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)

Exponential Definitions
I am discussing the equivalence of the definitions of exponential here:

http://forums.xkcd.com/viewtopic.php?f=17&t=80256

For anyone who has been following my progress or lack thereof on exponent combination laws/log laws etc, feel free to look on. --GFauxPas 16:59, 6 February 2012 (EST)

Does anyone have an idea how I can prove:


 * $\displaystyle \left({1 + \frac{x + y}{n}} + \frac {xy}{n^2}\right)^n \to \left({1 + \frac{x + y}{n}}\right)^n$ as $n \to +\infty$?--GFauxPas 17:45, 8 February 2012 (EST)


 * Observe:


 * $\displaystyle \frac{\left({1 + \frac{x + y}{n}} + \frac {xy}{n^2}\right)^n}{\left({1 + \frac{x + y}{n}}\right)^n} = \left({1 + \frac{xy}{n \left({n + x + y}\right)}}\right)^n = \sum_{k=0}^n \binom n k \left({\frac{xy}{n \left({n + x + y}\right)}}\right)^k$


 * which follows by algebra. Then use $\displaystyle\binom n k n^{-k} \le 1$ and prove $\displaystyle \lim_{n\to\infty} \sum_{k=1}^n \left({\frac{xy}{n + x + y}}\right)^k = 0$; this will follow from geometric series I think.


 * Note the splitting of of the term for $k = 0$; this will show that the limit of the quotient above equals $1$. Knowing that the bottom sequence converges, this gives the desired result (from Product Rule for Limits, I think). That should provide you enough to fill in the details (I am thinking on the fly here). --Lord_Farin 18:07, 8 February 2012 (EST)


 * Interesting. I don't understand how a geometric series would help though. --GFauxPas 08:20, 9 February 2012 (EST)
 * Edit: Oh, you meant that $=1$ not that it $=0$, right? --GFauxPas 08:23, 9 February 2012 (EST)


 * No, I tacitly made $k$ start at $1$ in the second line, effectively this is how one wants to prove that the limit with $k = 0$ included equals $1$; though that may also work directly, I don't know.


 * You can simply take one factor out of the sum, then apply geometric series to the remaining term, show that both the single term and the sum converge to zero and use the product of limits theorem. As said, a direct approach might be possible as well. --Lord_Farin 08:42, 9 February 2012 (EST)


 * Reading again what I wrote last night, I recall that the splitting of of the $k=0$ term was necessary to justify using the inequality (I think that maybe the squeeze is silently applied here). If you want, I will spell it out formally on my sandbox page. --Lord_Farin 08:44, 9 February 2012 (EST)

Okay I think I need to learn more about sums/series/sequences. I'm putting this on the back burner. --GFauxPas 10:41, 9 February 2012 (EST)
 * Oh, feel free to put up the proof if you want LF, I'm going to wait until I feel more comfortable with the material. --GFauxPas 11:11, 9 February 2012 (EST)
 * Ooh something just clicked. I shouldn't have copped out so quickly. I un-give up. --GFauxPas 14:59, 9 February 2012 (EST)

Proof
Let $\exp x$ be the exponential of $x$ as defined by the limit of a sequence:


 * $\exp x = \displaystyle \lim_{n \to +\infty} \left({1 + \frac x n}\right)^n$

Note that from Powers of Group Elements we can presuppose the exponent combination laws for natural number indices.

By definition:

Intuitively, the $\left({1 + \frac{x + y}{n}}\right)$ term is the most influential of the terms involved in the limit, and:

$\displaystyle \left({1 + \frac{x + y}{n}} + \frac {xy}{n^2}\right)^n \to \left({1 + \frac{x + y}{n}}\right)^n \ \text{as} \ n \to +\infty$

To formalize this claim:


 * $\exp \left({x + y}\right) = \exp x \cdot \exp y \iff \dfrac {\exp x \cdot \exp y} {\exp \left({x + y}\right)} = 1$

Now, as $n \to +\infty$, we use the Combination Theorem for Limits of Functions to investigate the behavior of this sequence.


 * The $1$ term trivially converges to $1$.


 * $\displaystyle \left| {\frac{n!}{k!\left({n-k}\right)!n^k}}\right| < 1$ and so that term converges to $0$ Power of a Number Less Than One.


 * For sufficiently large $n$, $\displaystyle \left\vert{ \frac{xy}{n + x + y}}\right\vert < 1$ and so the rightmost term converges, by Geometric Series.

Which means that the whole series converges to $1$.

Recall:


 * $\exp \left({x + y}\right) = \exp x \cdot \exp y \iff \dfrac {\exp x \cdot \exp y} {\exp \left({x + y}\right)} = 1$

Hence the result.

Like that, LF? --GFauxPas 16:48, 9 February 2012 (EST)