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@@ -375,10 +375,10 @@ <h2 class="anchored" data-anchor-id="recent-posts">Recent Posts</h2>
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-<p class="card-img-top"><img src="posts/clt-intuitive-derivation/images/normal_distribution.webp" alt="A diagram of the Normal Distribution bell curve. The central peak is at the mean mu. The curve shows that 68% of data falls within 1 standard deviation, 95% within 2, and 99.7% within 3." style="height: 150px;"  class="thumbnail-image card-img"/></p>
+<p class="card-img-top"><img src="posts/clt-intuitive-derivation/images/normal_distribution.png" alt="A diagram of the Normal Distribution bell curve. The central peak is at the mean mu. The curve shows that 68% of data falls within 1 standard deviation, 95% within 2, and 99.7% within 3." style="height: 150px;"  class="thumbnail-image card-img"/></p>
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 <h5 class="no-anchor card-title listing-title">
 The Bell Curve rising: An Intuitive CLT Derivation

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-<p class="card-img-top"><img src="posts/clt-intuitive-derivation/images/normal_distribution.webp" alt="A diagram of the Normal Distribution bell curve. The central peak is at the mean mu. The curve shows that 68% of data falls within 1 standard deviation, 95% within 2, and 99.7% within 3." style="height: 150px;"  class="thumbnail-image card-img"/></p>
+<p class="card-img-top"><img src="posts/clt-intuitive-derivation/images/normal_distribution.png" alt="A diagram of the Normal Distribution bell curve. The central peak is at the mean mu. The curve shows that 68% of data falls within 1 standard deviation, 95% within 2, and 99.7% within 3." style="height: 150px;"  class="thumbnail-image card-img"/></p>
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 The Bell Curve rising: An Intuitive CLT Derivation

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@@ -306,7 +306,7 @@ <h2 class="anchored" data-anchor-id="the-clt"><strong>The CLT</strong></h2>
 <p>The <strong>Central Limit Theorem (CLT)</strong> answers an important question: Why does the <strong>bell curve</strong> (or Normal Distribution) show up everywhere in the real world?</p>
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-<p><img src="images/normal_distribution.webp" class="img-fluid figure-img" alt="A diagram of the Normal Distribution bell curve. The central peak is at the mean mu. The curve shows that 68% of data falls within 1 standard deviation, 95% within 2, and 99.7% within 3."></p>
+<p><img src="images/normal_distribution.png" class="img-fluid figure-img" alt="A diagram of the Normal Distribution bell curve. The central peak is at the mean mu. The curve shows that 68% of data falls within 1 standard deviation, 95% within 2, and 99.7% within 3."></p>
 <figcaption>The Normal Distribution, showing the 68-95-99.7 rule.</figcaption>
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@@ -445,9 +445,9 @@ <h3 class="anchored" data-anchor-id="the-taylor-series-and-cumulants"><strong>Th
 <p>The rigorous answer for this choice lies in analyzing the magnitude of our integrand. The integral we must solve is <span class="math inline">\(\frac{1}{2\pi} \int \varphi_X(\theta)^m e^{-in\theta} d\theta\)</span>. The dominant term that dictates the behavior of the integral for large <span class="math inline">\(m\)</span> is <span class="math inline">\(\varphi_X(\theta)^m\)</span>. To analyze its magnitude, we use a key property of complex numbers: the magnitude of a number raised to a power is the magnitude of the number raised to that same power, i.e., <span class="math inline">\(|z^n|=|z|^n\)</span>. Therefore, the magnitude of our term is simply <span class="math inline">\(|\varphi_X(\theta)|^m\)</span>.</p>
 <p>Our first step is to prove that the magnitude of the characteristic function, <span class="math inline">\(|\varphi_X(\theta)|\)</span>, is at most 1. The intuition for this comes from visualizing the sum of complex numbers as vector addition—since complex numbers add component-wise, just like vectors, the process can be seen as joining vectors tip-to-tail. The <strong>triangle inequality</strong> (<span class="math inline">\(|\sum z_i| \le \sum |z_i|\)</span>) simply states that the length of the final resulting vector can never be greater than the sum of the lengths of all the individual vectors.</p>
 <p>By definition, <span class="math inline">\(\varphi_X(\theta) = \sum_k p_k e^{ik\theta}\)</span>. Applying the triangle inequality: <span class="math display">\[
-||\varphi_X(\theta)| = \left|\sum_k p_k e^{ik\theta}\right| \le \sum_k |p_k e^{ik\theta}| = \sum_k p_k |e^{ik\theta}|
+|\varphi_X(\theta)| = \left|\sum_k p_k e^{ik\theta}\right| \le \sum_k |p_k e^{ik\theta}| = \sum_k p_k |e^{ik\theta}|
 \]</span> Since <span class="math inline">\(|e^{ik\theta}|=1\)</span> for any real <span class="math inline">\(k\)</span> and <span class="math inline">\(\theta\)</span>, this simplifies to: <span class="math display">\[
-||\varphi_X(\theta)| \le \sum_k p_k = 1
+|\varphi_X(\theta)| \le \sum_k p_k = 1
 \]</span> The equality <span class="math inline">\(|\varphi_X(\theta)|=1\)</span> holds if and only if all the complex numbers <span class="math inline">\(e^{ik\theta}\)</span> (for which <span class="math inline">\(p_k&gt;0\)</span>) point in the same direction. For any non-trivial distribution (with at least two different outcomes), this only happens when <span class="math inline">\(\theta=0\)</span>. At <span class="math inline">\(\theta=0\)</span>, every term <span class="math inline">\(e^{ik\cdot 0}\)</span> is just 1. For any <span class="math inline">\(\theta \neq 0\)</span>, the different values of <span class="math inline">\(k\)</span> cause the terms to have different phases, so they are no longer perfectly aligned, and the magnitude of their sum is strictly less than 1. This means that the function <span class="math inline">\(|\varphi_X(\theta)|\)</span> has a unique, global maximum value of 1 at precisely <span class="math inline">\(\theta=0\)</span>.</p>
 <p>Now, consider what happens when we raise this to a large power, <span class="math inline">\(m\)</span>. The peak at <span class="math inline">\(\theta=0\)</span>, where the value is <span class="math inline">\(1^m = 1\)</span>, remains. However, for any other value of <span class="math inline">\(\theta\)</span> where <span class="math inline">\(|\varphi_X(\theta)| &lt; 1\)</span>, the value of <span class="math inline">\(|\varphi_X(\theta)|^m\)</span> plummets towards zero exponentially fast as <span class="math inline">\(m\)</span> grows. For instance, if at some point the magnitude is 0.99, for <span class="math inline">\(m=1000\)</span> it becomes <span class="math inline">\(0.99^{1000} \approx 4 \times 10^{-5}\)</span>.</p>
 <p>This creates an extremely sharp “spike” in the integrand’s magnitude, centered at <span class="math inline">\(\theta=0\)</span>. The result is that the only part of the integral that contributes significantly to the final value comes from an infinitesimally small neighborhood around <span class="math inline">\(\theta=0\)</span>. The contributions from all other regions are exponentially suppressed and become negligible. Therefore, to accurately approximate the integral for large <span class="math inline">\(m\)</span>, we <em>must</em> approximate the function <span class="math inline">\(g(\theta)\)</span> in this tiny, all-important region around the origin. The Taylor series is the fundamental mathematical tool for this. Expanding around <span class="math inline">\(\theta=0\)</span> is not a mere convenience; it is a mathematically necessary step.</p>