Migrate from 'jekyll-katex' to 'kramdown-math-sskatex'

1 files changed, 23 insertions, 22 deletions

diff --git a/source/know/concept/holomorphic-function/index.md b/source/know/concept/holomorphic-function/index.md
index e22799c..5dde240 100644
--- a/source/know/concept/holomorphic-function/index.md
+++ b/source/know/concept/holomorphic-function/index.md
@@ -8,7 +8,7 @@ categories:
 layout: "concept"
 ---
 
-In complex analysis, a complex function $f(z)$ of a complex variable $z$
+In complex analysis, a complex function $$f(z)$$ of a complex variable $$z$$
 is called **holomorphic** or **analytic** if it is complex differentiable in the
 neighbourhood of every point of its domain.
 This is a very strong condition.
@@ -17,9 +17,9 @@ As a result, holomorphic functions are infinitely differentiable and
 equal their Taylor expansion at every point. In physicists' terms,
 they are extremely "well-behaved" throughout their domain.
 
-More formally, a given function $f(z)$ is holomorphic in a certain region
-if the following limit exists for all $z$ in that region,
-and for all directions of $\Delta z$:
+More formally, a given function $$f(z)$$ is holomorphic in a certain region
+if the following limit exists for all $$z$$ in that region,
+and for all directions of $$\Delta z$$:
 
 $$\begin{aligned}
     \boxed{
@@ -27,13 +27,13 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-We decompose $f$ into the real functions $u$ and $v$ of real variables $x$ and $y$:
+We decompose $$f$$ into the real functions $$u$$ and $$v$$ of real variables $$x$$ and $$y$$:
 
 $$\begin{aligned}
     f(z) = f(x + i y) = u(x, y) + i v(x, y)
 \end{aligned}$$
 
-Since we are free to choose the direction of $\Delta z$, we choose $\Delta x$ and $\Delta y$:
+Since we are free to choose the direction of $$\Delta z$$, we choose $$\Delta x$$ and $$\Delta y$$:
 
 $$\begin{aligned}
     f'(z)
@@ -44,8 +44,8 @@ $$\begin{aligned}
     = \pdv{v}{y} - i \pdv{u}{y}
 \end{aligned}$$
 
-For $f(z)$ to be holomorphic, these two results must be equivalent.
-Because $u$ and $v$ are real by definition,
+For $$f(z)$$ to be holomorphic, these two results must be equivalent.
+Because $$u$$ and $$v$$ are real by definition,
 we thus arrive at the **Cauchy-Riemann equations**:
 
 $$\begin{aligned}
@@ -56,7 +56,7 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Therefore, a given function $f(z)$ is holomorphic if and only if its real
+Therefore, a given function $$f(z)$$ is holomorphic if and only if its real
 and imaginary parts satisfy these equations. This gives an idea of how
 strict the criteria are to qualify as holomorphic.
 
@@ -64,8 +64,8 @@ strict the criteria are to qualify as holomorphic.
 ## Integration formulas
 
 Holomorphic functions satisfy **Cauchy's integral theorem**, which states
-that the integral of $f(z)$ over any closed curve $C$ in the complex plane is zero,
-provided that $f(z)$ is holomorphic for all $z$ in the area enclosed by $C$:
+that the integral of $$f(z)$$ over any closed curve $$C$$ in the complex plane is zero,
+provided that $$f(z)$$ is holomorphic for all $$z$$ in the area enclosed by $$C$$:
 
 $$\begin{aligned}
     \boxed{
@@ -78,7 +78,7 @@ $$\begin{aligned}
 <label for="proof-int-theorem">Proof</label>
 <div class="hidden" markdown="1">
 <label for="proof-int-theorem">Proof.</label>
-Just like before, we decompose $f(z)$ into its real and imaginary parts:
+Just like before, we decompose $$f(z)$$ into its real and imaginary parts:
 
 $$\begin{aligned}
     \oint_C f(z) \dd{z}
@@ -88,21 +88,21 @@ $$\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 $$C$$:
 
 $$\begin{aligned}
     \oint_C f(z) \dd{z}
     &= - \iint_A \pdv{v}{x} + \pdv{u}{y} \dd{x} \dd{y} + i \iint_A \pdv{u}{x} - \pdv{v}{y} \dd{x} \dd{y}
 \end{aligned}$$
 
-Since $f(z)$ is holomorphic, $u$ and $v$ satisfy the Cauchy-Riemann
+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
-determined by its values on an arbitrary contour $C$ around $z_0$:
+states that the value of $$f(z)$$ at an arbitrary point $$z_0$$ is
+determined by its values on an arbitrary contour $$C$$ around $$z_0$$:
 
 $$\begin{aligned}
     \boxed{
@@ -116,8 +116,8 @@ $$\begin{aligned}
 <div class="hidden" markdown="1">
 <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
+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:
 
 $$\begin{aligned}
@@ -126,13 +126,14 @@ $$\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 $$z_0$$:
 
 $$\begin{aligned}
     \lim_{r \to 0}\:\: \frac{1}{2 \pi} \int_0^{2 \pi} f(z_0 + r e^{i \theta}) \dd{\theta}
     &= \frac{f(z_0)}{2 \pi} \int_0^{2 \pi} \dd{\theta}
     = f(z_0)
 \end{aligned}$$
+
 </div>
 </div>
 
@@ -153,7 +154,7 @@ $$\begin{aligned}
 <label for="proof-diff-formula">Proof</label>
 <div class="hidden" markdown="1">
 <label for="proof-diff-formula">Proof.</label>
-By definition, the first derivative $f'(z)$ of a
+By definition, the first derivative $$f'(z)$$ of a
 holomorphic function exists and is:
 
 $$\begin{aligned}
@@ -183,8 +184,8 @@ $$\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 $n$.
+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>