Early on in my hobby I came to the realization that cryptographic prowess has no viable market price point. More's the pity. Yet I think one day I may change that with my secrecy sauce.
"It seems to me now that mathematics is capable of an artistic excellence as great as that of any music, perhaps greater ; not because the pleasure it gives (although very pure) is comparable [...] to that of music [...]" – Bertrand Russell (1872–1970) #quote#mathematics#art#maths#math
Started listening to the audiobook edition of Everything Is Predictable. How Bayes' Remarkable Theorem Explains the World, written and read by Tom Chivers.
Il reprochait à #DidierRaoult des erreurs "niveau brevet des collèges" : accusé de diffamation, un prof de #maths relaxé
Attaqué en #justice par Didier Raoult pour diffamation et injure publique, Guillaume Limousin a été relaxé, mardi 14 mai [...]. Sur Twitter, ce professeur de mathématiques isérois, reprochait à l'infectiologue une série d'erreurs de "niveau brevet des collèges". Didier Raoult devra lui verser 2000€, au titre des frais de justice.
I just realized that all perfect squares mod 9 can only be 0, 1, 4, 7, but I can't find an easier proof than by exhaustion (square all numbers 0 to 8, mod 9). Is there a more elegant proof of this?
mod 11 has a wider choice (0, 1, 3, 4, 5, 9), but I wonder how good of a “perfect square detector” they can be together. Of course if either proof (by 9s and by 11s) fails, it's not a perfect square, but how many “not perfect square” are perfect squares both mod 9 and mod 11?
It's aimed at anyone who fits the aforementioned bill and is interesting in exploring the mathematical potential of their stories, objects and exhibitions, with participation in Maths Week Scotland in mind.
I have a question about the aperiodic spectre tile (or the hat/turtle).
I know that the proof of aperiodicity works by showing that the tiles must fit together in a hierarchical structure that eventually repeats itself at a larger scale. But the larger units aren't literally scaled copies of the spectre. I also know that there is some freedom as to how you draw the edges of the spectre.
Is there a way you can draw the edges that allows you to literally use spectres to cover a larger copy of themselves? If so, is this way of doing it unique?
(compared to python when it is forced to apply arbitrary functions with loops inside, element-wise to an array - that is, can't benefit from vectorised numpy functions)
this #maths experiment took about an hour in python and about 1 second in julia lang
sure my python isn't professional, but today was my first time with julia lang so that will be far from optimal either
"Mathematics must subdue the flights of our reason; they are the staff of the blind; no one can take a step without them; and to them and experience is due all that is certain in physics." – Voltaire (1694-1778) #quote#mathematics#math#maths
I bet that a lot of people in the Fediverse already know this very pretty pencil-based 3D art. But in case you haven’t, be prepared to marvel.
This sculpture is known as the hexastix and a variant series created by artist George Hart is titled 72 Pencils.
If you can get 72 unsharpened hexagonal pencils, and some flat rubber bands, you can attempt to create this. Search for a video by @standupmaths for a pseudo-tutorial.
"Numbers are free creations of the human mind, they serve as a means of apprehending more easily and more sharply the diversity of things." – Richard Dedekind (1831-1916) #quote#mathematics#math#maths#numbers
Imagine a circular wheel rolling, without skidding, on a flat, horizontal surface. The #locus of any given point on its #circumference is called a #cycloid. It is a #periodic#curve with #period over the #circle's circumference and has #cusps whenever the point is in contact with the surface (the two sides of the curve are tangentially vertical at that point).
Interestingly, it is also the curve that solves the #Brachistochrone problem, which means that starting at a cusp on the inverted curve (maximum height), a frictionless ball will roll under uniform gravity in minimum time from the start to any other point on the curve, even beating the straight line path.