Re: General--Exponential simplifications by default

• To: mathgroup at smc.vnet.net
• Subject: [mg69009] Re: General--Exponential simplifications by default
• From: p-valko at tamu.edu
• Date: Sat, 26 Aug 2006 02:04:57 -0400 (EDT)
• References: <ecmi96\$a1r\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```myform[expr_]:=Block[{num,den,fact},
num=Numerator[expr];
den=Denominator[expr];
fact=If[Count[den,E^a_]\[Equal]1,First[Cases[den,E^a_]],1];
Apart[num/fact]/Apart[den/fact]
]

E^(-2 eta y)(-theta + E^(eta y)(2+theta))/(2 alpha)  // myform

The output will look something like your preference:

(-theta E^(-2 eta y)+ E^(-eta y)(2+theta)) / (2alpha)

at least in the default OutputForm.

Peter

guillaume_evin at yahoo.fr wrote:
> Hi !
>
> I want to avoid simplifications when Mathematica integrates expressions with exponential terms. For example, I have :
>
> In[8]=      Espcond[y_] = Integrate[x*Densx[x, y], {x, 0, Infinity}, Assumptions -> alpha > 0]
> Out[8]=      E^(-2eta y)(-theta + E^(eta y)(2+theta))/(2alpha)
>
> I do not want to have a factorization by E^(-2eta y). More precisely I would like to have the following result:
> Out[8]=      (-theta E^(-2eta y)+ E^(-eta y)(2+theta))/(2alpha)
>
> I guess there is a way to tackle this problem with "ComplexityFunction" and "Simplify", but I tried different things such as "ComplexityFunction -> (Count[{#1}, Exp_, &#8734;] &)" in the "Simplify" function but no change appears.
>
> Is someone could give me some tricks on how tu use the "ComplexityFunction" ?
>