Re: question related to (-1)^(1/3)
- To: mathgroup at smc.vnet.net
- Subject: [mg96730] Re: [mg96711] question related to (-1)^(1/3)
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Mon, 23 Feb 2009 05:01:27 -0500 (EST)
- Reply-to: hanlonr at cox.net
(-1)^(1/3) == Exp[I*Pi/3] True Exp[I*Pi/3] // N 0.5+0.866025 I Exp[I*Pi/3]^3 -1 Sign[-1]*Abs[-1]^(1/3) -1 RecurrenceTable[{x[n + 1] == -Sign[x[n]]*Abs[x[n]]^(1/3), x[0] == 1}, x, {n, 1, 10}] {-1,1,-1,1,-1,1,-1,1,-1,1} Bob Hanlon ---- "=C3=92=C2=BB=C3=92=C2=B6=C3=96=C2=AA=C3=87=C3=AF" <lxguard-hw at yahoo.c= om.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