Introduce collapsible proofs to some articles

1 files changed, 45 insertions, 26 deletions

diff --git a/content/know/concept/holomorphic-function/index.pdc b/content/know/concept/holomorphic-function/index.pdc
index 1077060..1c2f092 100644
--- a/content/know/concept/holomorphic-function/index.pdc
+++ b/content/know/concept/holomorphic-function/index.pdc
@@ -77,8 +77,12 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-*__Proof__*.
-*Just like before, we decompose $f(z)$ into its real and imaginary parts:*
+<div class="accordion">
+<input type="checkbox" id="proof-int-theorem"/>
+<label for="proof-int-theorem">Proof</label>
+<div class="hidden">
+<label for="proof-int-theorem">Proof.</label>
+Just like before, we decompose $f(z)$ into its real and imaginary parts:
 
 $$\begin{aligned}
     \oint_C f(z) \:dz
@@ -88,16 +92,17 @@ $$\begin{aligned}
     &= \oint_C u \dd{x} - v \dd{y} + i \oint_C v \dd{x} + u \dd{y}
 \end{aligned}$$
 
-*Using Green's theorem, we integrate over the area $A$ enclosed by $C$:*
+Using Green's theorem, we integrate over the area $A$ enclosed by $C$:
 
 $$\begin{aligned}
     \oint_C f(z) \:dz
     &= - \iint_A \pdv{v}{x} + \pdv{u}{y} \dd{x} \dd{y} + i \iint_A \pdv{u}{x} - \pdv{v}{y} \dd{x} \dd{y}
 \end{aligned}$$
 
-*Since $f(z)$ is holomorphic, $u$ and $v$ satisfy the Cauchy-Riemann
-equations, such that the integrands disappear and the final result is zero.*
-*__Q.E.D.__*
+Since $f(z)$ is holomorphic, $u$ and $v$ satisfy the Cauchy-Riemann
+equations, such that the integrands disappear and the final result is zero.
+</div>
+</div>
 
 An interesting consequence is **Cauchy's integral formula**, which
 states that the value of $f(z)$ at an arbitrary point $z_0$ is
@@ -109,11 +114,15 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-*__Proof__*.
-*Thanks to the integral theorem, we know that the shape and size
+<div class="accordion">
+<input type="checkbox" id="proof-int-formula"/>
+<label for="proof-int-formula">Proof</label>
+<div class="hidden">
+<label for="proof-int-formula">Proof.</label>
+Thanks to the integral theorem, we know that the shape and size
 of $C$ is irrelevant. Therefore we choose it to be a circle with radius $r$,
 such that the integration variable becomes $z = z_0 + r e^{i \theta}$. Then
-we integrate by substitution:*
+we integrate by substitution:
 
 $$\begin{aligned}
     \frac{1}{2 \pi i} \oint_C \frac{f(z)}{z - z_0} \dd{z}
@@ -121,15 +130,15 @@ $$\begin{aligned}
     = \frac{1}{2 \pi} \int_0^{2 \pi} f(z_0 + r e^{i \theta}) \dd{\theta}
 \end{aligned}$$
 
-*We may choose an arbitrarily small radius $r$, such that the contour approaches $z_0$:*
+We may choose an arbitrarily small radius $r$, such that the contour approaches $z_0$:
 
 $$\begin{aligned}
     \lim_{r \to 0}\:\: \frac{1}{2 \pi} \int_0^{2 \pi} f(z_0 + r e^{i \theta}) \dd{\theta}
     &= \frac{f(z_0)}{2 \pi} \int_0^{2 \pi} \dd{\theta}
     = f(z_0)
 \end{aligned}$$
-
-*__Q.E.D.__*
+</div>
+</div>
 
 Similarly, **Cauchy's differentiation formula**,
 or **Cauchy's integral formula for derivatives**
@@ -143,16 +152,20 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-*__Proof__*.
-*By definition, the first derivative $f'(z)$ of a
-holomorphic function $f(z)$ exists and is given by:*
+<div class="accordion">
+<input type="checkbox" id="proof-diff-formula"/>
+<label for="proof-diff-formula">Proof</label>
+<div class="hidden">
+<label for="proof-diff-formula">Proof.</label>
+By definition, the first derivative $f'(z)$ of a
+holomorphic function exists and is:
 
 $$\begin{aligned}
     f'(z_0)
     = \lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0}
 \end{aligned}$$
 
-*We evaluate the numerator using Cauchy's integral theorem as follows:*
+We evaluate the numerator using Cauchy's integral theorem as follows:
 
 $$\begin{aligned}
     f'(z_0)
@@ -166,7 +179,7 @@ $$\begin{aligned}
     \oint_C \frac{f(\zeta) (z - z_0)}{(\zeta - z)(\zeta - z_0)} \dd{\zeta}
 \end{aligned}$$
 
-*This contour integral converges uniformly, so we may apply the limit on the inside:*
+This contour integral converges uniformly, so we may apply the limit on the inside:
 
 $$\begin{aligned}
     f'(z_0)
@@ -174,9 +187,10 @@ $$\begin{aligned}
     = \frac{1}{2 \pi i} \oint_C \frac{f(\zeta)}{(\zeta - z_0)^2} \dd{\zeta}
 \end{aligned}$$
 
-*Since the second-order derivative $f''(z)$ is simply the derivative of $f'(z)$,
-this proof works inductively for all higher orders $n$.*
-*__Q.E.D.__*
+Since the second-order derivative $f''(z)$ is simply the derivative of $f'(z)$,
+this proof works inductively for all higher orders $n$.
+</div>
+</div>
 
 
 ## Residue theorem
@@ -205,24 +219,29 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-*__Proof__*. *From the definition of a meromorphic function,
+<div class="accordion">
+<input type="checkbox" id="proof-res-theorem"/>
+<label for="proof-res-theorem">Proof</label>
+<div class="hidden">
+<label for="proof-res-theorem">Proof.</label>
+From the definition of a meromorphic function,
 we know that we can decompose $f(z)$ like so,
-where $h(z)$ is holomorphic and $p$ are all its poles:*
+where $h(z)$ is holomorphic and $p$ are all its poles:
 
 $$\begin{aligned}
     f(z) = h(z) + \sum_{p} \frac{R_p}{z - z_p}
 \end{aligned}$$
 
-*We integrate this over a contour $C$ which contains all poles, and apply
-both Cauchy's integral theorem and Cauchy's integral formula to get:*
+We integrate this over a contour $C$ which contains all poles, and apply
+both Cauchy's integral theorem and Cauchy's integral formula to get:
 
 $$\begin{aligned}
     \oint_C f(z) \dd{z}
     &= \oint_C h(z) \dd{z} + \sum_{p} R_p \oint_C \frac{1}{z - z_p} \dd{z}
     = \sum_{p} R_p \: 2 \pi i
 \end{aligned}$$
-
-*__Q.E.D.__*
+</div>
+</div>
 
 This theorem might not seem very useful,
 but in fact, thanks to some clever mathematical magic,