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Re: multiple of sum of fraction by common denominator keeps fractions in result

• To: mathgroup at smc.vnet.net
• Subject: [mg41631] Re: [mg41585] multiple of sum of fraction by common denominator keeps fractions in result
• From: Daniel Lichtblau <danl at wolfram.com>
• Date: Thu, 29 May 2003 08:13:58 -0400 (EDT)
• References: <200305280857.EAA09452@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

Friedrich Laher wrote:
> 
> In[8]:=
> \!\(\(1 - x\)\/\(5*\((1 - 2*x)\)\) - \(2 + x\)\/8 - 4 + \(2*x -
> 19\)\/80\ x\)
> 
> Out[8]=
> \!\(\(-4\) + 1\/8\ \((\(-2\) - x)\) + \(1 - x\)\/\(5\ \((1 - 2\ x)\)\) +
>      1\/80\ x\ \((\(-19\) + 2\ x)\)\)
> 
> In[17]:=
> Together[In[8]]
> 
> Out[17]=
> \!\(\(324 - 635\ x - 60\ x\^2 + 4\ x\^3\)\/\(80\ \((\(-1\) + 2\ x)\)\)\)
> 
> In[18]:=
> %*80*(-1+2x)
> 
> Out[18]=
> \!\(324 - 635\ x - 60\ x\^2 + 4\ x\^3\)
> 
> In[19]:=
> \!\(Expand[\((\(1 - x\)\/\(5*\((1 - 2*x)\)\) - \(2 + x\)\/8 -
>            4 + \(2*x - 19\)\/80\ x)\)*80*\((1 - 2*x)\)]\)
> 
> Out[19]=
> \!\(\(-340\) + 16\/\(1 - 2\ x\) + 651\ x - \(48\ x\)\/\(1 - 2\ x\) +
>      60\ x\^2 + \(32\ x\^2\)\/\(1 - 2\ x\) - 4\ x\^3\)
> 
> In[20]:=
> FullSimplify[Out[19]]
> 
> Out[20]=
> -324+x (635-4 (-15+x) x)
> 
> In[21]:=
> Expand[%]
> 
> Out[21]=
> \!\(\(-324\) + 635\ x + 60\ x\^2 - 4\ x\^3\)
> 
> of what use could Out[1] be, why not directly Out[21] ?

The input is a bit hard to parse by eye. Also you refer to Out[1] but
your first output shown is Out[8]. In any case I'm guessing you meant
Out[19].

It appears that you start with:

expr = -4 + (-2 - x)/8 + (1 - x)/(5*(1 - 2*x)) + (x*(-19 + 2*x))/80;

You do

Expand[expr*80*(-1+2*x)]

and want to know why the result is not the same as, say,

Expand[Together[expr*80*(-1+2*x)]]

If this is indeed your question, the answer is that Expand for rational
function input does just multiplication, addition, and expansion of
powers. It does not extract gcds or in other respects look for at
denominators of subexpressions (as Together would do). So in particular
it does it try to recombine subexpressions over common denominators.

Keeping in mind that (1-2*x) is a binomial rather than an "atomic"
entity, it then becomes clear why those factors of 1-2*x in some
denominators do not get cleared. Note that if the multiplier is made
atomic as below, they do in fact get cleared.

Expand[Expand[expr*a] /. a->80*(1-2*x)]


Daniel Lichtblau
Wolfram Research

• References:
  • multiple of sum of fraction by common denominator keeps fractions in result
    • From: Friedrich Laher <mathefritz@schmieder-laher.de>