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)

Thoughts on the picture, anyone? --GFauxPas 14:37, 9 November 2011 (CST)


 * Looks okay to me. I was going to get round to doing something similar myself at one point.


 * Mind, if you're planning on using it to illustrate sine and cosine, you might want to add the actual distances as lines. Oh, and purists may wince when they see x and y used to define the axes and the point on it, but I wouldn't be too fussed. --prime mover 14:47, 9 November 2011 (CST)

Try 2. I see I lost too many colors by saving it as a .gif, try 3 will be a .png or something.

File:Unitcirclev2.gif

\Let $P = (x,y)$ be a point on the unit circle centered at the origin.

Let $\theta$ be the angle formed by the arc $(1,0)$, $(x,y)$ subtending the origin, measured counterclockwise.

The unit circle definition of the trigonometric functions are $\cos \theta := x$

$\sin \theta := y$

That is, the directed distance between $P$ and the $x$-axis is the cosine, and the directed distance between $P$ and the $y$-axis is the sine.

Sources: khan academy "tau versus pi", wolfram mathworld "trigonometry"

After this is set up I can do a proof of the consistency between the right triangle definition and the circle definition. --GFauxPas 07:22, 17 November 2011 (CST)

Analysis
Let $f$ be a real function defined on some closed interval $[a..b]$.

Define the following sequence:

$S = \left \langle x_0, x_1 , x_2 , ... , x_{n-1}, x_n \right \rangle$

where $a = x_0$, $b = x_n$, and $x_0 < x_1 < x_2 < ... < x_n-1 < x_n$.

Let $\mathbb I_i$ be the $i$th sub-interval of $[a..b]$ of the form $[x_{n-1}..x_n]$, where $n$ is the $n$th term of $S$.

Further, let $c_i$ be an arbitrary constant in $\mathbb I_i$, and let $\Delta x_i$ be the width of $\mathbb I_i$.

Because $f$ is defined on $[a..b]$, the quantity $f(c_i) \cdot \Delta x_i$ exists in each $\mathbb I_i$.

The sum of all such quantities

$\displaystyle \sum_{i=1}^{n} f(c_i) \ \Delta x_i$, where $x_{i-1} < c_i < x_i$

is called a Riemann Sum of $f$ for the partition $\mathbb I$.

Note: I didn't yet go over what I wrote to make sure that it's accurate or precise, it's in sandbox-stage.

Geometric Interpretation
Rectangles. --GFauxPas 19:05, 24 November 2011 (CST)

Maybe we can have a page for Riemann sums in general, and then just point out the similarity in notation is not a coincidence? Just typing my thoughts here--GFauxPas 17:00, 24 November 2011 (CST)
 * Definition:Riemann Sum? Certainly, if you have a reliable source. --prime mover 17:04, 24 November 2011 (CST)
 * I have such a resource (lecture notes). It's in Dutch though, but can help to check for consistency.--Lord_Farin 17:14, 24 November 2011 (CST)
 * I'll dig around in my school's library next week. They're closed for Thanksgiving. --GFauxPas 17:20, 24 November 2011 (CST)

Third Axiom
(moved to Relative Frequency is a Probability Measure)

This proof doesn't seem quite right. I'm going to sleep on it and come back and look at it tomorrow. Feel free, PW-ers, to leave any comments you deem helpful. Induction isn't my strong point, and I have a feeling that I'm handwaving.--GFauxPas 22:25, 7 December 2011 (CST)
 * The critical error is that you prove something for every $n\in\N$, which is something different from having countably infinitely many experiments. I think you should use that you have a finite number of trials, and hence can have at most finitely many $\Pr(A_i)$ nonzero (from disjointness). Then I think you may reduce to this inductive proof. --Lord_Farin 02:53, 8 December 2011 (CST)
 * Thank you, Lord_Farin! How 'bout now? --GFauxPas 07:57, 8 December 2011 (CST)
 * You may have misunderstood. There could be infinitely many $A$, but only finitely many of them can assume a nonzero value (I think on PW this requires proof, though intuitively clear) as $n$ is finite and they are disjoint. So you would prove aforementioned observation, and then use above proof for all $A$ that have nonzero probability. Hope this formulation clears things up. --Lord_Farin 09:40, 8 December 2011 (CST)
 * I have to think about how to prove this, but as relative frequency is now defined, there definitely are only a finite number of $A$, am I wrong?. I think according to the second definition, defining RF as a limit, I may have to address infinitely many $A$, but I am not prepared to do that until I do sequences and series next semester. I'm not sure what you mean that only fnitely many of them are non-zero, I don't have that intuition here :(
 * The book I'm using has a proof, but I can't use it exactly as is because Gemignani has some of PW's axioms as theorems and vice versa. It also uses the second definition of RF, rather than the first. I will copy part of Gemignani's Induction step here for clarity, and I appreciate your patience with me. He is proving Axiom 3 for a probability measure in general.--GFauxPas 10:20, 8 December 2011 (CST)

We shall use finite induction. [base case] [induction hypothesis]

Let $A_1, A_2, ..., A_n, A_{n+1}$ be $n + 1$ events such that no two of these events can occur together. Set

$B = A_1 \cup A_2 \cup ... \cup A_n$.

Then $B \cap A_{n+1}$ is the empty set. For if $B \cap A_{n+1}$ contaned some element of $S$, [sample space] then $A_{n+1}$ and some $A_i, i = 1, ..., n,$ would have to contain some common element of $S$, which by assumption is impossible. Therefore $P(B \cap A_{n+1} )= 0$. But $A_1, A_2, ..., A_n, A_{n+1}$ is equivalent to $B \cup A_{n + 1}$. Therefore

$(12) \quad P(A_1, A_2, ..., A_n, A_{n+1}) = P(B \cap A_{n+1})$
 * $= P(B) + P(A_{n+1})$
 * $= P(A_1, A_2, ..., A_n) + P(A_{n+1})$.

However, $A_1, A_2, ..., A_n$ are $n$ events any two of which have an empty intersection... -Gemignani.

I misread a bit; I missed the 'finite' in '...in some finite reference class of events'. Certainly, with this restriction it is a probability measure. I however took the general $\sigma$-additivity allowing for countably infinitely many $A_i$. This restriction on the class of events means that the induction argument is valid without further ado, that is, the proof above this discussion now appears valid to me. --Lord_Farin 11:29, 8 December 2011 (CST)
 * Haha you didn't misread at all; I forgot to add "finite" in that page and your comments here made me realize my omission! I will have to research this more to see if the RF model can be extended to infinite reference classes--GFauxPas 11:36, 8 December 2011 (CST)