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)

What to Call this Theorem?
Hello friends, I'd like to add a proof for:


 * $\displaystyle \int \frac {1}{x^2 + a^2} \ \mathrm dx$

but I don't know what to call it. "Integral of a Rational Function"? "Integral Involving Arctangent"? "Integral of 1 Over (x^2 + a^2)"? What should I name the page? --GFauxPas 06:55, 15 December 2011 (CST)

Moved to Integral Involving Arctangent, please change the name if you think of something better. --GFauxPas 14:02, 15 December 2011 (CST)

where $a$ is a strictly positive constant and $a^2 > x^2$.

Moved to Integral Involving Arcsine --GFauxPas 08:26, 16 December 2011 (CST)


 * I appreciate the work you have been doing on these integrals. The only thing bothering me slightly is that you write equations like $\mathrm d x = a \cos\theta \mathrm d\theta$ while I suspect that you do not really understand what this means (the theory of differentials is really quite delicate and technical to deal with formally). Therefore, I suggest you stick with the substitution theorem instead. --Lord_Farin 08:38, 16 December 2011 (CST)
 * Just to muddy the waters, IMO it's okay to write $\mathrm d x = a \cos\theta \mathrm d\theta$ as long as the technique is defined as shorthand for the full derivatives. This would need to be done once on the page definition Integration by Substitution. This would remove the need for all this unsightly and unwieldy work every time it is invoked. Mind, I'm not a teacher (and never will be, officially, for legal reasons) so my view is subservient to LF's.
 * The same would apply to Integration by Parts. There's probably more work to be done on those pages so as to ensure the notation is appropraitely defiend (I havent checked because I'm doing other stuff atm). --prime mover 15:09, 16 December 2011 (CST)

Sure. I'm not sure exactly how you want me to write it though, what do you mean by the substitution theorem? --GFauxPas 08:45, 16 December 2011 (CST)
 * Well, it is quite easy to show that one may also use the substitution theorem (that is, Integration by Substitution) for indefinite integrals by plugging in values temporarily (this might need a separate page). Then you can just use this route instead of writing the equation with differentials (not that the equality is false; it is just a bit of an intuitive, physicist's shorthand which is hard to state formally). --Lord_Farin 08:58, 16 December 2011 (CST)
 * You seem to have bad experiences with physics :) . The other day in Calculus class we were challenged to solve an integral, I don't remember what it was exactly, that would have been simple were it not for a $ + 6$ floating around. So I said to my Calc II professor, "can't we assume it's negligible?" --GFauxPas 09:01, 16 December 2011 (CST)

Something like this? This looks... bizarre:

--GFauxPas 10:57, 16 December 2011 (CST)

Or maybe:

There's probably a more elegant way to do this ... --GFauxPas 11:25, 16 December 2011 (CST)


 * I'd mention Derivative of an Inverse Function to write down $a\cos\theta \frac{\mathrm d \theta}{\mathrm dx} = 1$ separately, and then just plug it in. Maybe by putting that equation in the middle, using $\implies$ to signify your conclusion. This would yield:


 * Hope that makes a bit more sense. --Lord_Farin 11:44, 16 December 2011 (CST)