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)

User:GFauxPas/Sandbox/Zeta2/lnxln1-x/existence

User:GFauxPas/Sandbox/Zeta2/lnxln1-x/integrand

User:GFauxPas/Sandbox/Zeta2/lnxln1-x/evaluation

User:GFauxPas/Sandbox/Zeta2/FourierSeries/

User:GFauxPas/Sandbox/Zeta2/Informal Proof

Primitive Solutions of Pythagorean Equation
The set of all primitive Pythagorean triples is generated by:
 * $\left({2 m n, m^2 - n^2, m^2 + n^2}\right)$

where:
 * $m, n \in \Z$ are positive integers
 * $m \perp n$, i.e. $m$ and $n$ are coprime
 * $m$ and $n$ are of opposite parity
 * $m > n$.

Proof 2
Let $\left({A,B,C}\right)$ be a Definition:Pythagorean Triple:


 * $A^2 + B^2 = C^2$

By the Pythagorean theorem, this equation describes the sides of a right triangle:

(Picture)

By the definitions of sine and cosine:

That is,

Next, we invoke Equiangular Triangles are Similar and Proportionality is Equivalence Relation to write:

Is there a prettier way to do this?

We restrict tangent as follows.

On the open interval $\left ({0 \,.\,.\, \pi} \right)$, $\tan \frac \theta 2$ can take all strictly positive real numbers.

Restrict $\tan \dfrac \theta 2$ on this interval, and then further restrict it so that its image is contained in the rationals:


 * $\tan \dfrac \theta 2 = \dfrac p q$

where $\dfrac p q$ is written in canonical form.

From the Double Angle Formulas:

Substituting this into the above proportion:

Is the eqn template unsuitable for this? --GFauxPas (talk) 21:08, 16 September 2014 (UTC)