When I type the following implicit equation into Mathematica I only get 'one quarter' of what the resulting plot should look like, i.e., I only get the upper-right quadrant.
ContourPlot[x^(2/3) + y^(2/3) == 1, {x, -1, 1}, {y, -1, 1}]
If you type $x^{\frac{2}{3}} + y^{\frac{2}{3}} = 1$ into this graphing calculator you will see the correct output which has a curve in each of the four quadrants.
So why is Mathematica giving an incorrect result, and is it possible to make it plot this implicit function correctly?
ContourPlot[Abs[x]^(2/3) + Abs[y]^(2/3) == 1, {x, -1, 1}, {y, -1, 1}]. Mathematica is giving the correct result; for instance, how to plot(-0.5)^(2/3)? – b.gates.you.know.what Aug 08 '18 at 05:30ContourPlot[CubeRoot[x^2]+CubeRoot[y^2]==1,{x,-1,1},{y,-1,1}]orContourPlot[(x^2)^(1/3)+(y^2)^(1/3)==1,{x,-1,1},{y,-1,1}]– matrix42 Aug 08 '18 at 06:08x^(a/b)as(x^a)^(1/b)whereas in Mathematica this is treated as(x^(1/b))^a. The two are of course not the same. The Mathematica treatment is a consequence of definingx^yas (in more customary notation)exp(y*log(x))– Daniel Lichtblau Aug 08 '18 at 14:28ContourPlot[(x^2)^(1/3) + (y^2)^(1/3) == 1, {x, -1, 1}, {y, -1, 1}]– Bob Hanlon Aug 08 '18 at 23:39