Nostr notes by Greg Egan

Here’s a piece of a quasicrystal with icosahedral symmetry. I ...

2026-05-12T23:55:02Z

In reply to nevent1q…9grw
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Here’s a piece of a quasicrystal with icosahedral symmetry.

I describe how to get this with the 3D version of the cut-and-project method here:

https://www.gregegan.net/APPLETS/12/deBruijnNotes.html#HD

In previous cases of automating mathematical activities, we’ve ...

2026-05-09T12:03:26Z

In reply to nevent1q…lggt
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In previous cases of automating mathematical activities, we’ve usually known very explicitly exactly what we were doing: computing trig tables, enumerating and checking graphs for the 4-colour map theorem, etc. This makes it relatively easy to understand the time and memory resources required, and how far they can be scaled up before hitting computational complexity limits.

It’s much less clear what LLMs are doing when arriving at these kinds of proofs, but it would be helpful if the people building and studying LLMs could make it equally transparent.

My guess is that the “algorithm” being effectively performed here (though of course not explicitly programmed as such) is something roughly along the lines of:

• extract a set P of relevant / related proofs from the literature
• break P down into a set of suitably fine-grained steps, S
• iterate through subsets of S looking for any collection of steps that will solve the problem

It’s impressive that it’s possible to automate these three things at all, even as imperfectly as LLMs do them, but if this description is more or less correct then it will eventually hit the same kind of scaling issues as any other combinatorial approach.

It might be possible to make this kind of smart-brute-force search more smart and less brutish in the manner of Alpha Go etc., but strategically pruning the decision trees for assembling valid proofs is likely to be much tougher than it was for Go.

I rarely used to weigh myself, but I recently started doing so ...

2026-05-05T06:04:51Z

I rarely used to weigh myself, but I recently started doing so daily, and I was surprised by how much the results fluctuated.

In this study:

https://pmc.ncbi.nlm.nih.gov/articles/PMC7192384/

they found:

“Within-week fluctuations of 0.35% were observed, characterised by weekend weight gain and weekday reduction which differed between all groups.”

But I seem to have peak-to-peak fluctuations closer to 4%, or ten times as much as the 0.35% seen in that study!

My food intake and exercise patterns are pretty constant, apart from doing a longish run every Sunday.

I’m curious as to what kind of fluctuations other people have measured on this timescale — especially when they’re not knowingly changing any of the relevant factors.

The Onion, gloriously on target as ever. ...

2026-03-19T12:44:03Z

The Onion, gloriously on target as ever.

https://theonion.com/the-onions-exclusive-interview-with-sam-altman/

“Add personality to your work with new editable shapes!” — ...

2026-01-29T04:30:34Z

“Add personality to your work with new editable shapes!”
— Apple, fondly imagining that this prospect is so alluring that I will pay them A$19.99/month for “Creator Studio” when the current apps I have (Pages, Numbers and Keynote) work perfectly well.

For crying out loud, ABC news, never send a line graph to do a ...

2026-01-21T10:22:16Z

For crying out loud, ABC news, never send a line graph to do a bar chart’s job.

Link: https://www.abc.net.au/news/2026-01-21/midnight-oil-rob-hirst-pancreatic-cancer-death/106252876

My non-AI web search suggests the theologian Paul Tillich.

2026-01-12T14:19:48Z

In reply to nevent1q…yed7
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My non-AI web search suggests the theologian Paul Tillich.

Gravity is the curvature of spacetime. So, how curved is the ...

2026-01-11T10:51:43Z

Gravity is the curvature of spacetime. So, how curved is the spacetime we are in right now?

Let’s start by looking at the effect of curvature in a very simple situation. Two meridians on a sphere, like the red and blue ones in the diagram, will come together and move apart as you follow them along their length. If their separation at the equator is s(0), after a distance d it will be:

s(d) ≈ s(0) cos(d/R)

where R is the radius of the sphere. The separation will repeat with a period of P = 2πR, so we can say that the radius of curvature of the space these geodesics are in is:

R = P / (2π)

“The polar ice sheet is moving at a rate of roughly 10 m per ...

2026-01-08T13:41:04Z

“The polar ice sheet is moving at a rate of roughly 10 m per year in a direction between 37° and 40° west of grid north, down towards the Weddell Sea.

The Geographic South Pole is marked by a stake in the ice alongside a small sign; these are repositioned each year in a ceremony on New Year's Day to compensate for the movement of the ice”

https://en.wikipedia.org/wiki/South_Pole

https://www.scientificamerican.com/article/the-south-pole-just-moved-heres-why/

Ha, I just looked this up and it sounds intriguing! I started ...

2026-01-07T23:27:53Z

In reply to nevent1q…jg0c
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Ha, I just looked this up and it sounds intriguing!

I started watching a (much more recent) Turkish series called “Hot Skull” with a broadly similar premise, but lost interest at some point as it was getting a bit too unfocused and meandering.

It’s fun (though no doubt very silly) to imagine taking this to ...

2026-01-07T13:33:54Z

In reply to nevent1q…3a7y
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It’s fun (though no doubt very silly) to imagine taking this to extremes. There are plenty of structures in modern mathematical physics that are as immutable as Euclidean space used to be as the stage on which Newtonian physics unfolded. What if every Hilbert space and Lie group was warped and distorted by the unruly behaviour of its constituents?

Commiserations! (I’m in Perth, so my own horrible days are not ...

2026-01-06T08:56:16Z

In reply to nevent1q…hszs
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Commiserations! (I’m in Perth, so my own horrible days are not in synch with the east coast, but we get our share.)

Every Australian politician who has dragged their feet on ...

2026-01-06T08:41:34Z

Every Australian politician who has dragged their feet on emissions reductions is cordially invited to spend the upcoming 44°C day in Sydney’s western suburbs in a two-bedroom flat without air conditioning.

Today in Rough Approximations For ArcSine: The Taylor series for ...

2026-01-05T12:32:43Z

Today in Rough Approximations For ArcSine:

The Taylor series for sin and cos start with:

sin θ ≈ θ - (1/6) θ^3
cos θ ≈ 1 - (1/2) θ^2

So:

(θ - sin θ) / (1 - cos θ) ≈ θ/3

Solving for θ:

θ ≈ 3 sin θ / (2 + cos θ)

or:

asin(x) ≈ 3 x / (2 + √[1-x^2])

This isn’t too bad for small x, and has the correct value and derivative at x = 0, but we can make it match those things at x = ±1 as well by tweaking the coefficients:

asin(x) ≈ π x / (2 + (π-2)√[1-x^2])

This has a maximum proportional error of 0.011.

That approximation is almost three orders of magnitude better ...

2026-01-05T00:01:11Z

In reply to nevent1q…ymrz
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That approximation is almost three orders of magnitude better than the ones I described as a(x) and b(x)!

Though they choose *not* to make it exactly match asin at the endpoints (as you can see from the error curve they plot for the function ψ), which leads to huge *proportional* errors close to x=0. If that was a problem, it would be easy to tweak it, making a_0 in their polynomial exactly π/2, which would roughly double its maximum absolute error elsewhere.

The original post: ...

2026-01-04T06:29:00Z

In reply to nevent1q…zd8h
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The original post:

Eniko Fox (npub193h…phpn)/115809815616652881">https://mathstodon.xyz/Eniko Fox (npub193h…phpn)/115809815616652881

A few days ago, I saw a nice approximation to the arcsine ...

2026-01-04T06:28:15Z

A few days ago, I saw a nice approximation to the arcsine function by Eniko Fox (npub193h…phpn)

asin(x) ≈ a(x) = x + (π/2-x)(1-√[1-x^2])^2

I won’t bother plotting a(x) against asin(x), because the two curves are indistinguishable to the eye, but the proportional error is plotted below.

I spent some time trying to figure out if there’s an underlying *geometric* reason why this works so well ... in the same sense that there is for, say:

asin(x) ≈ chord(x) = √[2(1-√[1-x^2])]

the length of the chord that subtends the angle asin(x).

There are some obvious nice features built into a(x): it clearly must agree with asin at x=0 and 1, and less obviously it will match derivatives at those points as well.

But surely there had to be some special geometric relationship as well?

If so, I never did find it. Maybe someone else will (or already has). But I found another approximation, roughly as simple and roughly as good:

asin(x) ≈ b(x) = ½(π-4)x^2 + x + 1 - √[1-x^2]

which also matches values and derivatives with asin(x) at x=0 and 1, and whose proportional error is the gold curve in the plot below.

That partly cured me of my conviction that there had to be a nice geometrical account for any approximation this good. Maybe all that’s really needed is a low-degree polynomial and one function, √[1-x^2], with an infinite derivative at x=1 the same as asin(x).

Of course They Might Be Giants have the perfect song to prepare ...

2025-12-31T13:24:09Z

Of course They Might Be Giants have the perfect song to prepare you for 2026:

https://www.youtube.com/watch?v=rBCVGshPGBk

Very nice! And if you change π/2 to sgn(x) π/2 you get an ...

2025-12-31T11:03:28Z

In reply to nevent1q…p66h
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Very nice!

And if you change π/2 to sgn(x) π/2 you get an equally good fit for all negative x values as well.

Sorry, my eyes glazed over immediately. If people have fun making ...

2025-12-29T11:26:41Z

In reply to nevent1q…5fpe
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Sorry, my eyes glazed over immediately. If people have fun making memes like this, that’s nice for them, but it has nothing to do with reality.

Three movies I’m looking forward to in 2026. (The dates here ...

2025-12-27T10:31:04Z

Three movies I’m looking forward to in 2026. (The dates here are when they open in my home town, at Perth’s Luna Cinemas.)

“No Other Choice” Jan 15
“The Secret Agent” Jan 22
“It Was Just An Accident” Jan 29

No, just a different directory.

2025-12-26T21:58:16Z

In reply to nevent1q…l834
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No, just a different directory.

This is the weirdest internet glitch I have encountered. When I ...

2025-12-26T08:25:33Z

This is the weirdest internet glitch I have encountered.

When I access any arxiv dot org abstract page via my home ISP, the page takes about 5 minutes to load. But the corresponding PDF for the same paper takes a second or two, so I’m not experiencing slow access to arxiv dot org in general.

And when I access the abstract pages via my mobile provider, there is no delay. So the problem isn’t anything on the arxiv servers.

Maybe this is Cloudflare, or some other content distribution network, interacting differently with my home and mobile ISPs. But it’s baffling that the (very small) abstract pages ever take so long to load.

Then maybe the Prime Field Product Theorem? It’s tough to pack ...

2025-12-24T14:14:13Z

In reply to nevent1q…er07
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Then maybe the Prime Field Product Theorem?

It’s tough to pack the entire contents of any theorem into a brief name, but (p-1)! mod p is just the product of all the non-zero elements in the field Z_p, and Wilson’s Theorem tells you this is -1.

You could call it the Paired Inverse theorem, from the way n ...

2025-12-24T12:44:16Z

In reply to nevent1q…geel
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You could call it the Paired Inverse theorem, from the way n being prime leads to the result if you pair up the distinct inverses in Z_n.

Chekhov would be ... pleased? #Pluribus

2025-12-24T12:20:43Z

Chekhov would be ... pleased?

#Pluribus

My Christmas was just saved from being fairly dire by D. C. ...

2025-12-24T10:19:00Z

My Christmas was just saved from being fairly dire by D. C. Stillson, inventor of the pipe wrench. My place is so old you can’t buy the square-plug tools for opening the drain access hatches anymore, but a pipe wrench did the trick, and now the kitchen sink ... empties again.

This is very nice, but I hope it won’t end up attracting the ...

2025-12-23T11:21:21Z

In reply to nevent1q…35u3
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This is very nice, but I hope it won’t end up attracting the awful kind of pop-sci coverage that makes it sound like the Halting Problem throws some shocking new spanner into the works of classical physics ... as if it had somehow been the case, before Turing and Gödel, that physicists had expected to be able to predict the behaviour of arbitrary systems at arbitrary future times with magical short-cuts that did not require them to calculate arbitrary numbers of intermediate states.

Larson is amusing, but maybe I should have said “Gold Age of ...

2025-12-20T21:57:52Z

In reply to nevent1q…54r3
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Larson is amusing, but maybe I should have said “Gold Age of animation” to make it clear what kind of work I meant.

https://en.wikipedia.org/wiki/Magical_Maestro

This sounds like something that would have appeared in the Golden ...

2025-12-20T11:01:06Z

In reply to nevent1q…gp8x
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This sounds like something that would have appeared in the Golden Age of cartoons: a farmyard full of animals frenetically plucking vegetables from the fields, carving them into instruments, and playing a symphony as hundreds of slapstick subplots unfolded around them.

A second season has been announced. I think Apple actually ...

2025-12-19T13:14:36Z

In reply to nevent1q…kw97
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A second season has been announced. I think Apple actually ordered two seasons when they bought it. I’m not sure how long the first season will be, but I guess it could be as short as 10 episodes (all I know for sure is that ep 9 is releasing on December 24)

Sadly, I have a feeling that most of the questions you ask are things that nobody involved in the show could answer, because no one has a sufficiently clear premise about the Hive Mind’s actual nature. I’d be delighted to be proved wrong, but on all the evidence so far it’s all just hand-waving to make whatever they want to happen, happen, rather than working out reasonable consequences of some clearly defined underlying reality.

Wikipedia’s article on sundogs claims: “It is possible to ...

2025-12-19T09:58:34Z

In reply to nevent1q…xfx6
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Wikipedia’s article on sundogs claims:

“It is possible to predict the forms of sun dogs as would be seen on other planets and moons. Mars might have sun dogs formed by both water-ice and CO2-ice. On the giant planets—Jupiter, Saturn, Uranus, and Neptune—other crystals form clouds of ammonia, methane, and other substances that can produce halos with four or more sun dogs.[9]”

Ref [9] was to a fairly old web site, https://www.atoptics.org.uk/halo/oworld.htm, which made several predictions, including:

“Halos from octahedral ammonia crystals as might exist in the cold high level clouds of Jupiter and Saturn. The 42° circular halo has four associated sundogs. The inner halos are produced by rays reflected within the crystal.”

That web site has no actual Martian haloes. However, we have seen a halo on Mars:

“Graphical Abstract

For the first time, a halo has been photographed from the surface of another planet. Its angular distance from the Sun indicates that the halo-generating Mars cloud consisted of water–ice crystals, not CO2 crystals.”

https://rmets.onlinelibrary.wiley.com/doi/10.1002/wea.4197 [Paywalled, alas]

I just got my annual COVID booster. The vaccine updated for ...

2025-12-18T02:00:45Z

I just got my annual COVID booster. The vaccine updated for LP.8.1 recently became available in Australia, so if you were waiting for that, now’s the time to get a pre-Christmas boost.

Time to try transporting a steamship over the Andes.

2025-12-16T12:14:43Z

In reply to nevent1q…6clq
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Time to try transporting a steamship over the Andes.

Until today I would have been hard-pressed to name a surface ...

2025-12-16T11:19:38Z

Until today I would have been hard-pressed to name a surface arising naturally in physics that’s described by a degree-6 polynomial, but there’s a family of them hiding in plain sight:

x^2 – k^2 (x^2 + y^2 + z^2)^3 = 0

These are the surfaces of equal electrostatic potential for a dipole pointing along the x-axis! The contours in the image come from a 2D slice through the axis.

But these surfaces also have two nice properties in Newtonian gravity.

[1] A point mass of m located *anywhere* on a k-surface has the same x-component for its gravitational attraction at the origin:

g_x = G k m

[2] If you fill the interior of a k-surface with matter of uniform density ρ, the gravitational attraction at the origin is *the maximum possible* for that density and total mass.

g(k) = 4πGρ/(5√k)

The volume of the interior of a k-surface is:

vol(k) = 4π/(15√[k^3])

In terms of total mass M and density ρ:

g_max(M,ρ) = G ((48/25)π^2 M ρ^2)^{1/3}

The attraction at the surface of a sphere is:

g_sph(M,ρ) = G ((16/9)π^2 M ρ^2)^{1/3}

So g_max/g_sph = (27/25)^{1/3} ≈ 1.026

The polar gravity of the “best” oblate spheroid can improve on a sphere *almost* as much:

g_oblate(ecc = 0.694464)/g_sphere ≈ 1.022

The max gravity result is proved in:

https://arxiv.org/abs/1412.5541

H/T Tom Lowe (npub1jj2…jlmk)

Orwell’s widow in 1974: “No, you may not make a musical ...

2025-12-15T23:55:27Z

Orwell’s widow in 1974: “No, you may *not* make a musical adaptation of 1984!” But “Diamond Dogs” was probably ten times better for needing to add and change enough to stand on its own.

Orwell’s estate in 2025: “Knock yourself out!”
Serkis: “Thank you! Here is my awful remake of ‘Chicken Run’”.

I’m pretty sure you’re right! In fact, if my numerical ...

2025-12-12T07:44:25Z

In reply to nevent1q…79p2
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I’m pretty sure you’re right!

In fact, if my numerical integrals can be trusted, for a cylinder of height 2 and radius 1.01, the gravitational attraction at the “equator” is 1.00455 times the gravitational attraction at the “poles”, despite the distance from the centre being greater at the equator. (This also kills off a potential alternative conjecture, where the maximum was achieved by the normal component of the gravity rather than the magnitude.)

Given a homogeneous, centrally symmetric convex body, has it been ...

2025-12-12T05:52:22Z

Given a homogeneous, centrally symmetric convex body, has it been proved that its gravitational acceleration on its surface attains its greatest magnitude at the closest point to the centre of mass, and its smallest magnitude at the farthest point?

Alternatively, are there known counterexamples?

This is certainly true for ellipsoids, and Newton more or less proved it for an oblate spheroid ( https://www.gregegan.net/SCIENCE/Ellipsoid/Ellipsoid.html#NPG ) but I can only get the ellipsoid result from an explicit calculation.

Just as a=b makes a spheroid special among ellipsoids, there is a ...

2025-12-11T12:06:47Z

In reply to nevent1q…cm6l
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Just as a=b makes a spheroid special among ellipsoids, there is a condition:

a^2 b^2 (B – A) / (a^2 – b^2) – C c^2 = 0

that makes an ellipsoid special, where A, B, C measure how fast the gravitational potential increases as you move away from the centre along the x, y, z axes.

Those coefficients A, B, C themselves depend on the semi-axes a, b, c, and are given by the values of certain integrals.

But the upshot is that we can find a curve in the space of eccentricities for two meridians of the ellipsoid, say e_xz and e_yz, in two orthogonal planes.

For any tri-axial ellipsoid with two meridians whose eccentricities lie on this curve, it is possible for *the same value of spin* to bring the corresponding meridians of the equipotential ellipsoids into alignment with the meridians of the body itself!

So, a shape like this is a possible shape for “sea level” in a spinning body of fluid.

Note that one of the meridians *must* have an eccentricity greater than a critical value, e_crit ≈ 0.812670. This means that the angular momentum must also exceed a threshold.

The second issue is whether the Jacobi ellipsoid has less total energy than the Maclaurin spheroid, for the same angular momentum.

The potential energy, U, of the Maclaurin spheroid is actually slightly *lower* than that of the Jacobi ellipsoid!

But the Jacobi ellipsoid, with a greater moment of inertia I, reaches the same angular momentum, L, with a smaller angular velocity ω … and ends up with a kinetic energy, K, sufficiently smaller to make its total energy, E, the lower of the two.

More at https://www.gregegan.net/SCIENCE/Ellipsoid/Ellipsoid.html

But how can this possibly work for a tri-axial ellipsoid? We only ...

2025-12-11T12:04:48Z

In reply to nevent1q…gwxv
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But how can this possibly work for a tri-axial ellipsoid? We only have one parameter to tweak, the rate of spin, so how can we match two *differently shaped* meridians?

The answer is that you *can’t* pull off this trick for a generic tri-axial ellipsoid, with any old a, b and c.

Given any oblate spheroid (a = b > c), made of rock, say, so ...

2025-12-11T12:03:44Z

In reply to nevent1q…t4yl
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Given any oblate spheroid (a = b > c), made of rock, say, so we can fix the shape, the surfaces of constant potential will also be spheroids — but if it’s not spinning they will not match the shape of the body.

However, by adding just enough spin the shapes can be made the same.

All the “equipotential ellipsoids” are just scaled versions of a single shape. Once you add enough spin to make one meridian of these ellipsoids the same shape as a meridian of the body itself, the problem is solved, since the rotational symmetry makes all meridians the same.

If a mass of incompressible fluid is floating in space, subject ...

2025-12-11T12:02:26Z

If a mass of incompressible fluid is floating in space, subject only to its own gravity and centrifugal force, what shape will it adopt as its angular momentum is increased?

Initially, it will form an oblate spheroid, with two equal semi-axes shorter than its axis of rotation.

However, past a certain value for the angular momentum, L, the fluid can have a lower total energy by breaking that symmetry and forming a “tri-axial” ellipsoid, with the semi-axes perpendicular to the axis, a and b, taking on different lengths!

The first configuration is known as a “Maclaurin spheroid”, after Colin Maclaurin, who studied it in 1742.

The second is called a “Jacobi ellipsoid”, after Carl Gustav Jacob Jacobi, who in 1834 realised that a fluid could be in equilibrium in this less symmetrical state.

There are two separate issues here.

One is the question of when the “level surfaces” of constant gravitational-plus-centrifugal potential for an ellipsoid can match the shape of the ellipsoid itself. What shapes can “sea level” take for a spinning mass of fluid?

Once a year, I need to print my French tax forms (in triplicate ...

2025-12-11T04:52:40Z

In reply to nevent1q…4sd3
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Once a year, I need to print my French tax forms (in triplicate for each publisher), so I buy a new ink cartridge and use the brief window of opportunity to print anything else I might conceivably want for the next 12 months, because the cartridges will dry up and become hopelessly clogged during any period of prolonged disuse.

For Dichronauts: Books-A-Million: ...

2025-12-11T03:44:44Z

In reply to nevent1q…m3hy
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For Dichronauts:

Books-A-Million:
http://www.booksamillion.com/p/9781597806053

Apple Books:
https://books.apple.com/us/book/dichronauts/id1171077019

Barnes and Noble:
https://www.barnesandnoble.com/w/?ean=9781597806053

Bookshop dot org
https://bookshop.org/a/1688/9781597806053

Kobo:
https://www.kobo.com/us/en/ebook/dichronauts?sId=8ac3153c-656f-4c4b-bba1-685c935ecd1b&ssId=3OYA7vbjgc7ujdoXQhfBA

(Those books where I have the US epub rights can be bought DRM-free from Smashwords and various other sites, but Dichronauts is one where the US publisher still has the US rights, so I can only sell it outside the US. All the books I self-publish are listed here:

https://www.gregegan.net/BIBLIOGRAPHY/Ebooks.html )

Euler?

2025-12-10T04:40:17Z

In reply to nevent1q…f5tm
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Euler?

Very nice!

2025-12-06T00:25:31Z

In reply to nevent1q…3w99
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Very nice!

Is there an elegant proof of this? I think I can kludge together ...

2025-12-05T13:20:35Z

In reply to nevent1q…jn2l
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Is there an elegant proof of this? I think I can kludge together a proof by induction that relies on looking at how many carries can occur each time you add 999 to the previous multiple, with some tedious book-keeping relating zeroes in the last 3 digits and 9's in the digits to the left of the last 3. But is there something nicer?

An embedding f:SO(3)->R^n needs to be a homeomorphism onto its ...

2025-12-05T12:59:38Z

In reply to nevent1q…vu3j
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An embedding f:SO(3)->R^n needs to be a homeomorphism onto its image, and that breaks down at the boundary of the ball in R^3, since antipodal points on the boundary are not identified in the ambient topology of R^3.

I think SO(3) can be embedded in R^5, but even that’s not ...

2025-12-05T11:15:14Z

In reply to nevent1q…e5mn
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I think SO(3) can be embedded in R^5, but even that’s not particularly useful or interesting for most purposes.

That’s so frustrating, because SO(3) is one of the easiest Lie ...

2025-12-05T10:52:03Z

In reply to nevent1q…nyju
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That’s so frustrating, because SO(3) is one of the easiest Lie groups by far to visualise without having to worry about higher dimensions!

And once you have that model of it — as a sphere with antipodal points identified — you can easily understand cool things like the Dirac belt trick, and why an electron needs to be rotated twice to return to its original state:

https://www.gregegan.net/APPLETS/21/21.html

I don’t recall how much emphasis was placed on the underlying ...

2025-12-05T00:04:36Z

In reply to nevent1q…7pay
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I don’t recall how much emphasis was placed on the underlying unity of these algorithms when I was in primary school, but if I was teaching kids myself it was something I would focus on.

Operations on decimal integers and decimals with non-integer parts are essentially identical. The only difference between operations on fractions and operations on integers is conversion to a common denominator, and understanding that geometrically (cutting up the same pie into fifths and quarters, etc., or overlaying two transparencies with different subdivisions of the number line) dispels any mystery and helps protect against errors.

My memory of early maths education is that it was so focused on ...

2025-12-04T23:09:12Z

In reply to nevent1q…a75q
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My memory of early maths education is that it was so focused on the number line — as the underlying reality to which all notation was pointing — that it would never have occurred to me that the integers and the rationals were anything other than particular subsets of the reals.

And FWIW, I think that’s by far the best starting point! Six-year-olds should not have to worry about Dedekind cuts.

A grim, but fascinating, unsolved mystery. A mostly peaceful and ...

2025-12-04T13:57:15Z

A grim, but fascinating, unsolved mystery. A mostly peaceful and prosperous farming culture known as LBK spread across Europe from Anatolia, from 11,000 to 7,000 years ago. When it eventually came to an end, in many places the transition appears to have been unremarkable, but in some sites there are vast mass graves, and other signs of strange tensions arising in various settlements.

https://www.science.org/content/article/headless-bodies-hint-why-europe-s-first-farmers-vanished

At the linked post, it says: “range of y coordinates is about ...

2025-12-04T13:26:01Z

In reply to nevent1q…cker
_________________________

At the linked post, it says:

“range of y coordinates is about +/-5537521000”

Great writeup from @npub1hh3…hqxd of an extraordinary ...

2025-12-01T08:34:40Z

Great writeup from StartsWithABang (npub1hh3…hqxd) of an extraordinary hypothesis: the reason the two sides of the moon are so different might come down to tidal locking when the moon was still forming, and chemical gradients in the proto-lunar material due to radiant heat from the Earth!

https://bigthink.com/starts-with-a-bang/moon-two-faces-different/

You might find this interesting. There probably isn’t a ...

2025-12-01T03:17:22Z

In reply to nevent1q…wgdy
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You might find this interesting. There probably isn’t a consensus, but the study discussed here suggests that some very large gas clouds might have directly collapsed into black holes at a very early time.

https://bigthink.com/starts-with-a-bang/first-supermassive-black-holes/

Sadly it’s not a joke, but if it’s any small consolation, ...

2025-11-28T06:22:36Z

In reply to nevent1q…6lwm
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Sadly it’s not a joke, but if it’s any small consolation, “Scientific Reports”, though it comes under the Nature umbrella, is not “Nature” as such.

Is there any word-producing thing in the universe that ...

2025-11-28T04:01:40Z

Is there any word-producing thing in the universe that understands even less about autism than RFK Jr and Andrew Wakefield?

Amazingly, yes: whatever 90% chatbot / 10% human chimera cranked out this article about, err, AIs diagnosing autism.

H/T feralrobots on BlueSky.

https://www.nature.com/articles/s41598-025-24662-9

No, they’re not that magical; you can only split them into ...

2025-11-27T08:01:16Z

In reply to nevent1q…ttmv
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No, they’re not that magical; you can only split them into measurable parts.

Black Friday Sale! Our 1,000-dimensional magic golden spheres[*] ...

2025-11-27T07:13:54Z

Black Friday Sale! Our 1,000-dimensional magic golden spheres[*] are discounted by a whopping 50%!

* Discounted 1,000-dimensional magic golden spheres are 99% the radius of normally priced spheres, which may result in a small decrease in magicness and total gold volume.

One thing that confuses me is that (unless I’m mistaken) the ...

2025-11-26T22:56:43Z

In reply to nevent1q…ll3j
_________________________

One thing that confuses me is that (unless I’m mistaken) the standard model gives accurate predictions of when and how leptons in the heavier generations decay into lighter versions, and yet the symmetry group underlying the generation structure remains unknown.

I thought the coupling that enables a decay like this tells you what gauge boson is involved, and if you know what gauge boson is involved, you know the symmetry group. Obviously I’m wrong about something major here! Can you tell me what I’m fundamentally wrong about?

By making the internet work again, I mean restore access to all ...

2025-11-26T22:23:08Z

In reply to nevent1q…ugvj
_________________________

By making the internet work again, I mean restore access to all the static resources that would still have been useful without any ongoing edits, e.g. the last pre-apocalypse version of Wikipedia, and various web sites covering technical issues that she would benefit greatly from learning something about. Potentially, she could also maintain direct contact with those other survivors she hadn’t completely alienated.

FFS, Carol, if the Hive Mind really will do anything you ask, ...

2025-11-26T13:11:14Z

FFS, Carol, if the Hive Mind really will do anything you ask, tell it to get the internet working again.

#Pluribus

For an account of how Newton reached this result (and also a ...

2025-11-26T11:35:06Z

In reply to nevent1q…46ln
_________________________

For an account of how Newton reached this result (and also a different way to obtain the same formula, using quadrupole moments):

https://www.gregegan.net/SCIENCE/Ellipsoid/Ellipsoid.html#NPG

St Peter: You studied Newtonian gravity, right? Me: Umm, a bit. ...

2025-11-25T13:11:37Z

St Peter: You studied Newtonian gravity, right?

Me: Umm, a bit.

SP: Explain to me, like I’m 5, why the ratio of polar gravity to equatorial for a uniform oblate spheroid is approximately 1 + 1/5 the ellipticity of the meridians.

Me: You let Isaac set the questions, didn’t you?

I’ve switched off all the explicitly AI nonsense in MacOS, but ...

2025-11-24T14:26:16Z

In reply to nevent1q…eqvx
_________________________

I’ve switched off all the explicitly AI nonsense in MacOS, but Preview still has some over-excitable “data detectors” that think any sequence of digits must be a phone number that needs to be added to your contacts.

One mouse click away from butt-dialling the Big Bang. ...

2025-11-24T07:38:21Z

One mouse click away from butt-dialling the Big Bang.

I’m still a bit curious as to why you mentioned credit cards! I ...

2025-11-24T04:28:43Z

In reply to nevent1q…4j8p
_________________________

I’m still a bit curious as to why you mentioned credit cards! I guess it can’t be a coincidence that they have exactly the same aspect ratio as a sheet of foolscap, and in the imperial measurements the specified sizes are exactly 4:1.

I’m used to all the cool relationships between the sizes of paper in the ISO 216 standard, but I never knew about the credit card / foolscap link.

I learned about “mirror spiders” from here (despite living in ...

2025-11-21T09:00:22Z

In reply to nevent1q…38c5
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I learned about “mirror spiders” from here (despite living in Australia I’d never heard of them before), and here’s an open access paper in Nature that dives into the mechanism:

“Animals precisely control the morphology and assembly of guanine crystals to produce diverse optical phenomena in coloration and vision.”

https://www.nature.com/articles/s41467-023-35894-6

A huge, beautiful library of tessellated materials in biology! ...

2025-11-21T03:59:04Z

A huge, beautiful library of tessellated materials in biology!

https://tessellated-materials.mpikg.mpg.de/collection

And an interview in Science with the creators of this library:

https://www.science.org/content/article/why-does-biology-keep-building-things-out-tiles

RE: https://mastodon.online/@tomstafford/115582077134907935 Some ...

2025-11-20T22:33:52Z

In reply to nevent1q…n6vf
_________________________

RE: https://mastodon.online/@tomstafford/115582077134907935

Some embarrassing Grok responses like this are vanishing from X right now. As of this moment, this one is still online, but it will probably be deleted eventually.

FWIW, here’s someone who independently verified the existence of the screenshotted Grok response earlier, in case you’re wondering if it was faked.

Tom Stafford (npub1m9f…m8nw)/115582077236839502">https://mathstodon.xyz/Tom Stafford (npub1m9f…m8nw)/115582077236839502

quoting
note1ktc…6qwm

fact check: real. For shame. https://x.com/grok/status/1990938698658218162

Who could have guessed that the end product of a fearless, ...

2025-11-20T12:34:06Z

Who could have guessed that the end product of a fearless, intellectually rigorous research program whose sole aim was to produce an LLM with the most reliable, objective and trustworthy responses possible would sound so much like a sycophantic courtier flattering a demented, narcissistic monarch?

I lived without a washing machine for about 20 years, ...

2025-11-20T12:08:40Z

In reply to nevent1q…mmmz
_________________________

I lived without a washing machine for about 20 years, soaking/handwashing everything. That started when I was living in tiny rented flats with no laundry, and I was too lazy and/or poor to go to a laundromat, but once I was used to it, it was just part of the daily routine.

I also found that puzzling. For some kinds of neuropharmaceutical ...

2025-11-18T04:10:58Z

In reply to nevent1q…s9zq
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I also found that puzzling. For some kinds of neuropharmaceutical testing in organoids, you would definitely want human neurons, but for biocomputing per se it seems irrelevant.

Maybe he’s decided that the future is human neurons in dishes. ...

2025-11-17T13:03:57Z

In reply to nevent1q…4w4u
_________________________

Maybe he’s decided that the future is human neurons in dishes.

“Biocomputing advocates claim that these systems could one day rival the capability of artificial intelligence and the potential of quantum computers.”

[There’s some genuinely interesting research here, but also a lot of ridiculous hype, like this quote.]

https://www.nature.com/articles/d41586-025-03633-0

This looks more or less identical to waving a magnet in front of ...

2025-11-16T12:36:25Z

In reply to nevent1q…hhs8
_________________________

This looks more or less identical to waving a magnet in front of a 1980s CRT television tuned to a station that is currently off the air.

Google’s “AI overview”, pointlessly appended to their usual ...

2025-11-15T14:28:40Z

Google’s “AI overview”, pointlessly appended to their usual weather forecast for my suburb:

“The weather in XXX is currently sunny with a temperature of about 17° C ... ”

The forecast already states the current temperature, but no, it is not “currently sunny” as it’s 10:23 pm.

The forecast by itself is perfectly understandable. It does not need a summary, an overview, an interpretation, a Reader’s Digest Condensed Version.

FFS, Google, is this what you’re building new data centres and buying billions of dollars worth of GPUs to do?

You can see the relationship of the focal conics to the surface ...

2025-11-12T23:33:50Z

In reply to nevent1q…h0en
_________________________

You can see the relationship of the focal conics to the surface in this construction:

https://mathstodon.xyz/@gregeganSF/115468415319176584

quoting
note1a93…qr75

Most people know how to draw an ellipse by pinning two ends of a string to a board and sweeping a pencil around inside the string, keeping it taut.

But what about the 3D equivalent?

Start with an ellipse (blue) and a hyperbola (green) in orthogonal planes, with each curve’s vertices (extreme points) being the other’s foci.

Pin one end of a string (red and black) to one focus of the hyperbola, and the other to the opposite focus of the ellipse, and then pull the string taut, while also requiring it to pass through whatever point on each curve minimises the distance. The point the string reaches in between will trace out an ellipsoid!

This construction was found in the late 1800s by the German mathematician Staude (Wikipedia says the famous physicist Maxwell also made a start on the problem a bit earlier). I haven’t read Staude’s proof, but you can read a modern treatment here:

https://arxiv.org/abs/1707.01233v1

The simplest definition is that two ellipsoids with the same ...

2025-11-12T23:30:59Z

In reply to nevent1q…lzqj
_________________________

The simplest definition is that two ellipsoids with the same centre, and with their axes aligned, are confocal if the squares of their semi-axes all differ by the same amount.

For an ellipse with semi-axes a and b, the distance f from the centre to the foci is given by:

f^2 = a^2 - b^2

so adding the same quantity to both a^2 and b^2 preserves the foci.

So algebraically this generalises that relationship: we add the same quantity to a^2, b^2, c^2 for the ellipsoid.

Geometrically, you can also identify “focal conics” of an ellipsoid that remain the same under this transformation.

https://en.wikipedia.org/wiki/Focal_conics

[3] The gravitational field at a given point outside an ellipsoid ...

2025-11-12T12:40:11Z

In reply to nevent1q…6qnj
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[3] The gravitational field at a given point outside an ellipsoid will be identical for any two ellipsoids that have the same mass, and that are “confocal” with each other.

Why? It turns out that if you map points between two confocal ellipsoids with a map T that rescales the three coordinates in proportion to the ratios between the ellipsoids’ axes, then for any points P and Q on the same ellipsoid you have:

dist(P, T(Q)) = dist(T(P), Q)

If we take two confocal ellipsoids with the same density and dissect them into rectangular rods, like AB and T(A)T(B) in the diagram, the component of the gravitational force those rods exert, in the direction of their own axes, at the points T(P) and P will depend only on the cross-sections of the rods and the distances of the endpoints of the rods from the point where the force is measured.

But the distances from the endpoints are the same! So we have a simple relationship between the force that the ellipsoid E_1 exerts on an exterior point T(P), and the force that the ellipsoid E_2 exerts on the interior point P.

Combined with the linearity of T and the linear force law in the interior, we end up with a force at the exterior point that is independent of the particular ellipsoid confocal with E_1 that we choose, and simply scales with its mass.

More details at: https://www.gregegan.net/SCIENCE/Ellipsoid/Ellipsoid.html

[2] Inside a solid ellipsoid, the components of the gravitational ...

2025-11-12T12:39:10Z

In reply to nevent1q…d770
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[2] Inside a solid ellipsoid, the components of the gravitational force along each of the three axes is proportional to the distance from the centre along the corresponding axis ... but unlike a spherical ball, the constants of proportionality will differ for the three axes:

g = –(A x, B y, C z)

We can prove this by looking at two double cones through a point P = (x, y, z). As mentioned upthread, the gravitational force from the material in a cone will be proportional to the length of the cone, and so the x-component of the force will be proportional to the *extent* of each cone along the x-axis, Δx. If the two double cones differ by flipping the y- and z-components of the direction of their axes, the sum of all four values for Δx will be the sum of both roots of two quadratic equations, and the flip in y and z will cause the final answer to simply be proportional to x, with no other dependence on the coordinates of P.

So when we cover all the material in the ellipsoid by integrating over all such cones, the total remains proportional to x, and we can reach the same kind of result for the y- and z-axes.

Three fun things about the gravity of ellipsoids. [1] There is no ...

2025-11-12T12:37:43Z

Three fun things about the gravity of ellipsoids.

[1] There is no gravitational force inside an ellipsoidal shell of uniform density ... but only if it’s the *right kind* of shell, known as a homoeoid. The inner and outer boundaries of such a shell must be concentric, axis-aligned, scaled copies of each other, rather than the shell having uniform thickness.

If you take a spherical shell whose boundaries are concentric spheres, it’s clear by symmetry that any straight line through a point P in the interior will intersect the shell along two segments, AB and CD, of equal length. But we can turn such a shell into a homoeoid with a linear transformation, and a linear transformation rescales all parallel line segments by the same factor, so AB and CD will also be equal for any homoeoid.

Now, consider a double cone with its apex at P. The amount of material in slices of each cone at a distance r grows with r^2, exactly cancelling the 1/r^2 diminution of gravity, so the attraction at P from the material will (in the limit of an infinitesimal cone) be proportional to the lengths AB and CD. And since AB and CD are equal, there will be equal and opposite forces at P, cancelling each other out.

OK, I’m just trying to understand exactly what is and isn’t ...

2025-11-10T11:44:58Z

In reply to nevent1q…96uz
_________________________

OK, I’m just trying to understand exactly what is and isn’t different in the two cultures. You seem to be saying that the thing that does NOT exist in Italy (setting aside the Fellini poster) is the practice of writing an integer and a fraction adjacent to each other without an explicit + sign. So it is common to write the equivalent expression with an explicit + sign?

If I solved a problem in primary school mathematics and needed to write down an *exact* answer of, say, 4 + 13/43, would the correct way to write that be “4 + 13/43”, or would I normally be expected to write “185/43”?

Are you talking about formal mathematics, or ordinary language ...

2025-11-10T11:14:22Z

In reply to nevent1q…a06k
_________________________

Are you talking about formal mathematics, or ordinary language and commerce? In Italy, if four friends share ten breadsticks, do they say “we have two and half each” or “we have five halves each”?

Mixed numbers certainly don’t seem to be entirely absent from Italian culture!

https://en.wikipedia.org/wiki/8%C2%BD

Colin the Mathmo (npub1s4h…qs84)

Every parallelepiped that you place around an ellipsoid whose ...

2025-11-06T13:17:16Z

In reply to nevent1q…jx04
_________________________

Every parallelepiped that you place around an ellipsoid whose faces are tangent to the ellipsoid at their centres has the same volume for a given ellipsoid: 8 a b c, where a, b and c are the semi-axes of the ellipsoid.

It’s not hard to prove that this generalises to n dimensions.

Suppose we have an n-ellipsoid given by the equation:

∑ᵢ₌₁ⁿ𝑥ᵢ²/𝑎ᵢ² = 1

where the 𝑥ᵢ are coordinates (x,y,z,...) and the 𝑎ᵢ are the semi-axes in each direction.

A vector normal to the surface can be found from the gradient of the left-hand side of this equation; since that function is constant on the surface, the gradient is orthogonal to the surface.

norm(𝑥) = (𝑥ᵢ/𝑎ᵢ²)

Suppose we choose n points on the surface of the ellipsoid, 𝑥ˢ, s=1,...n, such that:

𝑥ᵗ·norm(𝑥ˢ) = 0 when t≠s

We will also have:

𝑥ˢ·norm(𝑥ˢ) = ∑ᵢ₌₁ⁿ𝑥ˢᵢ²/𝑎ᵢ² = 1

If we define the matrix 𝑋 to have the vectors 𝑥ˢ as its rows, and the matrix 𝐴 to have diagonal elements 1/𝑎ᵢ² and zeroes elsewhere, then the dot products above correspond to:

𝑋 𝐴 𝑋ᵀ = 𝐼

If we take the determinants of both sides of this equation, we have:

det(𝑋)² det(𝐴) = 1

So det(𝑋) will always be the same, the product of all the semi-axes, and the volume of the parallelotope that encloses the ellipsoid will be 2ⁿ det(𝑋), with each of its faces formed by taking ±𝑥ˢ as the centre and adding ±𝑥ᵗ, for all the choices of t≠s. Since those 𝑥ᵗ will be orthogonal to norm(𝑥ˢ), each face will be tangent to the ellipsoid at its centre.

Every parallelogram that you draw around an ellipse whose sides ...

2025-11-06T13:15:32Z

Every parallelogram that you draw around an ellipse whose sides are tangent to the ellipse at their midpoints has the same area for a given ellipse: 4 a b, where a and b are the semi-axes of the ellipse.

I’m finding it mostly pretty silly so far, but Emma Thompson ...

2025-11-04T12:15:59Z

In reply to nevent1q…yh8f
_________________________

I’m finding it mostly pretty silly so far, but Emma Thompson plays an interesting character and I’ll probably stick with it and see if there’s enough entertainment value to compensate for the implausibility. (It’s based on a book by Mick Herron, who wrote the books on which the series “Slow Horses” is based, so if you enjoyed that you might like this as well ... though the intelligence agency portrayed in “Cemetery Road” seems to be ten times more cartoonish than the version of MI5 in “Slow Horses”.)

It’s old news that most screenwriters are clueless about ...

2025-11-04T05:56:21Z

It’s old news that most screenwriters are clueless about scientists and mathematicians, but you’d hope they'd be a *little* more familiar with what makes a good humanities student.

“Down Cemetery Road”: “She was brilliant! She memorised The Wasteland, including the footnotes!”

I liked the way they didn’t bludgeon you over the head with the ...

2025-11-04T02:23:43Z

In reply to nevent1q…64ty
_________________________

I liked the way they didn’t bludgeon you over the head with the school shootings thing, but went gleefully over the top / tongue-in-cheek with the Hansel-and-Gretel-on-steroids revenge, so you got some catharsis without making it all unbearably bleak.

It’s been a bit too long since I saw it to remember the ...

2025-11-04T00:54:53Z

In reply to nevent1q…448q
_________________________

It’s been a bit too long since I saw it to remember the progression of expectations I had. I was definitely surprised when the villain was revealed to be so old-school folkloric, but then that worked for me in the end as part of the whole fairy-tale-like allegory for a very grim aspect of modern American life.

It’s nice that an expert on aerosols has won the 2025 Prime ...

2025-11-03T12:41:20Z

It’s nice that an expert on aerosols has won the 2025 Prime Minister's Prize for Science … but the govt still can’t be bothered acting on her expertise.

"The reality is that there are no regulations for indoor air quality," she said.

https://www.abc.net.au/news/science/2025-11-03/lidia-morawska-prime-ministers-science-prize-air-quality/105959260

I liked it a lot! I was a bit afraid at first that we’d be ...

2025-11-03T08:00:30Z

In reply to nevent1q…7490
_________________________

I liked it a lot! I was a bit afraid at first that we’d be stuck with Julia Garner just spiralling into alcoholism as the town shunned her ever more cruelly, but once they unleashed ... no spoilers, so I’ll just say “the thing that comes next” ... I really started to enjoy it.

The portions of the string where it departs from the hyperbola or ...

2025-11-01T01:53:25Z

In reply to nevent1q…m6nr
_________________________

The portions of the string where it departs from the hyperbola or the ellipse both lie on cones whose axis is a tangent to the curve at that point, and which make an angle with the tangent that is the same as the adjacent segment of the string.

This is a geometrical way of saying that putting tension on the string won’t make it slide along these curves, because the sum of the distances of the adjoining segments is a local minimum, so it can’t offer up slack by moving the point.

Here is a version where the point on the ellipse is held still ...

2025-11-01T01:52:31Z

In reply to nevent1q…zjxz
_________________________

Here is a version where the point on the ellipse is held still while the point on the hyperbola is swept along it.

Most people know how to draw an ellipse by pinning two ends of a ...

2025-10-31T10:56:32Z

Most people know how to draw an ellipse by pinning two ends of a string to a board and sweeping a pencil around inside the string, keeping it taut.

But what about the 3D equivalent?

Start with an ellipse (blue) and a hyperbola (green) in orthogonal planes, with each curve’s vertices (extreme points) being the other’s foci.

Pin one end of a string (red and black) to one focus of the hyperbola, and the other to the opposite focus of the ellipse, and then pull the string taut, while also requiring it to pass through whatever point on each curve minimises the distance. The point the string reaches in between will trace out an ellipsoid!

This construction was found in the late 1800s by the German mathematician Staude (Wikipedia says the famous physicist Maxwell also made a start on the problem a bit earlier). I haven’t read Staude’s proof, but you can read a modern treatment here:

https://arxiv.org/abs/1707.01233v1

Yes, but more of what? The trouble with this kind of second-hand, ...

2025-10-30T23:56:25Z

In reply to nevent1q…xuzz
_________________________

Yes, but more of what?

The trouble with this kind of second-hand, context-free quote is that the speaker is not around to stop people twisting their words. I’m not accusing you of doing that, but honestly I hate the whole cottage industry of mining famous people’s utterances and dredging up fragments of speech or writing that they never intended to be treated as epigrams as if they are some kind of distillation of wisdom.

Since they’re expecting sunlight, I would imagine that their ...

2025-10-29T14:08:02Z

In reply to nevent1q…jmhp
_________________________

Since they’re expecting sunlight, I would imagine that their existing systems will take account of it.

If Reflect Orbital put their stupid fucking mirrors in orbit, ...

2025-10-29T13:55:55Z

In reply to nevent1q…54kc
_________________________

If Reflect Orbital put their stupid fucking mirrors in orbit, maybe we can crowd source a giant laser to push it off course with radiation pressure?

Since the Harvard Astronomer is getting excited about the Wow! ...

2025-10-28T22:15:21Z

Since the Harvard Astronomer is getting excited about the Wow! signal now, this is worth boosting:

“[...] We report the detection of narrowband signals (10 kHz) near the hydrogen line similar to the Wow! Signal, although two-orders of magnitude less intense and in multiple locations. Despite the similarities, these signals are easily identifiable as small interstellar clouds of cold hydrogen (HI) in the galaxy. We hypothesize that the Wow! Signal was caused by a sudden brightening of the hydrogen line in these clouds triggered by a strong transient radiation source, such as a magnetar flare or a soft gamma repeater (SGR). A maser flare or superradiance mechanisms can produce stimulated emission consistent with the Wow! Signal. Our hypothesis explains all observed properties of the Wow! Signal, proposes a new source of false positives in technosignature searches, and suggests that the Wow! Signal could be the first recorded event of an astronomical maser-like flare in the hydrogen line.”

https://arxiv.org/abs/2408.08513

Nostr event nevent1qqs0nvtzhnnh9w6du48wsynyxnczzmuq0kapdal0gvm6g5w9p82lgvgzyptrsggqfslyun60hmm55fjhmvefng88rfuzzjle0kctdjs9ldeaxhlh7xn

2025-10-28T13:50:42Z

In reply to nevent1q…vwag
_________________________

https://authorsguild.org/advocacy/artificial-intelligence/anthropic-settlement-faq/

“Does the settlement mean that Anthropic can now continue to use pirated books to train AI?

No. The settlement does not give Anthropic—or any AI company—permission to use pirated books going forward. It only resolves Anthropic’s liability for past use of books. In fact, the agreement requires Anthropic to destroy all copies in its possession.”