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)

Theorem
Let:

where:


 * $\forall i: 1 \le i \le m: \mathbf{a_i} = \begin{bmatrix} a_{1i} \\ a_{2i} \\ \vdots \\ a_{mi} \end{bmatrix} \in \R^m$.

Then:


 * $\left\{ \mathbf{a_1}, \mathbf{a_2}, \cdots, \mathbf{a_n} \right\}$ is a linearly independent set iff


 * $\operatorname{N}\left({\mathbf A}\right) = \left\{ \mathbf {0}_{n \times 1}\right\}$

Proof
Sufficient Condition:

Suppose $\left\{ \mathbf{a_1}, \mathbf{a_2}, \cdots, \mathbf{a_n} \right\}$ is independent.

Then by the definition of independence, $\forall k: 1 \le k \le n: x_k = 0 \iff \mathbf x = \mathbf 0_{n \times 1}$.

Then by the definition of null space, $\operatorname{N}\left({\mathbf A}\right) = \left\{ \mathbf {0}_{n \times 1}\right\}$.

Suppose $\operatorname{N}\left({\mathbf A}\right) = \left\{ \mathbf {0}_{n \times 1}\right\}$.

Then by the definition of null space, $\mathbf x = \mathbf {0}_{n \times 1} \implies \forall k: 1 \le k \le n: x_k = 0$, from which it follows that $\left\{ \mathbf{a_1}, \mathbf{a_2}, \cdots, \mathbf{a_n} \right\}$ is linearly independent.