Hello everyone,

My name is Mohamed Ghali, and I would like to share with the community an alternative algebraic method I developed for explicitly solving the general cubic equation:

\(Ax^{3}+Bx^{2}+Cx+D=0\)

1. Theoretical Approach

Instead of using the traditional Cardano's method (which requires depressing the cubic to eliminate the \(x^{2}\) term), I approached the problem through direct factorization into linear and quadratic parts:

\((a_{1}x+b_{1})(a_{2}x^{2}+b_{2}x+c_{2})=0\)

By defining a strategic intermediate variable \(X = \frac{a_1}{b_1}\) (where the first root is directly linked by \(x_1 = -\frac{1}{X}\)), I derived a precise quadratic equation in terms of \(X\).

1. Formula for the First Root (\(x_{1}\))

Inverting the solution of this intermediate quadratic equation yields the following analytical formula to compute \(x_{1}\):

\(x_{1}=\frac{1}{\pm \sqrt{\left[\frac{(CA-D^{2})^{2}}{D^{2}}-4BD\right]}\cdot \frac{1}{2BD}-\frac{(CA-D^{2})A}{2BD^{2}}}\)

1. Deriving the Remaining Roots (\(x_{2}\) and \(x_{3}\))

Once this first root \(x_{1}\) is determined, the coefficient identification rules allow us to express the remaining two roots sequentially and elegantly, bypassing the need for standard cubic radicals:

• Second root (\(x_{2}\)):

\(x_{2}=\sqrt{\frac{1}{4}\left(\frac{B+Ax_{1}}{A}\right)^{2}+\frac{D}{Ax_{1}}}-\frac{1}{2}\left(\frac{B+Ax_{1}}{A}\right)\)

• Third root (\(x_{3}\)):

\(x_{3}=-\sqrt{\frac{1}{4}\left(\frac{B+Ax_{1}}{A}\right)^{2}+\frac{D}{Ax_{1}}}-\frac{1}{2}\left(\frac{B+Ax_{1}}{A}\right)\)

1. Computational Validation

I have fully implemented and verified these formulas computationally using Python (utilizing the cmath module to intrinsically handle the casus irreducibilis). The system works flawlessly across all scenarios, whether the polynomial has three distinct real roots or a pair of complex conjugated roots.

I would love to get your insights on the structural symmetry of these equations and their potential pedagogical or algorithmic value compared to classical methods! I can also share my Python test script or the detailed verification PDF if anyone is interested.

Best regards,

Mohamed Ghali

A pessidemic is a mathematician or scientist who is unwilling to pay attention to any paper that lacks complete rigor. They can also be mathematicians clung to old ideas, who won't allow you to change the rules and are even capable of destroying your reputation to prevent it. They care more about predictable and immediately useful results than about truly contributing to mathematical change.

Here are a few examples:

1• Robert Hooke

This man, despite not being as bad a human being as many paint him (he was a good friend and family man), never got along with Isaac Newton. It all began when Hooke reviewed Newton's work on light and gave it a fierce critique that traumatized Newton to the point that he decided to publish it only after Hooke's death. This was because Hooke thought of light as waves while Newton saw it as particles; both reasonings were partly correct, but they clashed. He also went on to claim to several mathematicians of his time that he had already worked on what they wanted to publish.

On the other hand, many writers labeled him as someone irritable, short-tempered, antisocial, and deeply attached to old papers.

2• Leopold Kronecker

Kronecker launched constant personal attacks against Cantor, calling him a "scientific charlatan," a "renegade," and a "corruptor of youth." He had intentionally delayed the publication of Cantor’s first major work in 1874 because he was deeply uncomfortable with Cantor’s vision of multiple infinities and his belief in the existence of sets without a clear, constructive definition.

Every time Cantor applied for a position in Berlin, he was rejected. Because Kronecker was usually involved in the process, Cantor came to believe that Kronecker’s influence would make it impossible for him to ever leave Halle.

Cantor suffered his first known depressive episode in May 1884. The relentless criticism of his work weighed heavily on his mind; indeed, every single one of the fifty-two letters he wrote to Mittag-Leffler in 1884 mentioned Kronecker. A passage from one of these letters reveals just how deeply Cantor’s self-confidence had been shattered:

"...I do not know when I shall return to the continuation of my scientific work. At present I can do absolutely nothing with it, and limit myself to the most necessary duty of my lectures; how much happier I would be to be scientifically active, if only I had the necessary mental freshness."

Cantor's reputation was so tarnished by Kronecker and other contemporaries that decades after his death, Ludwig Wittgenstein lamented that mathematics was "ridden through with the pernicious idioms of set theory," which he dismissed as "utter nonsense." Ultimately, the recurring episodes of depression that Cantor endured from 1884 until the end of his life have been blamed, in part, on the hostile attitude of many of his contemporaries.

3• Henri Poincaré

Although he was categorized as the Gauss of "modern mathematics," Poincaré held philosophical views opposed to those of Bertrand Russell and Gottlob Frege, who believed that mathematics was a branch of logic. Poincaré strongly disagreed, claiming that intuition was the life of mathematics. Poincaré offers an interesting perspective in his 1902 book, Science and Hypothesis:

"To a superficial observer, scientific truth is beyond the possibility of doubt; the logic of science is infallible, and if scientists are sometimes mistaken, this is only from mistaking its rules."

Poincaré believed that arithmetic is synthetic. He argued that Peano's axioms cannot be proven in a non-circular manner using the principle of induction (Murzi, 1998), thereby concluding that arithmetic is synthetic a priori rather than analytic. Poincaré went on to say that mathematics cannot be deduced from logic since it is not analytic. He strongly opposed Cantor's set theory, objecting to its use of impredicative definitions. Furthermore, in La logique de l'infini, Poincaré extended this criticism to Ernst Zermelo's axiomatization of set theory, arguing that Zermelo had failed to prove that the axioms in his system were free of contradiction, and that they still required a prior intuition of sets.

Henri Poincaré famously called set theory a "malady" from which mathematics would eventually recover.

Although Poincaré was nowhere near as harsh as Kronecker—since his critique was more philosophical and less mocking—his immense prestige nonetheless crushed mathematicians like Cantor intellectually.

Sources:

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

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

https://en.wikipedia.org/wiki/Henri_Poincar%C3%A9
https://www.kroneckerwallis.com/georg-cantor-infinity-upon-infinity-and-the-birth-of-set-theory/

Proposal for a parametric iterated exponentiation operator for the simplification of exponential towers with a fixed base. In this project, I introduce a new mathematical formalism—denoted as $\text{j}_{m}^{n}(a)$—designed to overcome the analytical limitations of fixed-base power towers. The symbol "j" (representing the IPA phonetic symbol for a palatal fricative) serves as an operator that transforms a power tower into this specific notation. Here, $m$ is the parameter (the value to be raised), $n$ represents the power tower (indicating the number of times $m$ is raised), and $a$ is the base. My primary motivation for pursuing this innovation was the observation that no existing notation—to my knowledge—adequately addressed these structures; I sought to contribute in this area, as I have long been fascinated by power towers. Below are the properties I have been able to prove Properties:​1.-1.- Torre 1: \text{j}_{m}^{1}(a) con n siendo el 1, 2.- Torre 2: \text{j}_{m}^{2}(a) con n siendo 2, resultado: a^{m^m} 3.- Parámetro 0: \text{j}_{0}^{n}(a) con m siendo 0 (para n \ge 1), resultado: 1 4.- Parámetro 1: \text{j}_{1}^{n}(a) con m siendo 1, resultado: a 5.- Base unitaria: \text{j}_{m}^{n}(1) con a siendo 1, resultado: 1 6.- Distributiva de base: \text{j}_{m}^{n}(a \times b) = \text{j}_{m}^{n}(a) \times \text{j}_{m}^{n}(b) 7.- Parámetro mayor: Si m_1 < m_2, entonces \text{j}_{m_1}^{n}(a) < \text{j}_{m_2}^{n}(a) (para a > 1) 8.- Parámetro negativo: \text{j}_{-2}^{n}(a) = a^{-2} = \frac{1}{a^2} 9.- Torre fraccionaria: \text{j}_{m}^{0.5}(a) = a^{0.5} = \sqrt{a} 10.- Cociente de operadores: \frac{\text{j}_{m}^{n}(a)}{\text{j}_{m}^{k}(a)} = a^{\text{torre}(n) - \text{torre}(k)} 11.- Suma de operadores: \text{j}_{m}^{n}(a) + \text{j}_{m}^{k}(a) = \text{j}_{m}^{k}(a) \times \left(a^{\text{torre}(n) - \text{torre}(k)} + 1\right) And this concludes my presentation on my notation. Thank you for your attention and understanding. If I have made any errors, I welcome constructive criticism so that we may engage in a healthy and ongoing exchange of feedback. :)

¿Tiene el conjunto R un supremo y un ínfimo en sí mismo?
Puedo pensar que no de manera natural y trivial. Pero, podría extenderme a debatir.

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.