1

We use the notation $\mathbb N^{\gt 0} = \{1,2,\dots,n,\dots\}$.

If $m,n \in \mathbb N^{\gt 0}$ we can always apply Euclidean division to get a quotient - if $m \ge n$ we can call $m$ the dividend and $n$ the divisor, and if $m \lt n$ we can call $n$ the dividend and $m$ the divisor. This is a commutative binary operation, $\mathsf {EC}(m,n)$.

Examples: $\mathsf {EC}(3,5) = 1$,$\;\mathsf {EC}(11,11) = 1$ and $\mathsf {EC}(2,7) = 3$.

A mapping $f: \mathbb N^{\gt 0} \to \mathbb N^{\gt 0} $ is said to have $+\infty$ as a limit if for every $M \in \mathbb N^{\gt 0}$ there exist $N \in \mathbb N^{\gt 0}$ such that for every $n \ge N$ the image $f(n)$ is greater than or equal to $M$.

Let $f$ and $g$ both have $+\infty$ as a limit. We can define other mappings for each $k \in \mathbb N^{\gt 0}$,

$\tag 1 k \times \mathsf {EC}(f,g): \; n \mapsto \text{Max}[\;\mathsf {EC}(kf(n),g(n)),\,\mathsf {EC}(f(n),kg(n))\;]$

Definition: Two mappings $f$ and $g$ are said to approach $+\infty$ at the same rate if for every $k \in \mathbb N^{\gt 0}$, $k \times \mathsf {EC}(f,g)$ is eventually constant and equal to $k$.

Example: $f(n) = n^2 + 100n + 10000$ and $g(n) = n^2$ approach $+\infty$ at the same rate.

Question 1: Did this concept appear in Greek_mathematics or, for that matter, have these definitions ever been used in the mathematical literature?

I find it interesting that we can get the concept of a limit at a foundational level - our universe of discourse is restricted to only the natural numbers.

Question 2: Are there known results in mathematics that can be developed and expressed (perhaps in a watered down fashion) from this primitive platform?

CopyPasteIt
  • 11,366
  • 1
    Unless I'm misunderstanding, your definition is equivalent to (when we restrict attention to functions which do in fact approach $+\infty$) the simpler expression $$\lim_{n\rightarrow\infty}{f(n)\over g(n)}=1,$$ that is "$f$ and $g$ have the same order." And this - and related notions - is very well-studied; see e.g. here. – Noah Schweber Nov 04 '18 at 14:38
  • Presentation-wise, is there any particular reason you're keeping everything inside the integers? Note that this leads to the notation being much more cumbersome, since you can't write "${x\over y}$." – Noah Schweber Nov 04 '18 at 14:46
  • @NoahSchweber I am interested in the concept of unity, ratios, and proportions at a foundational level so that I can develop the theory of magnitudes that I've been working on. The central thesis is that a considerable percent of mathematics is naturally exposed by starting with just a few axioms and asking questions. Of course at some point of this 'exercise' , and all in due time, you will define the modern definition of a limit and deduce the rules for taking the limit of a quotient. – CopyPasteIt Nov 04 '18 at 15:04
  • @NoahSchweber Also, how about transformations $f$ and $g$ of $\mathbb N^{\gt 0}$ that can't be easily extended to the real numbers with $ \lim_{n\rightarrow\infty}{f(n)\over g(n)}=1$? Can we just rule them out? – CopyPasteIt Nov 04 '18 at 15:13
  • When I wrote "$\lim_{n\rightarrow\infty}$," I meant that the limit was being taken over the natural numbers - that is, we're looking at the limit of the sequence $({f(n)\over g(n)})_{n\in\mathbb{N}}$ (with the understanding that finitely many of its terms might be undefined). So there's no need to extend to reals. – Noah Schweber Nov 04 '18 at 15:19

1 Answers1

1

Yes, this is a very important and well-understood notion.

Your definition is more concisely written as $$\lim_{n\rightarrow\infty}{f(n)\over g(n)}=1,$$ which in turn is generally abbreviated "$f\sim g$." Here I'm taking the limit over naturals, so $f$ and $g$ are understood as functions with domain $\mathbb{N}$; we can just as easily define $\sim$ for functions on the rationals, reals, complexes, or etc. Some basic properties are described here, and you may also be interested in other varieties of asymptotic comparison.

As to its history, I'm unaware of any treatment of growth rates in "ancient" mathematics, Greek or otherwise - so far as I know, the earliest such investigations occurred in the $1800$s in the context of analytic number theory (e.g. the prime number theorem is a result in asymptotic analysis) - but it's hard to prove a negative, and I could very easily be wrong.

Noah Schweber
  • 245,398
  • Thanks for the links! Can you imagine how 'cool' it would be to prove the prime number theorem starting with just under, say 8 axioms, within 50 pages of logical deductive reasoning? (don't ask me to try this!). – CopyPasteIt Nov 04 '18 at 15:40
  • 2
    @CopyPasteIt Honestly, it doesn't grab me personally. It's easy to bootstrap from very basic axioms to the useful, more intricate definitions and their properties, and at that point why not use them? The usual "Peano road" - start with an infinite well-ordering with no (nonzero) limit elements, and get to $\mathbb{R}$ (and well beyond) - is quite straightforward and parsimonious. So I don't really see the point of avoiding introducing concepts when we have the machinery to do so. But ultimately this comes down to aesthetics, and what one wants to get out of an approach. – Noah Schweber Nov 04 '18 at 15:46
  • (And actually, rather than "infinite well-ordering with no nonzero limit elements," it's arguably better to say "well-ordering with no nonzero limit elements and no maximal element - then we don't have to start with a notion of "(in)finite." But again, aesthetic preference.) – Noah Schweber Nov 04 '18 at 15:48
  • It would be synchronicity indeed if you did not see my post https://math.stackexchange.com/questions/2630654/proof-involving-bootstrapping/2979224#2979224 And I still can't define the word bootstrap! – CopyPasteIt Nov 04 '18 at 15:52
  • Your word "parsimonious" caught my attention, but I'll probably continue with this a bit longer... – CopyPasteIt Nov 04 '18 at 15:53
  • 1
    @CopyPasteIt I've always understood it (outside of contexts where it has a precise technical meaning) to refer to seemingly "lifting oneself up by one's bootstraps" - basically, taking a really really really weak-seeming assumption (or definition or etc.) and building up from it to something impressive, the idea being that it looks like you've gotten something out of nothing. E.g. a well-ordering per se doesn't come equipped with any algebraic structure at all - but we can get "successor" from it, and then define addition and multiplication recursively, and then get $\mathbb{Q}$, ... – Noah Schweber Nov 04 '18 at 15:53
  • 1
    and then just show that there is precisely one Dedekind-complete ordered field (up to isomorphism) and that $\mathbb{Q}$ sits inside it in a unique way, and boom - $\mathbb{R}$! (Note that this actually reveals a principle which starting with algebraic structure can obscure: the recursion theorem (in the set-theoretic, not computability-theoretic, sense), which is how we get addition and multiplication. ) – Noah Schweber Nov 04 '18 at 15:56
  • Yes, that is what I am working on (like Tarski) - boom $\mathbb R$! – CopyPasteIt Nov 04 '18 at 15:57
  • @CopyPasteIt Well, what's wrong with the already-existing way I've referred to above? (Nothing need be wrong with it of course, you might prefer following a different line for aesthetic reasons, I'm just curious.) – Noah Schweber Nov 04 '18 at 15:58
  • 2
    @CopyPasteIt Depending on what you're looking for, you'll either be very satisfied with or very dissatisfied with the Metamath proof of the prime number theorem – Mark S. Nov 05 '18 at 01:36
  • @MarkS. Way cool! – CopyPasteIt Nov 05 '18 at 01:50
  • @NoahSchweber Funny that you mentioned the recursion theorem. As I go along with the algebra, I find myself having to awkwardly address it. Good point! – CopyPasteIt Nov 05 '18 at 01:57