"I can't draw" is almost (but not quite) as common a misconception as "I can't do #maths". Prove yourself wrong for one of these at Olivia's free drop-in #drawing workshop at @NPGLondon on 21st June:
A couple of weeks ago, I posted an #animation of a point on a circle generating a #cycloid.
If you turn the curve "upside down", you get the #BrachistochroneCurve. This curve provides the shortest travel time starting from one cusp to any other point on the curve for a ball rolling under uniform #gravity. It is always faster than the straight-line travel time.
Anyway, the #animation took a bit of thought as it requires a bit of #Mechanics, some #Integration and is made a bit more tricky as the curve is multi-valued and so you need to treat different branches separately. The #AnimatedGif was produce with #WxMaxima.
Here's something I just learned: the lucky numbers of Euler.
Euler's "lucky" numbers are positive integers n such that for all integers k with 1 ≤ k < n, the polynomial k² − k + n produces a prime number.
Leonhard Euler published the polynomial k² − k + 41 which produces prime numbers for all integer values of k from 1 to 40.
Only 6 lucky numbers of Euler exist, namely 2, 3, 5, 11, 17 and 41 (sequence A014556 in the OEIS).
The Heegner numbers 7, 11, 19, 43, 67, 163, yield prime generating functions of Euler's form for 2, 3, 5, 11, 17, 41; these latter numbers are called lucky numbers of Euler by F. Le Lionnais.
h/t John Carlos Baez
(@johncarlosbaez) for pointing this out.
The fascinating Heegner numbers [1] are so named for the amateur mathematician who proved Gauss' conjecture that the numbers {-1, -2, -3, -7, -11, -19, -43, -67,-163} are the only values of -d for which imaginary quadratic fields Q[√-d] are uniquely factorable into factors of the form a + b√-d (for a, b ∈ ℤ) (i.e., the field "splits" [2]). Today it is known that there are only nine Heegner numbers: -1, -2, -3, -7, -11, -19, -43, -67, and -163 [3].
Interestingly, the number 163 turns up in all kinds of surprising places, including the irrational constant e^{π√163} ≈ 262537412640768743.99999999999925... (≈ 2.6253741264×10^{17}), which is known as the Ramanujan Constant [4].
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
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.
Started listening to the audiobook edition of Everything Is Predictable. How Bayes' Remarkable Theorem Explains the World, written and read by Tom Chivers.
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?
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 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?