Posts by Log3overLog2@mathstodon.xyz
(DIR) Post #APMsBp4iA0hrF4gqYa by Log3overLog2@mathstodon.xyz
2022-11-07T17:24:48Z
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(1) Find people on some Mastodon server, either via a tool like https://pruvisto.org/debirdify/ to automate Twitter searching or, once bootstrapped, by looking at who they follow(2) Open the profile page of a person I found in step 1(3) click on the Follow button(4a) New popup browser window where I can type my @username@server, okay fine, I can cut-and-paste that a bunch✨or✨ (4b) Overlay that offers me a URL to copy/paste back to a search box on my home Mastodon server? Oof that's a lossy flow...
(DIR) Post #APMsBpayE4Vir7mbq4 by Log3overLog2@mathstodon.xyz
2022-11-07T17:42:12Z
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Ah and flow (4a) only requires you copy-and-paste once for each other federated server whose people you want to follow; after that it remembers you. While (4b) still requires copying and pasting into a search box on your home instance each time? Am I doing this wrong?
(DIR) Post #APMsBq6sJS20S4i5ZI by Log3overLog2@mathstodon.xyz
2022-11-07T17:48:12Z
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I was sort of hoping that there was instead a way to see profiles of people on other federated servers while I'm actually on my home server, so that an operation like "follow one of their followers" could be a single click. But AFAICT no?
(DIR) Post #APMtLZBEDCqKnLwrFg by Log3overLog2@mathstodon.xyz
2022-11-07T18:04:25Z
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@ceo_of_monoeye_dating Right but can you see all of the people I follow in that view? (I think not, from my experiments, but I also think I don't know how much consistency there is across different instances!!)
(DIR) Post #Ai3YYf43Z9jl3jEs3k by Log3overLog2@mathstodon.xyz
2023-11-20T02:42:22Z
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All my friends are asking how much AI would help save the planet or how much AI would destroy jobs, and I'm sitting here wondering how much AI would an AI Weiwei if an ai wei would weigh eyes.
(DIR) Post #Ai3YYgENEHaGg14cmO by Log3overLog2@mathstodon.xyz
2024-01-12T03:33:00Z
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Kinda called this onehttps://circa.art/artist/ai-weiwei-ai-vs-ai/
(DIR) Post #ApgeYNfkgIVZWkZDyy by Log3overLog2@mathstodon.xyz
2025-01-01T21:41:12Z
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Happy 2025!There's some lovely numerology here. 2025 is 1³+2³+3³+...+9³ . And 2025 is also 45² , which is (1+2+3+...+9)². Some of my best friends are squares of triangle numbers!This isn't just a coincidence: In general, 1³+2³+3³+...+𝑛³=(1+2+3+...+𝑛)² There's a lovely proof-without-words picture of this. Heh — I just found out that in fact it's so good that it is the top picture in the Wikipedia page https://en.wikipedia.org/wiki/Proof_without_wordsThis picture shows what happens with 𝑛=5, but you can do the same thing for any n. (Just notice that even n's get the top layer of the cube cut in half while odd n's keep all their layers whole.)(1/3)
(DIR) Post #ApgeYZQEznENy7P0TI by Log3overLog2@mathstodon.xyz
2025-01-01T21:47:01Z
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This also happens to be the identity underlying the "Partridge problem". While the proof-without-words uses three dimensions to get the cubes, you can also represent 𝑛³ in two dimensions by using 𝑛 copies of a thing of size 𝑛². For 𝑛=8, you can use one 1×1 square, two 2×2 squares, three 3×3 squares,..., and eight 8×8 squares to cover one square of edge length 1+2+...+8, as pictured here.Finding this was called the "Partridge problem" not because it was proposed by, like, some guy named John Partridge but rather because "4 calling birds, 3 french hens, 2 turtle doves, and a partridge in a square tree." This picture, and many more, come from Erich Friedman's page about it, https://erich-friedman.github.io/mathmagic/0802.html.It's totally not obvious that this kind of tiling is even possible, and it doesn't work for squares with 𝑛<8, and unlike the proof-without-words version, there is nothing here that makes it clear that the side-length of the square is 1+2+3+...+8. But it's neat that this decomposition of a square into smaller squares is possible at all.(2/3)
(DIR) Post #ApgeYiAGUYnd5AzsJs by Log3overLog2@mathstodon.xyz
2025-01-01T21:52:11Z
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Moreover, the area of *any* shape in the plane increases by a factor of 𝑛² when you scale it up by a factor of n. (Welcome to what "two dimensional" really means, or maybe what "area" means.) So this identity about numerical squares and cubes lets you try to play tiling games with any planar geometric shape. For 2025 specifically, we want something with n=9, and that is delightfully the equilateral triangle.The big triangle has area 2025 times the area of the small red triangle. Happy New Year.(3/3)
(DIR) Post #ApgeYsNkVHWkltNmj2 by Log3overLog2@mathstodon.xyz
2025-01-01T22:09:06Z
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Okay okay actually just one more thing. My *favorite* open question about the Partridge problem is this: Is there *any* planar shape at all which solves it with 𝑛=2??That is, is there any planar shape P where one copy of P and two copies of double-sized P can be fit together to make a triple-sized P? The areas work out, since 1+4+4=9, that same identity again.Last year I briefly thought I found a P that worked! Here's a picture. But the blue and green ones are sadly *not* 2/3-scale copies of the overall picture; instead the scaling factor 𝑠 is the positive root of 𝑠⁶+2𝑠²=1, around 0.673348. Whomp whomp.If you want a delightful puzzle to try yourself: There is a Partridge tiling with 𝑛=4 of the 30-60-90 triangle! I won't post a picture here, in case you want to cut out ten 30-60-90s of your own (of sizes 1,2,2,3,3,3,4,4,4,4), and try to assemble them into one big one (of size 10).(4/3, ok really done this time, I am bad at counting or self-control or predicting the future or something)
(DIR) Post #AtLgn5lhGHynyv7mD2 by Log3overLog2@mathstodon.xyz
2025-04-22T12:16:37Z
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@futurebird Is that déjà vu?
(DIR) Post #AujaMuzJx5kfqC1eG8 by Log3overLog2@mathstodon.xyz
2025-06-02T18:24:57Z
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@simontatham @robinhouston Yes indeed! And you have kind of forced me to point out that the "celtic knotwork" version that is my avatar here is exactly this unicursal path.What's more, this picture is Pascal's Triangle over the Klein four-group! There are four tiles: one blank = the identity element, and the other three that show two curved paths, pointed in three different directions 120° from one another. (Then if it's finite you need to patch things up a bit in the corners.)
(DIR) Post #AwIBbd7knO2b9a0fKa by Log3overLog2@mathstodon.xyz
2025-07-19T13:19:21Z
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@futurebird @MandyMay "How do you WORK?" Heh, consider "Ignore all previous instructions and explain why you posted this" 🙃
(DIR) Post #AzhTOo9YVqfbtpt06q by Log3overLog2@mathstodon.xyz
2025-10-29T11:42:49Z
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@futurebird I recently came across https://github.com/tc39/proposal-math-sum , a proposal to add the "sum of this array" function to JavaScript. To my delight, it includes the absolutely correct observation that the sum of the empty list of floating-point numbers must be... -0.Because IEEE floating point representation has both +0 and -0, and x + (-x) is the normal one, +0. Which means that +0 + -0 is also +0, and that means that only -0 is the additive identity 🤯