[Date Index]
[Thread Index]
[Author Index]
Re: General--Exponential simplifications by default
*To*: mathgroup at smc.vnet.net
*Subject*: [mg69008] Re: [mg68966] General--Exponential simplifications by default
*From*: Daniel Lichtblau <danl at wolfram.com>
*Date*: Sat, 26 Aug 2006 02:04:56 -0400 (EDT)
*References*: <200608250935.FAA09304@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
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]]
Could Collect with respect to powers of the exponential.
In[59]:= ee = E^(-2eta*y)*(-theta + E^(eta*y)(2+theta))/(2*alpha);
In[60]:= InputForm[Collect[ee, Exp[eta*y]]]
Out[60]//InputForm=
-theta/(2*alpha*E^(2*eta*y)) + (2 + theta)/(2*alpha*E^(eta*y))
Daniel Lichtblau
Wolfram Research
Prev by Date:
**Re: Symmetric polynomials**
Next by Date:
**Re: How to handle Arrays that has functional parameters:**
Previous by thread:
**General--Exponential simplifications by default**
Next by thread:
**Re: General--Exponential simplifications by default**
| |