Spirituality
19 Mar 07
22 Mar 07
1 edit
Originally posted by DoctorScribbles\lim_{n\to\infty}-\frac{1}{n}=0 and -\frac{1}{n} oscillates.
The limit you describe does not exist, because the function n(x)-(x/3) oscillates; it does not grow arbitrarily close to 0 as x increases.
and how do you know that it won't go to 0?
this is a serious question not some kind of a shot at you.
22 Mar 07
2 edits
Originally posted by adam warlockI think you have to use the Schnirelmann density, not asymptotic density.
\lim_{n\to\infty}-\frac{1}{n}=0 and -\frac{1}{n} oscillates.
and how do you that it won't go to 0?
this is a serious question not some kind of a shot at you.
As I said before, Scribbles asked you to describe what Leaky said (and you did that correctly), not to correct what he said.
PS: Yes, I looked it up in the Internet.
Edit2: So I'm probably wrong, anyway. 😛
22 Mar 07
1 edit
Originally posted by Palynkathe describe thing is a good point. and that's what i've tried to do.
I think you have to use the Schnirelmann density, not asymptotic density.
As I said before, Scribbles asked you to describe what Leaky said, not to correct what he said.
PS: Yes, I looked it up in the Internet.
but his point that the limit doesn't exist cause the function oscillates isn't true.
and by the way: never heard of the other density... 😕 😳
22 Mar 07
Originally posted by adam warlockThose are different limits than the one you described before.
\lim_{n\to\infty}-\frac{1}{n}=0 and -\frac{1}{n} oscillates.
and how do you that it won't go to 0?
this is a serious question not some kind of a shot at you.
The limit you described before was the limit as x approaches infinity of n(x)-(x/3).
Consider three cases:
Case 1: If x is divisible by three, n(x)-(x/3) = 0.
Case 2: If x+1 is divisible by three, n(x)-(x/3) = -2/3
Case 3: If x+2 is divisble by three, n(x)-(x/3) = -1/3
The values of the function in question are independent of the magnitude of x, and clearly oscillate between three values no matter how large x grows. In the language of epsilon-delta proofs, if you choose e to be .01, there exists no d such that x>d implies that |f - 0| < e.
22 Mar 07
Originally posted by DoctorScribblesok.that's all right. but you saying that the limit doesn't exist cause the function oscillates is wrong!
Those are different limits than the one you described before.
The limit you described before was the limit as x approaches infinity of n(x)-(x/3).
Consider three cases:
Case 1: If x is divisible by three, n(x)-(x/3) = 0.
Case 2: If x+1 is divisible by three, n(x)-(x/3) = -2/3
Case 3: If x+2 is divisble by three, n(x)-(x/3) = -1/3
The v ...[text shortened]... lta proofs, if you choose e to be .01, there exists no d such that x>d implies that |f - 0| < e.
22 Mar 07
Originally posted by DoctorScribblesmy definition of limit is the usual weierstrass one. i don't know that the limit exists. i just tried to formalize what chronicleaky said. i didn't payed much attention to correctness because i didn't think you wanted that.
Then what is your definition of limit, and what is your proof that it exists in this case?
in this case the function oscillates and the limit doesn't exist. but that's different than saying that because a function oscillates its limit doesn't exist.
and if the ratio is actually close to 1/3 that formalization has to be right.
22 Mar 07
2 edits
Originally posted by adam warlockFormalization does not consist in converting words to symbols.
i just tried to formalize what chronicleaky said. i didn't payed much attention to correctness because i didn't think you wanted that.
Formalization consists in taking an idea and making it a precise proposition subject to formal analysis.
What you have done is taken ChronicLeaky's words, rendered them as a symbolic proposition about a limit, found that the proposition was not sensible, and concluded that your formalization was correct and that his idea was incorrect. In actuality, it is your formalization that is incorrect.
22 Mar 07
1 edit
I don't know what I said. What I meant to say is:
Let S be a set of positive integers. Let S(x) = {n | n is in S and n < x}, for each positive real x. Then clearly |S(x)| increases (not necessarily strictly) with x. Also, |S(x)|/x is bounded above by 1 and by 0 below.
The asymptotic density of S in the integers is defined as the limit as x approaches infinity of |S(x)|/x.
The Schnirelmann density is the infimum of |S(n)|/n for natural numbers n.
In some cases, these agree. For example, the set of even numbers has asymptotic density 1/2 and Schnirelmann density 1/2. However, there are sets for which the Schnirelmann density exists and the asymptotic density does not, but none has been mentioned in this thread.
Finally, the asymptotic density can be a bit misleading. It's often better to find some real function f(x) such that |S(x)|/f(x) is bounded by a constant, or even better, approaches 1. This gives a more precise measurement. For example, the perfect squares and the primes both have asymptotic density 0, but for the former, f(x) = x^1/2 while for the latter, f(x) = x/log x. This shows there are "more" primes than squares in a way the asymptotic density does not.
Note I said the PROPORTION tends to 1/3. The limit you're talking about should be of N(x)/x - 1/3, not N(x) - x/3.
EDIT I'd like to know where the Spirituality is here.
22 Mar 07
Originally posted by ChronicLeakyright on the limit thing. my mistake ...but i guess you know what i meant 😳
I don't know what I said. What I meant to say is:
Let S be a set of positive integers. Let S(x) = {n | n is in S and n < x}, for each positive real x. Then clearly |S(x)| increases (not necessarily strictly) with x. Also, |S(x)|/x is bounded above by 1 and by 0 below.
The asymptotic density of S in the integers is defined as the limit as x app ...[text shortened]... of N(x)/x - 1/3, not N(x) - x/3.
EDIT I'd like to know where the Spirituality is here.
and if we make 1/3 three on my formalization everything is ok isnt it?
22 Mar 07
Originally posted by DoctorScribblessometimes formalization is just that....i never concluded his idea was incorrect just your oscillation point.
Formalization does not consist in converting words to symbols.
Formalization consists in taking an idea and making it a precise proposition subject to formal analysis.
What you have done is taken ChronicLeaky's words, rendered them as a symbolic proposition about a limit, found that the proposition was not sensible, and concluded that your form ...[text shortened]... ect and that his idea was incorrect. In actuality, it is your formalization that is incorrect.
22 Mar 07
Originally posted by DoctorScribblesHey Scribs, do you remember the riddle where there are infinite row of seeds being planted and two birds eating them at different speeds?
Hopefully the number of participants in this thread who have experienced mathematical humility in the presence of a higher power is greater than zero.
Can't we use Leaky's method of finding f(x) such that|S(x)|/ f(x) is bounded by a constant to choose between one of the two rows?