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]]