Math

This is a first draft of a TikZ picture illustraing this classical formula to be used for math help channels.

Adding \caption{for the picture} without “Figure 1:” requires \usepackage{caption} and wrapping with \begin{figure}. It also possible to use the primitive TeX command \par, but it would be complicated to use that with standalone. In the previous post, the SVG picture from the LaTeX table in an article has too much useless whitespace around the table. I don’t bother to learn other packages, as I need time for other more important stuff.

Motivation

I saw someone illustrating his/her solution with a “superior triangle”.

This reminds me the homothety about the centroid of factor −1/2.

The above picture \usetikzlibrary{calc} for computing coordinates from those of existing points. (A)!.25!(B) means $(A)+.25[(B)-(A)]$.

\begin{tikzpicture}[scale=2]
\coordinate (D) at (-0.7,1);
\coordinate (E) at (-1,0);
\coordinate (F) at (1,0);
\coordinate (A) at ($(E)!.5!(F)$);
\coordinate (B) at ($(F)!.5!(D)$);
\coordinate (C) at ($(D)!.5!(E)$);
\coordinate (G) at ($(D)!.5!(E)!1/3!(F)$);

\draw (A) -- (B) -- (C) -- cycle;
\draw (D) -- (E) -- (F) -- cycle;
\begin{scriptsize}
\fill (G) circle (0.5pt) node [left=2pt,anchor=north]{$G$};
\end{scriptsize}
\draw[->,-latex,dashed] (D) -- (A);
\draw[->,-latex,dashed] (E) -- (B);
\draw[->,-latex,dashed] (F) -- (C);
\end{tikzpicture}

Previous post

From the homothety between the nine-point circle and circumcircle about orthocenter with a factor of 2, we see that the nine-point center is the mid-point of orthocenter and circumcenter.

Motivation

Someone on Discord asked about the existence of the nine-point circle. It’s well-known that that can be proved by homothety.

Little reminder about homothety

Homothety preserves angles (and thus parallel lines). Homothetic polygons are similar, so the ratio of the corresponding sides is the same. Considering the radii of a circle under a homothety, we see that a homothety maps a circle to another circle.

Notation

• H: orthocenter
• G: centroid
• O: circumcenter
• ω: circumcircle
• H_A: feet of altitude with respect to A.
• M_A: midpoint of side a.
• E_A: Euler point with respect to A. (i.e. midpoint of A and H)

Problem

The second proof for nine-point circle on AoPS starts with a proved fact that the reflection of H about a and M_A lie on ω.

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.

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.

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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