Re: Re: Graphing Functions
- To: mathgroup at smc.vnet.net
- Subject: [mg22569] Re: [mg22543] Re: [mg22476] Graphing Functions
- From: Bojan Bistrovic <bojanb at python.physics.odu.edu>
- Date: Sat, 11 Mar 2000 17:52:42 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
> > As a relative novice in Mathematica, I need help with a very basic
> > problem involving the graphing of a certain function.
> >
> > The function in question is:
> >
> > x^(1/3) -
> > x^(2/3).
> >
> > The plotting function Plot[f, xmin, xmax] seems unable to deal with cube
> > roots of negative fractional real numbers.
> >
> > Please let me know how I can obtain a plot of the above function over
> > the real number line from say, from x = -10 to x = 10.
> >
> > Thank you.
> >
> > Julian
> >
> Using the identity x=Exp[Log[x]] you have
>
> x^(1/3) - x^(2/3)= Exp[1/3 Log[x] ] - Exp[2/3 Log[x]]
>
> which for negative numbers becomes
>
> Abs[x]^(1/3) Exp[I Pi/3] - Abs[x]^(2/3) Exp[2 I Pi/3]
>
> or
>
> Abs[x]^(1/3) (1 + I Sqrt[3])/2 - Abs[x]^(2/3) (-1 + I Sqrt[3])/2
>
> which in general ISN'T real for x<0 so you have to use 3D plots.
>
To comment more on this question, in the example above the 'first' brunch cut
was chosen leading to the phases (I Pi/3) for x^(1/3) and (2 I Pi/3) for
x^(2/3). In general, there is an infinite number of choices for the phase
(I Pi/3 * n) for for x^(1/3) and (2 I Pi/3 *n) for x^(2/3) where is integer.
However, only three give differet solutions since the rest differ by adding
(2 Pi I)*n to the phase: {I Pi/3, 2 I Pi/3, 3 I Pi/3= I Pi} for x^(1/3) and
{2 I Pi/3, 4 I Pi/3, 6 I Pi/3 = 1} for x^(2/3). All choices will still lead
to complex solutions for x<0 so simple 2D Plot won't suffice.
Bye, Bojan
--
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Bojan Bistrovic, bojanb at jlab.org
Old Dominion University, Norfolk VA & Jefferson Lab, Newport News, VA
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