herbert-quain https://personaljournal.ca/herbert-quain/ Sat, 06 Jun 2026 02:30:41 +0000 Testing MathJax https://personaljournal.ca/herbert-quain/testing-mathjax <![CDATA[Testing MathJax script src="https://yihui.name/js/math-code.js"/script !-- Just one possible MathJax CDN below. You may use others. -- <script async src="https://mathjax.rstudio.com/latest/MathJax.js?config=TeX-MML-AMCHTML" /script When $a \ne 0$, there are two solutions to \(ax^2 + bx + c = 0\) and they are: $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$ \\[ x = {-b \pm \sqrt{b^2-4ac} \over 2a} \\] div class="math" \begin{equation} \int0^\infty \frac{x^3}{e^x-1}\,dx = \frac{\pi^4}{15} \end{equation} /div This should be inline: \\( ax^2 + \sqrt{bx} + c = 0 \\) Hi $z = x + y$. $$a^2 + b^2 = c^2$$ `$$\begin{vmatrix}a & b\\ c & d \end{vmatrix}=ad-bc$$` e have \(x1 = 132\) and \(x2 = 370\) and so ... \\begin{array}{cc} a & b \\\\ c & c \\end{array} script type="text/x-mathjax-config" MathJax.Hub.Config({ TeX: { equationNumbers: { autoNumber: "AMS" } } }); /script When \(a \ne 0\), there are two solutions to \(ax^2 + bx + c = 0\) and they are $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$ <script type="text/javascript" async src="https://cdnjs.cloudflare.com/ajax/libs/mathjax/2.7.5/MathJax.js?config=TeX-MML-AM_CHTML" async /script /head body p When \(a \ne 0\), there are two solutions to \(ax^2 + bx + c = 0\) and they are $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$ /p /body ... when $x < y$ we have ...]]> Testing MathJax

When $a \ne 0$, there are two solutions to (ax^2 + bx + c = 0) and they are: $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$

\[ x = {-b \pm \sqrt{b^2-4ac} \over 2a} \]

\begin{equation} \int_0^\infty \frac{x^3}{e^x-1}\,dx = \frac{\pi^4}{15} \end{equation}

This should be inline: \( ax^2 + \sqrt{bx} + c = 0 \)

Hi $z = x + y$.

$$a^2 + b^2 = c^2$$

$$\begin{vmatrix}a & b\\ c & d \end{vmatrix}=ad-bc$$

e have \(x_1 = 132\) and \(x_2 = 370\) and so ...

\begin{array}{cc} a & b \\ c & c \end{array}

When (a \ne 0), there are two solutions to (ax^2 + bx + c = 0) and they are $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$

When (a \ne 0), there are two solutions to (ax^2 + bx + c = 0) and they are $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$

... when $x < y$ we have ...

]]> https://personaljournal.ca/herbert-quain/testing-mathjax Thu, 29 Aug 2019 19:49:05 +0000 Hello World https://personaljournal.ca/herbert-quain/hello-world <![CDATA[Hello! This is my first post. I am actually posting this from the command line through an API Client library I am working on for Write Freely.]]> Hello! This is my first post. I am actually posting this from the command line through an API Client library I am working on for Write Freely.

]]> https://personaljournal.ca/herbert-quain/hello-world Wed, 31 Jul 2019 03:02:56 +0000