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)

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:

$\displaystyle \left({1 + \frac{x + y}{n}} + \frac {xy}{n^2}\right)^n \to \left({1 + \frac{x + y}{n}}\right)^n$ 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, term by term.

As $1$ trivially converges to $1$, consider now the other terms of the sequence.

From Negative of Absolute Value:


 * $\displaystyle -\left\vert{\sum_{k=1}^n {n \choose k}n^{-k}\left({\frac{xy}{n + x + y} }\right)^k}\right\vert \le \sum_{k=1}^n {n \choose k}n^{-k}\left({\frac{xy}{n + x + y} }\right)^k \le +\left\vert{\sum_{k=1}^n {n \choose k}n^{-k}\left({\frac{xy}{n + x + y} }\right)^k}\right\vert$

From Triangle Inequality and from Absolute Value Bounded Below by Zero:


 * $\displaystyle 0 \le \left \vert{\sum_{k=1}^n {n \choose k}n^{-k}\left({\frac{xy}{n + x + y} }\right)^k}\right\vert \le \sum_{k=1}^n \left \vert{{n \choose k}n^{-k}\left({\frac{xy}{n + x + y} }\right)^k}\right\vert$

Observe that:


 * $\implies \displaystyle 0 \le \sum_{k=1}^n \left\vert{ {n \choose k}n^{-k}\left({\frac{xy}{n + x + y} }\right)^k}\right\vert \le\sum_{k=1}^n \left\vert{\frac{xy}{n + x + y} }\right\vert^k$


 * $0 \to 0$ as $n \to +\infty$, trivially.

From Sum of Infinite Geometric Progression, the expression in the absolute value in the right hand term converges to:

By the Squeeze Theorem for Sequences:


 * $\displaystyle \sum_{k=1}^n \left\vert{ {n \choose k}n^{-k}\left({\frac{xy}{n + x + y} }\right)^k}\right\vert$

also converges to $0$.

By the Squeeze Theorem for Sequences:


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

converges to $0$, as well.

Which means:


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

Which is equivalent to our hypothesis:

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

That was a lot longer than I expected it would be. But I suppose I deserve it, for being obstinate to prove the laws from this definition. Is this good, LF? Thanks for the help. --GFauxPas 22:50, 9 February 2012 (EST)


 * Looking at it, I'm wondering, perhaps we can do away with the whole absolute value business and shorten the prove considerably? The image of $e^x$ is $(0..+\infty)$, do we know that yet? --GFauxPas 00:24, 10 February 2012 (EST)