When applying the rule:
exp[c[x] + d[x] + e[x]] //. exp[a_[x] + b_[x]] -> exp[a[x]] + exp[b[x]] I would like to get
exp[c[x]] + exp[d[x]] + exp[e[x]] as a result. But instead I get the initial expression unchanged.
The rule only works for expressions with two additives:
exp[c[x] + d[x] + e[x]] //. exp[a_[x] + b_[x]] -> exp[a[x]] + exp[b[x]] How can I make it work for a general case?
The other answers do not explain why your rule doesn't work. They also don't capture the spirit of your question, in my opinion. I will use arbitrary functions so as not to be distracted by incidentals.
First, let's explain why f[c[x] + d[x] + e[x]] //. f[a_[x] + b_[x]] -> f[a[x]] + exp[b[x]] doesn't do what you want it to. You are wanting to use the Flat attribute of the Plus function to "break" f over addition, because you reason that since Plus[b, c, d]===Plus[b, Plus[c, d]], then f[Plus[b[x], c[x], d[x]]]/.f[p_[x] + q_[x]] -> f[b[x]]+f[c[x]+d[x]]. In fact, you may have noticed that it does work without the functions:
a+b+c/.x_+y_->{{x},{y}} {{a}, {b + c}}
The Blank pattern—and only the Blank pattern—is treated in a special way when used on Flat functions like Plus, because Blank is given a special meaning within Flat functions in the sense that it can match like f[__] (if it’s able to). In particular, a Blank pattern can match any sequence of arguments of f[a, b, c, ...] when f is flat. To drive home the point, we can use ReplaceList to list all of the different possible ways to match the pattern:
SetAttributes[f, Flat]; ReplaceList[f[b, c, d], f[x_, y_] -> {{x}, {y}}] {{{f[b]},{f[c,d]}}, {{f[b,c]},{f[d]}}, {{b},{f[c,d]}}, {{f[b,c]},{d}}}
However, p_[x], or any other pattern, is not endowed with this same special meaning within Flat functions, and so f[p_[x], q_[x], ...] has the effect of annihilating p_’s special abilities within Flat functions.
Here's the rule for non-Blank patterns (as of v11.3): If a pattern has to unflatten a flat function in order to match, then the pattern will not match. On the other hand, if a Blank pattern needs to unflatten a flat function in order to match, it will.
Here's the general case: Suppose g is any function and f is Flat. We are interested in transforming g[f[c[x], d[x], e[x]]] into f[g[c[x]], g[f[d[x], e[x]]] in a single step. Thus,
SetAttributes[f, Flat]; g[f[c[x], d[x], e[x]]] //. g[f[a_[x], b_]] :> f[g[a[x]], g[f[b]]] f[g[c[x]], g[d[x]], g[f[e[x]]]]
Notice that the final result has an inner f, because we have no rule for when g has only a single argument. In other words, it did exactly what we asked it, but not more.
PlusThere is nothing special about Plus. Again, for any function g, we want to transform g[c[x] + d[x] + e[x]] into g[c[x]] + g[d[x] + e[x]] in one step. Thus,
g[c[x] + d[x] + e[x]] //. g[a_[x] + b_] :> g[a[x]] + g[b] g[c[x]] + g[d[x]] + g[e[x]]
Many possibilities exist. One is
Exp[c[x] + d[x] + e[x]] /. Exp[z_] :> Total@Exp@(List @@ z) (* E^c[x] + E^d[x] + E^e[x] *) Another is
Exp[c[x] + d[x] + e[x]] /. Exp @ z_ :> Exp /@ z A slightly different implementation.
Exp[c[x] + d[x] + e[x]] /. Exp[HoldPattern@Plus[x__]] :> Plus @@ Exp[{x}] E^c[x] + E^d[x] + E^e[x]
Exp[c[x]]+Exp[d[x]+Exp[e[x]]orExp[c[x]]*Exp[d[x]*Exp[e[x]]? If byexpyou really meanExp, then the former is not the proper expansion while the latter is. $\endgroup$exp[c[x]]+exp[d[x]]+exp[e[x]].expwas supposed to mean "expected value" for some random distribution. Initially I used\[DoubleStruckE]for that function, but converted it to lower-caseexpto post my question here. $\endgroup$