4
$\begingroup$

In mathematica, I have a piecewise function similar to:

Piecewise[{ {x^2 + 2*x - 4, 0 <= x <= 1/4},
            {0, true}                    }]

I don't know how many pieces there will be, or what the internal ordering will look like, but I know that exactly one of the pieces will have x=0 as its left endpoint. How do I extract that piece in its general form (x^2 + 2*x - 4) without knowing the upper end of its domain (1/4)?

Improve this question
$\endgroup$

2 Answers 2

Reset to default
3
$\begingroup$

Given say:

expr = Piecewise[{ {x^2 + 2*x - 4, 0 <= x <= 2}, 
                   {0, True}}]

... you can convert your Piecewise expression back into a list form using:

 aa = Internal`FromPiecewise[expr, True]  // Transpose

{{x >= 0 && x <= 2, -4 + 2 x + x^2}, {x > 2 || x < 0, 0}}

Then, you can find the part that corresponds to x == 0 using:

Select[aa, (#[[1]] /. x -> 0) &]

{{x >= 0 && x <= 2, -4 + 2 x + x^2}}

Improve this answer
$\endgroup$
2
$\begingroup$

Similar to wolfies' approach - find the part for which the condition is True when x->0.

p = Piecewise[{{x^2 + 2*x - 4, 0 <= x <= 1/4}, {0, True}}];

Cases[First@p, {e_, c_} /; (c /. x -> 0) :> e]

(* {-4 + 2 x + x^2} *)
Improve this answer
$\endgroup$
1
  • $\begingroup$ Thanks for the alternative approach! I'm accepting wolfies' answer mainly for the reason that it deals with the default case better than this one. For example, Piecewise[{{x^2 + 2*x - 4, 1 <= x <= 2}}] for x=0 gives {{x > 2 || x < 1, 0}} for wolfies' (indicating the default of 0), but {} for yours (making it a special case to handle). $\endgroup$
    – Ponkadoodle
    Commented Apr 14, 2014 at 7:10

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.