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)

Derivative of a Series?
I'm learning convergence and divergence of sequences and series in my Calc Class, and Larson is implicitly using a theorem here, and it seems significant enough to warrant a page on PW if it's not already up, can someone explain what he did here?

The book is analyzing:


 * $\displaystyle \sum_{n=1}^{\infty} \ (-1)^n \frac {\sqrt{n}}{n + 1}$

He wants to use the Alternating Series Test, so to prove that $a_{n+1} \le a_n$, he creates a differentiable real function:


 * $f(x)= \displaystyle \frac {\sqrt{x}}{x + 1}$

and takes the derivative of that. The derivative is negative so $f$ is a decreasing function (for $x > 1$) but how does it prove that:


 * $\left \langle{ \displaystyle \frac {\sqrt{n}}{n + 1}}\right \rangle$

is decreasing? --GFauxPas 20:45, 4 February 2012 (EST)


 * Use FTIC and some result on inequality of integrals, together with the bounds $n, n+1$. I'm sure you'll figure from there. --Lord_Farin 17:14, 5 February 2012 (EST)


 * Oh, uh, hm. Does it use this guy?: Euler-Maclaurin Summation Formula. I don't know of any other things that tie together the limits of continuous functions with the limits of series. I guess I'll sleep on it. --GFauxPas 18:06, 5 February 2012 (EST)


 * I admit, I was testing you for the fun of it. You wanted to prove that mentioned sequence was decreasing; I merely tried to put up a sketch of an approach that could show that particular assertion. You want to investigate $\displaystyle f(n+1)-f(n) = \int_n^{n+1}f'(x)\ \mathrm dx <0$. That's what I was trying to say. --Lord_Farin 11:11, 6 February 2012 (EST)


 * Sorry to let you down. Are you doing what Larson is doing, or something different? --GFauxPas 11:35, 6 February 2012 (EST)


 * ah, and that's true for all $n \in \N$! Got it! thanks LF --GFauxPas 12:15, 6 February 2012 (EST)


 * LF thanks for the idea, I put it up: Monotonicity of Sequences. Do you think the proof needs induction, or it's okay the way it is? --GFauxPas 13:28, 7 February 2012 (EST)


 * It's okay as it stands. I think it might be a good idea to stipulate the domain of $f$ more explicitly, as to enable the statements of the derivative being positive and the other three to be more explicitly mentioned (i.e., $\forall x \in ...$). Otherwise, no big comments. --Lord_Farin 13:33, 7 February 2012 (EST)
 * Updated to house style. --Lord_Farin 14:11, 7 February 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)