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Topic: 0.999. (Read 106760 times) 

Icarus
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Re: 0.999.
« Reply #300 on: Feb 19^{th}, 2004, 4:19pm » 

I see that the problem I've getting on the last couple days has not been universal. However, towr has superbly answered the questions. To TenaliRamen, it's true that you are not likely to see any new insights into this problem in this thread again, and very little other discussion, as the large size of this thread makes responding to it a pain. But in terms of space used, I doubt it is very much by computer standards, and it does provide a place for those still interested &/or unclear about the truth here, such as Sameer was. Indeed, the discussion between Sameer, towr, & I demonstrates to me that this thread needs to be ongoing, as there will always be those for whom this topic is not fully settled. If we were to lock this thread, then I would want to immediately start a successor to continue the conversation, so that those new to this have a place where they can comment. Perhaps this would be a good idea, with the summaries of this thread leading off, since it would encourage new visitors to actually read it rather than simply skipping to the end and ignoring the previous conversations.


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SWF
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Re: 0.999. Tangent999.GIF
« Reply #301 on: Feb 19^{th}, 2004, 7:34pm » 

When x is less than or equal to 1: tan^{1}(x)=x  x^{3}/3 + x^{5}/5  x^{7}/7 + x^{9}/9  ... To facilitate the upcoming term by term comparison with the series for 0.999..., rearrange some of the terms to get: tan^{1}(x)= (xx^{3}/3) + (x^{5}/5x^{7}/7) + [sum] ( x^{4n+1}/(4n+1)  x^{8n5}/(8n5)  x^{8n1}/(8n1) ) where the summation is on n from 2 to infinity. If there is any doubt about convergence of this form for inverse tangent, see the attached graph comparing the exact value of inverse tangent with what is obtained with 30 terms of this series. Also, using standard convergence tests, it is not difficult to show the series converges for x less than or equal to 1. Inserting x=1 into this series and multiplying by 4/[pi] equals 1 since 4/[pi]*tan^{1}(1)=1. Compare term by term with 0.999... = 0.9+0.09+0.009+...: 4/[pi]*(11/3) < 0.9 4/[pi]*(1/51/7) < 0.09 4/[pi]*( x^{4n+1}/(4n+1)  x^{8n5}/(8n5)  x^{8n1}/(8n1) ) < 9*10^{N1} (for all n>=2) Therfore 1 < 0.999...

« Last Edit: Feb 19^{th}, 2004, 7:35pm by SWF » 
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Icarus
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Re: 0.999.
« Reply #302 on: Feb 20^{th}, 2004, 3:25pm » 

Very nice! Did you come up with it yourself, or find it somewhere? I know what is going on, but will leave it for someone else, other than giving this minor hint: It's has to do with something I've mentioned before in this thread.


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Sameer
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Re: 0.999.
« Reply #303 on: Feb 20^{th}, 2004, 4:14pm » 

on Feb 20^{th}, 2004, 3:25pm, Icarus wrote:Very nice! Did you come up with it yourself, or find it somewhere? I know what is going on, but will leave it for someone else, other than giving this minor hint: It's has to do with something I've mentioned before in this thread. 
 Hahah very nice hint indeed Icarus And I do agree with the earlier post. I do get confused sometimes with my own fundamentals and tend to project them wrong. Besides even though I have been on this forum for quite a while I still haven't been able to go through all of them. In fact I am just starting on medium. I havent even looked at hard (which I doubt would be just more than browsing after my grand show on page 12 of this thread). Tenali, my post on space was flowing with sarcasm too . I haven't been in school for a while and besides I am not working in the field I studied. That is why I have lost touch with lot of my math. Just the other day when I was flipping through one of the math books and I was like whoa, can't remember. This forum is the place where I can explore something I knew before and have forgotten or have had shaky foundations to begin with. I am simply honored to be in company of you all greats (the count is definitely higher than 5 so that is not the answer to that question in easy section )


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rmsgrey
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Re: 0.999.
« Reply #304 on: Feb 20^{th}, 2004, 4:56pm » 

The series for tan^{1} may converge, but it doesn't converge absolutely for x=1 (if you change each '' to '+', the result diverges). By observing that 8/3[pi]<8/9<.89, and substituting that for the first term in SWF's expansion of .999... you get that .9899999999999...>1 (using 8/9 rather than .89 gives .9888888...>1 but requires you to take the behaviour "at infinity" on faith; given the subject matter of this thread that may be unwise) [e]corrected equation (sorry Icarus)[/e]

« Last Edit: Feb 21^{st}, 2004, 8:06am by rmsgrey » 
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Icarus
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Re: 0.999.
« Reply #305 on: Feb 20^{th}, 2004, 8:41pm » 

on Feb 20^{th}, 2004, 4:56pm, rmsgrey wrote:The series for tan^{1} may converge, but it doesn't converge absolutely for x=1 (if you change each '' to '+', the result diverges). 
 And what does this tell you? Quote:By observing that 8/[pi]<8/9<.89 
 Whoa! Now, since 2 < 8/[pi] [approx] 2.55, we have 2 < 1! It's amazing what you can prove if your observation skills are a little rusty!


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rmsgrey
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Re: 0.999.
« Reply #306 on: Feb 21^{st}, 2004, 8:10am » 

on Feb 20^{th}, 2004, 8:41pm, Icarus wrote: And what does this tell you? 
 That rearranging the terms can change the limit. Quote: It's amazing what you can prove if your observation skills are a little rusty! 
 Particularly since I went through my entire post thinking I had put that pesky 3 in! (I've edited it in now)


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Icarus
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Re: 0.999.
« Reply #307 on: Feb 21^{st}, 2004, 8:58am » 

Exactly. When I was in college, one of my professors suggested as a research project to find conditions on rearrangements of a conditionally coverging series that would guarantee the rearranged series would converge to the same limit, and/or conditions guaranteeing it would converge at all. I couldn't think up any good ways of approaching the problem, however.


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Eigenray
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Re: 0.999.
« Reply #308 on: Feb 22^{nd}, 2004, 2:01am » 

If a (real) series converges, but does not converge absolutely, then it can be made to converge to any arbitrary (extended real) value, or diverge, by a suitable rearrangement of its terms. The proof is not hard; the key is that the subsequence of positive terms converges to 0, yet its sum must converge to infinity, and similarly for the negative terms. The converse is true as well: If a series converges absolutely, then any rearrangment of its terms converges to the same value.

« Last Edit: Feb 22^{nd}, 2004, 2:06am by Eigenray » 
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SWF
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Re: 0.999.
« Reply #309 on: Feb 22^{nd}, 2004, 5:48pm » 

on Feb 20^{th}, 2004, 3:25pm, Icarus wrote:Very nice! Did you come up with it yourself, or find it somewhere? 
 Thank you. I made this up for everyone's amusement, but I already knew that rearranging a conditionally convergent series can be made to give whatever one wants. I stayed away from using the standard example of the series for ln(2), hoping to make spotting the flaw more difficult, but that trick did not work. With all the pages in this thread, it takes quite a long time to post with a modem connection because the bottom of the screen seems to include all the previous posts.


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Icarus
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Re: 0.999.
« Reply #310 on: Feb 22^{nd}, 2004, 7:45pm » 

That you were familiar with the result was obvious to those of us who are also familiar with it. It was a nice change of pace from the recurring arguments, despite its tongueincheek nature. This thread is hard to reply to even on a cable modem. I've been thinking about taking Tenali Ramen's advice and locking it, but also replacing it with a new thread linking back, and perhaps starting out with copies of my thread summaries. To reiterate what I've said already, this thread serves still as a learning tool for those who have never confronted this issue, and for those who think they understand what is going on but still have misconceptions, such as Sameer did. For this reason I do not want to shut it down completely. But a shorter thread might make it more likely for such people to join in a meaningful conversation.


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Gatorshiz100
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Re: 0.999.
« Reply #311 on: Apr 1^{st}, 2004, 12:05pm » 

You crazy mathematicians. I'm an engineer, .9 = 1 is good enough for me.


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Avi Gavlovski
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Re: 0.999.
« Reply #312 on: Apr 8^{th}, 2004, 5:16pm » 

The real numbers are a complete metric space. Hence, every Cauchy sequence converges to a point in the set. Hence, since the sequence {Sn} = 9/10+9/10^2+...+9/10^n is Cauchy, it converges to a point in the set. Furthermore, the only valid definition of .999... is to define it as the limit of {Sn} as n>infinity. The nonvalid definitions that ive seen posted are all equivalent to: .999... = max{reals < 1} However, if this number were indeed less than 1, then .999... < (.999... + 1)/2 < 1 which would imply that its not a max. Further, it implies that there IS NO max. There is a sup: 1.


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hamby
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Re: 0.999.
« Reply #313 on: Apr 9^{th}, 2004, 7:20am » 

Does a hundred gabillion trillion zillion = infinity. No. .999... does not equal 1.


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Sameer
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Re: 0.999.
« Reply #314 on: Apr 9^{th}, 2004, 7:26am » 

on Apr 1^{st}, 2004, 12:05pm, Gatorshiz100 wrote:You crazy mathematicians. I'm an engineer, .9 = 1 is good enough for me. 
 I am an engineer too and I say you should know why .99... = 1


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rmsgrey
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Re: 0.999.
« Reply #315 on: Apr 19^{th}, 2004, 6:40am » 

on Apr 9^{th}, 2004, 7:20am, hamby wrote:Does a hundred gabillion trillion zillion = infinity. No. .999... does not equal 1. 
 "gabillion" and "zillion" are not terms I'm familiar with. A hundred trillion trillion trillion is a finite value, and can be expressed explicitly or generated with a finite number of steps. A trillion trillion trillion ... where the ... represents the limit of extending the number of "trillion"s by one each step cannot be generated by a finite process, and does equal infinity. Similarly, any finite string of nines (after the decimal point) gives a value less than one, but .999... cannot be generated by a finite process and does have the same value as 1.000...


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Grimbal
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Re: 0.999.
« Reply #316 on: Apr 29^{th}, 2004, 4:42pm » 

When you feel 0.999... should be < 1, it is because you never realized that the decimal writing of a number is not necessarily unique. You think that a number is identified by its decimal notation. It is not. 0.999... is the same number as 1.000..., just like 1/3 is the same number as 2/6. You never would say that 1/3 is not equal to 2/6 because it is written differently. To prove the equality, if x = 0.999..., you have x*109 = x. From there, x*9 = 9 and therefore x = 1.


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