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Re: Common factors in a list

  • To: mathgroup at
  • Subject: [mg58373] Re: Common factors in a list
  • From: Peter Pein <petsie at>
  • Date: Tue, 28 Jun 2005 21:57:05 -0400 (EDT)
  • References: <d9r50q$568$>
  • Sender: owner-wri-mathgroup at

John Reed schrieb:
> I'm working with a list (vector) that's composed of functions such as 
> complex exponentials and Bessel functions.  Some of the functions are common 
> to each element of the list.  I'm trying to find a way to do the following:
>    {a x, a y, a z}-> a *{x, y, z}
> In this expression, a, x, y, z are polynomials, exponentials and/or Bessel 
> functions.
> So far I haven't had any luck with this.  Does anyone have a simple 
> solution?
> John Reed 

Because of Mathematica's built-in "simplification" of
In[1]:= a*{x, y, z}
Out[1]={a*x, a*y, a*z},
I'll use \[CircleDot] as replacement for the multiplication character:
In[2]:= a \[CircleDot] {x, y, z}
Out[2]= a \[CircleDot] {x, y, z}

vecFactor[vec_List] :=
  Module[{ql, fl = FactorList[Plus @@ vec], mp, f2},
   ql = FoldList[Cancel[#1/Power @@ #2]& , vec, fl];
   f2 = ql[[ mp = Position[#1, Min[#1]]&[LeafCount /@ ql][[1, 1]]]];
   ql = Times @@ Power @@@ Take[fl, {1, mp - 1}];
   ql \[CircleDot] f2
 {6*BesselJ[0, x]* xp[-x]*(x^2 - 1), 4*(x - 1)*Exp[-x]*BesselJ[1, x],
  2*(1 - x)^2*BesselJ[2, x]*Exp[-x]}
((2*(-1 + x))/E^x) \[CircleDot]
  {3*(1 + x)*BesselJ[0, x], 2*BesselJ[1, x], (-1 + x)*BesselJ[2, x]}

Later you can use expr/.\[CircleDot]->Times to get back your product.
Maybe someone else has a better trick to avoid the distribution of the
common factor...

Peter Pein

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