Refurio Anachro

@RAnachro

Higher maths is cool – come and see invisible worlds with me!

Vrijeme pridruživanja: ožujak 2019.

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  1. prije 23 sata

    We can write this like so: C(a✕b,c) <=> C(a,c^b) where L = _✕b, and R = _^b. What I just said I got straight from Brendan Fong from this cool lecture where he and each give a complete lecture about adjunctions! 3/end

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  2. prije 23 sata

    The <=> means that there's a pair of arrows, one in each direction, that these two hom-sets are isomorphic. The first example given is currying: a function in two arguments is equivalent to a function of one argument returning another function in the remaining argument. 2/n

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  3. prije 23 sata

    An adjunction is a way to relate two objects a and c, but not directly: C(a,c) Instead, we get two maps, (L)eft and (R)ight, which allow to relate them: C(L(a),c) <=> A(a,R(c)) 1/n

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  4. proslijedio/la je Tweet
    30. sij
    Odgovor korisnicima

    I did some more experiments, and I think I know how to color by exit angle. When we naively color by exit angle, we see the colors repeated 2^i times for the band of points that escape after i iterations. Put differently, the i'th band's color index is folded by frac(2^i*x).

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  5. proslijedio/la je Tweet
    29. sij

    I’m back! Having a busy month moving and settling in , but this hasn’t totally kept me away from making cool pictures :) Here’s a depiction of an Anosov mapping class on a torus 1/n

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  6. 28. sij

    For clarification, I'm talking about 1/q, similar to A7237 (which only looks at squares q=n²), versus p/1, for which the series seems to vanish for p>12. Of course, we could also look at p/q in general, but that has two parameters, and wouldn't yield a nice 1d sequence.

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  7. 28. sij

    Herroneous! Heronian. Yours, Heroniac

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  8. 28. sij

    When p/q < 1 you get a ever more matches per run, otherwise there are only a few. 12/1 seems to give the last example, so it's all zeros from there on: 5 3 2 1 2 1 0 1 0 0 0 1 ... 4/4 (n=n, 4=4, etc.)

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  9. 28. sij

    But ..., shouldn't we also include square roots of integers? After all, computing an area requires taking a square root, so we should compare p*A² with q*P²! 3/4

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  10. 28. sij

    How many triangles with integer sides exist whose area is exactly twice their perimeter? What about n times? Of course, there's an entry in the online encyclopedia of integer sequences for that: 2/4

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  11. 28. sij

    There are 5 Heronic triangles whose area equals their perimeter. Of all things, five! Since area grows quadratically, and the perimeter only linearly, it's clear that there can only be finitely many such triangles. 1/4

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  12. 27. sij

    ... Shannon entropy! This seems to be a feature of operad algebras in general, a fact Tom apologizes or it to be not very intersting! The main result can be brought down to an axiomatic description of information-losing processes modelled as functions. Linearity.

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  13. 27. sij

    Tom describes operads, and gives simple examples (constant, monoidal, polynomials), and a probabilistic one. He then describes operad algebras, interconnecting operads. For probabilistic operads it leads to ...

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  14. proslijedio/la je Tweet
    27. sij

    The Mandelbrot set approximated by the 4096 first Fourier coefficients of its boundary (it would be nice to superimpose over a classical view but I don't know how to get Sage to align its coordinates exactly):

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  15. 24. sij
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  16. proslijedio/la je Tweet
    24. sij
    Odgovor korisnicima

    I haven't found out what a free critical orbit is, but I got a rabbit newton! Please ignore the shadows, that's due to the weird sign based coloring. It's based on gnofract4d's Mandel with this after "loop:" z = z - (sqr(sqr(z) + ) + )/(4*z*sqr(z) + 4*z*)

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  17. 23. sij

    That's it for now. Thanks again for the beautiful lecture series! I've picked on , because I tend to need to educate colleagues about programming in Ocaml. I have no students. I'm not even an academic. And I know even less amout the maths side :)

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  18. 23. sij

    Anyways, 'reasoning' means that the compiler can infer stuff. To improve the programming language. We want the compiler to do things for us! Therefore the type layer can't be another Turing complete language, because that would make automatic reasoning about it impossible.

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  19. 23. sij

    These two layers are really independent! Languages for computation need it the most, but logicians and model theorists seem like them, too.

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  20. 23. sij

    (2) There's two languages! Due to Gödel we know that we can't have a language that can reason about itself. Hence we need two languages: one about types, the other about computations. The type layer is for reasoning about the computation layer.

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