Are you interested in a federated alternative to Goodreads that doesn't use Amazon?

because I'm making a federated alternative to Goodreads that doesn't use Amazon

Information on creating marbled paper with LaTeX! https://people.csail.mit.edu/jaffer/Marbling/.

You can find the package documentation here: https://ctan.org/pkg/pst-marble .

Sir Thomas Urquhart was a 17th-century Scottish eccentric who tried to systematize a new language for trigonometry; the law of sines was abbreviated as “eproso”, which (if you know the system) encapsulates its meaning.

OMFG A FEDERATED SCHOLARLY COMMUNICATIONS PLATFORM!

I literally cannot wait to dig into this!! It's what I've been wanting!

#Accessibility

A color contrast checker that offers alternatives if your color combination has not enough contrast. It would be cool to chose if you want to change background or foreground though but it's still very nice to get suggestions

RT @DataVizSociety@twitter.com

#DataVisualization can help us make sense of and teach about the coronavirus, but the stakes are high. @abmakulec@twitter.com has 10 considerations to help you #VizResponsibly:

🐦🔗: https://twitter.com/DataVizSociety/status/1237803279486275585

Mathemagics

Article by Pierre Cartier

In collections: Notation and conventions, The act of doing maths

My thesis is:there is another way of doing mathematics, equally successful, and the two methods should supplement each other and not fight.

URL: http://ftp.gwdg.de/pub/misc/EMIS/journals/SLC/wpapers/s44cartier1.pdf

Entry: http://read.somethingorotherwhatever.com/entry/Mathemagics

The innovation here seems to be in the emphasis placed on the properties of the sum of the roots and the product of the roots.

And now, substituting into the formula for \(r\) and \(s\), we get \(r, s = -\frac{b}{2} \pm \sqrt{\frac{b^2}{4} - c}\).

But this is just the same as \(\frac{-b \pm \sqrt{b^2 - 4c}}{2}\), which is exactly what we would expect from the standard formula, given that \(a=1\)! The general derivation, for \(a \neq 1\), takes a few more steps, but is fairly straightforward.

Now, the only way we can have \(r+s = -b\) is if \(r = -\frac{b}{2} + z\) and \(s = -\frac{b}{2} - z\), for some \(z\).

Since \(rs = c\), this means that \(c = \frac{b^2}{4} - z^2\).

Rearranging, we get \(z = \pm \sqrt{\frac{b^2}{4} - c}\).

The trick is as follows: Assume, first of all, that \(a=1\), and let \(r\) and \(s\) be the (unknown) roots. Then we can write \(x^2 + bx + c = (x - r)(x - s)\).

Expanding the right-hand side, we get \(x^2 + bx + c = x^2 - (r+s)x + rs\).

So we have \(r + s = -b\), and \(rs = c\).

I've just discovered something incredible: Jim Fowler has compiled TikZ to WebAssembly! That means you can render TikZ diagrams in web pages, **on the fly**!!

I've made a demo page with an editor, so you can see it and believe it: https://tikzjax-demo.glitch.me

- Favourite Unicode math font
- Libertinus

- Currently studying
- Category theory; group theory

Joined Jul 2018