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Define m as an integer. I want to specify a Piecewise function f[x,n] that takes value n^3 x^2 except when x divides m(n+1). At these values, I want f[x,n]=1.

I can't figure out how to do this because I can't work out how to include the assumption that m is an integer. I have

f[x_, n_] := 
  Piecewise[{
    {1, x ∈ Divisors[m (n + 1)]},
    {x^2  n^(1/2), x ∉ Divisors[m (n + 1)]}
  }, 
    Assumptions -> m ∈ Integers];
table =Table[f[x, n], {x, 1, 5}, {n, 0, 5}]

This doesn't work. How do I do it?

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  • $\begingroup$ It's not clear what you're trying to do with m and the assumptions. Is m symbolic or numerical? If numerical, this is very simple: f[x_, n_] := Piecewise[{{n^3 *x^2, Mod[x, m (m + 1)] != 0}, {1, Mod[x, m (m + 1)] == 0}}] - but if m is symbolic, then this doesn't make any sense. $\endgroup$
    – flinty
    Commented Jul 13, 2020 at 16:22
  • $\begingroup$ Hi @flinty. The Mod trick works if I eliminate the use of m entirely - it was just a dummy variable to stand for "integer multiples of n+1". Thank you. If you could tweak and post it as an answer, I'll tick it. $\endgroup$
    – Richard Burke-Ward
    Commented Jul 13, 2020 at 18:08
  • $\begingroup$ Maybe ConditionalExpression is what you need? $\endgroup$
    – Sjoerd Smit
    Commented Jul 13, 2020 at 18:43

1 Answer 1

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As m is just an integer multiple of n+1, you can do this with Mod:

f[x_, n_] := Piecewise[{{1, Mod[x, n+1] == 0}}, n^3 *x^2]

f[4, 3] (* returns 1 *)
f[5, 3] (* returns 675 == 3^3 * 5^2 *)

table = Table[f[x, n], {x, 1, 5}, {n, 0, 5}]
(* returns: {{1, 1, 8, 27, 64, 125}, {1, 1, 32, 108, 256, 500}, {1, 9, 1, 243, 
  576, 1125}, {1, 1, 128, 1, 1024, 2000}, {1, 25, 200, 675, 1, 3125}} *)
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