This is probably a very simple question, but I couldn't find a duplicate.
As everybody knows, {x, y} + v gives {x + v, y + v}. But if I intend v to represent a vector, for example if I am going to substitute v -> {vx, vy} in the future, then the result {x + v, y + v} is meaningless.
How I can indicate to Mathematica that v is not a scalar and functions like Plus should not treat it as one? I tried setting $Assumptions = v ∈ Vectors[2] but that didn't help.



InactivatePlusbecause there are other computations you want to do? – march Dec 24 '15 at 03:51PiecewiseExpand[f[t, 0, 2]]for my Koch snowflake parametrization, because I couldn't figure out make the answers here work for it. But seeing all the different approaches inspired a solution of my own that did work out (though still not as elegant as I would like). – Dec 25 '15 at 02:51