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MathGroup Archive 2003

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Re: a first-time user question

  • To: mathgroup at smc.vnet.net
  • Subject: [mg39378] Re: [mg39352] a first-time user question
  • From: Dr Bob <drbob at bigfoot.com>
  • Date: Thu, 13 Feb 2003 04:53:16 -0500 (EST)
  • References: <200302120852.DAA14779@smc.vnet.net>
  • Reply-to: drbob at bigfoot.com
  • Sender: owner-wri-mathgroup at wolfram.com

Do you want the third root of x squared, or the square of the third root of 
x?

Here are the third roots of -1, the square of each, and finally that result 
to the reciprocal power:

{#, Conjugate@#, # # #} &@Exp[I 2 Pi/3] // ComplexExpand
%^2 // ComplexExpand
%^(3/2) // ComplexExpand

{-(1/2) + (I*Sqrt[3])/2,   -(1/2) - (I*Sqrt[3])/2, 1}
{-(1/2) - (I*Sqrt[3])/2,   -(1/2) + (I*Sqrt[3])/2, 1}
{                    -1,                       -1, 1}

The real root doesn't get us back to -1, as we would probably like.  In 
other words, (x^(2/3))^(3/2) is not x, if we insist on real roots.

If we take the other approach, squaring and then taking the third root, and 
then raising to the 3/2 power, we get the same answers.  (Take the real 
third root and then multiply by the three third roots of 1.)

((-1)^2)^(1/3){1, Exp[I 2Pi/3], Exp[I 4Pi/3]} // ComplexExpand
%^(3/2) // ComplexExpand

{1, -(1/2) + (I*Sqrt[3])/2,   -(1/2) - (I*Sqrt[3])/2}
{1, -1, -1}

Again, the real 2/3 power of x is positive, and raising it to the 3/2 power 
doesn't give x.  Obviously, the real 2/3 power isn't always the one you 
want, then!

Still, if you want to graph it, ask for it specifically:

Plot[(x^2)^(1/3), {x, -1, 0}]

Bobby

On Wed, 12 Feb 2003 03:52:57 -0500 (EST), Ye Hu <huye at wharton.upenn.edu> 
wrote:

> I have a simple (supposedly) question...
>
> I used the following command to draw a plot, but Mathematica could not 
> run
> properly.
>
> Plot[x^(2/3),{x,-1,0}]
>
>
> I know the problem is about calculating (-1)^(1/3) in mathematica. 
> Although
> (-1)^(1/3) = -1,
> mathematica tries to calculate it numerically and gives complex results
> 0.5+0.8i
> How to solve this problem and restrict the solution to be only real 
> numbers?
>
> Thanks very much.
>
>
>
>



-- 
majort at cox-internet.com
Bobby R. Treat



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