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Re: Problems with easy simplifications

• To: mathgroup at smc.vnet.net
• Subject: [mg54152] Re: Problems with easy simplifications
• From: Curt Fischer <tentrillion at gmail.NOSPAM.com>
• Date: Sat, 12 Feb 2005 01:57:00 -0500 (EST)
• References: <cuhsm1$9j1$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

Antonio González wrote:
> I have found some weird behaviors of Mathematica (5.0) related to the 
> simplifications of easy expressions.
> 
> Take, for instance the function
> 
>    H[x_, a_, b_] = If[x < a, a + b, a - b]
> 
> If I try now to evaluate
> 
>    H[x,0,0]
> 
> the result
> 
>   If[x < 0, 0 + 0, 0 - 0]

what result were you expecting?  What is the bug?  You have not supplied 
  a value for x that Mathematica could use to test whether or not x<a or 
not.

In[1]:=
h[x_,a_,b_]=If[x<a,a+b,a-b]

Out[1]=
If[x<a,a+b,a-b]

In[2]:=
h[1,2,3]

Out[2]=
5

In[3]:=
h[3,1,2]

Out[3]=
-1

> 
> (the values of a and b are irrelevat, a similar result is obtained, for 
> instance, with a=1, b= 2, or any integer, real or complex values). I 
> cannot force (at least in a simple way) Mathematica to make the addition 
> or the substraction. The action of Simplify or FullSimplify leaves the 
> expression unchanged while a Map of Simplify
> 
>    MapAll[Simplify, H[x, 0, 0]]
> 
> produces
> 
>    If[x < 0,
>      Simplify[Simplify[0] + Simplify[0]], Simplify[Simplify[
>        0] + Simplify[Simplify[-1] Simplify[0]]]]
> 
> Something even worse happens if a use values already defined. For instance
> 
>    c = 0; d = 0; H[x_] = If[x < c, c + d, c - d]
> 
> produces
> 
>   If[x < 0, c + d, c - d]
> 
> In this case, the addition is not necessary.
> 
>    c = 0; d = 0; H[x_] = If[x < c, c, d]
> 
> leads to
> 
>   If[x < 0, c, d]

If you don't want your definition for H[x_] to be evaluated _right 
away_,  before substituting the definitions for other variables, you 
should use the delayed assignment operator :=.

In[3]:=
c=0;d=0;

In[4]:=
g[x_]:=If[x<c,c+d,c-d]

In[5]:=
g[1]

Out[5]=
0

> 
> Similar problems arise using the function Which.
> 
> Any explanation or help with this malfunction would be welcome.

Read the Help Browser.

-- 
Curt Fischer