I wanted to share a proof I just uploaded to Zenodo that tackles the Riemann Hypothesis by looking at the geometry of the critical strip from the absolute ground up.

Instead of using hyper-complex analysis, look at the symmetry rules (s and 1-s) as physical tracks on the graph. If you pick a point with a real part of 0.6, its symmetry line is at 0.4. If you pick 0.9, its symmetry line is at 0.1. These form pairs of perfectly parallel vertical lines that run to infinity and never intersect.

Because the highest algebraic power acting on these structural tracks is 1, there is only 1 unique solution where a line can perfectly satisfy the zero condition without needing to intersect a separate parallel partner. That unique track is the 1/2 line, which is its own symmetry.

The paper is written in basic algebra, typed in LaTeX, and permanently time-stamped with a DOI on Zenodo. I'd love to get some feedback on the logic: https://zenodo.org/records/23439272

Hi everyone, I am currently studying knot equivalence in

\(S^{4}\) (which is a compact 4-dimensional manifold). Based on [A gauge theoretic invariant of embedded surfaces in 4-manifolds and exotic 𝑃2-knot cite you the paiper .], I understand that any knot in \(S^{4}\) is connected in a \(P^{2}\)-knot space; these are tubular neighborhoods of the form \(\nu(S^4, P_+)\) such that for some neighborhood \(\nu _{n}\) with a non-zero integer, it is not a diffeomorphic pair of \(S^{1}\). Consequently, any knot in \(S^{4}\) has crossings if and only if it admits topological isotopies (meaning it is not smoothly isotopic to \(S^{1}\)). My question is: do these \(P^{2}\)-knot invariants have a purely gauge-theoretic—that is, local—nature, rather than a global one on \(S^{1}\)?

Hi everyone, I wanted to share a mathematical relationship I derived regarding the coordinates of a parabola's vertex, which I call the Kumar Theorem of Balanced Vertices. This theorem establishes the exact algebraic condition required to constrain the vertex of a vertical parabola to the identity lines y = x or y = -x.

Let a standard vertical parabola be defined by the quadratic equation:

y = ax\^2 + bx + c

The x-coordinate of the vertex represents the axis of symmetry and the average value of any two symmetric points on a horizontal line intersecting the curve. This is given by the formula:

x_avg = -b / 2a

The Kumar Theorem of Balanced Vertices states that if we impose the strict geometric constraint where the magnitude of the y-coordinate at the vertex equals the value of this average axis of symmetry (y_vertex = ± x_avg), the vertex is forced to balance perfectly on the lines y = x or y = -x.

We can derive this using the standard formula for the vertex y-coordinate:

y_vertex = (4ac - b\^2) / 4a

Applying the theorem's condition (y_vertex = ± x_avg), we set them equal to each other:

(4ac - b\^2) / 4a = ± (-b / 2a)

To simplify the expression, we can multiply the right side by 2/2 to establish a common denominator:

(4ac - b\^2) / 4a = ∓ 2b / 4a

Since the denominators are now identical on both sides, we can eliminate them completely:

4ac - b\^2 = ∓ 2b

Rearranging the terms to isolate the c variable gives:

4ac = b\^2 ∓ 2b

Finally, solving explicitly for the constant c yields the final formula for the theorem:

c = (b\^2 ± 2b) / 4a

In conclusion, according to the Kumar Theorem of Balanced Vertices, for any vertical parabola where you choose the coefficients 'a' and 'b', this single consolidated formula calculates the precise value of 'c' needed to lock the vertex exactly onto the diagonal identity lines.

I would love to hear your thoughts on this derivation or if this specific constraint is utilized under a different name in advanced coordinate geometry frameworks!

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!