What is f(x) ≤ g(x) + O(1)? Inequalities With Asymptotics

(jamesoswald.dev)

24 points | by ibobev 3 days ago

7 comments

  • qsort 1 hour ago
    I think the confusion is because strictly speaking $f(x) = O(g(x))$ is an abuse of notation. $O(g(n)), \Theta(g(n))$ and friends are sets. We can't say that a function equals a set, or that a function "is less" than another function, but notoriously mathematics runs on javascript, so we try to do something instead of giving a type error.

    Here "is less" is interpreted as "eventually less for all values" and "plus a set" is interpreted as "plus any function of that set".

    I never liked this notation for asymptotics and I always preferred the $f(x) \in O(g(x))$ style, but it's just notation in the end.

    • sfpotter 10 minutes ago
      The reason it's preferred to use "=" instead of "\in" is because the way that Landau notation is generally used in practice is as a kind of ellipsis or placeholder. For example, the Taylor expansion e^x = 1 + x + O(x^2). I could just as well write e^x = 1 + x + ..., but the former conveys more meaning about what is hidden behind the ellipsis. It's an abuse of notation, but in the contexts that it's used, it's not clear what additional clarity using "\in" would bring over "=". Maybe also that big O is mainly used as a notation to facilitate doing calculations, less describing what family a function belongs to. Here are Knuth's thoughts, which I agree with: https://micromath.wordpress.com/2008/04/14/donald-knuth-calc...
    • ndriscoll 1 hour ago
      To me it seems similar to the + C on an antiderivative (or more generally, quotient objects). Technically, you are dealing with an equivalence class of functions, so a set. But it's usually counterproductive to think of it that way (and when you study this stuff properly, one of the first things you do is prove that you (usually) don't need to, and can instead use an arbitrary representative as a stand-in for the set), so you write F(x)+C.
      • qsort 53 minutes ago
        I think the Landau notation is a bit more finicky with the details. When it's really a quotient (like modular arithmetic) I'm with you, but here $O()$ morally means "at most this" and often you have to use the "direction of the inequality" to prove complexity bounds, so I'm more comfortable with the set notation. But again, it's just notation, I could use either.
      • ijustlovemath 33 minutes ago
        Huh, never thought about the potential connection between the set-containment operation and Stokes like that.
        • ndriscoll 19 minutes ago
          It's actually a linear (more generally, abstract) algebra thing. (All, Differentiable, Smooth, or all sorts of other sets of) functions form a vector space. The derivative is a linear operator (generalized matrix). If you have a linear equation Ax=b, then if you can find some solution X, the general solution set is X+kerA, where kerA (the kernel or nullspace) is the set of all v where Av=0. What's the kernel of the derivative operator (i.e. what has 0 derivative)? Constant functions. So the general solution is whatever particular antiderivative you find plus any constant function.
    • hyperpape 1 hour ago
      Although, when I learned foundations of mathematics, every function was a set, and if you wanted them, you'd get plenty of junk theorems from that fact.
      • qsort 1 hour ago
        "Everything is an object" is for boys, "everything is the empty set composed with copies of itself via the axiom of pairing" is for men ;)
    • foxes 1 hour ago
      I feel its not that bad an abuse of notation as kinda consistent with other areas of mathematics -

      A coset, quotients r + I, affine subspaces v + W, etc. Not literal sets but comparing some representative with a class label, and the `=, +` is defined not just on the actual objects but on some other structure used to make some comparison too.

  • josalhor 1 hour ago
    > computer science students should be familiar with the standard f(x)=O(g(x)) notation

    I have always thought that expressing it like that instead of f(x) ∈ O(g(x)) is very confusing. I understand the desire to apply arithmetic notation of summation to represent the factors, but "concluding" this notation with equality, when it's not an equality... Is grounds for confusion.

    • FartyMcFarter 1 hour ago
      Given this possible confusion, is it still valid to say the following two expressions are equivalent as the article does?

      f(x) = g(x) + O(1)

      f(x) - g(x) = O(1)

  • dataflow 32 minutes ago
    Maybe an easier explanation: just subtract g(x) from both sides.

    You get:

      f(x) - g(x) ≤ O(1)
    
    Now, if you already know that

      f(x) - g(x) = O(1)
    
    means "f and g eventually differ by no more than a constant", then

      f(x) - g(x) ≤ O(1)
    
    must mean "f eventually stops exceeding g by a constant".
  • bo1024 42 minutes ago
    The easiest way to read it is "there exists a function h in O(1) such that f(x) <= g(x) + h(x)."

    I think first we should teach "f in O(g)" notation, then teach the above, then observe that a special case of the above is the "abuse of notation" f(x) = O(g(x)).

  • edflsafoiewq 53 minutes ago
    O(1) just means "a bounded function (on a neighborhood of infinity)". Unlike f(x), which refers to some function by name, O(1) refers to some function by a property it has. It's the same principle at work in "even + odd = odd".

    Programmers wringing their hands over the meaning of f(x)=O(g(x)) never seem to have manipulated any expression more complex than f(x)=O(g(x)).

  • jheriko 1 hour ago
    [dead]