Why does $y=x^a$ flicker around the x-axis as you adjust $a$?

Question

I know raising to an even power makes things positive and raising to an odd power preserves sign, but if you graph fractional powers they seem to alternate. Why is this?

GIF demonstrating

Answer 1

The exponents shown on that image are decimals, but it helps if you write them as rational numbers. For example compare $x^{0.2}$, $x^{0.4}$, $x^{0.5}$, $x^{0.6}$, and $x^{0.75}$.

$x^{0.2}=x^{\frac15}=\sqrt[5]x$. This is defined for all $x$ and the function is odd, graph symmetric about origin.

$x^{0.4}=x^{\frac25}=(\sqrt[5]x)^2=\sqrt[5]{x^2}$. Defined for all $x$ but unlike the previous example, the function is even, all values are positive (except at $0$), graph is symmetric about $y$-axis.

$x^{0.5}=x^{\frac12}=\sqrt x$. Defined only for non-negative $x$, graph only in the right half-plane.

$x^{0.6}=x^{\frac35}=(\sqrt[5]x)^3=\sqrt[5]{x^3}$. Defined for all $x$ and, like in the first example, the function is odd, graph symmetric about origin.

$x^{0.75}=x^{\frac34}=(\sqrt[4]x)^3=\sqrt[4]{x^3}$. Defined only for non-negative $x$, graph only in the right half-plane.

You may try more examples on your own, like $x^{0.8}=x^{\frac45}$. But you will likely see three types of graphs (similar to the ones illustrated above):

(i) defined for all $x$ and graph symmetric about the origin (positive values for positive $x$, and negative values for negative $x$),

(ii) defined for all $x$ and graph symmetric about $y$-axis (positive values for all $x\not=0$),

(iii) defined only for $x\ge0$.

When you ask the computer to gradually vary the exponent $a$ within a certain range, then you should keep in mind that a computer could only work with approximations of real numbers, and these approximations are rational numbers, and in lowest terms you could have $a=\frac{\mathrm{odd}}{\mathrm{even}}$, or $a=\frac{\mathrm{even}}{\mathrm{odd}}$, or $a=\frac{\mathrm{odd}}{\mathrm{odd}}$. Try to match these three possibilities to (i), (ii), and (iii) listed earlier.

Answer 2

There is no good reason for this. The choice of branch for roots is rather arbitrary. You need to start looking at complex exponential to build any valuable intuition for this.

Answer 3

It's because of how your calculator chooses to define, or not define, exponentiation when $x$ is a negative number and $a$ is a real number.

In particular, what I suspect it is using is the definition that, for negative $x$,

$$x^a := \begin{cases} -\sqrt[q]{(-x)^p},\ \mbox{if $a = \frac{p}{q}$, rational in reduced form},\\ \mbox{undefined},\ \mbox{if $a$ is irrational} \end{cases}$$

This definition is basically one that generalizes how that

$$(-1)^{1/2}$$

is undefined on the real domain if thought of as a square root, because there is no real number who squares to -1, but

$$(-1)^{1/3}$$

can be thought of as a cube root, and, it turns out, -1 cubes to -1, thus we might be tempted (and a lot succumb to such temptation) to define

$$(-1)^{1/3} := -1$$

More generally, even roots are undefined for negatives, odd roots are defined. The problem with this definition is, of course, just what you've observed: it behaves pretty badly because the rationals densely intermix fractions with even and odd denominators, and looks really weerd (esp. try and contemplate graphing $y = (-2)^x$ ... lurkey ... VAAWQH!), while for positive $x$, things behave nicely.

To me, this is good reason I'd think to instead simply leave $x^a$ undefined when $x$ is a real number less than 0 and $a$ is a real number who is not an integer. (There are no problems defining when $a$ is an integer, of course.) And this doesn't mean you lose the ability to find the real solution of

$$x^3 = -1$$

and similar, it's just that you have to write it as

$$x = -(1^{1/3})$$

and not

$$x = (-1)^{1/3}$$

.

Sadly, the problem is convincing calculator manufacturers of this idea.

Cheers.