User:Neocapitalist





That which is not forbidden is compulsory. This is the law of human existence.


 * User:Neocapitalist/How to pick your nose
 * User:Neocapitalist/evolution debates
 * User:Neocapitalist/SDnet
 * http://en.wikipedia.org/wiki/Help:Formula
 * Thats: Help:Formula
 * User:Neocapitalist/He-man (Tribute to PA...)
 * User:Neocapitalist/boxes

other stuff
$$\mathbf{v} \in \mathbb{R}^n$$ $$\mathbf{v} = \langle v_1, v_2, v_3, ..., v_n \rangle$$

points on a sphere
The volume of a sphere is given by $$\frac{4\pi}{3}R^3$$; Then $$\frac{dV}{dt} = \frac{4\pi}{3}3r^2\frac{dR}{dt} = 4\pi r^2\frac{dr}{dt}$$. Since the infinitesimal distance on the surface of the sphere is given by $$d\theta\ ^2 + d\phi\ ^2 = dr^2$$, differentiating gives $$\frac{d\theta}{dt} d\theta + \frac{d\phi}{dt} d\phi = \frac{dr}{dt} dr = \left (\frac{1}{4\pi r^2} \right) \frac{dV}{dt} dr$$

points on a hypersphere
The volume of a hypercube should be given by $$ \int_{-R}^{R} \left (\int_{-y}^{y} \left (\sqrt{R^2-x^2-y^2}\right )^3 dx \right) dy $$ Then, if I have my numbers right,

Proof there exists a number of the form $$a^n$$ s.t. a, n are irrational, and $$a^n$$ is rational
Let $$k=n=\sqrt{2}$$. Then $$k^n=\sqrt{2}^\sqrt{2}$$. Now, $$\sqrt{2}^\sqrt{2}$$ is either rational or irrational. If it is rational, then the theorem is proven. If it is irrational, then we can choose a new $$a=(\sqrt{2}^\sqrt{2})$$. Then $$(\sqrt{2}^\sqrt{2})^\sqrt{2}=\sqrt{2}^2=2$$, which is rational.

Riemann
$$\zeta(z)=\sum_{n=0}^\infty \frac{1}{n^z}$$ $$\zeta(z)=0\wedge z\ne-2n$$$$\Rightarrow Re(\zeta(z))=\frac{1}{2}$$

Wiggy
$$\int \equiv $$