0
$\begingroup$

I have been disliking the theory of surreal numbers for a while, but let's test it.

So, we have a set of divergent improper integrals of continuous functions with the following ordering: $\int_0^\infty f(x)dx\ge\int_0^\infty g(x)dx$ if $\int_0^s (f(x)-g(x))dx\ge0$ for all $s$ greater than some large $S$. And all convergent integrals are considered equal to their sums.

Using surreal numbers can you give representations of such set of divergent integrals? Particularly, I wonder, what growth rate or divergent integral their $\omega$ corresponds to.

$\endgroup$
4
  • $\begingroup$ Do you identify divergent integrals with the corresponding functions $s \mapsto {\int_a}^s f(x)d x$? If that's the case, then you're looking at a way to represent certain real-valued functions as surreal numbers? Which ones exactly, every continuous real-valued function? $\endgroup$
    – nombre
    Jan 1, 2021 at 20:00
  • $\begingroup$ @nombre, well, yes, the growth rates. Say, the functions are monotonic and continuous. $\endgroup$
    – Anixx
    Jan 1, 2021 at 20:06
  • 1
    $\begingroup$ But convergent integrals are not identified with growth rates, but rather with the limit of said integral, is that it? In any case, surreal numbers form an ordered field, so if you want the representation to preserve the order and the arithmetic, then you must first select an ordered field of such growth rates. $\endgroup$
    – nombre
    Jan 1, 2021 at 20:50
  • $\begingroup$ Well, if you pick an infinite hyperreal number $N$, then comparing the hyperreals $\int_a^N f(x)\,dx$ is a way to compare growth rates, and if the integral converges then $\int_a^N f(x)\,dx$ is infinitely close to its value (so to make them equal, you could quotient the ring of appreciable hyperreals by the ideal of infinitesimals). And since the surreals are the universally embedding ordered field, any field of hyperreals can be embedded in the surreals. Are you looking for something more "natively surreal" than this? $\endgroup$ Jan 23, 2021 at 16:12

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.