Fifteen Most Famous Transcendental Numbers

(physics.wisc.edu)

35 points | by vismit2000 2 hours ago

5 comments

  • brianberns 1 hour ago
    I read this with pleasure, right up until the bit about the ants. Then I saw the note from myself at the end, which I had totally forgot writing seven years ago. I probably first encountered the article via HN back then as well. Thanks for publishing my thoughts!
  • nuancebydefault 51 minutes ago
    I would have expected more numbers originating from physics, like Reynolds number (bad example since it is not really constant though).

    The human-invented ones seem to be just a grasp of dozens man can come up with.

    i to the power of i is one I never heard of but is fascinating though!

  • barishnamazov 1 hour ago
    Don't want to be "that guy," but Euler's constant and Catalan's constant aren't proven to be transcendental yet.

    For context, a number is transcendental if it's not the root of any non-zero polynomial with rational coefficients. Essentially, it means the number cannot be constructed using a finite combination of integers and standard algebraic operations (addition, subtraction, multiplication, division, and integer roots). sqrt(2) is irrational but algebraic (it solves x^2 - 2 = 0); pi is transcendental.

    The reason we haven't been able to prove this for constants like Euler-Mascheroni (gamma) is that we currently lack the tools to even prove they are irrational. With numbers like e or pi, we found infinite series or continued fraction representations that allowed us to prove they cannot be expressed as a ratio of two integers.

    With gamma, we have no such "hook." It appears in many places (harmonics, gamma function derivatives), but we haven't found a relationship that forces a contradiction if we assume it is algebraic. For all we know right now, gamma could technically be a rational fraction with a denominator larger than the number of atoms in the universe, though most mathematicians would bet the house against it.

  • senfiaj 1 hour ago
    > Euler's constant, gamma = 0.577215 ... = lim n -> infinity > (1 + 1/2 + 1/3 + 1/4 + ... + 1/n - ln(n)) (Not proven to be transcendental, but generally believed to be by mathematicians.)

    So why bring some numbers here as transcendental if not proven?

    • auggierose 1 hour ago
      Because it still might be transcendental. Just because you don't know if the list is correct, doesn't mean it isn't.
      • loloquwowndueo 1 hour ago
        So it’s like “15 oldest actors to win an Oscar” and including someone who’s nominated this year but hasn’t actually won. But he might, right?

        No, my dudes. Just no. If it’s not proven transcendental, it’s not to be considered such.

        • chvid 32 minutes ago
          I think the Oscars should go to the algebraic numbers - think about it - they are far less common ...
  • zkmon 1 hour ago
    If a number system has a transcendental number as its base, would these numbers still be called transcendental in that number system?
    • moefh 1 hour ago
      Yes. A number is transcendental if it's not the root of a polynomial with integer coefficients; that's completely independent of how you represent it.
    • frutiger 1 hour ago
      I think the elements of the base need to be enumerable (proof needed but it feels natural), and transcendental numbers are not enumerable (proof also needed).
      • tocs3 7 minutes ago
      • JadeNB 5 minutes ago
        I think your parent comment was speaking of a "base-$\alpha$ representation", where $\alpha$ is a single transcendental number—no concerns about countability, though one must be quite careful about the "digits" in this base.

        (I'm not sure what "the elements of the base need to be enumerable" means—usually, as above, one speaks of a single base; while mixed-radix systems exist, the usual definition still has only one base per position, and only countably many positions. But the proof of countability of transcendental numbers is easy, since each is a root of a polynomial over $\mathbb Q$, there are only countably many such polynomials, and every polynomial has only finitely many roots.)

      • kinkyusa 1 hour ago
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    • kinkyusa 1 hour ago
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