@user21820 Wow, I had no idea. All the definitions I have come across so far are of the forms shown here. Some omit the absolute value bars, but all treat o and ω as the "strict" counterparts of O and Ω and only consider values of n that are greater than a threshold value like n_0 or N.

@user21820 Sorry, I had forgotten to upvote it. I'm sorry to hear about the downvotes, that sounds like harassment. This post says that serial voting can be reversed.

@user51462 Indeed, sometimes some of these trolls' votes get reversed, but the majority of them don't. Ironically, almost all of these votes are on my posts on logic. There is strong evidence that most of these trolls hate logic, and their anti-logic hate is not surprising at all.

@user51462 They are 'strict' counterparts, but that doesn't mean the inequality sign should be strict too. Most of the time, it doesn't matter. But in special cases like I mentioned, mine is better.

@user21820 Thanks for letting me know. I was not able to fully understand the justification you gave here, but I will add your definitions to my notes.

@user21820 Are you able to recommend any discrete maths books that are easier and more systematic than Concrete Maths?

@user51462 There's nothing we can do about trolls who downvote sparsely for no good reason and actual mathematicians who downvote wrong posts, because moderators (or SE staff) are not supposed to judge correctness. Anyway, let's not bother about this issue anymore.

@user51462 The point is that if x = 0 then of course ( x∈o(1) as t → c ), regardless of what t is. Nobody disputes this. But if x = 0 then ( exp(x) ∈ 1+x+o(x^2) as t → c ) is equivalent to ( exp(0) ∈ 1+0+o(0^2) as t → c ), which is equivalent to ( 0 ∈ o(0) as t → c ). If you reject this, then you must reject the theorem ( x∈o(1) as t → c ) ⇒ ( exp(x) ∈ 1+x+o(x^2) as t → c ), which is very useful!

We cannot just exclude special cases like x = 0, because there are an unlimited number of others. For example, if x = t·sin(1/t) when t ≠ 0, then ( x∈o(1) as t → 0 ), but ( exp(x) ∈ 1+x+o(x^2) as t → 0 ) is equivalent to ( exp(t·sin(1/t)) ∈ 1+t·sin(1/t)+o((t·sin(1/t))^2) as t → 0 ), which can be true only if you use my definition of little-o, because t·sin(1/t) = 0 infinitely many times as t → 0.

@user51462 Hmm unfortunately no. But just finish FOL+PA+ST per my exercises and you would be capable of learning practically everything in mathematics. I can guarantee you that... Because in the end what matters is that you have a fixed foundational system that you know how to use for all mathematics, which then allows you to mentally translate any mathematics that you read/hear into a 100% precise form that doesn't rely on unreliable intuition.

In any case, you can ask anything about Knuth's books in the Basic Math room.

So you can very well work through it concurrently with my course.