Re: Extracting a Function's Domain and Image
- To: mathgroup at smc.vnet.net
- Subject: [mg68050] Re: Extracting a Function's Domain and Image
- From: David Bailey <dave at Remove_Thisdbailey.co.uk>
- Date: Fri, 21 Jul 2006 05:37:42 -0400 (EDT)
- References: <e9ku8v$l1v$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Bruce Colletti wrote:
> Re Mathematica 5.2.2.
>
> Why does the last line of code return {x} and not {a,b,_}, while if the f[_]=0 is struck, the answer is {a,b}?
>
> I want to extract the domain and image of a user-defined function, and tech support recommended using DownValues (pretty handy). If you have an easier approach, please share it. Thanks.
>
> Bruce
>
> Remove@f;
> f[_]=0;
> f["a"]=1;
> f["b"]=2;
> Clear@x;
> h=DownValues@f
> Cases[h,f[x_]->x,Infinity]
>
> Out[52]=
> {HoldPattern[f[a]]:>1,HoldPattern[f[
> b]]:>2,HoldPattern[f[_]]:>0}
>
> Out[53]=
> {x}
>
Hello,
The problem here is that you are using an expression involving f (inside
your Cases command) which can itself evaluate using your definition of
f. This sort of slippery problem is often encountered when you
manipulate held expressions. One way round this is to replace f with
another function - ff, say - and then manipulate that:
Cases[h /. f -> ff, ff[x_] -> x, Infinity]
I am not very happy with your use of the terms domain and image for the
information that you can extract in this way. I hope you are using these
terms loosely, and not with their precise maths meaning.
David Bailey
http://www.dbaileyconsultancy.co.uk