Plot of (2 x^2 - x^3)^(1/3)
Bernard
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
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
--
Murray Eisenberg mur...@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
Matthias Bode
Table[{x, 2*x^2 - x^3, (2*x^2 - x^3)^
(1/3), Re[(2*x^2 - x^3)^(1/3)]},
{x, -4, 4}]
Best regards,
MATTHIAS BODE
S 17.35775=B0, W 066.14577=B0
> Date: Sat, 28 Aug 2010 07:01:34 -0400
> From: bernard.v...@edu.ge.ch
> Subject: Plot of (2 x^2 - x^3)^(1/3)
> To: math...@smc.vnet.net
>
> 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 -
Peter Breitfeld
The reason is, that Mathematica always uses complex arithmetic.
Especially for multi-valued roots it takes the main value e^x=Exp[e Log[b]]
f[x_]:=(2x^2-x^3)^(1/3)
f[3] --> (-1)^(1/3)*3^(2/3)
val=ComplexExpand[f[3]] --> (3/2)I*3^(1/6) + 3^(2/3)
So f[3] is a complex number with positive real part.
There are two ways to get your graph:
1. avoid non integer Powers, e.g. using ContourPlot
ContourPlot[2x^2-x^2==y^3, {x,-1,3},{y,-3,3}]
2. define a version of Power, that uses the real branch like this:
rprule=(b_?Negative)^Rational[m_,n_?OddQ]:>(-(-b)^(1/n))^m;
Attributes[realPower]={Listable,NumericFunction, OneIdentity};
realPower[b_?Negative,Rational[m_,n_?OddQ]]:=(-(-b)^(1/n))^m
realPower[x_,y_]:=x^y
realPower[x]:=x//.rprule
The last definition ensures, that realPower[(-8)^(1/3)] gives -2
Now Plot will work:
g[x_]:=realPower[2x^2-x^2,1/3]
Plot[g[x],{x,-1,3}]
--
_________________________________________________________________
Peter Breitfeld, Bad Saulgau, Germany -- http://www.pBreitfeld.de
Bernard
I conclude that Piskounov use another convention for the root z^(1/3).
He gives, for the graph of (2 x^2 - x^3)^(1/3) what I obtain with :
Plot[Piecewise[{{(2 x^2 - x^3)^(1/3), x <= 2}, {-Re[(2 x^2 -
x^3)^(1/3)], x > 2}}], {x, -1, 3}]
See : Calcul Differentiel et integral, N. Piskounov, Editions
MIR, Moscou 1970, on p. 210
Best Regards=
Alexei Boulbitch
but not in the second case, did not you?
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 !
--
Alexei Boulbitch, Dr. habil.
Senior Scientist
Material Development
IEE S.A.
ZAE Weiergewan
11, rue Edmond Reuter
L-5326 CONTERN
Luxembourg
Tel: +352 2454 2566
Fax: +352 2454 3566
Mobile: +49 (0) 151 52 40 66 44
e-mail: alexei.b...@iee.lu
--
This e-mail may contain trade secrets or privileged, undisclosed or
otherwise confidential information. If you are not the intended
recipient and have received this e-mail in error, you are hereby
notified that any review, copying or distribution of it is strictly
prohibited. Please inform us immediately and destroy the original
transmittal from your system. Thank you for your co-operation.
__________ Information from ESET NOD32 Antivirus, version of virus signature database 5407 (20100829) __________
The message was checked by ESET NOD32 Antivirus.
Peter
Hi Bernard,
maybe you want
Plot[Root[#^3-(2 x^2-x^3)&,1]//Re,{x,-1,4}]
?
Peter
chungyuandye
nsomnia.ber...@gmail.com> wrote:
> On 28 Aug., 13:02, Bernard <bernard.vuilleum...@edu.ge.ch> wrote:
>
s
> > 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 -
>
> > 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 !
>
> Hi Bernard,
>
> maybe you want
>
> Plot[Root[#^3-(2 x^2-x^3)&,1]//Re,{x,-1,4}]
>
> ?
>
> Peter
Unprotect[Power];
Power[x_?Negative, Rational[p_, q_?OddQ]] := (-(-x)^(1/q))^p;
Protect[Power];
Plot[(2 x^2 - x^3)^(1/3), {x, -1, 3}]