Re: branch of (-1)^(1/3)
- To: mathgroup at smc.vnet.net
- Subject: [mg94443] Re: branch of (-1)^(1/3)
- From: dimitris <dimmechan at yahoo.com>
- Date: Sun, 14 Dec 2008 07:33:06 -0500 (EST)
- References: <ghtjcj$r7f$1@smc.vnet.net>
On 12 =C4=E5=EA, 13:53, "slawek" <hu... at site.pl> wrote:
> Is a simple way to choose the branch of (-1)^(1/3) ?
>
> Mathematica gives a (correct) non-real answer. It is ok, but I need the (=
-1)
> as the output when I input (1)^(1/3) because I know that it is a solut=
ion
> of real-valued problem.
>
> Is any "standard way" to pick up a correct (i.e. arbitrary) root of
> (-1)^(1/n) instead the default?
>
> slawek
Say that exp is one an expression wrt x then you could something like
the following,
Solve[x^3 == 1, x]
exp /. %[[1]]
For n instead of 3 just change the exponent 3 above.
Another way is as follows,
In[14]:=
Unprotect[Power];
In[15]:=
FullForm[(-1)^(Rational[1, n_])] := 1
In[16]:=
Protect[Power];
In[17]:=
(-1)^(1/3)
Out[17]=
1
Note that in this way it might be side effects.
In[18]:=
Solve[x^3 == 1, x]
Out[18]=
{{x -> -1}, {x -> 1}, {x -> (-1)^(2/3)}}
Instead of
In[19]:=
Quit[]
In[1]:=
Solve[x^3 == 1, x]
Out[1]=
{{x -> 1}, {x -> -(-1)^(1/3)}, {x -> (-1)^(2/3)}}
which is the default. This may cause conflicts to some built in
functions.
Dimitris