For Pete’s sake. You quoted my entire post. In it I said explicitly that there was no difference. You’re obviously not even bothering to read anything we’re posting.
Prove me wrong.
Well, I don’t know which post your referring to, and I do not think you stated it so clearly if you did. All you had to do was state what I stated above?
I have read all the posts. Which one did you explcitly state this in?
You can click the arrow in a quote to go to the post it came from. Exapno Mapcase is referring to this post of yours, in which you quoted this post of theirs.
Yeah, sorry don’t see where you said that even fter reading it a 3rd time.
You may have said in some round about fashion…?
I like my answer better.
Allotrope, You’re really, really screwing up here. Indistinguishable’s knowledge of mathematics goes so far beyond Calculus I it’s not even funny.
Calculus I textbooks make a lot of assumptions that mathematicians in general do not consider themselves beholden to. Not because mathematicians think Calculus I textbooks are wrong, but because mathematicians are interested in many other, different mathematical systems than the ones found in Calculus I textbooks.
Really? I’ll believe that when he can read a limit equation properly as being evaluated AT infinity.
There is NO interpretation of calculus where ‘n’ would be read as a “truncated” value. That is simply bullshit. Sorry, but those are the facts.
Let me explaine why I say this. Your talking about me not follwoing the rules of infinity an what not… Which particular rule? I still see my argument as valid, except for the conclusion there was any contradiction.
I showed what I believe a valid means of showing the infintesimal value between 1 and .999… with the aid of Limits. I concluded that since in one answer we come up with 0.000…1 and limit says it should be 0.000…0 that there was a contradiction. There is none of course because in the end by the definiton of limits there is no difference between the 2 answers. I don’t see at all where you said that.
So you’re either from Harvard or from MIT.
[spoiler]It’s an old joke.
A guy enters the “10 items or fewer” line at a supermarket in Cambridge, Mass. with an overflowing basket containing many more than ten items. As he begins unloading his cart, the cashier says to the man, “Why, you must attend either Harvard or MIT.”
The man puffs up at this and replies, “Why yes I do, but how ever did you know?”
“Well, there’s a sign right in front of you that says ‘Ten items or fewer’ so either you attend MIT and can’t read, or attend Harvard and can’t count.”[/spoiler]
I am beginning to think what makes math so difficult is not the math but the mathematicians 
EM,
I was not trying solve Zeno’s paradox in the least. I was not even contending it is really a pardox.
I think it is you who have not read all the posts.
EM,
If your interested the quesiton/puzzle/(flase proof) starts here: http://boards.straightdope.com/sdmb/showpost.php?p=15353743&postcount=273
I was also sure a flaw would be found somewhere but it was a fun puzzle for me. Maybe for you, EM, it would be obvious form the onset.
This was the image you linked: ImageShack - Best place for all of your image hosting and image sharing needs
What I pointed out was:
It’s plain as day that the very first thing that the linked image does is say “0.999… = the limit, as n goes to infinity, of 0.999… truncated to n decimal places”. I am surprised that pointing this out should upset you so.
OK, why don’t you explain the importance of the sigma notation to me then.
Er, why should I? What importance? The notation isn’t important… it’s just a fancy way of saying “9/10^1 + 9/10^2 + … + 9/10^n”. Do you want to know the importance of that sum? That (finite) sum is the standard interpretation of the (finite) decimal string “0.99…9” (with n many 9s). That sum is also equal to (1 - 1/10^n). What of it?
None of this is in any contradiction to anything I said.
So when sigma says n = 1 to ∞, you think that’s a FINITE sum do you?
Well, good for you. Thanks for proving my point. :smack:
Read the image (I shall link it again, since you seem to have forgotten it). The Sigma is not marked from n = 1 to infinity. The Sigma is marked from k = 1 to n. And n is, presumably, meant to be taken as a natural number (ergo, finite).
I am sympathetic to taking n as an infinite value, if you like, as a consequence of the binding given by the limit as n goes to infinity. But this is not what pretty much any Calculus textbook (other than Keisler’s) will tell you is the interpretation of this notation… The party line is that limits, as n goes to infinity, are simply descriptions about the asymptotic behavior for finite n.
And, again, none of this is in any contradiction to anything I said.
As n approaches infinity. And then you evaluate the final version at infinity which you would only do if n were infinite.
How else do you think you get 1 - 0 = 1 in the last segment - hmm?
You get there because you evaluate 1/(10^n) as 1/(10^∞). This give you 1/∞ which = 0. Thus 1 - 0 = 1
Now show me how YOU get that 0 from the equation if n is a trucated value.
The “limit” notation, on the standard account (which needn’t be the only account, but is the one you will find in standard calculus textbooks), works like so:
To take the limit, as n goes to infinity, of 1/10^n, you do not have to look at anything other than the value of 1/10^n for large but finite natural numbers n. Specifically, it may turn out that there is some value L such that, for any given reciprocal natural number epsilon, there is some natural number delta such that the difference between L and 1/10^n is less than epsilon whenever n is greater than delta. In this case, we say that L is the limit, as n goes to infinity, of 1/10^n.
It turns out that L = 0 will satisfy this property, as, in order to make the difference between 0 and 1/10^n less than epsilon, it suffices to take delta to be, say, 1/epsilon (as 1/10^(1/epsilon) < epsilon, as 10^(1/epsilon) > 1/epsilon). Thus, the limit, as n goes to infinity, of 1/10^n is equal to 0.
Oh god, that was completely meaningless. You really went all out trying to camouflage the fact that n has to be evaluated AT infinity - which is precisely what I’ve been saying all along. But the epsilon delta stuff was a nice touch.