New answers tagged clifford-algebras
I did some original research using Mathematica, and this was the simplest form I could find. Given two axis angles A and B, with a=||A|| and b=||B|| The resulting vector has the length c=acos(cos(a)cos(b)-A.B/(a b)sin(a)sin(b)) and is in the direction E=A b cos(b)sin(a)+a B cos(a)sin(b)+A x B sin(a)sin(b) So C=c*E/||E|| Not very monstrous, but it ...
If you know the Hessian you can compute $\nabla^2$, but if you know only $\nabla^2$, you can't compute the Hessian, since you only know the trace and the antisymmetric parts of the Hessian. The traceless symmetric part is information that $\nabla^2$ does not have.
Top 50 recent answers are included