Adaptations for standalone documents:

• \usetikzlibrary{pattern} before \begin{document}
• \pgfplotsset{compat=1.6}

\begin{tikzpicture}
\begin{axis} [axis lines=center,legend style={at={(0.7,0.7)},anchor=south west}]
\addplot [domain=-3:3, thick, smooth, yellow] { 1/sqrt(2*pi)*exp(-x^2/2) };
\addlegendentry{$y = \tfrac{1}{\sqrt{2\pi}} e^{-x^2/2}$};
\addplot [dashed, yellow] coordinates {(1.5,0) (1.5,0.14)};
\addlegendentry{99th percentile};
\addplot[domain=-3:1.5, pattern=north east lines,draw=none, fill opacity=0.3]
{ 1/sqrt(2*pi)*exp(-x^2/2) } \closedcycle;
\end{axis}
\end{tikzpicture}

For secondary school students, I define cosine and sine as the x and y-components of the point A (cos θ, sin θ) on the unit circle x² + y² = 1, and the tangent function as the quotient of sine over cosine.

\begin{tikzpicture}[scale=3]
\coordinate (O) at (0,0);
\coordinate (H) at (0.6,0);
\coordinate (A) at (0.6,0.8);
\coordinate (E) at (1,0);
\coordinate (T) at (1,0.8/0.6);
\draw (O) circle (1);
\draw[->] (-1.3,0) -- (1.3,0) node [right]{$x$};
\draw[->] (0,-1.3) -- (0,1.3) node [above]{$y$};
\begin{scope}[thick]
\draw (O) node [below left] {$O$}
    -- (H) node [below right] {$H$}
    node [below, midway] {$\cos \theta$}
    -- (A) node [below, midway, sloped] {$\sin \theta$}
    node [above=5pt] {$A$}
    -- cycle node [above left, midway] {$1$};
\begin{scope}
    \clip (O) -- (A) -- (H) -- cycle;
    \draw (O) circle (0.1) node[right=7pt, above=5pt, anchor=west] {\small $\theta$};
\end{scope}
\draw (H) rectangle ++(-0.1,0.1);
\draw (E) rectangle ++(-0.1,0.1);
\draw (E) node [below right] {$E$}
    -- (T) node [below, midway, sloped] {$\tan \theta$}
    node [above, right] {$T$}
    -- (A);
\end{scope}
\end{tikzpicture}

Here’s the $\LaTeX$ code of my diagram for matrix diagonalisation to be used on Discord.

Why do matrix diagonalisation on square matrix $P$?

If we can find a diagonal matrix $D$ and a square matrix $Q$ such that $P = QDQ^{-1}$, then we can easily compute $(P + \lambda I)^n$ for any scalar $\lambda$ and integer $n$ because $D^n$ is easy to compute.

\[\begin{tikzcd}
    {{}} & {{}} & \cdots & {} \\
    {{}} & {{}} & \cdots & {{}}
    \arrow["P", from=1-1, to=1-2]
    \arrow["{Q^{-1}}"', from=1-1, to=2-1]
    \arrow["D"', from=2-1, to=2-2]
    \arrow["Q"', from=2-2, to=1-2]
    \arrow["P", from=1-2, to=1-3]
    \arrow["D"', from=2-2, to=2-3]
    \arrow["P", from=1-3, to=1-4]
    \arrow["D"', from=2-3, to=2-4]
    \arrow["Q"', from=2-4, to=1-4]
\end{tikzcd}\]

After viewing the power of the Discord bot $\TeX{}$it, which renders $\LaTeX$ code on Discord, I gave up spending more time on exploring more functionalities of $\KaTeX$ (say, commutative diagrams) because Discord and $\LaTeX$ spread math knowledge much better than a static blog for basic math: the former allows instant feedback from the reader. The later is better for taking notes. To display more complicated graphics, I can compile to PDF first, then use dvisvgm with -P for --pdf. (The small -p selects --page=ranges.)

Source code:

Background

I’m doing exercise 4.9 of Think Julia, which asks for a function for a polar rose using Luxor’s turtle graphics.

Difficulties

1. Work out the geometric structure of the family of polar roses. The key is to construct some auxiliary isoceles triangles and work out the angles between them. One sees that they are parametrized by two varaibles n and k.
   • n: number of petals
   • k: petal increment
   • constraint: k ≠ n ÷ 2
2. Handle the case when gcd(n, k) > 1, i.e. more than one closed loop.
3. The positive x direction goes to the right; the positive y direction goes down.

Attempt

1. Use ThinkJulia.Reposition(t::Turtle, x, y) to reposition the turtle.
2. Use turn(t::Turtle, θ) to turn t
3. Use ThinkJulia.Orientation(t::Turtle, θ) to restore the turtle’s orientation after the move.

Code

I spend three days writing and testing this function.

Motivation

To test PRs on the upstream of a Hugo theme by setting up a testing branch.

Goal

To deploy a forked GitHub repo for a Hugo theme with exampleSite to GitHub Pages using GitHub Actions.

The whole article is based on my fork of Hugo Future Imperfect Slim.

References

1. GitHub Actions for Hugo
2. A Stack Overflow question showing pwd in GitHub Actions
3. A Hugo Discourse post about testing exampleSite

Difficulties

I had failed for about ten times before I got the job done.

Goal

To build an ASP.NET Core 5 MVC web app linked with a PostgreSQL.

Motivation

1. SQL Server is proprietary.
2. SQLite used in Microsoft’s ASP.NET Core 5 MVC tutorial isn’t made for web apps.
3. MySQL doesn’t perform well with concurrent read-writes. It’s dual-licensed like GitLab.
4. Some users find PostgreSQL cost-effective.

Useful tutorials

1. MS’s tutorial in item 2 above.
2. Wes Doyle’s YouTube video goes through the steps
3. MS’s tutorial for Razor Pages with EF Core migrations

Steps

0. Create a superuser in the database.

   [Read More]

A backup of a deleted PSQ : https://math.stackexchange.com/q/3955443/290189

OP : irfanmat

It has a detailed answer by Atticus Stonestrom. It’s pity that his post got deleted. As there’s no reason for undeletion, I’m posting it here so as to preserve the contents.

Question body

Is there a semi-simple and projective but not injective module? I will be glad if you help.

Response(s)

In the non-commutative case, the answer is yes. Consider $R$ the ring of upper triangular $2\times 2$ matrices over a field $F$, and denote by $e_{ij}$ the element of $R$ with the $ij$-th entry equal to $1$ and all other entries equal to $0$. We can decompose $R$ as a direct sum of left ideals $$Re_{11}\oplus (Re_{12}+Re_{22})=Re_{11}\oplus Re_{22},$$ so let $M=Re_{11}$. Then $M$ is clearly simple, and – as a direct summand of the free module $R$ – is also projective. However, $M$ is not injective; consider the map $f:Re_{11}\oplus Re_{12}\to M$ taking $e_{11}$ and $e_{12}$ to $e_{11}$. $Re_{11}\oplus Re_{12}$ is a left ideal of $R$, but there is no way to extend $f$ to a map $R\rightarrow M$, so this gives the desired example.

[Read More]

I wanted to post the following answer to a question on double integral on Math.SE, but someone had submitted his work before I finished typing. As a result, I’m posting this on my personal blog.

Let $r = \sqrt{x^2+4y^2}$ and $t = \begin{cases} \tan^{-1}(2y/x) &\text{ if } x > 0 \\ \pi/2 &\text{ if } x = 0. \end{cases}$ Then $\begin{cases} x &= r \cos t \\ y &= (r \sin t)/2 \end{cases}$ and $D = { (r,t) \mid r \ge 0, t \in [\pi/4, \pi/2] }$. Calculate the Jacobian

I wanted to start a meta question, but I don’t see a point of that after viewing some related posts listed at the end of the next subsection.

Intended question

You may vote on your preferred way.

Related questions: