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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
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