User:MarSch

The joys of good math notation

If e is a bivector, then we have ${\displaystyle e^{ix}=-e^{xi}\;}$ or equivalently ${\displaystyle e^{ij}=-e^{ji}\;}$. In quaternions much the same formula is true. We have ${\displaystyle \mathrm {e} ^{\mathrm {i} \mathrm {j} }=\mathrm {e} ^{-\mathrm {j} \mathrm {i} }=\mathrm {e} ^{\mathrm {k} }\;}$, but ${\displaystyle \mathrm {e} ^{\mathrm {i} x}=\mathrm {e} ^{-x\mathrm {i} }\;}$ holds only if x is purely quaternionic: ${\displaystyle x=j\mathrm {j} +k\mathrm {k} \;}$ for real j and k.

Now let x, p, d and μ be real numbers and e a bivector on a compact 2-manifold R. Then we can calculate the integral ${\displaystyle \int _{R}expd\mu \in \mathbf {R} }$ and it is a real number. On the other hand it is much easier to simply calculate ${\displaystyle \int _{\mathbf {R} }\exp \;\mathrm {d\mu } =\infty }$, because it is infinity. Unfortunately Greek letters are autoitalicized.

This stuff is all pretty simple. Let's get into some deeper waters.

Let ${\displaystyle i:=({}_{\mathrm {i} }^{0}{}_{0}^{-\mathrm {i} })\;}$. Then ${\displaystyle i^{2}=1\;}$ so we have ${\displaystyle \mathrm {e} ^{ix}=\cosh x+i\sinh x\;}$. On the other hand ${\displaystyle \mathrm {e} ^{\mathrm {i} x}=\cos x+\mathrm {i} \sin x\;}$.

Misc hacking

take closer look at User:Korath/autovfd.js

Illegal numbers

I am in support of fully stating illegal primes, AACS encryption keys and other numbers and facts while they pose no legal risk to Wikipedia and while no official decision from the operators of Wikipedia has been made to censor such numbers.

Wikipedia:Babel
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