Comment #3 on issue 1905 by abro...@verizon.net: Differentiation and
Geometric Algebra
http://code.google.com/p/sympy/issues/detail?id=1905
Also in your case gamma_x**2 = -1 so that exp(x*gamma_x) =
cos(x)+sin(x)*gamma_x and
d/dx(exp(x*gamma_x)) = -sin(x)+cos(x)*gamma_x. If gamma_x**2 = 1 then
exp(x*gamma_x) = cosh(x)+sinh(x)*gamma_x and
d/dx(exp(x*gamma_x)) = sinh(x)+cosh(x)*gamma_x. The actual formula for
exp(A) if A is
a multivector such that A*A is a scalar would be:
exp(A) = cosh(sqrt((A*A)()))+sinh(sqrt((A*A)()))A
you have to use (A*A)() to extract the scalar part of the scalar
multivector A*A as a
sympy symbol.
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