March 10
Show that the following theorem is equivalent to the completeness axiom (i.e. that every bounded subset of the real numbers has a least upper bound):
- Any nondecreasing sequence has a limit.
Going in the direction "limit => completeness axiom", any proof that I can think of will show that the set has a least upper bound, and therefore that any countable subset of the real numbers has a least upper bound. How do you extend this to uncountable subsets? Widener (talk) 00:34, 10 March 2012 (UTC)[reply]
- Hint: a set bounded above but without a least upper bound will generate a nondecreasing sequence. Rschwieb (talk) 02:13, 10 March 2012 (UTC)[reply]
- Use the density of the rationals.--121.74.121.82 (talk) 05:09, 10 March 2012 (UTC)[reply]
- Am I missing something? a nondecreasing sequence like an=n does not have a limit. Shouldn't that be Any bounded nondecreasing sequence has a limit? 84.197.178.75 (talk) 18:21, 10 March 2012 (UTC)[reply]
- Yes, that is what I meant. (talk) 10:47, 11 March 2012 (UTC)[reply]
- I think it's true that any uncountable bounded subset of can be formed by taking a countable subset of with least upper bound and adding an uncountable number of reals to that subset. As long as every real number added is less than or equal to , will again be the least upper bound. Widener (talk) 11:51, 11 March 2012 (UTC)[reply]
- Hmm interesting question, had me thinking for a while. Like one of the above posters said, the separability of the real numbers (i.e. density of rationals) is essentially here. Let A be any set with upper bound. If it had a greatest element it would have an sup, so suppose it has no greatest element. Take the set K of rationals in A; obviously for each x in A there's some y in K such that x is less than y. Now enumerate K as x1,x2..., and define a new sequence y1,y2... as follows. Let y1 be any point of K, and inductively let yn be any point of K such that yn is bigger than y(n-1) and x1,...,x(n-1) (I'll let you verify why such a yn always exist). The sequence y1,y2,... is non decreasing, bounded, and for every x in A there's some yn that's bigger than x. The limit of this sequence is the sup of A.
- Just wondering, can anyone give a totally ordered set such that it doesn't have the least upper bound property but every non decreasing sequence converges? Money is tight (talk) 14:37, 11 March 2012 (UTC)[reply]
- .--130.195.2.100 (talk) 03:01, 12 March 2012 (UTC)[reply]
- This isn't necessarily obvious though. Is there a proof of this? Widener (talk) 05:16, 12 March 2012 (UTC)[reply]
What is the eccentricity of the following conic section?
(666*x)^2-(666*y)^2+(x+y+666^2)=2012 123.24.112.17 (talk) 16:56, 10 March 2012 (UTC)[reply]
- From the cartesian coordinates representation in Conic section article, I get A=666^2, B=0 and C=-666^2; so B*B-4*A*C > 0 making it a hyperbola, and since A+C=0 its a rectangular one. And Hyperbola mentions that for a rectangular hyperbola, the eccentricity is . But I'm not familiar with the subject so I can't tell if this is correct... 84.197.178.75 (talk) 18:49, 10 March 2012 (UTC)[reply]
When I'm writing out a power series in a single variable, I use the big O notation for the truncated terms, don't I? For example:
Does the point of evaluation need to be mentioned/taken into account for non-entire functions? — Fly by Night (talk) 19:00, 10 March 2012 (UTC)[reply]
- Yes, you'd use O here. The function is O(x3) for x → 0, but not o(x3).
- The notation O(x3) is only meaningful if "for x → 0" or something similar is specified or understood from context. Of course, if you were dealing with x → a for some nonzero a, then O(x3) would mean the same as O(1), so there would be no point using O(x3), and it's unlikely you'd be referring to a nonzero a here. These remarks are applicable to any function with an asymptotic expansion, entire or not. 96.46.204.126 (talk) 19:49, 10 March 2012 (UTC)[reply]
- It's unlikely to use for with , but it's very likely to use it for . -- Meni Rosenfeld (talk) 21:23, 10 March 2012 (UTC)[reply]
- Thank you. I made a mistake in not considering the case a=±∞. 96.46.204.126 (talk) 19:38, 12 March 2012 (UTC)[reply]
- See Big O notation section Usage: Infinitesimal asymptotics for this exact example. 84.197.178.75 (talk) 19:56, 10 March 2012 (UTC)[reply]