Improve knowledge baseHEADmaster

1 files changed, 9 insertions, 9 deletions

diff --git a/source/know/concept/heaviside-step-function/index.md b/source/know/concept/heaviside-step-function/index.md
index 9f5d4ec..6412914 100644
--- a/source/know/concept/heaviside-step-function/index.md
+++ b/source/know/concept/heaviside-step-function/index.md
@@ -45,15 +45,15 @@ $$\begin{aligned}
 \end{aligned}$$
 
 The [Fourier transform](/know/concept/fourier-transform/)
-of $$\Theta(t)$$ is as follows,
-where $$\pv{}$$ is the Cauchy principal value,
+of $$\Theta(t)$$ is as follows, where $$\mathcal{P}$$
+is the [Cauchy principal value](/know/concept/cauchy-principal-value/),
 $$A$$ and $$s$$ are constants from the FT's definition,
 and $$\mathrm{sgn}$$ is the signum function:
 
 $$\begin{aligned}
     \boxed{
         \tilde{\Theta}(\omega)
-        = \frac{A}{|s|} \Big( \pi \delta(\omega) + i \: \mathrm{sgn}(s) \pv{\frac{1}{\omega}} \Big)
+        = \frac{A}{|s|} \Big( \pi \delta(\omega) + i \:\mathrm{sgn}(s) \:\mathcal{P} \frac{1}{\omega} \Big)
     }
 \end{aligned}$$
 
@@ -77,18 +77,18 @@ $$\begin{aligned}
 \end{aligned}$$
 
 The first term is proportional to the Dirac delta function.
-The second integral is problematic, so we take the Cauchy principal value $$\pv{}$$
-and look up the integral:
+The second integral is problematic, so we take
+the Cauchy principal value $$\mathcal{P}$$ and look up the integral:
 
 $$\begin{aligned}
     \tilde{\Theta}(\omega)
-    &= A \pi \delta(s \omega) + \frac{A}{2} \pv{\int_{-\infty}^\infty \mathrm{sgn}(t) \exp(i s \omega t) \dd{t}}
-    = \frac{A}{|s|} \pi \delta(\omega) + i \frac{A}{s} \pv{\frac{1}{\omega}}
+    &= A \pi \delta(s \omega) + \frac{A}{2} \:\mathcal{P}\! \int_{-\infty}^\infty \mathrm{sgn}(t) \exp(i s \omega t) \dd{t}
+    = \frac{A}{|s|} \pi \delta(\omega) + i \frac{A}{s} \:\mathcal{P} \frac{1}{\omega}
 \end{aligned}$$
 {% include proof/end.html id="proof-fourier" %}
 
 
-The use of $$\pv{}$$ without an integral is an abuse of notation,
+The use of $$\mathcal{P}$$ without an integral is an abuse of notation,
 and means that this result only makes sense when wrapped in an integral.
-Formally, $$\pv{\{1 / \omega\}}$$ is a [Schwartz distribution](/know/concept/schwartz-distribution/).
+Formally, $$\mathcal{P}\{1 / \omega\}$$ is a [Schwartz distribution](/know/concept/schwartz-distribution/).