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)

Normal Distributions
There are a bunch of definitions and theorems I plan on putting up regarding normal distributions. However, as Brace/Brace's book does not assume knowledge of Calculus, I'm going to have to make some calls. Regarding the definite integral:


 * $\displaystyle \int_a^b \frac 1 {\sqrt{2\pi}} \exp \left({-\frac 1 2x^2 }\right) \, \mathrm dx$

I can approach it two ways. As a function of two variables $a,b$ or as a function of intervals of the form $[a..b]$. The first option is I think easier, but the second option might closer represent the underlying sample space of all real intervals. Or am I arguing semantics? Also, if the second way, what's the best way to represent "the set of all real intervals"?

Also, it's often used as a real function:


 * $x \mapsto \displaystyle \int_{\to -\infty}^x \frac 1 {\sqrt{2\pi}} \exp \left({-\frac 1 2t^2 }\right) \, \mathrm dt$

Should I give that guy a transclusion, or put them both on one page? --GFauxPas (talk) 15:10, 12 December 2012 (UTC)


 * The first stage is for the raw indefinite integral itself to be extracted into a separate page (it might already be, as the Gaussian integral) - not trivial to solve of course.


 * You can then treat the def int just as a standard instance of a definite integral: I would treat it as a function of two variables a and b because the interval-ness of it is completely subsumed by the integral-ness of it. Once you've established it's a d.i. you don't need to establish that it's an interval - you know it is because it's an integral.


 * The second is then a function of one variable, $x$. Give these two babes separate pages to start with, both linking to the same indef int (see above) and then if it's appropriate we can merge them with a transclusion. Maybe transclude both def ints into the one main indef int page. That will set us a precedent for all calculus pages: make the indefinite integral the main page and transclude the def int as a subpage. I confess I haven't addressed integral calculus since this new paradigm evolved, so none of this has been even thought about. Feel free to play. --prime mover (talk) 16:24, 12 December 2012 (UTC)


 * I hunted around a bit and have found no sources that deal with the above as an indef integral - all sources skip right to the definite integral. I'll deal with the definite first, because I'd like to put up published theorems and definitions before I wax adventurous. --GFauxPas (talk) 13:52, 20 December 2012 (UTC)


 * IIRC the indefinite integral usually goes by the name "error function", $\operatorname{erf}$. --Lord_Farin (talk) 14:24, 20 December 2012 (UTC)

Continuous Random Variables
Boo, there's not enough foundation for continuous random variables to do stuff on normal distributions. That means more work for me -_-


 * $\displaystyle \operatorname{var} \left({X}\right) = \int_{x \in \Omega_X} \left({x - \mu}\right)^2 \operatorname{pdf}\left({x}\right) \, \mathrm dx$

Do we have stuff up about density functions? I couldn't find anything. --GFauxPas (talk) 15:21, 13 December 2012 (UTC)


 * The relevant articles are Definition:Probability Distribution and Definition:Probability Mass Function. Most of the abstract measure-theoretic foundation for analysis has been laid down, but indeed the application to random variables is still (largely) missing. --Lord_Farin (talk) 15:31, 13 December 2012 (UTC)

To do
I'll need this theorem in order to solidify some results with the definite integrals above.

Let $f$ be a differentiable real function. Then:


 * $\dfrac {f(x+h) - f(x-h)} {2h} \to f\,'(x)$ as $h \to 0$ --GFauxPas (talk) 22:28, 19 December 2012 (UTC)


 * Is the "2" in the denominator extraneous? --GFauxPas (talk) 22:31, 19 December 2012 (UTC)


 * Not unless $f'(x) = 0$. Try multiplying both sides with $2$. --Lord_Farin (talk) 23:30, 19 December 2012 (UTC)


 * Oh, okay, right. So to prove this I'm going to use that $f\,'_+ = f\,'_-$, from Limit iff Limits from Left and Right, right? And then what's the next step, can I have a hint?


 * It may be worthwhile to add definitions for "derivative from the left" and "derivative from the right". This also allows for neater structuring of Definition:Differentiable/Real Function/Interval.


 * Furthermore, it appears that the following equivalent formulation of differentiability is not covered yet (it's rather trivial that they're equivalent but needs proof nonetheless):


 * Let $x \in \R$ be a real number.


 * Let $f$ be a real function whose domain is a neighborhood of $x$.


 * Then $f$ is said to be differentiable at $x$ iff the following limit exists:


 * $f' \left({x}\right) := \displaystyle \lim_{h \to 0} \frac {f \left({x + h}\right) - f \left({x}\right)} {h}$


 * If this is the case, $f' \left({x}\right)$ is called the derivative of $f$ at $x$.


 * With this definition, we observe that your expression is (one half times) $\displaystyle \lim_{h \to 0} \frac {f \left({x + h}\right) - f \left({x}\right)} {h} + \frac {f \left({x - h}\right) - f \left({x}\right)} {-h} = 2 f' \left({x}\right)$, using that $-h \to 0$ iff $h \to 0$. --Lord_Farin (talk) 08:16, 20 December 2012 (UTC)


 * Equivalence of Definitions of the Derivative, Definition:Right-Hand Derivative, Definition:Left-Hand Derivative --GFauxPas (talk) 12:10, 20 December 2012 (UTC)

Is there a such thing as a one-sided (real valued) definite integral? --GFauxPas (talk) 13:54, 20 December 2012 (UTC)


 * The closest thing I can think of is Definition:Stieltjes Function of Measure on Real Numbers. --Lord_Farin (talk) 14:24, 20 December 2012 (UTC)

Continuity of Inverse Functions
Larson, appendix, A12

Theorem
Let $f: \mathbb I \to \mathbb J$ be a real function, where $\mathbb I$ is some real interval.

Let $f$ admit an inverse.

Then, if $f$ is continuous on its domain, then $f^{-1}$ is continuous on its domain.

Comment: yowzers. That's ... a bit of a mess. First off, the part two should be separate from the part one, and they can be linked together corollaryishly. Nextishwise, refactor: Is there anything else? --Dfeuer (talk) 03:35, 1 February 2013 (UTC) Ah yes, one more:
 * A strictly monotone bijection between tosets is an order isomorphism.
 * An order isomorphism between linearly ordered spaces is a homeomorphism.
 * The order topology on an interval in a linearly ordered space is the same as its subspace topology. --Dfeuer (talk) 03:37, 1 February 2013 (UTC)


 * Well I know it's messy, I'm just trying to make the proof work before I make it neat. Unfortunately I don't know enough to generalize results to everything you have said, as I have not learned order theory or topology. --GFauxPas (talk) 03:41, 1 February 2013 (UTC)


 * The order topology on a totally ordered set is typically one of the first ones introduced in a topology text. You could probably get through those basics in a few hours. The usual topology on the reals is the order topology, so the ideas are very similar. The basic idea is that a set is "open" iff it is a union of "basic" open sets, which take one of three forms: open rays to the left, open rays to the right, and (bounded) open intervals. The last bit, about the order topology on an interval being the subspace topology, means you don't have to make yourself crazy about the endpoints. You could probably express all these ideas directly in terms of real numbers &hellip; somehow. --Dfeuer (talk) 03:50, 1 February 2013 (UTC)


 * And what I'm calling an interval here is called a Convex Set. --Dfeuer (talk) 03:55, 1 February 2013 (UTC)

Proof
Let $f$ be invertible on $\mathbb I$.

Part 1:


 * $f$ is a bijection from Bijection iff Inverse is Bijection.


 * $f$ is then strictly monotone from Continuous Bijection of Interval is Strictly Monotone.

Part 2:

Because $f$ is a bijection, $\mathbb J = f\left({\mathbb I}\right)$.

From Image of Interval by Continuous Function is Interval, $\mathbb J$ is an interval.


 * $\downarrow$ PROVE THIS $\downarrow$

Let $a$ be an interior point of $\mathbb J$. Then $f^{-1} \left({a}\right)$ is an interior point of $\mathbb I$.


 * $\uparrow$ PROVE THIS $\uparrow$

Let $\epsilon > 0$. From Interval Defined by Betweenness there exists some $0 < \epsilon_1 < \epsilon$ such that:


 * $\mathbb I\,' = \left ({f^{-1}\left({a}\right) - \epsilon_1 \,.\,.\, f^{-1}\left({a}\right) + \epsilon_1} \right)$

is a subinterval of $\mathbb I$.


 * $\downarrow$ does this need proof? $\downarrow$

Because $f$ is strictly monotone, $f\left({\mathbb I\,'}\right) \subseteq \mathbb J$.


 * $\uparrow$ Does this need proof? $\uparrow$

From Interval Defined by Betweenness there exists some $\delta > 0$ such that:


 * $\left ({a - \delta \,.\,.\, a + \delta} \right)$

is a subinterval of $f\left({\mathbb I\,'}\right)$.

Note that if $y \in \left ({a - \delta \,.\,.\, a + \delta} \right)$, then $\left \vert {x - a} \right \vert < \delta$.

This means that:


 * $\left \vert {x - a} \right \vert < \delta \implies \left \vert {f^{-1}\left({y}\right) - f^{-1}\left({a}\right)} \right \vert < \epsilon_1 < \epsilon$.

Thus $f^{-1}$ is continuous at $a$.

As $a$ was an arbitrary interior point, all that remains to be proven is to address the endpoints of $\mathbb J$, if $\mathbb J$ is half-open or closed.

--GFauxPas (talk) 03:24, 1 February 2013 (UTC)