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
(for Relative Frequency is a Probability Measure)

Basis for the Induction
The case $n = 2$ is verified as follows:

Let $A, B$ be two pairwise disjoint events.

Let $p$ and $q$ be the amount of times $A$ and $B$ have been observed, respectively. Let $n$ be the total number of trials observed.

By the definition of pairwise disjoint, $A$ and $B$ never happened at the same time. So in all $n$ observations, $A$ happened $p$ times, $B$ happened $q$ times, and $A \lor B$ happened $p + q$ times. By hypothesis:

$\Pr \left({A \cup B} \right) = \dfrac {p + q} n$


 * $= \dfrac p n + \dfrac q n$


 * $= \Pr\left({A}\right) + \Pr\left({B}\right)$

This is the basis for the induction.

Induction Hypothesis
Let $A_1, A_2, A_3,..., A_j$ be $j$ pairwise disjoint events.

Assume $\Pr \left({\displaystyle \bigcup_{i=1}^j A_i}\right) = \Pr\left({A_1}\right) + \Pr\left({A_2}\right) + \Pr\left({A_3}\right)+ ... + \Pr\left({A_j}\right)$.

This is our induction hypothesis.

Induction Step
This is our induction step:

Let $A_1, A_2, A_3,..., A_j, A_{j+1}$ be $j+1$ pairwise disjoint events.

Define $C = A_1, A_2, A_3,..., A_j$.

Then $C$ and $A_{j+1}$ are also pairwise disjoint.

By the base case,

$\Pr \left({A \cup C} \right) = \Pr\left({A}\right) + \Pr\left({C}\right)$

The result follows by the Principle of Mathematical Induction.

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)