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