User:Neocapitalist

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7h15 u53r 15 31337.

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$\frac{d}{dx} \int_{a}^{g(x)}f(t)dt$

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This user is a Capitalist. |}

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

User:Neocapitalist/How to pick your nose

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http://en.wikipedia.org/wiki/Help:Formula

Thats: Help:Formula

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1 Math

[edit] Math

[edit] other stuff

$\mathbf{v} \in \mathbb{R}^n$

$\mathbf{v} = \langle v_1, v_2, v_3, ..., v_n \rangle$

[edit] 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 $\frac{d\theta}{dt} d\theta + \frac{d\phi}{dt} d\phi = \frac{dr}{dt} dr = \left (\frac{1}{4\pi r^2} \right) \frac{dV}{dt} dr$

[edit] 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,

[edit] 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.