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MathGroup Archive 2004

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Re: Re: Log[4]==2*Log[2]

  • To: mathgroup at smc.vnet.net
  • Subject: [mg50611] Re: [mg50557] Re: [mg50520] Log[4]==2*Log[2]
  • From: DrBob <drbob at bigfoot.com>
  • Date: Sun, 12 Sep 2004 04:42:27 -0400 (EDT)
  • References: <200409090917.FAA19334@smc.vnet.net> <200409100805.EAA04777@smc.vnet.net> <opsd4ehiyriz9bcq@monster.cox-internet.com> <2E4E6EFA-03F0-11D9-99E9-000A95B4967A@akikoz.net> <opsd59t8ufiz9bcq@monster.cox-internet.com> <598B2886-043B-11D9-99E9-000A95B4967A@akikoz.net> <746B4F54-043E-11D9-99E9-000A95B4967A@akikoz.net>
  • Reply-to: drbob at bigfoot.com
  • Sender: owner-wri-mathgroup at wolfram.com

As is often the case with a CAS, the trick is to manage our expectations.

Bobby

On Sun, 12 Sep 2004 07:03:47 +0900, Andrzej Kozlowski <andrzej at akikoz.net> wrote:

> *This message was transferred with a trial version of CommuniGate(tm) Pro*
> More fully and correctly (taking the rist of sounding very  pedantic):
> in the case when the quantities compared are not equal, even if they
> are exact, Mathematica can "settle" the question only if the difference
> is "large enough".
> In the case of exact input, as in the posted example, he important
> thing is, that for equality the "required precision" is Infinity. So,
> an equality between exact numbers, when it holds, will be settled
> without applying Simplify or FullSimplify only if it can be settled by
> using exact numbers and a few very simple transformation rules used by
> the Evaluator. When equality does not hold it will be settled in cases
> when the difference between the two quantities is large enough, and in
> the remaining cases it will require increasing $MaxExtraPrecision.
>
> In general with exact input it seems preferable to apply Simplify or
> FullSimplify without prior numerical evaluation, although of course
> there are many cases when simplification will not succeed while
> numerical comparison will, and cases when neither simplification nor
> numerical check can settle the question of equality.
>
>
> Andrzej
>
>
>
> On 12 Sep 2004, at 06:41, Andrzej Kozlowski wrote:
>
>> Yes, I agree. Indeed, what it does it tries to settle "exact equality"
>> using inexact numbers, and it of course it can't be done. However,  as
>> you correctly point out, had the quantities entered not been equal
>> Mathematica would have settled the question. This is of course the
>> reason why it does anything at all except just simply returning the
>> input.
>>
>> Andrzej
>>
>>
>> On 12 Sep 2004, at 02:07, DrBob wrote:
>>
>>> *This message was transferred with a trial version of CommuniGate(tm)
>>> Pro*
>>>>> Mathematica could not determine anything because it tries to
>>>>> compare the numbers "numerically" without using approximate
>>>>> numerical
>>>>> values, which can't be done.
>>>
>>> I agree it can't be done; so why is Mathematica trying to do that?
>>> Documentation says:
>>>
>>>>> For exact numeric quantities, Equal internally uses numerical
>>>>> approximations to establish inequality.
>>>
>>> Inequality usually can be established this way, but equality never
>>> can be. I suppose that's what the error message "intends to mean".
>>> Equal tried to establish INEQUALITY and failed, yet is unwilling to
>>> declare EQUALITY.
>>>
>>> Hence the test could never be True for exact quantities. But...
>>>
>>> Log[2]==Log[2]
>>>
>>> True
>>>
>>> Sigh....
>>>
>>> Bobby
>>>
>>> On Sat, 11 Sep 2004 21:43:28 +0900, Andrzej Kozlowski
>>> <andrzej at akikoz.net> wrote:
>>>
>>>> *This message was transferred with a trial version of
>>>> CommuniGate(tm) Pro*
>>>> Actually, I don't think Mathematica does any real "determining" since
>>>> it does not replace the exact values given in the input by
>>>> numerical approximations. The message issued is, I think, purely
>>>> formal. Mathematica could not determine anything because it tries to
>>>> compare the numbers "numerically" without using approximate numerical
>>>> values, which can't be done. (You have to apply N for it to use
>>>> numerical values). That't what I meant by "not surprisingly". I don't
>>>> think I  really understand your point?
>>>>
>>>> ANdrzej
>>>>
>>>>
>>>> On 11 Sep 2004, at 01:52, DrBob wrote:
>>>>
>>>>> *This message was transferred with a trial version of
>>>>> CommuniGate(tm)
>>>>> Pro*
>>>>>>> Mathematica does not apply any simplification rules but justtries
>>>>>>> to
>>>>>>> evaluate the expression numerically and, not
>>>>>>> surprisingly, it can't determine if the LHS is zero or not
>>>>>>> up to the required precision.
>>>>>
>>>>> On the contrary, I think the error message itself clearly indicates
>>>>> the difference IS zero to "the required precision". If 50 digits
>>>>> extra
>>>>> precision isn't enough to determine that the difference ISN'T zero,
>>>>> why doesn't Equal return True?
>>>>>
>>>>> Bobby
>>>>>
>>>>> On Fri, 10 Sep 2004 04:05:56 -0400 (EDT), Andrzej Kozlowski
>>>>> <andrzej at akikoz.net> wrote:
>>>>>
>>>>>> On 9 Sep 2004, at 18:17, Andreas Stahel wrote:
>>>>>>
>>>>>>>
>>>>>>> To whom it may concern
>>>>>>>
>>>>>>> the following answer of Mathematica 5.0 puzzeled me
>>>>>>>
>>>>>>> Log[4]==2*Log[2]
>>>>>>> leads to
>>>>>>>
>>>>>>> N::meprec: Internal precision limit $MaxExtraPrecision = 50.`
>>>>>>> reached
>>>>>>> while \
>>>>>>> evaluating -2\Log[2]+Log[4]
>>>>>>>
>>>>>>> with the inputs given as answer. But the input
>>>>>>>
>>>>>>> Log[4.0]==2*Log[2]
>>>>>>>
>>>>>>> leads to a sound "True"
>>>>>>>
>>>>>>> Simplify[Log[4]-2*Log[2]]
>>>>>>> leads to the correct 0, but
>>>>>>> Simplify[Log[4]-2*Log[2]==0]
>>>>>>> yields no result
>>>>>>>
>>>>>>> There must be some systematic behind thid surprising behaviour.
>>>>>>> Could somebody give me a hint please
>>>>>>>
>>>>>>> With best regards
>>>>>>>
>>>>>>> Andreas
>>>>>>> --
>>>>>>> Andreas Stahel       E-Mail: Andreas.Stahel at [ANTI-SPAM]hti.bfh.ch
>>>>>>> Mathematics, HTI     Phone: ++41 +32 32 16 258
>>>>>>> Quellgasse 21        Fax:   ++41 +32 321 500
>>>>>>> CH-2501 Biel         WWW:   www.hta-bi.bfh.ch/~sha
>>>>>>> Switzerland
>>>>>>>
>>>>>>>
>>>>>>
>>>>>> When you enter
>>>>>>
>>>>>> Log[4] - 2*Log[2] == 0
>>>>>>
>>>>>> Mathematica does not apply any simplification rules but just tries
>>>>>> to
>>>>>> evaluate the expression numerically and, not surprisingly, it can't
>>>>>> determine if the LHS is zero or not up to the required precision.
>>>>>>
>>>>>> If you use
>>>>>>
>>>>>> Simplify[Log[4] - 2*Log[2] == 0]
>>>>>>
>>>>>> Mathematica first tries to evaluate the argument of Simplify and
>>>>>> the
>>>>>> same thig happens as above, but then it actually applies Simplify
>>>>>> to
>>>>>> the output and gets the right answer True.
>>>>>>
>>>>>> The best thing to do is:
>>>>>>
>>>>>>
>>>>>> Simplify[Unevaluated[Log[4]-2*Log[2]==0]]
>>>>>>
>>>>>>
>>>>>> True
>>>>>>
>>>>>> which avoids evaluation of the argument and instead uses Simplify
>>>>>> on
>>>>>> the unevaluated input.
>>>>>>
>>>>>>
>>>>>>
>>>>>> Andrzej Kozlowski
>>>>>> Chiba, Japan
>>>>>> http://www.akikoz.net/~andrzej/
>>>>>> http://www.mimuw.edu.pl/~akoz/
>>>>>>
>>>>>>
>>>>>>
>>>>>
>>>>>
>>>>>
>>>>> --
>>>>> DrBob at bigfoot.com
>>>>> www.eclecticdreams.net
>>>>>
>>>>
>>>>
>>>>
>>>
>>>
>>>
>>> --
>>> DrBob at bigfoot.com
>>> www.eclecticdreams.net
>>>
>>
>
>
>



-- 
DrBob at bigfoot.com
www.eclecticdreams.net


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