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MathGroup Archive 2005

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Re: Modifying displayed form of an expression?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg57139] Re: [mg57127] Modifying displayed form of an expression?
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Thu, 19 May 2005 03:08:06 -0400 (EDT)
  • References: <200505170135.VAA23181@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

On 17 May 2005, at 10:35, AES wrote:

> I have a number of symbolic results which emerge from a Mathematica
> calculation in the form
>
>                 a  Exp[b]
>   f  =  ----------------------------------
>               Exp[c] - Exp[d]
>
> where a is "other stuff" and b, c and d are simple sums like b = b1 + 
> b2
> + b3.  By default Mathematica seems to shuffle the elements of b, c and
> d in these results such that there will be only positive signs in the
> exponents b, c and d.
>
> I want to force Mathematica to display each of these expressions in the
> form
>
>                   a Exp[b - c]
>      -------------------------------------------
>                1 -  Exp[d - c]
>
> so that the leading term in the denominator is always one (which will 
> in
> fact mean that these expressions all have the same common denominator).
> How might I do this?
>
> Note: I only want to force this display of each expression _once_, when
> I first calculate it, and I'm willing to do some handwork on each
> expression to accomplish this.  Thus, I don't need some sophisticated
> general tool, just a way to do this "by hand".  I've been trying Inputs
> like
>
>       (Exp[-c] Numerator[f]) / (Exp[-c] Denominator[f])
>
> and not getting anywhere.
>
> Ideas?
>
>

Your "by hand " way does not work simply because you do not force 
Mathematica to multiply-out the terms in the denominator, so 
immediately on evaluation  the whole expression is put back into the 
original form. All you need to do is:


(Exp[-c]*Numerator[f]/Expand[Exp[-c]*Denominator[f]])

If you do not wish to do so much typing a slightly quicker approach is:


f /. Exp[x_] :> Exp[x - c]


(a*E^(b - c))/(1 - E^(d - c))


Andrzej Kozlowski


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