Solve the following equation f[f[a+b]]==f[2a]+2f[b]?

Questions
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1

how to Solve this eq in mathematica

f[f[a+b]]==f[2a]+2f[b]
2

f[x_] := 2 x

In[93]:= f[f[a + b]] // Simplify

Out[93]= 4 (a + b)

In[94]:= f[2 a] + 2 f[b] // Simplify

Out[94]= 4 (a + b)
3

How can I know
f[x]=2x
by any kind of mathematica tool for example LIKE
Rsolve ,

4

Although mathematica is an extraordinary powerful tool I am afraid that will not work. You have to use your brain.

5

Thanks for HELP
I’m thinking is there a general rule TO SOLVE This kind of equation
for example
f[f[x]]==f[x]

f[f[f[x]]]==g[f[x]]

ReSolve is a good tool put Cant solve a nest case
ill try to make a small code to solve it put first i need general rule

6

I am afraid there is no general rule to cope with problems of this kind.

7

One approach is to make a parametrized guess and then use Reduce. For example, try

g[c0_,c1_,c2_][x_]:=c0+c1*x+c2*x^2

and then use Reduce on a small set of examples.

FullSimplify[Reduce[ Flatten[Table[f[f[a+b]]==f[2a]+2f[b],{a,0,5},{b,0,5}]]/.f->g[c0,c1,c2], {c0,c1,c2}]]
(* c2 == 0 && ((c0 == 0 && c1 == 0) || c1 == 2) *)

I added FullSimplify so it would get rid of redundant results which can be confusing.

It is telling us that either f is identically zero or f[x]==2x+constant

Note: if this is a homework question for a math class, it should be noted that there is some subtlety with the domain and other complications arise if f is not continuous. In general, use the powerful functions like Reduce and Solve, but you are also supposed to exper