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Joined 2 years ago
Cake day: June 12th, 2023
  • Kogasa@programming.devtoScience Memes@mander.xyz•ManifoldsEnglish
    27·
    1 month ago

    Manifolds and differential forms are foundational concepts of differential topology, and connections are a foundational concept of differential geometry. They are mathematical building blocks used in modern physics, essentially enabling the transfer of multivariable calculus to arbitrary curved surfaces (without relying on an explicit embedding into Euclidean space). I think the joke is that physics students don’t typically learn the details of these building blocks, rather just the relevant results, and get confused when they’re emphasized.

    For a tl;dr about the concepts mentioned:

    • A manifold is a curve, surface, or higher-dimensional object which locally resembles Euclidean space around each point (e.g. the surface of a sphere is a 2D manifold; tiny person standing on a big sphere perceives the area around them to resemble a flat 2D plane).

    • Differential forms are “things that can be integrated over a manifold of the corresponding dimension.” In ordinary calculus of 1 variable, that’s a suitably regular function (e.g. a continuous function), and we view such a function f(x) as a differential form by writing it as “f(x) dx.”

    • A connection is a way of translating local tangent vectors from one point on a manifold to another in a parallel manner, i.e. literally connecting the local geometries of different points on the manifold. The existence of a connection on a manifold enables one to reason consistently about geometric concepts on the whole manifold.

  • Kogasa@programming.devtoProgrammer Humor@programming.dev•*Permanently Deleted*
    15·
    2 months ago

    Let f(x) = 1/((x-1)^(2)). Given an integer n, compute the nth derivative of f as f^((n))(x) = (-1)(n)(n+1)!/((x-1)(n+2)), which lets us write f as the Taylor series about x=0 whose nth coefficient is f^((n))(0)/n! = (-1)^(-2)(n+1)!/n! = n+1. We now compute the nth coefficient with a simple recursion. To show this process works, we make an inductive argument: the 0th coefficient is f(0) = 1, and the nth coefficient is (f(x) - (1 + 2x + 3x^(2) + … + nx(n-1)))/x(n) evaluated at x=0. Note that each coefficient appearing in the previous expression is an integer between 0 and n, so by inductive hypothesis we can represent it by incrementing 0 repeatedly. Unfortunately, the expression we’ve written isn’t well-defined at x=0 since we can’t divide by 0, but as we’d expect, the limit as x->0 is defined and equal to n+1 (exercise: prove this). To compute the limit, we can evaluate at a sufficiently small value of x and argue by monotonicity or squeezing that n+1 is the nearest integer. (exercise: determine an upper bound for |x| that makes this argument work and fill in the details). Finally, evaluate our expression at the appropriate value of x for each k from 1 to n, using each result to compute the next, until we are able to write each coefficient. Evaluate one more time and conclude by rounding to the value of n+1. This increments n.