Introduce collapsible proofs to some articles

1 files changed, 49 insertions, 23 deletions

diff --git a/content/know/concept/binomial-distribution/index.pdc b/content/know/concept/binomial-distribution/index.pdc
index 70cc897..e644164 100644
--- a/content/know/concept/binomial-distribution/index.pdc
+++ b/content/know/concept/binomial-distribution/index.pdc
@@ -22,7 +22,7 @@ that $n$ out of the $N$ trials succeed:
 
 $$\begin{aligned}
     \boxed{
-        P_N(n) = \binom{N}{n} \: p^n (1 - p)^{N - n}
+        P_N(n) = \binom{N}{n} \: p^n q^{N - n}
     }
 \end{aligned}$$
 
@@ -41,8 +41,20 @@ $$\begin{aligned}
 The remaining factor $p^n (1 - p)^{N - n}$ is then just the
 probability of attaining each microstate.
 
-To find the mean number of successes $\mu$,
-the trick is to treat $p$ and $q$ as independent:
+The expected or mean number of successes $\mu$ after $N$ trials is as follows:
+
+$$\begin{aligned}
+    \boxed{
+        \mu = N p
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-mean"/>
+<label for="proof-mean">Proof</label>
+<div class="hidden">
+<label for="proof-mean">Proof.</label>
+The trick is to treat $p$ and $q$ as independent until the last moment:
 
 $$\begin{aligned}
     \mu
@@ -54,16 +66,26 @@ $$\begin{aligned}
     = N p (p + q)^{N - 1}
 \end{aligned}$$
 
-By inserting $q = 1 - p$, we find the following expression for the mean:
+Inserting $q = 1 - p$ then gives the desired result.
+</div>
+</div>
+
+Meanwhile, we find the following variance $\sigma^2$,
+with $\sigma$ being the standard deviation:
 
 $$\begin{aligned}
     \boxed{
-        \mu = N p
+        \sigma^2 = N p q
     }
 \end{aligned}$$
 
-Next, we use the same trick to calculate $\overline{n^2}$
-(the mean of the squared number of successes):
+<div class="accordion">
+<input type="checkbox" id="proof-var"/>
+<label for="proof-var">Proof</label>
+<div class="hidden">
+<label for="proof-var">Proof.</label>
+We use the same trick to calculate $\overline{n^2}$
+(the mean squared number of successes):
 
 $$\begin{aligned}
     \overline{n^2}
@@ -79,7 +101,7 @@ $$\begin{aligned}
     &= N p + N^2 p^2 - N p^2
 \end{aligned}$$
 
-Using this and the earlier expression for $\mu$, we find the variance $\sigma^2$:
+Using this and the earlier expression $\mu = N p$, we find the variance $\sigma^2$:
 
 $$\begin{aligned}
     \sigma^2
@@ -88,18 +110,26 @@ $$\begin{aligned}
     = N p (1 - p)
 \end{aligned}$$
 
-Once again, by inserting $q = 1 - p$, we find the following expression for the variance:
+By inserting $q = 1 - p$, we arrive at the desired expression.
+</div>
+</div>
+
+As $N \to \infty$, the binomial distribution
+turns into the continuous normal distribution:
 
 $$\begin{aligned}
     \boxed{
-        \sigma^2 = N p q
+        \lim_{N \to \infty} P_N(n) = \frac{1}{\sqrt{2 \pi \sigma^2}} \exp\!\Big(\!-\!\frac{(n - \mu)^2}{2 \sigma^2} \Big)
     }
 \end{aligned}$$
 
-As $N$ grows to infinity, the binomial distribution
-turns into the continuous normal distribution.
-We demonstrate this by taking the Taylor expansion of its
-natural logarithm $\ln\!\big(P_N(n)\big)$ around the mean $\mu = Np$:
+<div class="accordion">
+<input type="checkbox" id="proof-normal"/>
+<label for="proof-normal">Proof</label>
+<div class="hidden">
+<label for="proof-normal">Proof.</label>
+We take the Taylor expansion of $\ln\!\big(P_N(n)\big)$
+around the mean $\mu = Np$:
 
 $$\begin{aligned}
     \ln\!\big(P_N(n)\big)
@@ -108,7 +138,7 @@ $$\begin{aligned}
     D_m(n) = \dv[m]{\ln\!\big(P_N(n)\big)}{n}
 \end{aligned}$$
 
-We use Stirling's approximation to calculate all these factorials:
+We use Stirling's approximation to calculate the factorials in $D_m$:
 
 $$\begin{aligned}
     \ln\!\big(P_N(n)\big)
@@ -163,7 +193,7 @@ $$\begin{aligned}
     = - \frac{1}{\sigma^2}
 \end{aligned}$$
 
-The higher-order derivatives tend to zero for large $N$, so we discard them:
+The higher-order derivatives tend to zero for $N \to \infty$, so we discard them:
 
 $$\begin{aligned}
     D_3(n)
@@ -184,13 +214,9 @@ $$\begin{aligned}
     = \ln\!\Big( \frac{1}{\sqrt{2\pi \sigma^2}} \Big) - \frac{(n - \mu)^2}{2 \sigma^2}
 \end{aligned}$$
 
-Thus, as $N$ goes to infinity, the binomial distribution becomes a Gaussian:
-
-$$\begin{aligned}
-    \boxed{
-        \lim_{N \to \infty} P_N(n) = \frac{1}{\sqrt{2 \pi \sigma^2}} \exp\!\Big(\!-\!\frac{(n - \mu)^2}{2 \sigma^2} \Big)
-    }
-\end{aligned}$$
+Taking $\exp$ of this expression then yields a normalized Gaussian distribution.
+</div>
+</div>
 
 
 ## References