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Re: Getting rid of annoying zeroes in algebraic expressions
*To*: mathgroup at smc.vnet.net
*Subject*: [mg15249] Re: [mg15228] Getting rid of annoying zeroes in algebraic expressions
*From*: jtischer at col2.telecom.com.co (Cl-Jurgen Tischer)
*Date*: Sun, 27 Dec 1998 03:58:40 -0500
*Organization*: Universidad del Valle
*Sender*: owner-wri-mathgroup at wolfram.com
Sean,
first, on my computer your two versions where equivalent.
In[1]:= xx=0. + (a*f)/(0. + f) + (b*f)/(0. + f) + (c*f)/(0. + f);
In[2]:= xx /. 0. + (q_) -> q
Out[2]= (a*f)/(0. + f) + (b*f)/(0. + f) + (c*f)/(0. + f)
In[3]:= 0. + (a*f)/(0. + f) + (b*f)/(0. + f) + (c*f)/(0. + f) /.
0. + (q_) -> q
Out[3]= (a*f)/(0. + f) + (b*f)/(0. + f) + (c*f)/(0. + f)
Second, it's easy to get what you want:
In[4]:= xx //. 0. + (q_) -> q
Out[4]= a+b+c
or if you like it more
In[5]:= Chop[xx]
Out[5]= a+b+c
Jurgen
"Ross, Sean" wrote:
>
> Let xx be the result of some algebraic manipulations which, for some
> reason, mathematica thinks the real zeroes ought to be kept and I think
> they ought to be dropped.
>
> xx=0. + (a*f)/(0. + f) + (b*f)/(0. + f) + (c*f)/(0. + f)
>
> xx/.Plus[0.,q_]->q
>
> returns
>
> 0. + (a*f)/(0. + f) + (b*f)/(0. + f) + (c*f)/(0. + f)
>
> while
>
> 0. + (a*f)/(0. + f) + (b*f)/(0. + f) + (c*f)/(0. + f)/.Plus[0.,q_->q]
>
> returns the expected
>
> a+b+c.
>
> xx is supposed to be equivalent to the expression, but in this case, it
> is not.
> I can't write a function to automatically drop zeroes if expressions
> behave differently when they are alone on an input line and when they
> are set equal to a symbol. Does anyone know how to eliminate this kind
> of behavior? I am sure it has something to do with some obscure step
> in the "standard evaluation cycle" and that there is probably a long,
> convoluted explanation as to why someone would want this behavior, but
> I don't and I would appreciate it if someone could tell me how to get
> rid of it or circumvent it.
>
> Thanks,
>
> Sean Ross
>
> Please reply to rosss at plk.af.mil as I no longer subscribe to the
> mathgroup.
>
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