Why does Expand work in one case but not the other?

Question

I was playing with the Expand command and found something that puzzled me. Using Trace did not clarify the situation.

expr=Exp[x^2] (-2 Exp[-x^2] + 4 Exp[-x^2] x^2);
Expand[expr, Exp[x^2]]
(* expr is left unchanged *)
Expand[expr, Exp[-x^2]]
(* - 2 + 4 x^2 - it worked! *)
Expand[-2 Exp[-x^2] + 4 Exp[-x^2] x^2, Exp[-x^2]]
(* -2 Exp[-x^2] + 4 Exp[-x^2] x^2 is left unchanged *)

What exactly happened when the second call to Expand above successfully simplified expr?

Note: Simplify also does the job, but I am trying to figure out Expand.

Thanks for the help.

I would like to clarify my question.

I reasoned that Mathematica would evaluate the second call to Expand above in two steps:

Step 1) Leave Exp[x^2] aside and Expand the remainder of expr with argument Exp[-x^2].

Step 2) Evaluate the resulting expression.

However, the third call above to Expand does not change its argument, and Step 1 is identical to the third call. So Step 1 should leave expr unchanged and therefore in Step 2 expr would be evaluated. But expr evaluates to itself, so Step 2 also should not change expr. So neither of the two steps in the second call should change expr. But the second call does change expr. That is what I do not understand. My hypothesis of a two-step evaluation must be wrong, but I do not know why.

Answer 1

Expand[expr, patt], according to the documentation, works as follows:

leaves unexpanded any parts of expr that are free of the pattern patt.

Consider first an example from the docs:

(* Leave parts free of 1+x unexpanded: *)
Expand[(1 + x)^2 + (2 + x)^2, 1 + x]

1 + 2 x + x^2 + (2 + x)^2

That means "if there is 1+x, expand; if there is no 1+x, don't expand".

Now,

Expand[expr, Exp[x^2]]

means "if there is Exp[x^2], expand; leave the rest unchanged". Exp[x^2], as it is, is present only outside of the bracket - it has no way to be expanded further (like there is no way to further expand a single x); the bracket is free of it, so it's left unexpanded.

On the other hand,

Expand[expr, Exp[-x^2]]

says "if there is Exp[-x^2], expand; leave the rest unchanged". This occurs in the bracket, and what's outside the bracket is free of it. Hence, obviously Exp[x^2] is undexpanded (again: there's even no way to expand it by itself), but the bracket isn't free of Exp[-x^2], hence it is expanded.

Finally, as a summary and final example, consider

Expand[expr, y]

which is left unexpanded, because the whole expr is free of y.