En algún nivel de agregación de la materia, para un observador con el intelecto y los instrumentos de medición adecuados, yo soy (por extensión toda mi genialidad y perversión) indistinguible de una máquina, un objeto, un proceso inanimado (sin alma).
In this era of artificial intelligence, it is becoming urgent to distinguish human art from what machines produce. There is an ontological difference, even before an aesthetic one, between art and what a machine can generate through statistical calculation based on millions of images created by others. Algorithms lack the spark of humanity. For this reason, the Church wishes to renew an alliance with artists and cultural institutions to safeguard our humanity.
Today we’re introducing Gemini 4 Argon.
It delivers frontier performance in complex workflows across real-world software engineering, knowledge work, and cybersecurity defense with an industry-leading 1M token output limit.
We’re sharing a solution to the Navier-Stokes Millennium Prize Problem, one of the deepest problems at the frontier of mathematics.
The proof was produced by a group of agents, using an OpenAI next-generation model significantly more capable than GPT-6 Astra.
The problem concerns whether the description of smooth three-dimensional fluid motion modeled by the Navier-Stokes equations can break down. It has remained unresolved for roughly 90 years.
GPT-6 Astra has the best result yet on the Bach Benchmark. Its chorale contains no voice-leading errors, and its harmonic palette is sophisticated enough to include a Neapolitan sixth chord. More importantly, it is the first model to ever write passing tones on this benchmark, a significant leap in musical understanding.
Extra High effort. Prompt below:
“In LilyPond (version 2.24), write a 4-part chorale in the style of Bach, 3/4 time, G minor. Use two staves -- soprano and alto on the top staff, tenor and bass on the bottom staff. Respond with only the code block.”
hello there the jacobian conjecture is false thanx to my close friend akhil for asking about it and my other close friend fable for working during the world cup final
((1+xy)^3 z + y^2 (1+xy) (4+3xy), y + 3 x (1+xy)^2 z + 3 x y^2 (4+3xy), 2 x - 3 x^2 y - x^3 z): \C^3\to \C^3, has jacobian determinant -2, and sends (0, 0, -1/4), (1, -3/2, 13/2), and (-1, 3/2, 13/2) to (-1/4, 0, 0)