Optional argument as a correction of a main argument

Question

Let's consider a function

f[a_,b___]

I want that if I input only one argument I get

f[a]=a

but if I input 2 arguments I get

 f[a,b]=2a+b

How can I implement that?

I thought I could use If:

f[a_,b___]:=If[b===Null,a,2a+b]

but reading this previous post Are UnsamQ and Not[SameQ] the same function? I think it's not the correct way.

I could use 2 different definition for f[a_] and f[a_,b_] but I'd like to have only one definition. In fact, my problem arises because if f has only one argument, f performs analytic calculation, but with two argument (where the second argument is a "correction" of the first one) f needs, in only one part, a numerical calculations; anyway the computation that f performs is the same in both cases.

Answer 1

Two definitions, as recommended by m_goldberg and Michael E2, really is the preferred solution:

f[a_]     := a
f[a_, b_] := 2 a + b

This is both clear and efficient.

If you prefer an If statement then perhaps:

ClearAll[f]

f[a_, b___] := If[Hold[b] === Hold[], a, 2 a + b]

I am using Hold for generality, as in some cases one would not wish for the arguments to evaluate. That may not matter if f does not itself have a hold attribute but I felt it better to illustrate this than to omit it.

If you want to be clever we can exploit the specific definition of the example and write:

ClearAll[f]

f[a_: 0, b_] := 2 a + b

f[x]
f[x, y]

x
2 x + y

This works by making parameter a Optional with a default value of zero. When it is not provided the term 2 a therefore vanishes from the sum. If this kind of trickery pleases you see also what I call "vanishing patterns."