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Prithu biswas
Prithu biswas
12:20 PM
@user21820 Well , I actually didn't understood much of your last message.I also don't know why I am struggling so much in understanding what I am even doing.
Maybe I haven't learned the basics of AA (Asymptotic Analysis) properly yet. Maybe I should learn it thoroughly before even trying to prove lim(1/f)=1/lim(f).
I know that you have spent a lot of time teaching me AA , but it seems like I understood only 10% of what you said in our past messages. If I understood all of it , then I wouldn't struggle at all.
When I am learning something (non rigorous or rigorous) , I try to understand all of the rules behind it. If I don't understand the rules behind it , then I try my best to reverse engineer the rules myself.
This often drives me insane. It is kind of like playing chess without knowing the rules.
To me , it seems like AA is a intuitive method to find a rigorous proof:
Asymptotic Proof ⇒ [Some process] ⇒ Non-Asymptotic proof
But I can't seem to find all of the rules to apply AA properly.
Here is what I have noted so far about AA from our previous conversation: 
(*) Here "As d∈ℝ → 0" means "As real d eventually stays close enough to but different from 0".
(*) And "≈" means "eventually stays close enough to" (allows equality).
(*) And "o(E)" means "a set of values that for every c∈ℝ[>0] eventually has absolute value staying less than c·abs(E)".
(This works for ℂ-valued limits as well, but you can ignore that for now.)
(*) Addition between subsets of ℝ is what you expect: S+T = { x+y : x∈S ∧ y∈T } for each S,T⊆ℝ.
(*) Note that "→" and "≈" and "o(...)" are only meaningful under the "As d∈ℝ→0".
But It seems like I can't really apply AA with 100% confidence at all.Maybe because I haven't learned the axioms of AA yet for both o(x) and O(x).
Or maybe I am trying to learn too much too quickly , so my knowledge about this is foggy.
I hope you are not upset about me. I really love the way you teach , but this time I can't seem to learn the same rate you are teaching about AA.
In Short: I want to start all over again from the basics of AA.
 
Prithu biswas
Prithu biswas
12:39 PM
Note : I want to set aside lim(fg)=lim(f)+lim(g) and lim(1/f)=1/lim(f) for now.
And I have also read the Non Asymptotic Proof
Which is quite transparent because of the ST notation. Especially the B(r) set =)
 
MaxH
MaxH
(Q1) ¬∀x∈S ( P(x) ) ⇒ ∃x∈S ( ¬P(x) ).

	If not for all x in S (P(x)):
		If not (there exists x in S (not P(x))):
			Given y in S: ( I cant write "Given x in S here, right? Meaning, I have to use a new variable.)
				y in S.
				If not P(y):
					there exists x in S (not P(y)).
					not (there exists x in S (not P(x))).
					contradiction.
				P(y).
			For all x in S (P(x)).
			not (for all x in S (P(x)).
			contradiction.
		there exists x in S (not P(x)).
	not (for all x in S(P(x))) implies there exists x in S (not P(x)).
(Q2) ¬∃x∈S ( P(x) ) ⇒ ∀x∈S ( ¬P(x) ).

	If not (exists x in S (P(x)):
		If not (for all x in S (not P(x))):
			given y in S:
				If P(y):
					exists y in S (P(y)).
					exists x in S (P(x)).
					contradiction.
				not P(y).
			for all y in S (not P(y)).
		contradiction.
	for all x in S (not P(x)).
@user21820 I tried those two so far, are they correct?
 
 
9 hours later…
user21820
user21820
10:10 PM
@Prithubiswas Yes, it's true that I didn't give you a complete axiomatization for asymptotic analysis. It's kind of troublesome to have to list all the basic properties out; people who use it simply do it via intuition. Also, we would of course need to know the asymptotic expansions for every function involved in an expression.
@Prithubiswas In theory, all you need to be able to do is this, so it's good that you find it transparent.
But in practice, asymptotic analysis is always much easier to use, as you can see from the links on my profile under "Asymptotic expansions".
@Prithubiswas If abs(x)<min(1/2,ε), then abs(1/(1+x)−1) = abs(x)/abs(1+x) ≤ abs(x)/(abs(1)−abs(x)) < ε/(1−1/2). I'm pretty sure I told you this before. So proving that limit property doesn't need any asymptotic analysis.
@MaxH No. Read the definition of "used variable" properly. You can do "Given x in S:" at line 3. Also, your ∀intro step is wrong; follow the rules.
@MaxH Wrong again, because you do not have a contradiction. You're not following the rules at all.
 
MaxH
MaxH
11:05 PM
(Q1) ¬∀x∈S ( P(x) ) ⇒ ∃x∈S ( ¬P(x) ).

	If not for all x in S (P(x)):
		If not (there exists x in S (not P(x))):
			Given y in S:
				y in S.
				If not P(y):
					there exists x in S (not P(y)).
					not (there exists x in S (not P(x))).
					contradiction.
				P(y).
			For all y in S (P(y)).
			For all x in S (P(x)).
			not (for all x in S (P(x)).
			contradiction.
		there exists x in S (not P(x)).
	not (for all x in S(P(x))) implies there exists x in S (not P(x)).
(Q2) ¬∃x∈S ( P(x) ) ⇒ ∀x∈S ( ¬P(x) ).

	If not (exists x in S (P(x)):
		If not (for all x in S (not P(x))):
			given y in S:
				If P(y):
					exists y in S (P(y)).
					exists x in S (P(x)).
					not (exists x in S (P(x)).
					contradiction.
				not P(y).
			for all y in S (not P(y)).
			for all x in S (not P(x)).
			not (for all x in S (not P(x))).
			contradiction.
		not (not (for all x in S (not P(x))).
		for all x in S (not P(x))).
	not (exists x in S (P(x)) implies (for all x in S ( not P(x))).
@user21820 Would those be fixes?
 

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