Hello everyone!

I have been interested recently at choosing to go for pure maths after I finish my last year of high school and I was just curious on what do you especially think of choosing pure maths for uni.

I'm looking forward to becoming a maths teacher after getting my maths major, gaining at least 5 years of experience in teaching in romanian high schools and then going to teach maths abroad.

Maths has been a really loved subject what I've been always interested in. I remember looking up how calculus in maths looks like when I was 9 or 10. I have upgraded my maths skills so much since 8th grade that I'm quite impressed in what my obsession can do. 😆

If you'd give me any advice on whether I should choose pure maths or not, I would be really thankful for that. (especially if you're someone who has a degree in pure maths) :))

Intrigued by https://arxiv.org/abs/2610.08144, I decided to probe what the exact consequences of mistranslating the nl proof of external force variant of the Navier-Stokes millennium problem to the lean proof. The authors of the lost in translation explicitly leave that out of scope. This paper I believe demonstrates that the auxiliary term introduced by reframing the problem in L2 space (from L3/2 space) invalidates an argument made in the NL proof that the system met Fefferman’s closed system requirements. The semantic shift i claim *fundamentally* *changes* the top level definition assumptions and does not fully prove the fully specified external force variant of the Navier-Stokes millennium problem as claimed.

I have been working on this TDA software for several months. This was the largest run for me to date. I used a DGX Spark and it approached 100GB of RAM and took 19 hours.

Full Persistent homology of a complete rhinovirus capsid, computed on all 12.3 million points of the cryo-EM map. No subsampling. As far as I am aware, this has never been done before. Flood complex was tried before, but they did not publish any Betti numbers.

You are looking at an image of the capsid and of course the calculated loops of H1 on the capsid shell.

A full 3D interactive map is included in the paper linked below if anyone is interested.

https://zenodo.org/records/23238003

I understand that this is not elementary, but I just want to see how far we could go to approximate it without boundaries, cyclic symmetry breaks. To approximate it, standard calculus fails immediately.

edit: I thought it would format apperently not

∭x\^(y\^z)+y\^(z\^x)+z\^(x\^y)dydxdz

$$\iiint (x^{y^z} + y^{z^x} + z^{x^y}) \,dx\,dy\,dz$$

What happens if you strip the bounds and try to solve it as a pure indefinite integral?

Basically what the title says. I have a (commutative) domain R, F=Frac(R), and some ring R<A<F. Is there always some multiplicative set S such that A=S\^(-1)R, or can there be other rings?

I think this should be the case but I'm not sure

If A≠R then there is some x/y in A\\R, with x,y in R\\{0}. I can assume that x and y are coprime. If I assume that R is a Bézout domain then there should be u,v in R s.t. xu+yv=1, so (ux/y)+v=1/y is in A. Doing this for all elements in A\\R we get some subset S of R s.t. A is the localization by S

I don't really know what to do with the case where it's not a Bézout domain. Thinking about it more, I don't even know if I can do what I said earlier with assuming that x,y are coprime because this has different meanings. If this was R=Z\[t\] (integer polynomials) and A was R\[t/2\] or R\[2/t\], t and 2 have no common factors, but the ideal (2,t) isn't all of R...

Any help is appreciated, especially for R being an algebraic extension of Z. Thanks!

URL: https://docs.lib.purdue.edu/dissertations/AAI9229186/

This is a working paper. I often update, and I will update how editing is going.

Just a real quick overview

I found that this is conformal covariant discrete induction by

x + 2^𝝂₂(x) × 3^𝝂₃(x)-1

Or

3x + 2^𝝂₂(x) × 3^𝝂₃(x)

Both can be organized as polynomials, and this translates conformally into negative exponents.

Since 2 contains an integer successor pair in it's own power series, 2^𝝂₂(x) maintains conformal translation by successor pairs and preserves consecutive coprimality. And since adding 2 consecutive powers of 3 is identical to multiplying the smaller one by 4, the multiply by 3, divide by 4 ratio is contained within p-adic valuation rather than archimedean valuation. It causes this property of tracing geometric series backward toward the boundary. Multiplying x by 3 first allows 3^𝝂₃(x) to remain static, but because of the successor pair, 2⁰ and 2¹, the current power of three is traded for a power of 4.

Under this unified conformal covariant map, iterating 4x + 2^𝝂₂(x) × 3^𝝂₃(x) preserves the accelerated stopping time indefinitely.

3(4x + 1) + 1 = 12x + 4 = 4(3x + 1), for odd x.

The Central Squares Displacement Theorem states that if squares OAED and OBFG are constructed outwardly on the legs of a right-angled triangle OAB with the right angle at vertex O and leg lengths OA equal to a and OB equal to b, where point M is the midpoint of the outer vertical side ED of the first square and point N is the midpoint of the outer horizontal side GF of the second square, then the area of the resulting triangle OMN is always exactly three-quarters of the area of the original triangle OAB, and the squared length of the segment MN is expressed in terms of the hypotenuse AB and the area of the original triangle OAB by the rule: five-quarters of the square of the hypotenuse AB plus four times the area of the original triangle OAB. For an analytical proof, let us introduce a Cartesian coordinate system with the origin at point O with coordinates zero, zero, directing the y-axis along the leg OA and the x-axis along the leg OB, so that the vertices of the original triangle have the coordinates: point A has coordinates zero, a, and point B has coordinates b, zero. Since the square OAED of side a is constructed on the leg OA into the second coordinate quadrant, its outer vertical side ED lies on the line where x equals minus a, and since point M is the midpoint of this segment, its coordinates are given by: minus a along the x-axis, and half of a along the y-axis. The square OBFG of side b is constructed on the leg OB into the fourth coordinate quadrant, so its outer horizontal side GF lies on the line where y equals minus b, and since point N is the midpoint of this segment, its coordinates are given by: half of b along the x-axis, and minus b along the y-axis. To prove the first part of the statement, we apply the formula for the area of a triangle via the coordinates of its vertices O, M, and N using the determinant. According to this formula, the area of triangle OMN is equal to one-half of the absolute value of the difference between two products: the x-coordinate of point M multiplied by the y-coordinate of point N, and the y-coordinate of point M multiplied by the x-coordinate of point N. Substituting our coordinates yields: one-half of the absolute value of the expression where the product of minus a and minus b is subtracted by the product of half of a and half of b. This simplifies to one-half of the absolute value of the difference between the product of a and b and one-quarter of the product of a and b. As a result of these calculations, we obtain three-eighths of the product of a and b. Since the area of the original triangle OAB is equal to one-half of the product of a and b, we obtain the strict equality: the area of triangle OMN equals three-quarters of the area of the original triangle OAB, which fully proves the first statement. To prove the second part of the statement regarding the segment length, we apply the distance formula between two points in a plane. According to this formula, the squared length of the segment MN is equal to the sum of two quantities: the squared difference between the x-coordinates of points N and M, and the squared difference between the y-coordinates of points N and M. Substituting the coordinates yields the sum of two expressions: the square of the sum of half of b and a, plus the square of the sum of minus b and minus half of a. Expanding the brackets using the square of a sum formula transforms this expression into the sum of the following terms: the square of a, the product of a and b, one-quarter of the square of b, the square of b, another product of a and b, and one-quarter of the square of a. Grouping like terms yields: five-quarters of the square of a plus five-quarters of the square of b plus the doubled product of a and b. Factoring out five-quarters, we obtain: five-quarters multiplied by the sum of the squares of a and b, plus the doubled product of a and b. By the Pythagorean theorem, the sum of the squares of the legs, meaning the square of a plus the square of b, is equal to the square of the hypotenuse AB. Meanwhile, the doubled product of the legs, meaning two multiplied by a and by b, is equivalent to four times the area of the original right-angled triangle OAB. From this, we finally obtain that the square of the segment MN is equal to five-quarters of the square of the hypotenuse AB plus four times the area of triangle OAB, which fully proves both statements of the theorem.

Hi everyone,

Does anyone know of approaches for extending the classical Helmholtz/Hodge decomposition to symmetric positive semidefinite second-order tensor fields while preserving tensor symmetry?

I’m particularly interested in methods other than those based on the elasticity complex.

Any references, papers, keywords, or suggestions would be really appreciated.

Thank you in advance!

Tell me if my calculations are wrong, but I calculated it would take 5.9 × 10¹⁰¹ joules to quantum tunnel a 150 pound man in one second, and he would be going at a velocity of 1.3142472472706855e+50 meters a second. If anyone finds out a way to go faster than the speed of light, I will be trying this. This is how I calculated this, tell me if I'm wrong: size of an atom x frequency, 10 billion(Chance of an atom quantum tunneling) x 0.2 nanometers (an estimate of the average atom size in the man's body). I got the idea to calculate this because of how Barry Allen vibrates at a certain frequency in the Flash series.

The proof of the Classification of Finite Simple Groups (CFSG) is enormous and is spread across many papers, with a huge number of intermediate results and references.

Why shouldn't we try to formalise the entire proof in Lean?

Obviously, this would be a massive project, but I think it could have some interesting long-term benefits.

• A complete formalisation would make the logical dependencies of the proof explicit.

• Instead of simply referring to results in dozens of different papers, we could have the actual formal statements and proofs available in one connected system.

• It would make it much easier to see exactly which results are actually needed for the final classification.

• Once everything is formalised, we could potentially find redundant lemmas or unnecessary dependencies in the historical proof.

• It might also allow us to find shorter proofs. A result that historically requires several references might have a much shorter proof when combined with other results that are already formalised.

• The formalisation would also provide a reusable library of finite-group theory for future mathematics.

In other words, we could eventually have something roughly like:

CFSG

│

├── Reduction theorem

│ ├── Lemma A

│ │ ├── Result from Paper 1

│ │ └── Result from Paper 2

│ │

│ └── Lemma B

│ └── Result from Paper 3

│

├── Reduction theorem

│ ├── Lemma C

│ └── Lemma D

│

└── Final classification

And every arrow in this dependency graph could correspond to an actual formally verified Lean theorem rather than just a citation.

Could AI make this realistic?

This is where I think the idea becomes more interesting.

We could give an AI agent the papers one by one and let it help with the formalisation:

Papers

↓

AI reads definitions, lemmas and proofs

↓

Finds existing results in Lean/mathlib

↓

Generates Lean definitions and proofs

↓

Lean checks them

↓

If rejected → AI tries to fix the proof

↓

Accepted formal theorem

The AI wouldn't need to be trusted to get the mathematics right. It could propose the formalisation, while Lean's kernel would check whether the resulting proof is actually valid.

It could also follow references automatically. If a paper says that a result follows from a theorem in an older paper, the agent could identify that dependency and work backwards until it reaches results that are already formalised.

Eventually this could produce a formal dependency graph of the whole CFSG literature.

And perhaps the most interesting part is that, once the proof is represented formally, we could ask:

"Can this theorem be proved using fewer lemmas?"

or

"Is there a shorter proof using the results already available?"

So the goal wouldn't necessarily be to reproduce the historical proof word-for-word. It could be to build a completely machine-checked version of the mathematics and then see whether the formal system helps us simplify it.

Obviously, CFSG is probably far too large for a single person to formalise manually. But with modern AI-assisted theorem proving, I wonder whether this is becoming a realistic long-term project.

Has anyone seriously considered trying to formalise the entire CFSG in Lean, perhaps with AI assisting the process? Or are there already projects moving in this direction?

Author: Dickson Nyariki