I would like to have a series of functions that depend one another, indexed by a number, and that are defined via a Do loop. However, Mathematica seems unable to replace the index in the loop. My trial code:

  If[i == 1,
    ff[i, x_, y_] := x + y;,
    ff[i, x_, y_] := (x - y)*ff[i - 1, x, y];
  , {i, 1, 2}];
ff[2, x, y]

Output: (x - y) ff[-1 + i, x, y]

Desired output is however: (x - y)(x + y)

This thing surprises me since if I manually define

ff[2, x_, y_] := (x - y)*ff[1, x, y];

then I get the desired output. It seems that the Do loop is not replacing inside the function.

Is this normal? :-o


1 Answer 1


In your Do loop use Set rather than SetDelayed for the RHS to be evaluated.


Using RSolve to directly obtain a closed-form solution for the nth function

ff[n_, x_, y_] = 
 f[n] /. RSolve[{f[1] == x + y, f[n] == (x - y)*f[n - 1]}, f[n], n][[1]]

(* (x - y)^(-1 + n) (x + y) *)

Alternatively, using explicit recursion

f2[1] = x + y;
f2[n_] := f2[n] = (x - y)*f2[n - 1]

Generating a sequence from the recursion

seq = f2 /@ Range[5]

(* {x + y, (x - y) (x + y), (x - y)^2 (x + y), (x - y)^3 (x + y), 
    (x - y)^4 (x + y)} *)

Then using FindSequenceFunction to generalize from the sequence, the result is identical to that provided by RSolve

FindSequenceFunction[seq, n] === ff[n, x, y]

(* True *)
  • $\begingroup$ Thanks. Nice solution but I think it's too much writing for such a simple thing. All this is really weird in Mathematica (IMHO). In any standard programming language the set-delayed involves only the parameters of a function (the ones with underscore, here). The rest is replaced instantaneously. Sometimes Evaluate[..] helps, but here Evaluate[i-1] does not. I found a simplest solution with: Do[ If[i == 1, ff[i, x_, y_] := x + y;, pp = i; ff[i, x_, y_] := (x - y)*ff[pp - 1, x, y]; ]; , {i, 1, 2}]; ff[2, x, y] which I think it forces the i-replacement. $\endgroup$
    – Wizzerad
    May 4, 2021 at 17:57
  • $\begingroup$ I don't understand your objection. ff[n_, x_, y_] = f[n] /. RSolve[{f[1] == x + y, f[n] == (x - y)*f[n - 1]}, f[n], n][[1]] is shorter than your code, works for arbitrary n, and is more efficient. The remainder of the post was just to demonstrate an alternate approach. $\endgroup$
    – Bob Hanlon
    May 4, 2021 at 18:09

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