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)

3 implies 4
Now someone prove that the limit exists as $n \to +\infty$. Any takers? --GFauxPas 12:26, 7 February 2012 (EST)

This might start you off:


 * $\displaystyle \frac {\left({n-1}\right) \left({n-2}\right) \ldots \left({n-m}\right)} {n^m} = \frac {\left({n-1}\right)} n \frac {\left({n-2}\right)} n \cdots \frac {\left({n-m}\right)} n$

Each factor is less than $1$ and we already know (from somewhere) that the sum without these factors converges. Use the squeeze. --prime mover 16:25, 7 February 2012 (EST)


 * More explicit would be to show that the difference $e^x - \left({1 + \frac x n}\right)^n$ converges to zero. It gives the value of the limit for free. --Lord_Farin 16:27, 7 February 2012 (EST)


 * O yes of course, it's $n$ that's going to $\infty$. --prime mover 16:31, 7 February 2012 (EST)


 * But LF, we're defining $e^x$, how can we use it as something other than the sequence if we're assuming the limit definition? --GFauxPas 17:30, 7 February 2012 (EST)


 * We want to show 3 implies 4. I probably should have said that this $e^x$ is the power series definition ($\exp x$, if you wish). --Lord_Farin 03:11, 8 February 2012 (EST)


 * Aaah I get it, I was thinking about it in a weird way, you explained it fine. --GFauxPas 06:09, 8 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)