I need to find the nthComposition of a function. I thought there could be something like nthComposition[f,x,5] means five times composition of f with x as input parameter. I don't want to repeat f in my code as there is no fixed number for my composition, it could change every time. Assume it is my function:

f [va_] := ( va*2);

Composition[f, f][s]
4 s

Composition[f, f, f][s]
8 s

3 Answers 3


Depending on exactly how you want to use this, you might want to bundle this up into its own function (or function overload).

f[nestLevel_, va_] := Nest[f, va, nestLevel]


compf[nestLevel_] := Composition @@ ConstantArray[f, nestLevel]


f[2, x]
(* f[f[x]] *)
(* or 4 x using your sample function *)

(* f@*f@*f *)

(* f[f[f[x]]] *)
(* or 8 x using your sample function*)
  • $\begingroup$ thanks but I'm a bit confused about the first solution : f[nestLevel_, va_] := Nest[f, va, nestLevel] . we are defining f on left side but giving it to Nest as input on right side. how it doesn't show any conflict? $\endgroup$
    – Azzurro94
    Jun 6, 2022 at 14:55
  • 1
    $\begingroup$ f is just a symbol. We're not defining f directly, so to speak, we're specifying DownValues (replacement rules for a particular form). There is no infinite recursion, because once the Nest has been performed, we have forms with f that have only a single argument. So, the evaluator will use the one-argument definition to resolve those (if it exists). You could make this more explicit/safe if you define a new function instead of overloading f: nestf[nestLevel_, va_] := Nest[f, va, nestLevel]. $\endgroup$
    – lericr
    Jun 6, 2022 at 15:21

As answered well in the comment, we can find it by

Nest[f, x, 5] 

For some functions f, you can use NestList and FindSequenceFunction to solve the problem more generally

f[va_] := (va*2);

seq = Rest@NestList[f, x, 6]

(* {2 x, 4 x, 8 x, 16 x, 32 x, 64 x} *)


f[x_, n_ : 1] = FindSequenceFunction[seq, n]

(* 2^n x *)


(* 2 s *)

f[s, 5]

(* 32 s *)

f[s, n]

(* 2^n s *)

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