User:MarSch

Medicine patents

The joys of good math notation

If e is a bivector, then we have $e^{ix}=-e^{xi}\;$ or equivalently $e^{ij}=-e^{ji}\;$ . In quaternions much the same formula is true. We have $\mathrm {e} ^{\mathrm {i} \mathrm {j} }=\mathrm {e} ^{-\mathrm {j} \mathrm {i} }=\mathrm {e} ^{\mathrm {k} }\;$ , but $\mathrm {e} ^{\mathrm {i} x}=\mathrm {e} ^{-x\mathrm {i} }\;$ holds only if x is purely quaternionic: $x=j\mathrm {j} +k\mathrm {k} \;$ 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 $\int _{R}expd\mu \in \mathbf {R}$ and it is a real number. On the other hand it is much easier to simply calculate $\int _{\mathbf {R} }\exp \;\mathrm {d\mu } =\infty$ , because it is infinity. Unfortunately

Greek letters

are autoitalicized.

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

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

Main page subdivision hacking

Misc hacking

user:MarSch/sandbox
user:MarSch/templatelab

/deleteproposal
user:MarSch/sectionbox
user:MarSch/portalskeleton

take closer look at User:Korath/autovfd.js

Articles significantly edited

Mathematics, manifold, topological manifold, differentiable manifold, Laplace operator, scalar, scalar field, bijection, injection and surjection

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