other stuff
edit
v
∈
R
n
{\displaystyle \mathbf {v} \in \mathbb {R} ^{n}}
v
=
⟨
v
1
,
v
2
,
v
3
,
.
.
.
,
v
n
⟩
{\displaystyle \mathbf {v} =\langle v_{1},v_{2},v_{3},...,v_{n}\rangle }
points on a sphere
edit
The volume of a sphere is given by
4
π
3
R
3
{\displaystyle {\frac {4\pi }{3}}R^{3}}
Then
d
V
d
t
=
4
π
3
3
r
2
d
R
d
t
=
4
π
r
2
d
r
d
t
{\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
d
θ
2
+
d
ϕ
2
=
d
r
2
{\displaystyle d\theta \ ^{2}+d\phi \ ^{2}=dr^{2}}
differentiating gives
d
θ
d
t
d
θ
+
d
ϕ
d
t
d
ϕ
=
d
r
d
t
d
r
=
(
1
4
π
r
2
)
d
V
d
t
d
r
{\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
edit
The volume of a hypercube should be given by
∫
−
R
R
(
∫
−
y
y
(
R
2
−
x
2
−
y
2
)
3
d
x
)
d
y
{\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
a
n
{\displaystyle a^{n}}
a
n
{\displaystyle a^{n}}
edit
Let
k
=
n
=
2
{\displaystyle k=n={\sqrt {2}}}
k
n
=
2
2
{\displaystyle k^{n}={\sqrt {2}}^{\sqrt {2}}}
2
2
{\displaystyle {\sqrt {2}}^{\sqrt {2}}}
a
=
(
2
2
)
{\displaystyle a=({\sqrt {2}}^{\sqrt {2}})}
(
2
2
)
2
=
2
2
=
2
{\displaystyle ({\sqrt {2}}^{\sqrt {2}})^{\sqrt {2}}={\sqrt {2}}^{2}=2}
ζ
(
z
)
=
∑
n
=
0
∞
1
n
z
{\displaystyle \zeta (z)=\sum _{n=0}^{\infty }{\frac {1}{n^{z}}}}
ζ
(
z
)
=
0
∧
z
≠
−
2
n
{\displaystyle \zeta (z)=0\wedge z\neq -2n}
⇒
R
e
(
ζ
(
z
)
)
=
1
2
{\displaystyle \Rightarrow Re(\zeta (z))={\frac {1}{2}}}
∫
≡
{\displaystyle \int \equiv }
