Re: question related to (-1)^(1/3)
- To: mathgroup at smc.vnet.net
- Subject: [mg96738] Re: question related to (-1)^(1/3)
- From: Bill Rowe <readnews at sbcglobal.net>
- Date: Mon, 23 Feb 2009 05:02:54 -0500 (EST)
On 2/22/09 at 12:33 AM, lxguard-hw at yahoo.com.cn
(=C3=92=C2=BB=C3=92=C2=B6=C3=96=C2=AA=C3=87=C3=AF) 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, ... }
By default, Mathematica does not restrict solutions to real
numbers. Specifically,
In[21]:= NSolve[x^3 == -1, x]
Out[21]= {{x->-1.},{x->0.5-0.866025 I},{x->0.5+0.866025 I}}
and
In[23]:= RSolve[{x[n + 1] == -x[n]^(1/3), x[0] == 1}, x[n], n]
Out[23]= {{x[n] -> E^((-(1/2))*I*Pi*3^(1 - n)*(1 - 3^n))},
{x[n] -> E^(-((1/2)*I*Pi*3^(1 - n)*(1 - 3^n)) -
(2*I*Pi)/3^n)},
{x[n] -> E^((2*I*Pi)/3^n - (1/2)*I*Pi*3^(1 - n)*
(1 - 3^n))}}
Consequently, the list retuned by RecurrenceTable also includes
complex values.