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)

Proof 3 of derivative of exponential function
source: khan academy, larson book

Unit circle def'n
Just getting the skeleton of the definition out, I'll work on it later.

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

$\sin \theta := y$

$\cos \theta := x$



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

Sum of Squares of Sine and Cosine
This proof depends on the unit circle definition of sine and cosine.

Every point $(x,y)$ on the unit circle is $(\cos \theta,\sin \theta)$.

By definition, the graph of the unit circle is the locus of

$x^2 + y^2 = 1$

$\implies$

$\cos^2 \theta + \sin^2 \theta = 1$

Derivative of Arcsecant
Most of proof moved to here

$ |\dfrac {dy} {dx}| = \frac{1}{|x|\sqrt {x^2 - 1}}$

I'm stuck here. I'm trying to find a way to get rid of the absolute value sign on the left hand side. I'm going to think about it more, hopefully it's not a dead end... --GFauxPas 13:31, 28 October 2011 (CDT)
 * In such cases, it is probably a good idea to do a case distinction on the sign of $\frac{dy}{dx}$ (i.e., handle cases $<0$, $=0$ and $>0$ separately). Also, when $-1 0$, this simplifies to

$ \dfrac {dy} {dx} = \frac{1}{|x|\sqrt {x^2 - 1}}$

Because $\frac{1}{|x|\sqrt {x^2 - 1}}$ is never non-positive, we have exhausted all the cases. ???

--GFauxPas 10:49, 1 November 2011 (CDT)


 * Well, not exactly. The idea is to read off the sign of $\dfrac{dy}{dx}$ from the sign of $\dfrac{dx}{dy}$ (they are the same), depending on $x$. Most generally, one investigates when the absolute value does, and when it does not, coincide with the identity, and separates those cases. In the present case, when the domain is corrected to something where $\operatorname{arcsec}$ in fact takes real values, one has to be incredibly careful to make the statement precise. I suggest you read some more, eg. at MathWorld. --Lord_Farin 14:00, 1 November 2011 (CDT)

Thanks for your help Lord_Farin, my proof-writing skills or lack thereof are self-taught so I'm not very good.

How about this?

Since $\dfrac {dy} {dx} = \frac{1}{sec y \ tan y}$ the sign of $\dfrac {dy} {dx}$ is the same as the sign of $secy tan y$.

Writing $sec y tan y$ as $\frac{siny}{cos^2y}$ it is evident that the sign of $\dfrac {dy} {dx}$ is the same as the sign of $siny$.

From Sine and Cosine are Periodic on Reals $siny$ is never negative on its domain ($y \in [0..\pi] \land y \ne \pi/2$). Thus our absolute value is unnecessary and we have

$ \dfrac {dy} {dx} = \frac{1}{|x|\sqrt {x^2 - 1}}$

Which is what we desired to prove?

??? Thanks for your patience. --GFauxPas 11:02, 2 November 2011 (CDT)


 * I wrote it out today, and arrived at a similar formula. Actually, I even used about the same reasoning. Autodidactism is not a problem, it might just conflict with some standard notions in the literature. I recommend you read a lot of proofs here on PW (where the proof style is somewhat more rigorous than in most literature), and also read a lot of books on subjects you deem interesting. --Lord_Farin 12:29, 2 November 2011 (CDT)

Thanks for the help Lord_Farin, I'll do that. For now, Derivative of arcsecant function. --GFauxPas 13:28, 2 November 2011 (CDT)


 * Furthermore, I prefer these: ,  above the spacing you choose (  ) for the eqn template. It enhances readability. --Lord_Farin 11:38, 7 November 2011 (CST)
 * One more thing: You could use single equality signs to effectively add a header level. This will help keep this page structured (cf. this). --Lord_Farin 11:39, 7 November 2011 (CST)