User:Neocapitalist

other stuff

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

${\displaystyle \mathbf {v} =\langle v_{1},v_{2},v_{3},...,v_{n}\rangle }$

points on a sphere

The volume of a sphere is given by

${\displaystyle {\frac {4\pi }{3}}R^{3}}$;

Then

${\displaystyle {\frac {dV}{dt}}={\frac {4\pi }{3}}3r^{2}{\frac {dR}{dt}}=4\pi r^{2}{\frac {dr}{dt}}}$.

Since the infinitesimal distance on the surface of the sphere is given by

${\displaystyle d\theta \ ^{2}+d\phi \ ^{2}=dr^{2}}$,

differentiating gives

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

points on a hypersphere

The volume of a hypercube should be given by

${\displaystyle \int _{-R}^{R}\left(\int _{-y}^{y}\left({\sqrt {R^{2}-x^{2}-y^{2}}}\right)^{3}dx\right)dy}$

Then, if I have my numbers right,

Proof there exists a number of the form ${\displaystyle a^{n}}$ s.t. a, n are irrational, and ${\displaystyle a^{n}}$ is rational

Let ${\displaystyle k=n={\sqrt {2}}}$. Then ${\displaystyle k^{n}={\sqrt {2}}^{\sqrt {2}}}$. Now, ${\displaystyle {\sqrt {2}}^{\sqrt {2}}}$ is either rational or irrational. If it is rational, then the theorem is proven. If it is irrational, then we can choose a new ${\displaystyle a=({\sqrt {2}}^{\sqrt {2}})}$. Then ${\displaystyle ({\sqrt {2}}^{\sqrt {2}})^{\sqrt {2}}={\sqrt {2}}^{2}=2}$, which is rational.

Riemann

${\displaystyle \zeta (z)=\sum _{n=0}^{\infty }{\frac {1}{n^{z}}}}$ ${\displaystyle \zeta (z)=0\wedge z\neq -2n}$${\displaystyle \Rightarrow Re(\zeta (z))={\frac {1}{2}}}$

Wiggy

${\displaystyle \int \equiv }$