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Re: piecewise vs which

Bradley Stoll wrote:
> Consider defining a function in Mathematica (v. 5.2) in two different
> ways:  f[x_]=Piecewise[{{x^2,x<2},{3x,x>2}}] and
> g[x_]=Which[x<2,x^2,x>2,3x].  Notice that 2 is not in the domain of
> either function.  However, if I ask for f[2], Mathematica returns 0 and if I ask
> for g[2] Mathematica (correctly) returns nothing.  Is this a bug with
> Mathematica (that Mathematica returns 0 for f[2]), since 2 is not in the domain?

Hi Bradley,

Mathematica is "correct" by returning 0 for f[2] since this is the 
expected behavior as documented in the documentation for *Piecewise*

"Piecewise[{{val_1, cond_1}, ...}, val] uses default value _val_ if none 
of the cond_i apply. *The default for _val_ is 0.*"

So it may be better to define your function in the following way

f[x_] = Piecewise[{{x^2, x < 2}, {3*x, x > 2}}, "Undefined"];


Best regards,

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