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I am trying to solve a difference equation using RecurrenceTable as follows.

xhat0 = {0.00000001, 256};
boxedx[x_] := Piecewise[{{-2, x < -2}, {2, 2 < x}}, x]
SetAttributes[boxedx, Listable];
test = RecurrenceTable[{xhat[n] == boxedx[xhat[n - 1]],xhat[0] == xhat0}, {xhat}, {n, 1, 4}]

If I remove the boxedx function acting on the right hand side, or replace it by a built-in function such as Sqrt, the code works. I tried the definition

boxedx[x_List] := Map[Piecewise[{{-2, # < -2}, {2, 2 < #}}, #] &, x]

which also resulted in a numerical exception. And if I write the equality as 

xhat[n] == xhat[n - 1] + {1, 1}

a numerical exception happens again. I would appreciate any help!

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    $\begingroup$ Does it work if you use Clip[x, {-2, 2}] instead of boxedx? $\endgroup$
    – bill s
    Commented Feb 12, 2018 at 17:06
  • $\begingroup$ It does, I hadn't thought of it! I still don't understand why - and Attributes of Clip don't include listable - but your suggestions solves my problem. Thanks! $\endgroup$
    – Pait
    Commented Feb 12, 2018 at 17:54

1 Answer 1

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You can use Clip -- it acts on the elements of a list

Clip[Range[-5, 5], {-2, 2}]
{-2, -2, -2, -2, -1, 0, 1, 2, 2, 2, 2}
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  • $\begingroup$ Thank you very much! Do you have any idea of why defining a function doesn't work? I'm pretty sure I'll need something more complex sooner or later, so any pointer would be helpful. $\endgroup$
    – Pait
    Commented Feb 13, 2018 at 1:06
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    $\begingroup$ Sorry -- I'm not sure why -- I didn't really understand what you are trying to do in the RecurrenceTable. $\endgroup$
    – bill s
    Commented Feb 13, 2018 at 4:45
  • $\begingroup$ It was a simplified version of my problem, I removed complications in the RecurrenceTable to identify the source of the error messages. Your suggestion of using Clip solved my problem, though I still don't understand what was causing the errors. Thanks! $\endgroup$
    – Pait
    Commented Feb 14, 2018 at 16:51

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