Vector substitution

Question

I have a huge dynamic system in wxmaxima, and I need to do some vector substitution but it ends up just with some crazy results. This is what I need to do:

forces:[
    F1=[x1,y1,z1],
    F2=[x2,y2,z2]
];
equations:[F3=-F2];
subst(forces,subst(equations,F1+F3));

the result which I'm seeking is just a simple [x1+x2,y1+y2,z1+z2], but I got instead: [[x1-x2,x1-y2,x1-z2],[y1-x2,y1-y2,y1-z2],[z1-x2,z1-y2,z1-z2]]

any suggestions?

What is the value of `F3`? – Robert Dodier Oct 24 '19 at 21:40 — Robert Dodier, Oct 24 '19 at 21:40

score 2 · Accepted Answer · answered Oct 27 '19 at 17:48

OK, that is pretty puzzling, although I see now what's going on.

subst is serial (one by one) substitution, so subst([F1 = ..., F2 = ...], ...) is equivalent to subst(F2 = ..., subst(F1 = ..., ...)). That is, substitute for F1 first and then substitute F2 into the result of that.

However the result of subst(F1 = [x1, y1, z1], F1 - F2) is [x1 - F2, y1 - F2, z1 - F2]. You can see now what's going to happen if you substitute F2 into that -- you'll get the messy nested list result.

I think if you try psubst (parallel substitution) you'll get the expected result.

(%i2) forces:[
    F1=[x1,y1,z1],
    F2=[x2,y2,z2]
];
(%o2)        [F1 = [x1, y1, z1], F2 = [x2, y2, z2]]
(%i3) equations:[F3=-F2];
(%o3)                      [F3 = - F2]
(%i4) subst(equations, F1 + F3);
(%o4)                        F1 - F2
(%i5) psubst (forces, %o4);
(%o5)              [x1 - x2, y1 - y2, z1 - z2]
(%i6) psubst(forces, subst(equations, F1 + F3));
(%o6)              [x1 - x2, y1 - y2, z1 - z2]

Vector substitution

1 Answers1