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)

Tangent Line
Let $f$ be a real function that is continuous on some closed interval $[a..b]$ and differentiable and concave up on some open interval $(a..b)$.

Then all the tangent lines to $f$ are below the graph of $f$.

Proof
Let $\mathcal T$ be the tangent line to $f$ at some point $\left(c, f\left(c\right)\right)$, $c \in \mathbb I$.

From the gradient-intercept form of a line, given any point $\left({x_1, y_1}\right)$ and the gradient $m$, the equation of such a line is:


 * $y - y_1 = m \left({x - x_1}\right)$

$\implies y = m\left(x - x_1\right) + y_1$

For $\mathcal T$:


 * $y = \mathcal T \left(x\right)$


 * $y_1 = f\left(c\right)$


 * $m = f^\prime \left(c\right)$


 * $x = x$


 * $x_1 = c$

so


 * $\mathcal T \left(x\right) = f^\prime \left(c\right)\left(x - c\right) + f\left(c\right)$

Consider the graph of $f$ to the right of $\left(c, f\left(c\right)\right)$, that is, any $x$ in $(c..b)$.

Let $d$ be the directed vertical distance from the graph of $f$ to $\mathcal T$. That is, if $\mathcal T$ is below $f$, then $d < 0$. If $\mathcal T$ is above $f$, then $d > 0$.

From the diagram, it is apparent that $\mathcal T$ is below $f$, but seeking a contradiction, assume it's above, i.e. assume that $d > 0$.

$d$ can be evaluated by

$d = f\left(x\right) - \mathcal T \left(x\right)$


 * $= f\left(x\right) - f^\prime \left(c\right)\left(x - c\right) - f\left(c\right)$