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This is my code:

ClearAll[f, g] f[x_] := x^3 g[h_, x_] := h[x] - h[x + 1] g[f, x] (* x^3 - (1 + x)^3 *) g[g[f, x], x] (* (x^3 - (1 + x)^3)[x] - (x^3 - (1 + x)^3)[1 + x] *) r[x_] = g[f, x] (* x^3 - (1 + x)^3 *) g[r, x] (* x^3 - 2 (1 + x)^3 + (2 + x)^3 *)

g[f,x] outputs what I'd expect. However, if I try g[g[f,x],x], I don't get the expected result.

I think I see why; g[f,x] is producing an equation, whereas I believe I need a function head?

The last two lines gives the correct result ( x^3 - 2 (1 + x)^3 + (2 + x)^3) by defining r[x] as the output of g[f,x] and then calculating g[r,x], but I am looking to reperform this recursively and so this solution doesn't easily work, and in any case is clunky. I'm sure there must be a neat elegant solution to allow me to use something like g[g[f,x],x] or NestList, but I'm not seeing it.

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5 Answers 5

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This is just an extension of @Domen's answer.

The semantics of what you're doing is Nest. You're nesting to create function composition that you can apply to an argument, but it's still nesting. You're wanting to nest your g function twice with an initial argument of f. So, as Domen showed, what you want is to get this expression to work:

Nest[g, f, 2][x]

Sometimes it's helpful to see the structure of things using dummy symbols, so look at this (or just clear all your symbols and look at what the above produces):

Nest[gg, ff, 2][x] (* gg[gg[ff]][x] *)

So, what we need gg to be is a function that takes a function and returns another function. Again, Domen showed you two ways to do this, but another way that parallels directly the expression above is to use SubValues:

gg[fn_][arg_] := fn[arg] - fn[arg + 1]

Now evaluate that expression above again:

Nest[gg, ff, 2][x] (* ff[x] - 2 ff[1 + x] + ff[2 + x] *)

And with your f:

f[x_] := x^3; Nest[gg, f, 2][x] (* x^3 - 2*(1 + x)^3 + (2 + x)^3 *)

But, you can also use CurryApplied to make your original g into the kind of function you need (so no need to redefine your functions at all):

g[h_, x_] := h[x] - h[x + 1]; f[x_] := x^3; Nest[CurryApplied[g, 2], f, 2][x] (* x^3 - 2*(1 + x)^3 + (2 + x)^3 *)

It might be instructive to look at

Nest[CurryApplied[g, 2], f, 2] (* CurryApplied[g, 2][CurryApplied[g, 2][f]] *)
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You have correctly identified the problem. In the first iteration, you give f as the first argument to g, and then you can apply f to x. However, the result is now an expression containing x, which cannot be applied to the argument anymore.

Here are two possible ways to redefine your functions.

  1. Using ReplaceAll.
ClearAll[f, g] f[x_] := x^3; g[h_, x_] := h - (h /. x -> x + 1); g[f[x], x] (* x^3 - (1 + x)^3 *) g[g[f[x], x], x] (* x^3 - 2 (1 + x)^3 + (2 + x)^3 *) Nest[g[#, x] &, f[x], 5] (* x^3 - 5 (1 + x)^3 + 10 (2 + x)^3 - 10 (3 + x)^3 + 5 (4 + x)^3 - (5 + x)^3 *)
  1. Using pure functions.
ClearAll[f, g] f[x_] := x^3; g[h_] := Function[Evaluate[(h[#] - h[# + 1])]]; g[f][x] (* x^3 - (1 + x)^3 *) g[g[f]][x] (* x^3 - 2 (1 + x)^3 + (2 + x)^3 *) Nest[g, f, 5][x] (* x^3 - 5 (1 + x)^3 + 10 (2 + x)^3 - 10 (3 + x)^3 + 5 (4 + x)^3 - (5 + x)^3 *)
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Or, maybe, as a convenient workaround:

g[g[f, #] &, #] &[x] (* x^3 - 2 (1 + x)^3 + (2 + x)^3 *)

A 'slot free' version

(CurryApplied[g, 2]@CurryApplied[g, 2][f])[x] (* x^3 - 2 (1 + x)^3 + (2 + x)^3 *)

Or, as already pointed out by lericr, with Nest

Nest[CurryApplied[g, 2], f, 2][x] (* x^3 - 2 (1 + x)^3 + (2 + x)^3 *)

And

Nest[CurryApplied[g, 2], f, #][x] & /@ Range[5] (* { x^3 - (1 + x)^3, x^3 - 2 (1 + x)^3 + (2 + x)^3, x^3 - 3 (1 + x)^3 + 3 (2 + x)^3 - (3 + x)^3, x^3 - 4 (1 + x)^3 + 6 (2 + x)^3 - 4 (3 + x)^3 + (4 + x)^3, x^3 - 5 (1 + x)^3 + 10 (2 + x)^3 - 10 (3 + x)^3 + 5 (4 + x)^3 - (5 + x)^3 } *)
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  • Define g as an operator T which act on the function h.
Clear["Global`*"]; f[x_] := x^3; T[h_][x_] := Minus@DifferenceDelta[h[x], x]; list = NestList[T, f, 3] Through[list@x]

{f, T[f], T[T[f]], T[T[T[f]]]}

{x^3, -1 - 3 x - 3 x^2, 6 + 6 x, -6}

  • Since the definition is h[x]-h[x+1] instead of h[x+1]-h[x], it can not directly use DifferenceDelta[h[x],{x,n}] for n.
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Another workaround is to use Composition as follows:

f[x_] := x^3; g[h_] := Function[x, h[x] - h[x + 1]]; (g@*g)[f][x]

x^3 - 2 (1 + x)^3 + (2 + x)^3

Alternatively, use Fold with your definitions for f and g:

f[x_] := x^3; g[h_, x_] := h[x] - h[x + 1]; G = Function[h, Function[x, g[h, x]]]; Fold[G, f, Array[g &, 2]][x]

x^3 - 2 (1 + x)^3 + (2 + x)^3

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