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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




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