1 files changed, 16 insertions, 15 deletions
diff --git a/content/know/concept/binomial-distribution/index.pdc b/content/know/concept/binomial-distribution/index.pdc
index e644164..183e7e9 100644
--- a/content/know/concept/binomial-distribution/index.pdc
+++ b/content/know/concept/binomial-distribution/index.pdc
@@ -115,7 +115,8 @@ By inserting $q = 1 - p$, we arrive at the desired expression.
 </div>
 
 As $N \to \infty$, the binomial distribution
-turns into the continuous normal distribution:
+turns into the continuous normal distribution,
+a fact that is sometimes called the **de Moivre-Laplace theorem**:
 
 $$\begin{aligned}
     \boxed{
@@ -142,9 +143,9 @@ We use Stirling's approximation to calculate the factorials in $D_m$:
 
 $$\begin{aligned}
     \ln\!\big(P_N(n)\big)
-    &= \ln(N!) - \ln(n!) - \ln\!\big((N - n)!\big) + n \ln(p) + (N - n) \ln(q)
+    &= \ln\!(N!) - \ln\!(n!) - \ln\!\big((N - n)!\big) + n \ln\!(p) + (N - n) \ln\!(q)
     \\
-    &\approx \ln(N!) - n \big( \ln(n)\!-\!\ln(p)\!-\!1 \big) - (N\!-\!n) \big( \ln(N\!-\!n)\!-\!\ln(q)\!-\!1 \big)
+    &\approx \ln\!(N!) - n \big( \ln\!(n)\!-\!\ln\!(p)\!-\!1 \big) - (N\!-\!n) \big( \ln\!(N\!-\!n)\!-\!\ln\!(q)\!-\!1 \big)
 \end{aligned}$$
 
 For $D_0(\mu)$, we need to use a stronger version of Stirling's approximation
@@ -152,16 +153,16 @@ to get a non-zero result. We take advantage of $N - N p = N q$:
 
 $$\begin{aligned}
     D_0(\mu)
-    &= \ln(N!) - \ln\!\big((N p)!\big) - \ln\!\big((N q)!\big) + N p \ln(p) + N q \ln(q)
+    &= \ln\!(N!) - \ln\!\big((N p)!\big) - \ln\!\big((N q)!\big) + N p \ln\!(p) + N q \ln\!(q)
     \\
-    &= \Big( N \ln(N) - N + \frac{1}{2} \ln(2\pi N) \Big)
-    - \Big( N p \ln(N p) - N p + \frac{1}{2} \ln(2\pi N p) \Big) \\
-    &\qquad - \Big( N q \ln(N q) - N q + \frac{1}{2} \ln(2\pi N q) \Big)
-    + N p \ln(p) + N q \ln(q)
+    &= \Big( N \ln\!(N) - N + \frac{1}{2} \ln\!(2\pi N) \Big)
+    - \Big( N p \ln\!(N p) - N p + \frac{1}{2} \ln\!(2\pi N p) \Big) \\
+    &\qquad - \Big( N q \ln\!(N q) - N q + \frac{1}{2} \ln\!(2\pi N q) \Big)
+    + N p \ln\!(p) + N q \ln\!(q)
     \\
-    &= N \ln(N) - N (p + q) \ln(N) + N (p + q) - N - \frac{1}{2} \ln(2\pi N p q)
+    &= N \ln\!(N) - N (p + q) \ln\!(N) + N (p + q) - N - \frac{1}{2} \ln\!(2\pi N p q)
     \\
-    &= - \frac{1}{2} \ln(2\pi N p q)
+    &= - \frac{1}{2} \ln\!(2\pi N p q)
     = \ln\!\Big( \frac{1}{\sqrt{2\pi \sigma^2}} \Big)
 \end{aligned}$$
 
@@ -170,17 +171,17 @@ This is indeed the case:
 
 $$\begin{aligned}
     D_1(n)
-    &= - \big( \ln(n)\!-\!\ln(p)\!-\!1 \big) + \big( \ln(N\!-\!n)\!-\!\ln(q)\!-\!1 \big) - 1 + 1
+    &= - \big( \ln\!(n)\!-\!\ln\!(p)\!-\!1 \big) + \big( \ln\!(N\!-\!n)\!-\!\ln\!(q)\!-\!1 \big) - 1 + 1
     \\
-    &= - \ln(n) + \ln(N - n) + \ln(p) - \ln(q)
+    &= - \ln\!(n) + \ln\!(N - n) + \ln\!(p) - \ln\!(q)
     \\
     D_1(\mu)
-    &= \ln(N q) - \ln(N p) + \ln(p) - \ln(q)
-    = \ln(N p q) - \ln(N p q) 
+    &= \ln\!(N q) - \ln\!(N p) + \ln\!(p) - \ln\!(q)
+    = \ln\!(N p q) - \ln\!(N p q)
     = 0
 \end{aligned}$$
 
-For the same reason, we expect that $D_2(\mu)$ is negative
+For the same reason, we expect that $D_2(\mu)$ is negative.
 We find the following expression:
 
 $$\begin{aligned}