This is related to Simplifying cross product expressions
Consider the following:
F2[R + 2 a] = F2[R] + 2 a D[F2[R], R]
F3[R + 3 a] = F3[R] + 3 a D[F3[R], R]
F[R + 4 a] = F[R] + 4 a D[F4[R], R]
FullSimplify[TensorExpand[Cross[F3[R + 3 a], F[R + 4 a] - F2[R + 2 a]]]]
(*=> F[R]\[Cross]F3[R] - 3 F[R]\[Cross](a Derivative[1][F3][R]) +
F2[R]\[Cross]F3[R] + 3 F2[R]\[Cross](a Derivative[1][F3][R]) -
2 F3[R]\[Cross](a Derivative[1][F2][R]) +
4 F3[R]\[Cross](a Derivative[1][F4][R]) +
6 (a Derivative[1][F2][R])\[Cross](a Derivative[1][F3][R]) +
12 (a Derivative[1][F3][R])\[Cross](a Derivative[1][F4][R])
a
is scalar, but Mathematica doesn't know it, so it isn't moved outside the cross product.
I tried
FullSimplify[TensorExpand[Cross[F3[R + 3 a], F[R + 4 a] - F2[R + 2 a]]],
Assumptions -> {F[R], F2[R], F3[R], D[F[R], R], D[F3[R], R], D[F2[R], R]} \[Epsilon] Vectors[3]]
but the result is the same. How can I specifically declare that a
is a scalar?
\[Epsilon]
is not the same as\[Element]
! $\endgroup${F[R], F2[R], F3[R], D[F[R], R], D[F3[R], R], D[F2[R], R]} \[Element] Vectors[3]]
does not express what you think it expresses; it says that the single vector{F[R], F2[R], F3[R], D[F[R], R], D[F3[R], R], D[F2[R], R]}
is inVectors[3]
and thus evaluates toFalse
. Rather, you need a way of saying they're all inVectors[3]
, e.g.# \[Element] Vectors[3] & /@ {F[R], F2[R], F3[R], D[F[R], R], D[F3[R], R], D[F2[R], R]}
. But for this application, all that's needed is information ona
. $\endgroup$