Real Analysis Real Analysis Math Forum

April 12th, 2016, 11:31 AM   #21
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 Originally Posted by Mathbound Thank you for taking the time, but it just doesn't make sense to me to start two discussions on a specific topic simultaneously with two different people. From other help forums that I have seen, only one person takes on a student, at least until somebody else sees something wrong with how that person is helping the student or just wants to add some fresh ideas if the other method is obviously not working. And of course, one may jump in if the other just simply gives up on that student. We are in the "Wild West" of forum etiquette. There is no known universal way to behave on these forums yet, at least not that I can tell. And worst of all, the communication relayed is not quite complete as there is no body communication. For example, if we were all at a coffee shop discussing this, and you see my eyes locked on v8archie's as he/she tries to help me, you may get the impression that extra help is not necessary. But if you saw me huffing and puffing and looking up at the ceiling, and there is a pause, you may just instinctually know to jump in and help guide me into what v8archie is trying to convey. So please, trust me, people are grateful for your help, and nobody wants to be rude to someone like you who wants to help.
Noted.

I can see that your principal 'conversation' is with v8archie.
You should definitely listen to his words, they are usually wise.
Furthermore I both acknowledged and tried to build on what archie said, when saw that you were facing the same difficulties that beset so many studying this subject.
That is why I limited my input, whilst acknowledging his.

However sometimes a change of viewpoint helps and this is what I offered you.
The new view hopefully offers enlightenment, after which you can return to the formal path.

That does not require a long conversation, but an acknowledgement is appreciated.

April 12th, 2016, 11:54 AM   #22
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 Originally Posted by studiot Noted. I can see that your principal 'conversation' is with v8archie. You should definitely listen to his words, they are usually wise. Furthermore I both acknowledged and tried to build on what archie said, when saw that you were facing the same difficulties that beset so many studying this subject. That is why I limited my input, whilst acknowledging his. However sometimes a change of viewpoint helps and this is what I offered you. The new view hopefully offers enlightenment, after which you can return to the formal path. That does not require a long conversation, but an acknowledgement is appreciated.

April 12th, 2016, 12:04 PM   #23
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This subject is very tricky and has beeen debated and discussed for several hundred years.

So anytime you need amplification, just holler.

It is always best if you can figure the details out for yourself, once pointed in the right direction.

April 12th, 2016, 01:44 PM   #24
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 Originally Posted by Mathbound Interesting, but I think I am just going to have to accept it unless I can prove that there is a contradiction, and that is something that I am currently not able to properly show.
Do you know the difference between potential infinities (as used in real analysis) and the actual infinities of set-theory (e.g. $\aleph_0$, $\omega$, etc.).

April 12th, 2016, 08:20 PM   #25
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 Originally Posted by v8archie Do you know the difference between potential infinities (as used in real analysis) and the actual infinities of set-theory (e.g. $\aleph_0$, $\omega$, etc.).
I do not know anything about that. Is this a competing theory to set theory?

 April 12th, 2016, 09:04 PM #26 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,649 Thanks: 2630 Math Focus: Mainly analysis and algebra No. Potential infinities are used in analysis - in particular in limit expressions such as $\sum \limits_{n=0}^{\infty} {1 \over 2^n}$. Although the rigorous definition avoids the use of any infinity entirely (as outlined about), we talk about the limit as $n$ goes to infinity. This is a potential infinity because there is no suggestion that $n$ actually attains any infinite value. One reason it never does so is because there are no infinite natural numbers and in the context of an infinite sum, $n$ is a natural number. Another reason is that the definition of a limit never actually uses the value of the expression at that limit. (It is this allows us to define the continuity of a function in terms of limits). The actual infinities of set theory do, however exist (for some definition of existence). There are an infinite number of natural numbers and we denote this quantity as $\aleph_0$. Then there are the ordinal numbers which are particularly applicable to the infinite series we've been looking at. Unfortunately, I'm far from an expert on this stuff, but we can describe the length of the series above as being $\omega$. This doesn't mean that it has $\omega$ terms (it has $\aleph_0$ terms). But $\omega$ is called a "limit ordinal" because it describes the length of the series in the limit. We can write the sum as $\sum \limits_{n = 0}^{\infty} {1 \over 2^{2n}} + \sum \limits_{n = 0}^{\infty} {1 \over 2^{2n+1}}$. This is just a re-ordering of the terms (all those with an even index, followed by all those with an odd index). And, because the original series is absolutely convergent, we should get the same result. \begin{aligned}\sum \limits_{n = 0}^{\infty} {1 \over 2^{2n}} + \sum \limits_{n = 0}^{\infty} {1 \over 2^{2n+1}} &= \sum \limits_{n = 0}^{\infty} {1 \over 2^{2n}} + \frac12 \sum \limits_{n = 0}^{\infty} {1 \over 2^{2n}} \\ &= \sum \limits_{n = 0}^{\infty} {1 \over \left(2^2\right)^n} + \frac12 \sum \limits_{n = 0}^{\infty} {1 \over \left(2^2\right)^n} \\ &= \sum \limits_{n = 0}^{\infty} {1 \over 4^n} + \frac12 \sum \limits_{n = 0}^{\infty} {1 \over 4^n} \\ &= {1 \over 1 - \frac14} + \frac12 \cdot {1 \over 1 - \frac14} \\ &= \frac43 + \frac12 \cdot \frac43 = \frac63 = 2\end{aligned} But looking back at that derivation, we have two infinite sums. There are still $\aleph_0$ terms (the number of terms doesn't change just because we write them in a different order) but now the length of the sum is $\omega + \omega = \omega \cdot 2$. In other words, it is twice as long as before. Freaky. So, all these infinities do work together in a logical way, but each has its own domain and its own meaning. And you can't mix them. Analysis doesn't need (or want) to get bogged down in the mechanics of the sum it is calculating, especially when things get freaky and apparently contradictory. Instead, it's much less bother to avoid the actual infinities all together. We also avoid them because you can't get the these infinities by adding another terms to the sum which is what we do when we say $\sum \limits_{n=0}^{\infty} {1 \over 2^n} = \lim \limits_{n \to \infty} \sum \limits_{k = 0}^n {1 \over 2^k}$. They only exist "in the limit" and the limit expression doesn't actually use that limit at "infinity". Another reason is that neither $\aleph_0$ (the quantity of terms) nor $\omega$ (the length of the sum) are numbers that can be manipulated like $n$ in the sum. $\aleph_0$ doesn't submit to any normal rules of arithmetic and $\omega$ isn't commutative under addition or multiplication, so we definitely don't want them in our series anywhere. Probably I've just confused you more. It's difficult stuff to grasp, and very strange, but also has a practical sort of logic to it. I apologise for any errors in the above, it's my interpretation of what I've been reading. As I said, this is not in my field. Thanks from Mathbound Last edited by v8archie; April 12th, 2016 at 09:49 PM.
 April 12th, 2016, 10:39 PM #27 Member   Joined: Jun 2014 From: Alberta Posts: 56 Thanks: 2 This is just cool; I don't have any other way to describe it. I feel like I am fortunate enough to know that I am confused. My Background is just first-year calculus, first-year linear algebra, and one semester of analysis. But I am obsessed with this stuff anyways.

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