From f5105dc7b183fd540006fb4f21039d8b2d126621 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Fri, 7 Oct 2022 19:43:33 +0200
Subject: Expand knowledge base

---
 content/know/concept/central-limit-theorem/index.pdc | 14 +++++++-------
 1 file changed, 7 insertions(+), 7 deletions(-)

(limited to 'content/know/concept/central-limit-theorem/index.pdc')

diff --git a/content/know/concept/central-limit-theorem/index.pdc b/content/know/concept/central-limit-theorem/index.pdc
index a957736..518c224 100644
--- a/content/know/concept/central-limit-theorem/index.pdc
+++ b/content/know/concept/central-limit-theorem/index.pdc
@@ -53,7 +53,7 @@ $$\begin{aligned}
     \phi(k) = \int_{-\infty}^\infty p(x) \exp(i k x) \dd{x}
 \end{aligned}$$
 
-Note that $\phi(k)$ can be interpreted as the average of $\exp(i k x)$.
+Note that $\phi(k)$ can be interpreted as the average of $\exp\!(i k x)$.
 We take its Taylor expansion in two separate ways,
 where an overline denotes the mean:
 
@@ -62,7 +62,7 @@ $$\begin{aligned}
     = \sum_{n = 0}^\infty \frac{k^n}{n!} \: \phi^{(n)}(0)
     \qquad
     \phi(k)
-    = \overline{\exp(i k x)} = \sum_{n = 0}^\infty \frac{(ik)^n}{n!} \overline{x^n}
+    = \overline{\exp\!(i k x)} = \sum_{n = 0}^\infty \frac{(ik)^n}{n!} \overline{x^n}
 \end{aligned}$$
 
 By comparing the coefficients of these two power series,
@@ -88,12 +88,12 @@ using our earlier relation:
 $$\begin{aligned}
     C^{(1)}
     &= - i \dv{k} \Big(\ln\!\big(\phi(k)\big)\Big) \big|_{k = 0}
-    = - i \frac{\phi'(0)}{\exp(0)}
+    = - i \frac{\phi'(0)}{\exp\!(0)}
     = \overline{x}
     \\
     C^{(2)}
     &= - \dv[2]{k} \Big(\ln\!\big(\phi(k)\big)\Big) \big|_{k = 0}
-    = \frac{\big(\phi'(0)\big)^2}{\exp(0)^2} - \frac{\phi''(0)}{\exp(0)}
+    = \frac{\big(\phi'(0)\big)^2}{\exp\!(0)^2} - \frac{\phi''(0)}{\exp\!(0)}
     = - \overline{x}^2 + \overline{x^2} = \sigma^2
 \end{aligned}$$
 
@@ -153,7 +153,7 @@ with $\sqrt{N}$ appearing in the arguments of $\phi_n$:
 $$\begin{aligned}
     \phi_z(k)
     &= \idotsint
-    \Big( \prod_{n = 1}^N p(x_n) \Big) \: \delta\Big( z - \frac{1}{\sqrt{N}} \sum_{n = 1}^N x_n \Big) \exp(i k z)
+    \Big( \prod_{n = 1}^N p(x_n) \Big) \: \delta\Big( z - \frac{1}{\sqrt{N}} \sum_{n = 1}^N x_n \Big) \exp\!(i k z)
     \dd{x_1} \cdots \dd{x_N}
     \\
     &= \idotsint
@@ -184,7 +184,7 @@ $$\begin{aligned}
     = i k \overline{z} - \frac{k^2}{2} \sigma_z^2
     \\
     \phi_z(k)
-    &\approx \exp(i k \overline{z}) \exp\!(- k^2 \sigma_z^2 / 2)
+    &\approx \exp\!(i k \overline{z}) \exp\!(- k^2 \sigma_z^2 / 2)
 \end{aligned}$$
 
 We take its inverse Fourier transform to get the density $p(z)$,
@@ -194,7 +194,7 @@ which is even already normalized:
 $$\begin{aligned}
     p(z)
     = \hat{\mathcal{F}}^{-1} \{\phi_z(k)\}
-    &= \frac{1}{2 \pi} \int_{-\infty}^\infty \exp\!\big(\!-\! i k (z - \overline{z})\big) \exp(- k^2 \sigma_z^2 / 2) \dd{k}
+    &= \frac{1}{2 \pi} \int_{-\infty}^\infty \exp\!\big(\!-\! i k (z - \overline{z})\big) \exp\!(- k^2 \sigma_z^2 / 2) \dd{k}
     \\
     &= \frac{1}{\sqrt{2 \pi \sigma_z^2}} \exp\!\Big(\!-\! \frac{(z - \overline{z})^2}{2 \sigma_z^2} \Big)
 \end{aligned}$$
-- 
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