2
$\begingroup$

How can I ask Mathematica to find all polynomial $ f \! $s such that, for a constant $ k $:

$$ f(x) \ f(y) = f(x) + f(y) + f(xy) + k $$

I've solved it on paper and want to check my results. I've tried the following, none of which work:

  • Solve[f[x] f[y] == f[x] + f[y] + f[x y] + k, f]
  • Solve[f[x] f[y] == f[x] + f[y] + f[x y] + k, f[x]]
  • Solve[f[x] f[y] == f[x] + f[y] + f[x y] + k, f[z]]
  • Reduce[f[x] f[y] == f[x] + f[y] + f[x y] + k, f]
  • f[x_] := Sum[Subscript[c, i] x^i, {i, 0, n}]; Solve[f[x] f[y] == f[x] + f[y] + f[x y] + k, Subscript[c, 0]]
$\endgroup$
2
  • $\begingroup$ Take a look at this quite analogous problem: Solving functional equations in Mathematica $\endgroup$
    – Artes
    Feb 20, 2020 at 2:20
  • $\begingroup$ I would consider polynomials of increasing degree, and write them in terms of their coefficients. Then solve for the coefficients $\endgroup$
    – mikado
    Feb 20, 2020 at 6:50

1 Answer 1

1
$\begingroup$

It is better to use SolveAlways and to consider a specific value of n. Here is $n=4$:

f[x_] := Sum[Subscript[c, i] x^i, {i, 0, 4}]
{k, f[x]} /. SolveAlways[f[x] f[y] == f[x] + f[y] + f[x y] + k, {x, y}]
(* {{{-2, 1}, {-2, 1 + x^4}, {-2, 1 + x^3}, {-2, 1 + x^2}, {-2, 1 + x}} *)
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.