<?xml version="1.0" encoding="utf-8" standalone="yes"?><rss version="2.0" xmlns:atom="http://www.w3.org/2005/Atom"><channel><title>KaTeX on SAN == 0 ?</title><link>https://lr-ch.netlify.app/tags/katex/</link><description>Recent content in KaTeX on SAN == 0 ?</description><generator>Hugo -- gohugo.io</generator><language>zh-tw</language><lastBuildDate>Fri, 12 Aug 2022 09:00:00 +0000</lastBuildDate><atom:link href="https://lr-ch.netlify.app/tags/katex/index.xml" rel="self" type="application/rss+xml"/><item><title>使用數學符號</title><link>https://lr-ch.netlify.app/p/%E4%BD%BF%E7%94%A8%E6%95%B8%E5%AD%B8%E7%AC%A6%E8%99%9F/</link><pubDate>Fri, 12 Aug 2022 09:00:00 +0000</pubDate><guid>https://lr-ch.netlify.app/p/%E4%BD%BF%E7%94%A8%E6%95%B8%E5%AD%B8%E7%AC%A6%E8%99%9F/</guid><description>&lt;h2 id="使用katex">使用KaTeX&lt;/h2>
&lt;h3 id="行內區塊">行內、區塊&lt;/h3>
&lt;p>When \(a \ne 0\), there are two solutions to \(ax^2 + bx + c = 0\) and they are&lt;/p>
&lt;p>$$
x = {-b \pm \sqrt{b^2-4ac} \over 2a}
$$&lt;/p>
&lt;h3 id="羅倫茲方程">羅倫茲方程&lt;/h3>
&lt;p>$$
\begin{aligned}
\dot{x} &amp;amp; = \sigma(y-x) \\
\dot{y} &amp;amp; = \rho x - y - xz \\
\dot{z} &amp;amp; = -\beta z + xy
\end{aligned}
$$&lt;/p>
&lt;h3 id="柯西史瓦茲不等式">柯西－史瓦茲不等式&lt;/h3>
&lt;p>$$
\left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)
$$&lt;/p>
&lt;h3 id="向量外積">向量外積&lt;/h3>
&lt;p>$$
\mathbf{V}_1 \times \mathbf{V}_2 = \left| \begin{matrix}
\mathbf{i} &amp;amp; \mathbf{j} &amp;amp; \mathbf{k} \\[0.3em]
\frac{\partial X}{\partial u} &amp;amp; \frac{\partial Y}{\partial u} &amp;amp; 0 \\[0.5em]
\frac{\partial X}{\partial v} &amp;amp; \frac{\partial Y}{\partial v} &amp;amp; 0
\end{matrix} \right|
$$&lt;/p>
&lt;h3 id="投擲-n-個硬幣得到-k-個人頭的機率">投擲 &lt;em>n&lt;/em> 個硬幣得到 &lt;em>k&lt;/em> 個人頭的機率&lt;/h3>
&lt;p>$$
P(E) = {n \choose k} p^k (1-p)^{ n-k}
$$&lt;/p>
&lt;h3 id="an-identity-of-ramanujan">An Identity of Ramanujan&lt;/h3>
&lt;p>$$
\frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} =
1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}}
{1+\frac{e^{-8\pi}} {1+\ldots} } } }
$$&lt;/p>
&lt;h3 id="a-rogers-ramanujan-identity">A Rogers-Ramanujan Identity&lt;/h3>
&lt;p>$$
1 + \frac{q^2}{(1-q)}+\frac{q^6}{(1-q)(1-q^2)}+\cdots =
\prod_{j=0}^{\infty}\frac{1}{(1-q^{5j+2})(1-q^{5j+3})},
\quad\quad \text{for $|q|&amp;lt;1$}.
$$&lt;/p>
&lt;h3 id="馬克士威方程">馬克士威方程&lt;/h3>
&lt;p>$$
\begin{aligned}
\nabla \times \vec{\mathbf{B}} -\ \frac1c\ \frac{\partial\vec{\mathbf{E}}}{\partial t} &amp;amp; = \frac{4\pi}{c}\vec{\mathbf{j}} \\[1.0em]
\nabla \cdot \vec{\mathbf{E}} &amp;amp; = 4 \pi \rho \\[0.5em]
\nabla \times \vec{\mathbf{E}}\ +\ \frac1c\ \frac{\partial\vec{\mathbf{B}}}{\partial t} &amp;amp; = \vec{\mathbf{0}} \\[1.0em]
\nabla \cdot \vec{\mathbf{B}} &amp;amp; = 0 \end{aligned}
$$&lt;/p>
&lt;p>Inline math: \(\varphi = \dfrac{1+\sqrt5}{2}= 1.6180339887…\)&lt;/p>
&lt;p>Block math:
$$
\varphi = 1+\frac{1} {1+\frac{1} {1+\frac{1} {1+\cdots} } }
$$&lt;/p></description></item></channel></rss>