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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_, ∞] &)" in the "Simplify" function but no change appears.
>
> Is someone could give me some tricks on how tu use the "ComplexityFunction" ?
>
> Thank you in advance.
>
> Guillaume
>
> Link to the forum page for this post:
> http://www.mathematica-users.org/webMathematica/wiki/wiki.jsp?pageName=Special:Forum_ViewTopic&pid=12974#p12974
> Posted through http://www.mathematica-users.org [[postId=12974]]
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