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Re: branch of (-1)^(1/3)

  • To: mathgroup at smc.vnet.net
  • Subject: [mg94508] Re: branch of (-1)^(1/3)
  • From: "slawek" <human at site.pl>
  • Date: Mon, 15 Dec 2008 07:44:55 -0500 (EST)
  • References: <200812121154.GAA27892@smc.vnet.net> <gi2uug$a6m$1@smc.vnet.net>

U¿ytkownik "Carl Woll" <carlw at wolfram.com> napisa³ w wiadomo¶ci 
news:gi2uug$a6m$1 at smc.vnet.net...
> I assume you mean a simple way to choose the branch of a^(1/3), where a
> is real. If so, you can use:
>
> Root[#^3-a&, 1]


It doesn't work, because this appoach may be used in this way

In[24]:= (-1)^(1/3) /. a_^(1/3) -> Root[#^3 - a &, 2]
Out[24]= 1/2 (1 - I Sqrt[3])

nevertheless is completly unusable in this example

In[25]:= (-Sin[x])^(1/3) /. a_^(1/3) -> Root[#^3 - a &, 2]
Out[25]= Root[Sin[x] + #1^3 &, 2]

You can see, that Root works only for "numbers" - whereas a simple Sin[x] is 
enouch to stop evaluating Root[ ].

Obviously, I still can make the calculation by the pencil... and maybe my 
old log ruler. Is Mathematica suitable for?

And there is no help if

something /. a_^(1/3) -> Abs[a]^(1/3)

because the ^(1/3) still will fail recognize that the Abs[a] is real and 
that most obvious is a real result. (Because there are many things that are 
really real - for example taxes - if in any country the tax will be computed 
as (income/factor)^(-1/3) it would mean that the tax is imaginary!!! :) )

Regarding the numbering of roots: the system is arbitrary, it may be any 
permutation and the mathematics will remain the same.
The real - non-real is real: we are in R or in C, quite different sets.
The "branch number" is artifical similary as artifical are plate numbers on 
cars.

slawek
 



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