Re: General--Exponential simplifications by default

*To*: mathgroup at smc.vnet.net*Subject*: [mg69156] Re: General--Exponential simplifications by default*From*: p-valko at tamu.edu*Date*: Thu, 31 Aug 2006 04:39:16 -0400 (EDT)*References*: <2219B7076306D04492C4696EC5C9401C680001@PE-MAIL.pe.tamu.edu> <ed3sle$s89$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

You simply neglected my last example and went back to your example (the value of 2/3). However, your example is not interesting, because in it the result of Numerator[] is not contradictory at all. In my example, the value (in your words) of the input expression 1/Sqrt[2] is still 1/Sqrt[2] but the result of applying Numerator[] to it gives a strikingly different than anticipated result. Numerator[] does something else in between and we do not know what she is doing. If you use Trace, you can see that everything strange is happening inside the Numerator[] that is In[2]:= Map[InputForm,Trace[Numerator[1/Sqrt[2]]]] Out[2]= {{{{HoldForm[Sqrt[2]], HoldForm[Sqrt[2]]}, HoldForm[Sqrt[2]^(-1)], HoldForm[1/Sqrt[2]]}, HoldForm[1/Sqrt[2]], HoldForm[1/Sqrt[2]]},HoldForm[Numerator[1/Sqrt[2]]],HoldForm[Sqrt[2]]} The rersult of Numerator[1/Sqrt[2]] is Sqrt[2] and "all of this is quite not standard staff and cannot be found in every book on functional languages..." Andrzej Kozlowski wrote: > The problem is that you are using the word "evaluate" in a different > sense than the normal sense used in the context of functional > programming languages and computer algebra systems. In this normal > usage all "simplifications" (your words) performed by "the > Evaluator" are "evaluations" and their results are "values" of the > input expressions. Simplify is, something quite different - a > function applied by the user with its own semantics and its own > evaluation sequence, quite separate from what the "Evaluator" does. > > For example, the expression 2/3 when entered into Mathematica has > FullForm > > > FullForm[Hold[2/3]] > > Hold[Times[2,Power[3,-1]]] > > but the Evaluator evaluates this to: > > FullForm[2/3] > > Rational[2,3] > > This is normally described by saying that the value of Times[2,Power > [3,-1]] in Mathematica is Rational[2,3]. There are many such cases in > Mathematica and what you have come across are just some of them. > This is completely analogous to what you find in other functional > languages (like Lisp) where you have to carefully distinguish between > unevaluated expressions and their "values". (In Lisp you use the > quote sign to make this distinction). All of this is standard staff > and can be found in every book on functional languages etc. > > Andrzej Kozlowski > > > > > > On 29 Aug 2006, at 18:27, Valko, Peter wrote: > > > Andrzej: > > > > You tried to convince me that Denominator[] and Numerator[] evaluates > > its argument in a standard way and then they find the positive and > > negative powers. > > I tried to convince you, that Denominator[] and Numerator[] do > > something > > more: practically intermitting a Simplify[] operation before really > > getting to do the real job. > > Proof: > > > > If I write > > 1/Sqrt[2] > > or I write > > Evaluate[1/Sqrt[2]] > > I still get > > 1/Sqrt[2] (in input form) or Power[2,Rational[-1,2]] (in full form). > > > > However, when I ask for > > Numerator[1/Sqrt[2]] > > I get > > Sqrt[2] (in input form) or Power[2, Rational[1, 2]] (in full form) > > so "it is pretty clear" that Numerator[] and Denominator[] are doing > > something secret after evaluating the argument and before really > > finding > > the Numerator and Denominator. > > > > Your "solution" suggesting to use > > Numerator[Unevaluated[1/Sqrt[2]]] > > does something else than just preventing the evaluation of > > 1/Sqrt[2] > > (because we already agreed that the evaluation does not do anything to > > the expression 1/Sqrt[2]). > > What it really prevents is the simplification of the unevaluated or > > the > > evaluated expression, because (I guess) an Unevaluated[expression] is > > left intact by Simplify[]. > > > > > > Or am I missing something? > > > > > > > > > > -----Original Message----- > > From: Andrzej Kozlowski [mailto:akoz at mimuw.edu.pl] To: mathgroup at smc.vnet.net > > Subject: [mg69156] Re: Re: General--Exponential simplifications by > > default > > > > Pro* The statement about FullForm you are quoting is "true", at best, > > only in a "certain sense": > > FullForm does "not affect" evaluation "inside" FullForm, as in: > > > > FullForm[2^2]//InputForm > > > > FullForm[4] > > > > But of course it does affect evaluation "outside" FullForm, as in: > > > > > > FullForm[2]^2//InputForm > > > > > > FullForm[2]^2 > > > > This is the reason why you also get: > > > > > > {Numerator[#],Denominator[#]}&@FullForm[2/3]//InputForm > > > > > > {FullForm[2/3], 1} > > > > which I think makes pretty clear what happened also in your example. > > > > Andrzej Kozlowski > > > > > > > > On 29 Aug 2006, at 08:26, p-valko at tamu.edu wrote: > > > >> Thanks and basically I agree. However, a funny side effect is the > >> following: > >> > >> Denominator[Exp[x] /. x -> -2 ] > >> gives > >> E^2 > >> > >> but > >> > >> Denominator[FullForm[Exp[x]] /. x -> -2 ] gives > >> 1 > >> > >> In other words, wrapping the FullForm around an expression changes > >> its > > > >> "meaning". > >> This contradicts the statement that > >>>> FullForm acts as a "wrapper", which affects printing, but not > >>>> evaluation.<< > >> Or am I missing something? > >> > >> > >> > >> Andrzej Kozlowski wrote: > >>> I do not consider any of these examples "a contradiction". What > >>> needs > > > >>> to be understood is the difference between certain Mathematica > >>> expressions before and after they are evaluated. In the case of your > >>> examples consider: > >>> > >>> > >>> Numerator[Unevaluated[Exp[-x]]] > >>> > >>> E^(-x) > >>> > >>> vs > >>> > >>> Numerator[Exp[-x]] > >>> > >>> 1 > >>> > >>> > >>> Numerator[Unevaluated[1/Sqrt[2]]] > >>> > >>> 1 > >>> > >>> vs > >>> > >>> Numerator[1/Sqrt[2]] > >>> > >>> Sqrt[2] > >>> > >>> The cause of the apparent problem is that Numerator does not hold > >>> its > > > >>> argument, so whiteout Unevaluated you are getting the numerator of > >>> the whatever your expression evaluates to, not of your actual input. > >>> TO get the latter you simply need to use Unevaluated. This is no > >>> different from: > >>> > >>> > >>> Numerator[3/6] > >>> > >>> 1 > >>> > >>> Numerator[Unevaluated[3/6]] > >>> > >>> 3 > >>> > >>> If this is not a contradiction than neither are the other examples. > >>> Understanding the process of evaluation is perhaps the most > >>> important > > > >>> thing when using functional languages (not just Mathematica, the > >>> same > > > >>> sort of things occur, perhaps in an even more striking way, in > >>> languages like Lisp ). > >>> > >>> Andrzej Kozlowski > >>> > >>> > >>> > >>> On 27 Aug 2006, at 07:24, p-valko at tamu.edu wrote: > >>> > >>>> 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 > >>>> > >> > >

**RE: "Anti-Comments"?**

**RE: Can't Remove[] variable "K"? (Really strange behavior . . . )**

**Re: Re: General--Exponential simplifications by default**

**nullspaces**