       Re: Plot of (2 x^2 - x^3)^(1/3)

• To: mathgroup at smc.vnet.net
• Subject: [mg112080] Re: Plot of (2 x^2 - x^3)^(1/3)
• From: Murray Eisenberg <murray at math.umass.edu>
• Date: Sun, 29 Aug 2010 02:50:44 -0400 (EDT)

```Why do you think the second Plot should give the same result: you're
clearly looking at a different function. In the first Plot, you forced
taking negatives of the real parts; in the second Plot, you do not.

Take, for example, x->3:

(2 x^2 - x^3)^(1/3) /. x -> 3 // ComplexExpand // InputForm
((3*I)/2)*3^(1/6) + 3^(2/3)/2

Its real part is positive, which is exactly what the second Plot is showing.

I presume you realize that the function z^(1/3) gives the principal cube
root. And since

2 x^2 - x^3 /. x -> 3
-9

Arg[-9]
Pi

then the value of (2 x^2 - x^3)^(1/3) for x->3 returned by Mathematica
is what the definition of principal cube root says it should be, namely,
the same as the value of:

Exp[Log[Abs[-9]]/3] Exp[I Arg[-9]/3] // ComplexExpand

On 8/28/2010 7:01 AM, Bernard wrote:
> I see in Calcul Diff=E9rentiel et int=E9gral, N. Piskounov, Editions
> MIR, Moscou 1970, on p. 210 the graph of (2 x^2 - x^3)^(1/3) with
> negative values for x>  2, something like :
>
> Plot[Piecewise[{{(2 x^2 - x^3)^(1/3), x<= 2}, {-Re[(2 x^2 -
> x^3)^(1/3)],  x>  2}}], {x, -1, 3}]
>
> When I plot this function with :
>
> Plot[(2 x^2 - x^3)^(1/3) // Re, {x, -1, 3}]
>
> I obtain positives values for x>  3. I don't understand why.
> Thank you very much for your help !
>

--
Murray Eisenberg                     murray at math.umass.edu
Mathematics & Statistics Dept.
Lederle Graduate Research Tower      phone 413 549-1020 (H)
University of Massachusetts                413 545-2859 (W)
710 North Pleasant Street            fax   413 545-1801
Amherst, MA 01003-9305

```

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