# why the two images are different? Are the functions same?

why the second image just has the positive part? Is the problem in the domain of definition? 0.0

The code is here:

{Plot[1-Power[t^2, (3)^-1],{t,-1,1}],Plot[1-t^(2/3),{t,-1,1}]}


• Use CubeRoot or Surd instead of 1/3 Jul 22, 2021 at 2:46

As others have pointed out alternative methods to get the desired result, I'll address the why of the matter.

The difference stems from a two-step cascade, the first being computational and the second, mathematical. We can see the first step with the use of Trace.

Power[t^2, (3)^-1] /. t -> -0.5 // Trace
t^(2/3) /. t -> -0.5 // Trace


{...,{(-0.5)^2, 0.25}, 0.25^(1/3), 0.629961}

{..., (-0.5)^(2/3), -0.31498+0.545562 I}

These results show that the fundamental difference in evaluation is that the first procedure first computes the square which yields a positive number and is thus in the principal branch of the following cube root, giving a real number which Mathematica can plot (as opposed to a non-real). This is not the case with the direct evaluation in the second procedure. The shortcomings of using Power are highlighted in the "Possible Issues" documentation of CubeRoot.

• Thanks for your explaination. It's really helpful! (/▽＼) Jul 22, 2021 at 8:31
{Plot[1 - Power[t^2, (3)^-1], {t, -1, 1}],
Plot[1 - Surd[t, 3]^2, {t, -1, 1}]}


Or

{Plot[1 - Power[t^2, (3)^-1], {t, -1, 1}],
Plot[1 - CubeRoot[t]^2, {t, -1, 1}]}