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.