Re: Re: Mathematica language issues

*To*: mathgroup at smc.vnet.net*Subject*: [mg53021] Re: [mg53011] Re: Mathematica language issues*From*: DrBob <drbob at bigfoot.com>*Date*: Mon, 20 Dec 2004 06:34:40 -0500 (EST)*References*: <200412171020.FAA16185@smc.vnet.net> <cq0tm1$2m2$1@smc.vnet.net> <200412191115.GAA18073@smc.vnet.net>*Reply-to*: drbob at bigfoot.com*Sender*: owner-wri-mathgroup at wolfram.com

Ditto, Maxim. Anyway, an explanation of this behavior (if we had one) doesn't equate to a REASON for it. Bobby On Sun, 19 Dec 2004 06:15:22 -0500 (EST), Maxim <ab_def at prontomail.com> wrote: > On Sat, 18 Dec 2004 09:36:01 +0000 (UTC), Andrzej Kozlowski > <akoz at mimuw.edu.pl> wrote: > >> >> On 17 Dec 2004, at 19:20, Maxim wrote: >> >>> In[5]:= >>> Unevaluated[1 + 1]*2 >>> 2*Unevaluated[1 + 1] >>> >>> Out[5]= >>> 4 >>> >>> Out[6]= >>> 2*Unevaluated[1 + 1] >>> >> >> This is not a glitch but works exactly as one woudl expect. You can >> see the difference and the reason by looking at Trace in both cases >> (although there is no need for that, if you understand Unevaluated you >> can see it right away). >> >> First: >> 2*Unevaluated[1+1]//Trace >> >> >> {2 (1+1),2 Unevaluated[1+1]} >> >> >> First Unevaluated is stipped away and Mathematica attempts ot evaluate >> 2*(1+1). Since it knows no rule to apply and the expression has not >> changed Unevaluated is restored and evaluation is completed with the >> output you see. >> >> >> >> Unevaluated[1+1]*2//Trace >> >> {(1+1) 2,2 (1+1),{1+1,2},2 2,4} >> >> As before, first Unevaluated is stripped away and Mathematica tires to >> evaluate 2*(1+1). It now knows a rule to apply, which is given by the >> Orderless attribute and the canonical ordering, so it converts the >> expression into the form 2 (1+1). But now Unevaluated is not restored >> because the expression has changed so evaluation continues with 1+1 >> evaluationg to 2 and finally you obtain 4. >> >> Now, I have honstly considered this case only because I could see at >> once what what was going on. I do not knwo if any of the others are >> glitches but jusdging by my experience with the past "language >> glitches" you have reported (unlike the more serious problems desribed >> in your last posting) I rather doubt it. However I have no time to >> spend on this just to prove a point (again). >> >> >> >> Andrzej Kozlowski >> Chiba, Japan >> http://www.akikoz.net/~andrzej/ >> http://www.mimuw.edu.pl/~akoz/ >> > > I do not agree. Suppose we evaluate z*Unevaluated[1 + 1]; according to > your explanation, after the reordering of the factors Unevaluated should > disappear from the final result. However, the expression evaluates to > Unevaluated[1 + 1]*z. Further, suppose we take Unevaluated[1 > + 1]*Sin[Pi/4]: Sin[Pi/4] evaluates to 1/Sqrt[2], so in this case an > evaluation step definitely takes place; however, the output is > Unevaluated[1 + 1]/Sqrt[2]. Your theory simply doesn't work. But even if > it did, there is another problem: suppose I use Sin[Pi/8] instead of > Sin[Pi/4] -- then first you would need to know whether Mathematica has a > built-in rule for Sin[Pi/8] to arrive at any conclusion as to how it might > work with Unevaluated (that is, what will count as an evaluation step?). > So to apply your explanation we would have to search through all the > built-in rules of Mathematica. > > Maxim Rytin > m.r at inbox.ru > > > > -- DrBob at bigfoot.com www.eclecticdreams.net

**References**:**Mathematica language issues***From:*ab_def@prontomail.com (Maxim)

**Re: Mathematica language issues***From:*Maxim <ab_def@prontomail.com>