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Re: Interesting failure of Collect
*To*: mathgroup at smc.vnet.net
*Subject*: [mg61337] Re: Interesting failure of Collect
*From*: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
*Date*: Sun, 16 Oct 2005 00:17:53 -0400 (EDT)
*Organization*: The Open University, Milton Keynes, UK
*References*: <dipqve$hdt$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Blimbaum, Jerry CIV NSWC PC wrote:
> Given the expr = Sin[p]/(4 Pi a) + Sin[p]/(4 Pi b)
>
> and then applying the command
>
>
> Collect[expr, Sin[p]/(4 Pi)] works as it should, however,
>
> expr2 = Sin[p]/(4 Pi a) - Sin[p]/(4 Pi b)
>
> leaves the expression untouched. (I realize I could use Simplify but
> this problem occurred on a much longer expression and this is intended
> just to convey the basic idea)....for longer expressions the Collect
> process works for all quantities with a + sign but the one with a minus
> sign will not be collected.....strikes me as a "bug".....
>
>
> thanks....jerry blimbaum
>
>
Hi Jerry,
I have got the same behavior with Mathematica 5.2 for Windows XP (see
below). One workaround is to add -- or more precisely to replace the
negative sign by -- an additional parameter, say c, as in the following
example, and to use a replacement rule to change the parameter c by -1.
*Collect* seems to work correctly in this case.
In[8]:=
expr = Sin[p]/(4*Pi*a) + Sin[p]/(4*Pi*b)
Out[8]=
Sin[p]/(4*a*Pi) + Sin[p]/(4*b*Pi)
In[9]:=
Collect[expr, Sin[p]/(4*Pi)]
Out[9]=
((1/a + 1/b)*Sin[p])/(4*Pi)
In[10]:=
expr2 = Sin[p]/(4*Pi*a) - Sin[p]/(4*Pi*b)
Out[10]=
Sin[p]/(4*a*Pi) - Sin[p]/(4*b*Pi)
In[11]:=
Collect[expr2, Sin[p]/(4*Pi)]
Out[11]=
Sin[p]/(4*a*Pi) - Sin[p]/(4*b*Pi)
In[12]:=
expr3 = Sin[p]/(4*Pi*a) + c*(Sin[p]/(4*Pi*b))
Out[12]=
Sin[p]/(4*a*Pi) + (c*Sin[p])/(4*b*Pi)
In[13]:=
Collect[expr3, Sin[p]/(4*Pi)] /. c -> -1
Out[13]=
((1/a - 1/b)*Sin[p])/(4*Pi)
In[14]:=
$Version
Out[14]=
"5.2 for Microsoft Windows (June 20, 2005)"
Best regards,
/J.M.
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