Re: Re: Re: question related to (-1)^(1/3)
- To: mathgroup at smc.vnet.net
- Subject: [mg96787] Re: [mg96766] Re: [mg96749] Re: question related to (-1)^(1/3)
- From: Murray Eisenberg <murray at math.umass.edu>
- Date: Wed, 25 Feb 2009 04:01:33 -0500 (EST)
- Organization: Mathematics & Statistics, Univ. of Mass./Amherst
- References: <gnqo3c$c95$1@smc.vnet.net> <200902231004.FAA28251@smc.vnet.net> <200902241046.FAA23618@smc.vnet.net>
- Reply-to: murray at math.umass.edu
You COULD use the legacy RealOnly package:
<<Miscellaneous`RealOnly`
RecurrenceTable[{x[n+1]==-x[n]^(1/3), x[0]==1},x, {n,0,200}]//N
You'll get what you wanted. Note that you have to start the iterator n
at 0, though, not 1; otherwise, you'll have as output {-1.,1.,-1.,...}.
The Mathematica paradigm is different from another system. There's
no direct analog of "domain".
patapon wrote:
> I just want to get the real solution of (-1)^(1/3).
> BTW, I have read the tutorial but I am still confused with its explanation.
> But that is not important, I just feel it is inconvenient for the following
> problem:
>
> RecurrenceTable[{x[n + 1] == -x[n]^(1/3), x[0] == 1}, x, {n, 1, 200}] // N
>
> I consider there should be a way to limit the output in the domain real.
> for example, in MAXIMA
>
> (%i1) domain;
> (%o1) real
> (%i2) (-1)^(1/3);
> (%o2) -1
> (%i3) domain:complex;
> (%o3) complex
> (%i4) (-1)^(1/3),numer;
> (%o4) 0.86602540378444*%i+0.5
>
> On Mon, Feb 23, 2009 at 6:04 PM, Sjoerd C. de Vries <
> sjoerd.c.devries at gmail.com> wrote:
>
>> This misunderstanding pops up over and over again in this group.
>> Please type tutorial/FunctionsThatDoNotHaveUniqueValues in the search
>> bar of the mathematica doc centre.
>>
>> Cheers -- Sjoerd
>>
>> On Feb 22, 7:33 am, "=D2=BB=D2=B6=D6=AA=C7=EF" <lxguard... at yahoo.com.cn> =
>> wrote:
>>> I have tried expression:
>>> RecurrenceTable[{x[n + 1] == -x[n]^(1/3), x[0] == 1},
>>> x, {n, 1, 200}] // N
>>>
>>> Mathematica produce
>>> {-1., -0.5 - 0.866025 I, -0.766044 + 0.642788 I, -0.686242 -
>>> 0.727374 I, -0.71393 + 0.700217 I, -0.704818 - 0.709389 I,
>>> ...
>>>
>>> But it should be {1, -1, 1, -1, ... }
>>>
>>> If you try (-1)^(1/3)
>>>
>>> In[10]:= (-1)^(1/3)
>>>
>>> Out[10]= (-1)^(1/3)
>>>
>>> In[11]:= % // N
>>>
>>> Out[11]= 0.5 + 0.866025 I
>>
>>
>
--
Murray Eisenberg murray at math.umass.edu
Mathematics & Statistics Dept.
Lederle Graduate Research Tower phone 413 549-1020 (H)
University of Massachusetts 413 545-2859 (W)
710 North Pleasant Street fax 413 545-1801
Amherst, MA 01003-9305
- References:
- Re: question related to (-1)^(1/3)
- From: "Sjoerd C. de Vries" <sjoerd.c.devries@gmail.com>
- Re: Re: question related to (-1)^(1/3)
- From: patapon <lxguard-hw@yahoo.com.cn>
- Re: question related to (-1)^(1/3)