Re: piecewise vs which
- To: mathgroup at smc.vnet.net
- Subject: [mg60120] Re: piecewise vs which
- From: Peter Pein <petsie at dordos.net>
- Date: Sat, 3 Sep 2005 02:06:08 -0400 (EDT)
- References: <df9437$620$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Bradley Stoll schrieb:
> 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?
> While I have eyes, there is another question regarding limits. Is it
> the case that Limit[f[x],x->2] defaulted as
> Limit[f[x],x->2,Direction->-1] (a right hand limit)? Both return 6 in
> the above example. I'm using Mathematica in my calculus class and would
> like to explain why Mathematica does certain things. It doesn't seem
> that it would've been too difficult to just have two subroutines (a
> right and left hand limit) to determine whether a 'full' limit actually
> existed.
> Thanks!
>
> Bradley
>
Hi Bradley,
if you have a look at the documentation, you'll see that Piecewise has
a default value 0. If you want f[2] to return Null (as g[2] does, enter
In[1]:=
f[x_] = Piecewise[{{x^2, x < 2}, {3*x, x > 2}},Null];
g[x_] = Which[x < 2, x^2, x > 2, 3*x];
In[3]:=
f[2] == g[2]
Out[3]=
True
--
Peter Pein, Berlin
GnuPG Key ID: 0xA34C5A82
http://people.freenet.de/Peter_Berlin/