From 16555851b6514a736c5c9d8e73de7da7fc9b6288 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Thu, 20 Oct 2022 18:25:31 +0200
Subject: Migrate from 'jekyll-katex' to 'kramdown-math-sskatex'

---
 source/know/concept/laplace-transform/index.md | 37 +++++++++++++-------------
 1 file changed, 19 insertions(+), 18 deletions(-)

(limited to 'source/know/concept/laplace-transform')

diff --git a/source/know/concept/laplace-transform/index.md b/source/know/concept/laplace-transform/index.md
index 5b834c3..c7f352a 100644
--- a/source/know/concept/laplace-transform/index.md
+++ b/source/know/concept/laplace-transform/index.md
@@ -9,9 +9,9 @@ layout: "concept"
 ---
 
 The **Laplace transform** is an integral transform
-that losslessly converts a function $f(t)$ of a real variable $t$,
-into a function $\tilde{f}(s)$ of a complex variable $s$,
-where $s$ is sometimes called the **complex frequency**,
+that losslessly converts a function $$f(t)$$ of a real variable $$t$$,
+into a function $$\tilde{f}(s)$$ of a complex variable $$s$$,
+where $$s$$ is sometimes called the **complex frequency**,
 analogously to the [Fourier transform](/know/concept/fourier-transform/).
 The transform is defined as follows:
 
@@ -23,14 +23,14 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Depending on $f(t)$, this integral may diverge.
-This is solved by restricting the domain of $\tilde{f}(s)$
-to $s$ where $\mathrm{Re}\{s\} > s_0$,
-for an $s_0$ large enough to compensate for the growth of $f(t)$.
+Depending on $$f(t)$$, this integral may diverge.
+This is solved by restricting the domain of $$\tilde{f}(s)$$
+to $$s$$ where $$\mathrm{Re}\{s\} > s_0$$,
+for an $$s_0$$ large enough to compensate for the growth of $$f(t)$$.
 
-The **inverse Laplace transform** $\hat{\mathcal{L}}{}^{-1}$ involves complex integration,
+The **inverse Laplace transform** $$\hat{\mathcal{L}}{}^{-1}$$ involves complex integration,
 and is therefore a lot more difficult to calculate.
-Fortunately, it is usually avoidable by rewriting a given $s$-space expression
+Fortunately, it is usually avoidable by rewriting a given $$s$$-space expression
 using [partial fraction decomposition](/know/concept/partial-fraction-decomposition/),
 and then looking up the individual terms.
 
@@ -47,7 +47,7 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-This property generalizes nicely to higher-order derivatives of $s$, so:
+This property generalizes nicely to higher-order derivatives of $$s$$, so:
 
 $$\begin{aligned}
     \boxed{
@@ -60,7 +60,7 @@ $$\begin{aligned}
 <label for="proof-dv-s">Proof</label>
 <div class="hidden" markdown="1">
 <label for="proof-dv-s">Proof.</label>
-The exponential $\exp(- s t)$ is the only thing that depends on $s$ here:
+The exponential $$\exp(- s t)$$ is the only thing that depends on $$s$$ here:
 
 $$\begin{aligned}
     \dvn{n}{\tilde{f}}{s}
@@ -69,11 +69,12 @@ $$\begin{aligned}
     &= \int_0^\infty (-t)^n f(t) \exp(- s t) \dd{t}
     = (-1)^n \hat{\mathcal{L}}\{t^n f(t)\}
 \end{aligned}$$
+
 </div>
 </div>
 
 The Laplace transform of a derivative introduces the initial conditions into the result.
-Notice that $f(0)$ is the initial value in the original $t$-domain:
+Notice that $$f(0)$$ is the initial value in the original $$t$$-domain:
 
 $$\begin{aligned}
     \boxed{
@@ -92,17 +93,17 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Where $f^{(n)}(t)$ is shorthand for the $n$th derivative of $f(t)$,
-and $f^{(0)}(t) = f(t)$.
-As an example, $\hat{\mathcal{L}}\{f'''(t)\}$ becomes
-$- f''(0) - s f'(0) - s^2 f(0) + s^3 \tilde{f}(s)$.
+Where $$f^{(n)}(t)$$ is shorthand for the $$n$$th derivative of $$f(t)$$,
+and $$f^{(0)}(t) = f(t)$$.
+As an example, $$\hat{\mathcal{L}}\{f'''(t)\}$$ becomes
+$$- f''(0) - s f'(0) - s^2 f(0) + s^3 \tilde{f}(s)$$.
 
 <div class="accordion">
 <input type="checkbox" id="proof-dv-t"/>
 <label for="proof-dv-t">Proof</label>
 <div class="hidden" markdown="1">
 <label for="proof-dv-t">Proof.</label>
-We integrate by parts and use the fact that $\lim_{x \to \infty} \exp(-x) = 0$:
+We integrate by parts and use the fact that $$\lim_{x \to \infty} \exp(-x) = 0$$:
 
 $$\begin{aligned}
     \hat{\mathcal{L}} \big\{ f^{(n)}(t) \big\}
@@ -114,7 +115,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 And so on.
-By partially integrating $n$ times in total we arrive at the conclusion.
+By partially integrating $$n$$ times in total we arrive at the conclusion.
 </div>
 </div>
 
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