@johncarlosbaez looks up your street:
[2605.08224] Uniqueness on a Continuum: Quantifying Tonal Ambiguity Using Information Theory
https://arxiv.org/abs/2605.08224

Does anyone know how I would find a translated (into English) copy of this 17th Century mathematical treatise: Theoremata de centro grauitatis partium circulis et ellipsis?

It's on the Internet Archive here:
https://archive.org/details/ita-bnc-mag-00001352-001

Boosts appreciated!
#librarian #library #latin #translation

What maths-based fiction books do teachers have on reading lists for secondary / high school pupils? #teaching #mtbos

Help me identify this shape. We have a cookie cutter which makes these. I can't tell what it's supposed to be. #baking

@VergaraLautaro I'm looking to read a book or two which tells the story of how the advent of quantum mechanics impacted Chemistry. Specifically, how did the theory of orbitals and filling rules develop, what experiments confirmed them, and when. When and how were covalent and ionic bonds understood, etc. All of that. Everyone more than welcome to respond! Boosts appreciated.

@johncarlosbaez this looks like it might interest you: https://arxiv.org/abs/2408.14551

Very happy day at the second hand bookstore :)

Can anyone help me write some manim code? I want to produce an animation where I have some coloured squares arranged in a rectangular grid, and sometimes adjacent squares swap positions. It's this swapping which I am hoping manim will animate for me. Please boost or share with people you know who could help. Thanks! #manim

This looks interesting: The Penrose Tiling is a Quantum Error-Correcting Code https://arxiv.org/abs/2311.13040

Nothing terribly deep or exciting for #combinatorics -- just very good recreational fun for me!

In my paper https://arxiv.org/abs/2311.01569 which appeared today I find a formal (non-convergent) generating function to count bubbles by their size. I also find that configurations consisting of a single bubble of size 2n are discontinuous in number compared to their smaller counterparts. Removing them, one obtains a smooth asymptotic distribution in the \(n \to \infty\) limit.

Here we have one bubble of size 4:

I've been having fun thinking about gadgets I call bubbles. Bubbles are sequences of cards delimited by these adjacent matches. For example below we have two bubbles, each of size two (the two sequences on either side of the paired 3's).

Kreweras and Poupard answered this back in 1978. The answer is that the mean number of adjacent matches is 1, no matter the value of n. So if we shuffled the full deck, and defined a match by value and colour, we would have n=26, and would still expect just one adjacent match on average.

G. Kreweras and Y. Poupard, Sur les partitions en paires d’un ensemble fini totalement
ordonné, Publications de l’Institut de Statistique de l’Université de Paris 23 (1978),
57–74.

When you have a random list of n matching pairs of cards, what is the probability that a card and its match will be found side by side? Here is an example with n=3, where we take the card values (not suits) to define the matches. Here we have found the 3 beside its match.

I'm hoping the #mathstodon community can help me feel better about some basic linear algebra.

I'm thinking about a \(2\times 2\) matrix \(M\) and its invariant lines.

Let \(M\) have two distinct eigenvalues \(e_1\neq1\) and \(e_2\neq1\), with associated eigenvectors \(\bf{v_1}\) and \(\bf{v_2}\), respectively. Then there are exactly two invariant lines: \(\lambda \bf{v_1}\) and \(\lambda \bf{v_2}\), and they both pass through the origin.

If we want to have an invariant line which does not pass through the origin, we must have 1 as one of our eigenvalues. Let \(e_1=1\), and \(e_2\neq 1\). Of course \(\lambda\bf{v_1}\) is one invariant line, but consider the line \(\lambda \bf{v_2} + \bf{r}\) where \(\bf{r}\) is a general vector.

Since we have distinct eigenvalues, the eigenvectors span the plane, and so \(\bf{r}=\alpha \bf{v_1} + \beta \bf{ v_2}\). Now
\[ M ( \lambda \bf{v_2} + \bf{r}) = ((\lambda+\beta) e_2 - \beta)\bf{v_2} + \bf{r}\]
which means the line is invariant.

So far, so good. But I get uncomfortable in the case \(e_1=e_2=1\). Then there is one eigenvector \(\bf{v}\). I find it weird that not only \(\lambda\bf{v}\), but \(\lambda\bf{v}+\bf{r}\), for any \(\bf{r}\), is also invariant, even though there is no second eigenvector to "generate" this family of lines which don't pass through the origin. The implication is that \((M-\mathbb{I})\bf{r}\propto\bf{v}\) for all vectors \(\bf{r}\).

I'd like to construct an \(M\) where I can start with \(e_1=1,e_2\neq1\) and then smoothly deform it to the \(e_1=e_2=1\) case, to somehow see that the infinite family of invariant lines persists even after the eigenvector that generated them has become equal to the other eigenvector.

Can you help?

Someone should design a pc or laptop or operating system or all three, designed to minimize two things: machine boot up time, application launch time.

Anyone else get the "test for Ricardo" notification?

This looks like it could be interesting to Category theory folks @johncarlosbaez https://arxiv.org/abs/2307.03831

I've recently started reading this:
https://books.google.com/books/about/The_Crest_of_the_Peacock.html?id=ymud91nTc9YC#v=onepage&q&f=false

and learned some lovely things about multiplication in ancient Egypt. They did theirs by producing a table of doubles of one multiplicand, then representing the other in binary. The entries in the table of doubles corresponding to binary ones were then summed!