[Date Index]
[Thread Index]
[Author Index]
Re: General--Exponential simplifications by default
*To*: mathgroup at smc.vnet.net
*Subject*: [mg69039] Re: General--Exponential simplifications by default
*From*: p-valko at tamu.edu
*Date*: Sun, 27 Aug 2006 01:24:22 -0400 (EDT)
*References*: <200608250935.FAA09304@smc.vnet.net><ecoqnr$3bb$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Daniel,
Could you tell me why the contradiction?:
In[14]:=Exp[-x]//InputForm
Out[14]//InputForm=E^(-x)
In[15]:=Numerator[Exp[-x]]//InputForm
Out[15]//InputForm=1
Mathematica automatically turns it into Power[E, Times[-1, x]], but the user
beleives that the exponent is in the numerator.
What you see is not what you get!
My favorite contradiction is the following:
In[24]:=Numerator[1/Sqrt[2]]//InputForm
Out[24]//InputForm=Sqrt[2]
In[25]:=Denominator[1/Sqrt[2]]//InputForm
Out[25]//InputForm=2
I think "Numerator" and "Denominator" should not be allowed to do
whatever they want. Too much liberty here...
Peter
Daniel Lichtblau wrote:
> 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: Selecting Lists Without Null Results**
Next by Date:
**Re: General--Difficulties in Understanding Mathematica Syntax**
Previous by thread:
**Re: General--Exponential simplifications by default**
Next by thread:
**Re: Re: General--Exponential simplifications by default**
| |