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Re: piecewise vs which
*To*: mathgroup at smc.vnet.net
*Subject*: [mg60118] Re: [mg60101] piecewise vs which
*From*: Curtis Osterhoudt <gardyloo at mail.wsu.edu>
*Date*: Sat, 3 Sep 2005 02:06:06 -0400 (EDT)
*References*: <200509020833.EAA05912@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Hi, Bradley,
Some similar questions -- but with limits of +/- Infinity -- have
been bantered about on this group recently regarding limits.
The Piecewise function returns 0 as a default value when a domain
isn't given for an argument. I found this from the command ?Piecewise,
but it's also in the fuller Help Index documentation, of course. If you
make a larger "gap", say,
f[x_] := Piecewise[{{x^2, x < 2}, {3*x, x > 10}}]
then Piecewise will return 0 for values for 2<= x <= 10.
For the students' sake you might define f to return Null at
"unspecified" arguments, with
f[x_] := Piecewise[{{x^2, x < 2}, {3*x, x > 10}}, Null]
This should plot and behave more as you expect. Also, the Limit function
now works as you'd expect, including directed limits: try
Limit[f[x], x -> 2, Direction -> 1]
vs.
Limit[f[x], x -> 2].
Regards,
Curtis O.
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?
>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
>
>
>
>
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