Re: Not all points plot on my graph...
- To: mathgroup at smc.vnet.net
- Subject: [mg105191] Re: [mg105158] Not all points plot on my graph...
- From: Murray Eisenberg <murray at math.umass.edu>
- Date: Tue, 24 Nov 2009 05:46:41 -0500 (EST)
- Organization: Mathematics & Statistics, Univ. of Mass./Amherst
- References: <200911231149.GAA21721@smc.vnet.net>
- Reply-to: murray at math.umass.edu
Because Mathematica uses the principal cube root, hence the principal 2/3 power; when a number is negative, the principal cube root is not the negative real number you might expect, but rather a non-real complex number. For example: cuberoot = ComplexExpand[(-3)^(1/3)] 3^(1/3)/2 + (I/2)*3^(5/6) Expand[cuberoot^2] ((3*I)/2)*3^(1/6) - 3^(2/3)/2 This is often an unwanted and, for Mathematica novices, an unexpected behavior. But Mathematica wants things to be complex when they can be! There's a very easy fix here, since the 2/3 power is the cube-root of a square, and any square is nonnegative: f[x_] := ((x + 2)^2)^(1/3) davef wrote: > Why is it when I write this... > > Clear[f, x] > f[x_] = (x + 2)^(2/3) > Plot[f[x], {x, -3, 0}, PlotRange -> {{-3, 0}, {-4, 4}}] > f[-3] > > .. I get no values show in the graph for x<-2 even though f[-3] > evaluates to (-1)^(2/3). > > (-1)^(2/3) evaluates as f[-3]=1 so why wouldn't the point (-3,1) plot > on my graph? > > -- 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
- References:
- Not all points plot on my graph...
- From: davef <davidfrick2003@yahoo.com>
- Not all points plot on my graph...