Installing CheckStyle plugin for Eclipse

Tue, 16 May 2017 23:59:39 +0000

I recently discovered a tool for checking Java code formatting and other coding style related issues -- CheckStyle. It features a list of style checks such as

...
* Restricts nested boolean operators (&&, ||, &, | and ^) to a specified depth (default = 3).
* Checks for empty blocks.
* Checks for long source files.
...

There's also an Eclipse plugin for it: https://checkstyle.org/eclipse-cs/#!/. The Eclispe update URL is http://eclipse-cs.sf.net/update/. As of at least Eclipse 4.8.0 (Aug., 2018), this update URL no longer works.

There is a caveat at the time I install it:

the plugin's homepage http://eclipse-cs.sourceforge.net/ gives http://sevntu-checkstyle.github.com/sevntu.checkstyle/update-site/ as the installation URL, but it does not work. However on http://eclipse-cs.sourceforge.net/downloads.html a different installation URL http://eclipse-cs.sf.net/update/ is given, which turns out to work correctly. It seems someone has not updated the Web pages consistently. And it has certainly confused some: http://stackoverflow.com/questions/11865025/error-while-installing-check-style-plugin-for-eclipse.

MathJax rendering test

Tue, 16 May 2017 23:59:39 +0000

Testing math content rendering with MathJax. Test cases are originally copy from Gautam Iyer's test cases, and then gradually modified and suplemented.

Inline math

This is an equation: $E = m c²$.

Theorem

Theorem (Mean Value Property). Let $\Omega \subset \mathbb{R}^3$ be a domain, and $u$ is harmonic in $\Omega$ (i.e. $u = 0$ in $\Omega$). Suppose $B$ is a ball of radius $R$ and center $x_0$ that is completely contained in $\Omega$. Then $$ u(x_0) = \frac{1}{4 \pi R²} \int_{\partial B} u \, dS $$

Characters

Subscripts

Subscripts: ${{a} {1}}^{{b} {1}}$

Curly braces

The curly brace is a special character in LaTeX. Therefore, to display it, it needs to be escaped by a backslash. The backslash that is not part of the original LaTeX keyword such as \right needs to be escaped. So, at the end you need \\} to display a closing curly brace:

\left \\{X_1, X_2, \cdots, X_N \right \\} displays as $\left \{X_1, X_2, \cdots, X_N \right \}$.

Multi-line equations

$$ \begin{equation} \label{eq1} \begin{split} A & = \frac{\pi r²}{2} \\ & = \frac{1}{2} \pi r² \end{split} \end{equation} $$

$$ \begin{align} 2x - 5y &= 8 \\ 3x + 9y &= -12 \end{align} $$

$$ \begin{align} x&=y & w &=z & a&=b+c\\ 2x&=-y & 3w&=\frac{1}{2}z & a&=b\\ -4 + 5x&=2+y & w+2&=-1+w & ab&=cb \end{align} $$

Rotational Symmetry of the Laplacian

Let $T:\Real² \to \Real²$ be a rotation, and $u$ be any function. Then $$ \Laplace (u \circ T) = (\Laplace u) \circ T $$ This is also true in higher dimensions.

Given that the Laplacian has this rotational symmetry, one might expect a nice cancellation / compact formula for the Laplacian in polar coordinates. We try this next.

The Laplacian in Polar Coordinates

Let $\hat x$, $\hat y$ be the unit vectors in the $x$ and $y$ direction respectively, and let $\hat r$ and $\hat \theta$ be the unit vectors in the $r$ and $\theta$ direction respectively. Explicitly, $$ \hat x = \begin{pmatrix}1\cr 0\end{pmatrix},\quad \hat y = \begin{pmatrix}0\cr1\end{pmatrix},\quad \hat r = \frac{1}{r}\begin{pmatrix}x\cr y\end{pmatrix},\quad \hat \theta = \frac{1}{r}\begin{pmatrix}-y\cr x\end{pmatrix}. $$

First we compute that $$ \Grad u = \partial_x u \hat x + \partial_y u \hat y = \partial_r u \hat r + \frac{1}{r} \partial\theta u \hat \theta. $$ Now we compute $$ \Laplace u = \Div \Grad u = (\Grad \partial_r u) \cdot \hat r + \partial_r u (\Div \hat r) + (\Grad \frac{1}{r} \partial\theta u) \cdot \hat \theta + 0 = \partial_r² u + \partial_r u (\Div \hat r) + \frac{1}{r²} \partial_\theta² u. $$

To finish the calculation, we only need to compute $\Div \hat r$. We do this as follows $$ \Div \hat r = \Div \left( \frac{1}{r} \begin{pmatrix} x \cr y \end{pmatrix} \right) = \frac{2}{r} + \Grad\left( \frac{1}{r} \right) \cdot \begin{pmatrix} x \cr y \end{pmatrix} = \frac{2}{r} - \frac{1}{r} = \frac{1}{r}. $$ Substituting back gives $$ \Laplace u = \partial_r² u + \frac{1}{r} \partial_r u + \frac{1}{r²} \partial_\theta² u. $$

Basic tests to check interference with markdown.

Sub and super scripts.

Here's a superscript without math: a^b, a_b. Here they are with math:

One backslash: ( a^b, a_b ).
Two backslashes: $ a^b, a_b $.
Three backslashes: $ a^b, a_b $.

Smileys

{x} produces in normal mode. In math mode it will mess things up! However, smileys need whitespace before and after to be recognized. So while $ {x} $ will mess things up, ${x}$ will be OK and produce ${x}$.

Checking if * and _ mess up math

The code

$$
a * b + b * c \qquad
a_b + b_c, \qquad
a^{b + c}
$$

produces $$ a * b + b * c \qquad a_b + b_c, \qquad a^{b + c} $$

MathJAX equation reference test

Here is a labeled equation: \begin{equation}x+1\over\sqrt{1-x²}\label{ref1}\end{equation} with a reference to ref1: \ref{ref1}, and another numbered one with no label: $$x+1\over\sqrt{1-x²}$$ This one uses \nonumber: \begin{equation}x+1\over\sqrt{1-x²}\nonumber\end{equation}

Here's one using the equation environment: \begin{equation} x+1\over\sqrt{1-x²} \end{equation} and one with equation environment: \begin{equation} x+1\over\sqrt{1-x²} \end{equation*}

This is a forward reference [\ref{ref2}] and another \eqref{ref2} for the following equation: \begin{equation}x+1\over\sqrt{1-x²}\label{ref2}\end{equation} More math: $$x+1\over\sqrt{1-x²}$$ Here is a ref inside math: (\ref{ref2}+1) and text after it.

\begin{align} x& = y_1-y_2+y_3-y_5+y_8-\dots && \text{by \eqref{ref1}}\cr & = y'\circ y^* && \text{(by \eqref{ref3})}\cr & = y(0) y' && \text {by Axiom 1.} \end{align}

Here's a bad ref [\ref{ref4}] to a nonexistent label.

An alignment: \begin{align} a&=b\label{ref3}\cr &=c+d \end{align} and a starred one: \begin{align} a&=b\cr &=c+d \end{align}

Custom macro tests

This tests some of my predefined macros.

Inequalities

$$ a < b, \quad a \leq b \quad a \geq b \quad a > b. $$

Limits

Inline limits: $\displaystyle \lim {x \to a} \frac{1}{x}$, $\displaystyle \max {0, 1}$.

Displayed \displaystyle {\operatorname{ess} \sup} _{x \in \Real} produces $$ \displaystyle {\operatorname{ess} \sup} _{x \in \Real} $$

Topics

Subscript `_{xxx}` needs to be escaped `\_{xxx}`

Since pairs of _ denotes emphasis in Markdown syntax, it needs to be escaped:

$$
\begin{align}
\hat\sigma^2\_\mathrm{N} 
&= \frac{1}{N} \sum\_{i=1}^N \left( X\_i - \bar{X}\_N \right)^2 \\\\
&= \frac{1}{N} \sum\_{i=1}^N \left( X\_i^2 - \bar{X}\_N^2 \right)~,
\end{align}
$$

$$ \begin{align} \hat\sigma²_\mathrm{N} &= \frac{1}{N} \sum_{i=1}^N \left( X_i - \bar{X}_N \right)² \\ &= \frac{1}{N} \sum_{i=1}^N \left( X_i² - \bar{X}_N² \right)~, \end{align} $$

Or, use a white space in front of _:

$$
\begin{align}
\hat\sigma^2\_\mathrm{N} 
&= \frac{1}{N} \sum\_{i=1}^N \left( X\_i - \bar{X}\_N \right)^2 \\\\
&= \frac{1}{N} \sum\_{i=1}^N \left( X\_i^2 - \bar{X}\_N^2 \right)~,
\end{align}
$$

$$ \begin{align} \hat\sigma² \mathrm{N} &= \frac{1}{N} \sum {i=1}^N \left( X i - \bar{X} N \right)² \\ &= \frac{1}{N} \sum {i=1}^N \left( X i² - \bar{X} _N² \right)~, \end{align} $$

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