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Re: Simplify
*To*: mathgroup at smc.vnet.net
*Subject*: [mg98703] Re: [mg98659] Simplify
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Thu, 16 Apr 2009 04:18:08 -0400 (EDT)
*References*: <200904150902.FAA08151@smc.vnet.net> <ACCF1343-E098-469B-A599-4CA4F9E1A5E0@mimuw.edu.pl> <49E5CA92.7030005@metrohm.com> <7379A405-70A4-404B-8ACB-B4367F9A5A2A@mimuw.edu.pl> <49E5EA19.7030004@metrohm.com>
Yes, you are quite right. I underestimated WRI ;-)
I have version 7.01 and it does behave this way. The example I used I
took from my past posts to the MathGroup: I wrote about it at least
two times or more, before Mathematica 6, I think. But Simplify has
been improved and is much cleverer now.(I should have expected that
and checked). What has happened is that it now "knows" about this
problem and treats it as a special case. Thus, although the difference
between
2*6^z - 6^z
2^(z + 1) 3^z - 6^z
and
2*5^z - 5^z
5^z
still remains, when you apply Simplify you get the same answer.
However, although the "unpleasant consequence" that I wrote about has
been fixed, my main point remains unchanged. You can never Simplify
2^(z + 1)*3^z
to
2*6^z
even though
LeafCount[Hold[2*6^z]]
6
while
LeafCount[2^(z + 1)*3^z]
9
Andrzej Kozlowski
On 15 Apr 2009, at 23:07, dh wrote:
> Hi Andrzej,
> which version of Mathematica do you have? In version 7.0.1 Simplify seems to
> have changed.
> Your example:
>
> > FullSimplify[2*6^z - 6^z]
> > 2^(z + 1)*3^z - 6^z
>
> in my version:
> Simplify[2*6^z - 6^z]
> 6^z
>
> Daniel
>
> Andrzej Kozlowski wrote:
>> Yes, you are right but I still think it is beside the point.
>> Consider this rather more striking (in my opinion) example.
>> let
>> expr = 2^(z + 1)*3^z
>> You can easily see that this can be re-written as 2*6^z and that:
>> LeafCount[Hold[2*6^z]]
>> 6
>> while
>> LeafCount[expr]
>> 9
>> so 2*6^z has a much smaller LeafCount than expr, but you can never
>> Simplify expr to 2*6^z because the Evaluator will always rewrite it
>> again as expr. This has a rather unpleasant consequence:
>> FullSimplify[2*6^z - 6^z]
>> 2^(z + 1)*3^z - 6^z
>> Simplify cannot see that the simplest answer if 6^z (no matter what
>> ComplexityFunction you use) because immediately on evaluation 2*6^z
>> is converted to expr above and then it is too late; Simplify can't
>> see that the first term is simply twice the second term. Yet
>> Simplify has no problem noticing that:
>> Simplify[2*6^z - 6^z == 6^z]
>> True
>> The reason for the problem is the following "canonical form":
>> expressions like (a^k1*b^k2*c^k3...)^z *(a*l1*b^l2*...) where a,
>> b, c are positive are re-written as a^(k1+l1)*b^(k2+l2)*....
>> For example
>> (2^u*5^v)*(2*3^w*5^z)
>> 2^(u + 1) 3^w 5^(v + z)
>> (actually the rule used is note general, but I just want to give an
>> example of a "canonical form" that causes trouble).
>> Since these reductions are made by the Evaluator, Simplify has no
>> effect on them. But because of them we get the following
>> inconsistent behaviour:
>> 2*5^z - 5^z
>> 5^z
>> but
>> 2*6^z - 6^z
>> 2^(z + 1)*3^z - 6^z
>> This, of course, cannot be changed by applying Simplify with any
>> ComplexityFunction because canonical forms can't be changed by
>> Simplify (unless you apply Hold).
>> Andrzej
>> On 15 Apr 2009, at 20:52, dh wrote:
>>> Hi Andrzej,
>>> thanks for your helpfull explanations.
>>> Let me just mention that there is an additional reason, internally
>>> the expression x^2/y^2 is rewritten and has the the same leaf
>>> count as (x/y)^2. Consider:
>>> FullForm[Hold[x^2/y^2]]
>>> FullForm[x^2/y^2]
>>> Daniel
>>>
>>>
>>> Andrzej Kozlowski wrote:
>>>> Actually, this point has been explained many times (by me ;-))
>>>> (I like to think of Mathematica's evaluation process in terms of
>>>> something called "The Evaluator", which I think I first found in
>>>> David Wagner's book "Power Programming with Mathematica". I think
>>>> it is only an abstraction, along with "the Parser", "the
>>>> Typesetter" etc, but a convenient one when one is thinking about
>>>> the evaluation process. )
>>>> The "Evaluator" always evaluates
>>>> (x/y)^2
>>>> to
>>>> x^2/y^2
>>>> This happens before Simplify takes any effect. Even if Simplify
>>>> converted x^2/y^2 to (x/y)^2 the Evaluator would kick in and
>>>> again convert it back to x^2/y^2. Since the Evaluator always
>>>> overrides Simplify there is no way to get (x/y)^2 as the output
>>>> without using Hold.
>>>> Perhaps you are asking why Mathematica (or the "Evaluator")
>>>> automatically converts (x/y)^2 to x^2/y^2. It's because of
>>>> something called "canonical forms" or "standard forms".
>>>> Basically, in order to optimize performance in computer algebra
>>>> systems one want to reduce to 0 as quickly as possible as many
>>>> expressions that are actually equal. The earlier you do this the
>>>> better the performance. If you allow expressions to contain a
>>>> large number of subexpressions that are actually 0 until you
>>>> apply Simplify, it may very seriously impair performance.
>>>> "Canonical forms" (or "normal forms", there is a slight
>>>> difference between them but I shall ignore it) are certain unique
>>>> forms to which various expressions are reduced automatically by
>>>> the Evaluator (before applying Simplify). This has the effect
>>>> causing cancellations to occur early. Also, these "canonical
>>>> forms" have to be independent of any particular
>>>> ComplexityFunction used, hence the reduction has to be performed
>>>> outside Simplify. The advantage of using canonical forms
>>>> independent of ComplexityFunction is that they often enable
>>>> Mathematica to identify two expressions as equal even if Simplify
>>>> can't fine a sequence of Complexity reducing transformations that
>>>> will convert one expression into the other.
>>>> Not surprisingly, using "canonical forms" can sometimes produce
>>>> undesirable side-effects and this is one of them (a rather minor
>>>> one, worse ones do occur).
>>>> Andrzej Kozlowski
>>>> Andrzej Kozlowski
>>>> On 15 Apr 2009, at 18:02, dh wrote:
>>>>>
>>>>>
>>>>> Hi,
>>>>>
>>>>> can somebody explain, why
>>>>>
>>>>> Simplify[x^2/y^2,ComplexityFunction->LeafCount]
>>>>>
>>>>> does not simplify to (x/y)^2, although the LeafCount is:
>>>>>
>>>>> LeafCount[Hold[x^2/y^2]] gives 10
>>>>>
>>>>> and
>>>>>
>>>>> LeafCount[Hold[(x/y)^2]] gives 8
>>>>>
>>>>>
>>>>>
>>>>> Daniel
>>>>>
>>>>>
>>>>>
>>>
>>>
>>> --
>>>
>>> Daniel Huber
>>> Metrohm Ltd.
>>> Oberdorfstr. 68
>>> CH-9100 Herisau
>>> Tel. +41 71 353 8585, Fax +41 71 353 8907
>>> E-Mail:<mailto:dh at metrohm.com>
>>> Internet:<http://www.metrohm.com>
>>>
>
>
> --
>
> Daniel Huber
> Metrohm Ltd.
> Oberdorfstr. 68
> CH-9100 Herisau
> Tel. +41 71 353 8585, Fax +41 71 353 8907
> E-Mail:<mailto:dh at metrohm.com>
> Internet:<http://www.metrohm.com>
>
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