Why do Replace and ReplaceAll give different resul

Question

Maybe someone can explain to me why Replace gives a different answer than ReplaceAll, even though I am using one rule, and, I think, I have one expression.

According to the documentation:

ReplaceAll looks at each part of expr, tries all the rules on it, and then goes on to 
the next part of expr. The first rule that applies to a particular part is used; no 
further rules are tried on that part, or on any of its subparts

and for Replace

A list of rules can be given. The rules are tried in order. The result of the first 
one that applies is returned. If none of the rules apply, the original expr is 
returned.

I have this expression:

z/(0.5 -1.4 z+z^2)

and this one rule

z -> Exp[I*w]

The following input

Clear[z]
hz = z/(0.5 - 1.4 z + z^2);
Replace[hz, z -> Exp[I*w]]
ReplaceAll[hz, z -> Exp[I*w]]

yields this result:

z/(0.5 -1.4 z+z^2)                    (*from Replace*)
E^(I w)/(0.5 -1.4 E^(I w)+E^(2 I w))  (*from ReplaceAll*)

So, ReplaceAll did the job I expected, but not Replace. The only thing I think I am missing here, is what makes Parts of an expression. So maybe hz is not 'one' expression? Ok, fine, but it should then have been able to apply the rule at least to one part of it. But it did not do any replacement anywhere

FullForm[hz]
Times[z,Power[Plus[0.5`,Times[-1.4`,z],Power[z,2]],-1]]

Isn't hz one expression? If not, how many parts does it have? Is it the number of leaves at the end of the graph generated by TreeForm[hz]? If so, then I see 3 'z' in there, and Replace[] should then have replaced one of them

I always used /. without thinking about it, which is the same as ReplaceAll, but now I used Replace[] thinking is was the same as /., I found this and noticed the different functions. (so many functions to learn, so little time :)

Answer 1

In fact, Replace and ReplaceAll are not the same. You can think of Replace as a more precise version, since with Replace you can specify the levels in expression on which you want the replacements to happen. The difference here is somewhat similar to that between Map and MapAll, except the subtlety that ReplaceAll does not work depth-first (see below). Another difference, also rather subtle, is that Replace takes the Heads option, while ReplaceAll does not, which makes Replace yet more precise than ReplaceAll.

By default, Replace works only at level 0, which is, entire expression. Your z is deeper however:

In[220]:= Clear[z]
hz = z/(0.5 - 1.4 z + z^2);


Position[hz, z]

Out[222]= {{1}, {2, 1, 2, 2}, {2, 1, 3, 1}}

If you use the level specification for Replace, you can achieve an effect similar to, but not always the same as that of ReplaceAll:

In[223]:= Replace[hz,z->Exp[I*w],{0,Infinity}]

Out[223]= E^(I w)/(0.5\[VeryThinSpace]-1.4 E^(I w)+E^(2 I w))

The difference between Replace with lev.spec {0,Infinity} and ReplaceAll is that the former acts depth-first, sub-expressions before expressions, while the latter works from larger expressions to their parts. It is discussed in more detail e.g. here. One example where this difference was used to one's advantage can be found in this post.

Coming back to default behavior of Replace, which operates on entire expression: it is very useful when you want to transform only the entire expression, but none of its parts (which may accidentally match the pattern in your rule). One example of such application of Replace is here.

Answer 2

I think the relevant bit of the documentation for Replace is: "applies a rule or list of rules in an attempt to transform the entire expression expr. " i.e., to transform the entire expression. Thus,

Replace[z, z -> Exp[I*w]]

would transform z to Exp[I*w], but your example fails because the rule does not match the entire expression.

Note that Replace takes an argument level spec; so to operate on the end leaves of your tree, try

Replace[hz, z -> Exp[I*w], -1]

giving

E^(I w)/(0.5 -1.4 E^(I w)+E^(2 I w))