s)}
 \end{aligned}$$
 
-We prove this by expanding the left-hand side. Note that the lower
-integration limit is 0 instead of $-\infty$, because we set both $f(t)$
-and $g(t)$ to zero for $t < 0$:
+<div class="accordion">
+<input type="checkbox" id="proof-laplace"/>
+<label for="proof-laplace">Proof</label>
+<div class="hidden">
+<label for="proof-laplace">Proof.</label>
+We expand the left-hand side.
+Note that the lower integration limit is 0 instead of $-\infty$,
+because we set both $f(t)$ and $g(t)$ to zero for $t < 0$:
 
 $$\begin{aligned}
     \hat{\mathcal{L}}\{(f * g)(t)\}
@@ -98,6 +110,8 @@ $$\begin{aligned}
     &= \int_0^\infty \tilde{f}(s) g(t') \exp(- s t') \dd{t'}
     = \tilde{f}(s) \: \tilde{g}(s)
 \end{aligned}$$
+</div>
+</div>