Can substitution rules be used in function definitions?

Question

Simple input like this

f[x_]:=2x
f[2]

has the predicted output of 4. However, this

r={a->2x};
f[x_] := a/.r
f[2]

Produces the output "2 x". I would like it to produce 4. More generally, is there a way I can incorporate rules into a function definition?

Answer 1

It's not a problem with using rules in the def; it's a problem of whether x literally appears in the rhs or not.

When applying a definition where the pattern x_ on the lhs has been matched to an argument (e.g. 2), Mathematica looks to see if the pattern name x appears anywhere in the definition's rhs, substitutes the matched value for any x's it finds, then evaluates. Here, it looks at f[2] and matches 2 to x_, then looks at the rhs of the definition, and only sees a, /., and r (and no x). So, it doesn't substitute anything, and finishes its application of the definition. All further evaluation (of a /. r) "doesn't know about" the definition anymore. (In particular, it doesn't know about he value 2 that got matched to x, and x is now "just a symbol again".)

Further, when you use :=, the rhs you typed in is unevaluated, and the definition Mathematica learns is the replacement rule f[x_] :> a /. r. However, when you use =, the rhs is evaluated first, and then learned as a definition. So we start with f[x_] = a /. r, a /. r becomes 2x, then the definition that Mathematica learns is the rule f[x_] :> 2x. That's why cvgmt's solution in the comments (f[x_] = a /. r) works. See this for more about the difference between := and = :)

(Note: I omitted some of the technicalities when describing what rule Mathematica actually learns and how it uses the rule, but all in all, it behaves roughly like :>.)