User:GFauxPas/Sandbox

Welcome to my sandbox, you are free to play here as long as you don't track sand onto the main wiki. --GFauxPas 09:28, 7 November 2011 (CST)


 * $u \ v \ \mathsf{u} \ \mathsf{v} \ \nu \ \upsilon$

Anyone else have a hard time distinguishing between $u$ and $v$? I would like it to look more like this, does it confuse anyone else? It seems PW doesn't have the upgreek package. --GFauxPas 07:49, 27 January 2012 (EST)


 * Nope. Multiple years of extensive TeX writing and reading have trained my eye. I agree that referenced $v$ looks more distinguished, but imagine it is hard to implement. --Lord_Farin 08:08, 27 January 2012 (EST)

Convergence
We have that $\displaystyle \lim_{n \to +\infty}a_n = 0$, by hypothesis.

To show that $\displaystyle \sum_{k=1}^n{n \choose k}\frac {{a_n}^{k-1}} {n^k}$ converges, observe that, for $n$ large enough:

Recall that $a_n \to 0$, by hypothesis.

This means that $0 \le \left\vert{a_n}\right\vert \le 1$ for sufficiently large $n$, because $a_n$ is Cauchy.

Then, we can say:


 * $\displaystyle \left \vert{ \frac{ 1 - {a_n }^k } {1 - a_n} }\right\vert \le \left \vert{ \frac{ 1 - a_n } {1 - a_n} }\right\vert = 1$ as $n \to +\infty$, because $k \ge 1$.

Hence $\displaystyle \left \vert{\sum_{k=1}^n{n \choose k}\frac { {a_n}^{k-1} } {n^k} }\right\vert$ converges, by the Comparison Test.

That means that $\displaystyle \sum_{k=1}^n{n \choose k}\frac { {a_n}^{k-1} } {n^k}$ converges as well, because Absolutely Convergent Series is Convergent.

Is this okay? --GFauxPas 08:54, 6 March 2012 (EST)


 * Unfortunately, no. I think I got carried away when I said that the argument generalized to arbitrary complex sequences. However, it functions in the sense that we have demonstrated that the sequence is bounded. This means that we can estimate the (modulus of the) product of this thing with $a_n$ by a fixed number times the modulus of $a_n$, which then converges to zero. This will suffice to prove that the limit of the whole expression is zero because $|a_n|\to0$ does imply $a_n\to0$ (unlike when $0$ is replaced by another complex number). These tedious considerations are necessary because, as mentioned before, the sequence is not in the form of a series, hence results for series can't be applied. I hope you grasp the sketchy adapted approach. --Lord_Farin 13:56, 6 March 2012 (EST)
 * But we don't care what $\displaystyle \sum_{k=1}^n{n \choose k}\frac {{a_n}^{k-1}} {n^k}$ converges to, just that it converges to some number... Just to try to separate out what I know and what I don't know, is my proof legitimate for $a_n$ being a real sequence? I'm trying to find a way to deal with this strange not-a-series sequence that kind of looks like a series but it isn't but it maybe it is just a little. I'll leave complex analysis for another time. --GFauxPas 14:08, 6 March 2012 (EST)


 * I'm not so sure anymore. The proof as it stands now seems to only be guaranteed to work when all $a_n$ are positive real. However, the approach sketched for the complex case works; a page establishing $a_n\to0, |b_n|\le B\implies a_nb_n \to 0$ should be established (if it isn't already). That's the theorem we want to invoke. --Lord_Farin 18:04, 6 March 2012 (EST)

R is subset of C?
I notice that in general we're treating all real numbers as complex numbers. But isn't that a bit incorrect? Sure, if you build $(\R^2,+,\cdot ) = \C$ then every element of $\C$ such that $\Im(z) = 0$ acts like a real number. But isn't there a difference between $2 \in \R$ and $(2,0) \in \C$? one's a 1-tuple or whatever you want to call it, one's a 2-tuple, no? --GFauxPas 12:57, 9 March 2012 (EST)


 * For all intents and purposes, the canonical embedding $\iota: \R \to \R^2, x \mapsto \left({x, 0}\right)$ is an isomorphism. Hence we needn't bother unless we are relying on the explicit set structure. This is on the same level as regarding different definitions of the natural numbers to be the same. Generally, it doesn't pose a problem and often the applying of natural isomorphisms and their inverses is suppressed from the notation. It's good that you noticed though. --Lord_Farin 13:06, 9 March 2012 (EST)


 * "I notice that in general we're treating all real numbers as complex numbers" - I'm fighting that tendency. However, there's a progressive element which is extending the scope of all the hard fought real-analysis results to just extend them to the complex plane, without a thought to the possible implications on the chain of proof, however much the message goes out that the complex result needs to be stated and proved as a completely separate object. --prime mover 13:16, 9 March 2012 (EST)


 * I see your point there; eventually results will need deeper, genuinely different proofs (or, for a small amount, easier) to be generalized to the complex plane. For me, this is where it is justified to add a section 'Proof for Reals' or something like that. However, if the proof is identical, I deem it unnecessary to have the proof twice. Rather, one could say 'let $x \in \Bbb \F \in \{\R,\C\}$' (borrowed that phrase from Conway) or 'let $x \in \R$ or $\C$' to emphasize that the result is also valid for the reals. --Lord_Farin 13:21, 9 March 2012 (EST)


 * One case in particular where the proof is identical can stand as it is, but it still needs to be stated as to which number fields it applies. It's particularly important to be fussy about this where a proof applies to $\Q, \R, \C$ but not $\Z$, or for $\R$ and $\C$ but not $\Q$, and even when they can be extended to $\mathbb H$ and so on. So this is not just an idle me-being-fussy. Lots to be done still. --prime mover 14:40, 9 March 2012 (EST)

I hope I didn't somehow put the message that you were being fussy. In fact, I conclude that we agree on what needs and what needn't be done. Case dismissed, let's get to business. --Lord_Farin 19:28, 9 March 2012 (EST)


 * No worries, it was a light, self-deprecating comment in response to any tendency (not shown by you) to gloss over not-obviously-important details. --prime mover 04:07, 10 March 2012 (EST)