Wikipedia:Reference desk/Archives/Mathematics/2012 March 10

= March 10 =

Extending proof to uncountable subsets of R
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 $$\{a_n\}$$ has a limit.

Going in the direction "limit => completeness axiom", any proof that I can think of will show that the set $$\{x \in \mathbb{R} : \exists n \in \mathbb{N} : x=a_n\}$$ 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)
 * Hint: a set bounded above but without a least upper bound will generate a nondecreasing sequence. Rschwieb (talk) 02:13, 10 March 2012 (UTC)
 * Use the density of the rationals.--121.74.121.82 (talk) 05:09, 10 March 2012 (UTC)
 * Am I missing something? a nondecreasing sequence like an=n does not have a limit. Shouldn't that be Any bounded nondecreasing sequence $$\{a_n\}$$ has a limit? 84.197.178.75 (talk) 18:21, 10 March 2012 (UTC)
 * Yes, that is what I meant. (talk) 10:47, 11 March 2012 (UTC)
 * I think it's true that any uncountable bounded subset of $$\mathbb{R}$$ can be formed by taking a countable subset of $$\mathbb{R}$$ with least upper bound $$L$$ and adding an uncountable number of reals to that subset. As long as every real number added is less than or equal to $$L$$, $$L$$ will again be the least upper bound. Widener (talk) 11:51, 11 March 2012 (UTC)
 * 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)
 * $$\omega_1 + \omega^*$$.--130.195.2.100 (talk) 03:01, 12 March 2012 (UTC)
 * This isn't necessarily obvious though. Is there a proof of this? Widener (talk) 05:16, 12 March 2012 (UTC)

Eccentricity of a conic section
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)
 * 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 $$\sqrt{2}$$. 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)

Big or Little
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:
 * $$ e^x = 1 + x + \frac{x^2}{2} + O(x^3) \, . $$

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)
 * Yes, you'd use O here. The function $$ e^x - 1 - x - \frac{x^2}{2}$$ 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)
 * It's unlikely to use $$O(x^3)$$ for $$x\to a$$ with $$0<a<\infty$$, but it's very likely to use it for $$x\to\infty$$. -- Meni Rosenfeld (talk) 21:23, 10 March 2012 (UTC)
 * Thank you. I made a mistake in not considering the case a=±∞. 96.46.204.126 (talk) 19:38, 12 March 2012 (UTC)
 * See Big O notation section Usage: Infinitesimal asymptotics for this exact example. 84.197.178.75 (talk) 19:56, 10 March 2012 (UTC)