Re: Re: Why can't Mathematica tell when something is algebraically
- To: mathgroup at smc.vnet.net
- Subject: [mg108193] Re: [mg108127] Re: Why can't Mathematica tell when something is algebraically
- From: Adam Strzebonski <adams at wolfram.com>
- Date: Wed, 10 Mar 2010 01:45:24 -0500 (EST)
- References: <hn2ltj$3kt$1@smc.vnet.net> <201003091119.GAA06777@smc.vnet.net> <AF2BE895-2E51-4F9C-9410-8CB5CB418899@mimuw.edu.pl> <0F45FC76-FF73-4401-B3E6-011F8C6AFCDA@mimuw.edu.pl>
- Reply-to: adams at wolfram.com
All that Assuming does is locally changing the value of $Assumptions.
Assuming[assum, expr]
is a shorthand for
Block[{$Assumptions = $Assumptions && assum}, expr]
All functions that take the Assumptions option, with the exception of
PowerExpand, have the default option setting of
Assuptions :> $Assumptions
This means that they will automatically use the value of $Assumptions,
whether it has been set with Assuming, or directly with
$Assumptions = assum
PowerExpand is an exception, its default option setting is
In[1]:= Options[PowerExpand]
Out[1]= {Assumptions -> Automatic}
where Automatic means that PowerExpand is allowed to assume that
all "simple" power and log expansion transformations are valid.
This is the expected default behaviour of PowerExpand.
In[2]:= PowerExpand[(a b)^c]
c c
Out[2]= a b
If Assumptions was set to $Assumptions, which by default evaluates
to True, PowerExpand would add terms that make the expansion valid
without any assumptions and the expected default behaviour would not
be preserved.
In[3]:= PowerExpand[(a b)^c, Assumptions :> $Assumptions]
c c (2 I) c Pi Floor[1/2 - Arg[a]/(2 Pi) - Arg[b]/(2 Pi)]
Out[3]= a b E
Best Regards,
Adam Strzebonski
Wolfram Research
Andrzej Kozlowski wrote:
> In fact, I was wrong, and I forgot of something I once new (or assumed that it had been "fixed"). Assuming has no effect on PowerExpand and the answer returned is completely misleading. For example:
>
> Assuming[Im[r] != 0,
> PowerExpand[r^2 Sqrt[(r^3 + r + 2)/r] - Sqrt[r^3 (r^3 + r + 2)]]]
> 0
>
> which is of course wrong.
>
> PowerExpand only works with Assumptions passed view the Assumptions -> ... method and then the answer is very complicated:
>
> PowerExpand[
> r^2*Sqrt[(r^3 + r + 2)/r] - Sqrt[r^3*(r^3 + r + 2)],
> Assumptions -> Element[r, Reals]]
>
> r^(3/2)*Sqrt[r^3 + r + 2]*
> E^(I*Pi*Floor[Arg[r]/(2*Pi) - Arg[r^3 + r + 2]/
> (2*Pi) + 1/2]) - r^(3/2)*Sqrt[r^3 + r + 2]*
> E^(I*Pi*Floor[-((3*Arg[r])/(2*Pi)) -
> Arg[r^3 + r + 2]/(2*Pi) + 1/2])
>
> But, what is rather strange, this works:
>
> PowerExpand[r^2 Sqrt[(r^3 + r + 2)/r] - Sqrt[r^3 (r^3 + r + 2)],
> Assumptions -> r <= 0]
>
> even though it does not work with Simplify and Reduce
>
> Reduce[r^2 Sqrt[(r^3 + r + 2)/r] - Sqrt[r^3 (r^3 + r + 2)] == 0 &&
> r < 0, r]
> Reduce::cpow: Reduce was unable to prove that a radical of an expression containing only real variables and parameters is real valued. If you are interested only in solutions for which all radicals contained in the input are real valued, use Reduce with domain argument Reals. >>
>
> The question is then, does PowerExpand really do that correctly? It would be reassuring if someone from WRI would confirm it. (Getting the right answer with PowerExpand itself is not very reassuring...).
>
> I remember a while ago there was already a discussion here concerning the fact that PowerExpand is probably the only function that admits Assumptions only in the form Assumptions -> and not via Assuming. I though that was going to be changed but it remains as it used to be. If changing this behaviour is really impossible I think there ought to be some sort of warning message that appears when one attempts to use
> Assuming[something,PowerExpand[expr]], for obvious reasons.
>
> Andrzej Kozlowski
>
>
> On 9 Mar 2010, at 14:22, Andrzej Kozlowski wrote:
>
>> It seems better to use:
>>
>> Assuming[Element[r, Reals],
>> PowerExpand[r^2 Sqrt[(r^3 + r + 2)/r] - Sqrt[r^3 (r^3 + r + 2)]]]
>>
>> 0
>>
>> PowerExpand with Assumptions ought to be as reliable as Simplify (which does not work in this case), whereas if you use it without assumptions it can return answers that will be false for some values of the variables (it makes use of the maximal assumptions that permit the expansion). The fact that Simplify works here with the assumption r>=0 but not r<0 appears to be a bug.
>>
>> Andrzej Kozlowski
>>
>>
>>
>>
>> On 9 Mar 2010, at 12:19, dh wrote:
>>
>>> Hi,
>>> Sqrt is a multivalued function. Therefore, your expression is not
>>> necessarily zero. If you are sure that all your variables are real and
>>> that you only want the main branch of Power, then you can use PowerExpand:
>>> r^2 Sqrt[(r^3 + r + 2)/r] - Sqrt[r^3 (r^3 + r + 2)] // PowerExpand
>>> Daniel
>>>
>>> On 08.03.2010 12:09, mmdanziger wrote:
>>>> This isn't the first time that I've encountered something like this in
>>>> Mathematica but in my calculations I got a term like this:
>>>>
>>>> r^2 Sqrt[(r^3 + r + 2)/r] - Sqrt[r^3 (r^3 + r + 2)]
>>>>
>>>> Which is obviously identically zero. For some reason Simplify or even
>>>> FullSimplify can't figure this out. Once you get dependent on
>>>> Mathematica these things are pretty disturbing...you forget about your
>>>> own knowledge because the program tells you that things are
>>>> different. Then you sit there like an idiot checking an algebraic
>>>> identity that any beginning precalc student should be able to solve no
>>>> problem.
>>>>
>>>> Is there any way to get Mathematica to "wake up" to these things? It
>>>> has such a powerful algebraic engine for most things, why can't it see
>>>> something simple like the above? Do you really have to manually
>>>> override and tell the program when things should be zero?
>>>>
>>>> For the time being I'll just sift through and test things by hand but
>>>> I can't believe that there isn't a better way.
>>>>
>>>> Best,
>>>> md
>>>>
>>>
>>> --
>>>
>>> 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>
>>>
>>>
>
- Follow-Ups:
- Re: Why can't Mathematica tell when something is algebraically
- From: Andrzej Kozlowski <akoz@mimuw.edu.pl>
- Re: Why can't Mathematica tell when something is algebraically
- References:
- Re: Why can't Mathematica tell when something is algebraically zero?
- From: dh <dh@metrohm.com>
- Re: Why can't Mathematica tell when something is algebraically zero?