Fractional/decimal powers

[DaveC426913]

TL;DR Summary

Apparently some values are not defined in y=x^z where z is a real number

It has been many decades since high school math, so be patient with my fumblings.

I was lying in bed last night trying to do some calcs on the ceiling - the nature of which I have now forgotten - when I tried to picture the transition from x^2 to x^3. I broadened it to the general question y=x^z here z is a real number and got some weird results. I sat up to check on my phone and find that apparently some values are just not defined. I asked a chatbot, and it provided a partial answer (it only solved for integer z's) - but it didn't explain why there are some undefined values.

When I tried to do it in my head, I got weird results. Probably because I'm doing it wrong:

	X	Z		Y
	2	2	2x2	4
	2	3	2x2x2	8
	2	2.5	2x2x1	4
	2	2.25	2x2x.5	2
	2	2.75	2x2x1.75	7

I see I'm doing it wrong. Apparently, 2^2.5 should be arranged as a fraction: 2^5 / 2^2?
But that is 32/4=8, which is the same as 2^3, so that must be wrong.
I punched it into a calculator and got 5.65685424949. So now I'm even more confused.

Still, what is mystifying me is that the parabola of x^2 is entirely positive (no solutions below y=0), yet x^1.9999 and x^2.0001 has a negative component (full set of solutions in lower left quadrant). The 3D graph seems to have discontinuities at every even value of x that are infinitesimally thin planes.


So now I'm looking to graph the whole thing, say, from x=-2 to x=2 and from z=-1 to z=3 where z is a real number. Is this the kind of thing that can be done online? I tried Wolfram Alpha many years ago but it was too complex for me to get very far. Is that still a thing?

 

[erobz]

TL;DR Summary: Apparently some values are not defined in y=x^z where z is a real number

It has been many decades since high school math, so be patient with my fumblings.

I was lying in bed last night trying to do some calcs on the ceiling - the nature of which I have now forgotten - when I tried to picture the transition from x^2 to x^3. I broadened it to the general question y=x^z here z is a real number and got some weird results. I sat up to check on my phone and find that apparently some values are just not defined. I asked a chatbot, and it provided a partial answer (it only solved for integer z's) - but it didn't explain why there are some undefined values.

When I tried to do it in my head, I got weird results. Probably because I'm doing it wrong:

	X	Z		Y
	2	2	2x2	4
	2	3	2x2x2	8
	2	2.5	2x2x1	4
	2	2.25	2x2x.5	2
	2	2.75	2x2x1.75	7

I see I'm doing it wrong. Apparently, 2^2.5 should be arranged as a fraction: 2^5 / 2^2?
But that is 32/4=8, which is the same as 2^3, so that must be wrong.
I punched it into a calculator and got 5.65685424949. So now I'm even more confused.

$2^{2.5} = 2^{2 + \frac{1}{2}} = 2^{2} \cdot 2^{\frac{1}{2}} = 2^{\left( \frac{5}{2}\right)} = \left( 2^5\right)^{\frac{1}{2}} = \sqrt{32} = 4 \sqrt{2} \approx 5.656...$

You get these by rules for exponents.

 

[DaveC426913]

Thanks.

 

[erobz]

Thanks.

You are getting mixed up on what the definition of a fractional exponent is. It's read like an inverse argument. $2 ^{\frac{1}{2}}$ is read like "this number is the number that when it is multiplied by itself two times gives you $2$" (the base). The answer being $\sqrt{2}$ for the example.

 

[DaveC426913]

No, I get that a number raised to 1/2 is the root.

But I see where I made my mistake.
2^1/2 is not 2¹/ 2², (which would be 1/4).
It's the root of 2¹, which is root 2.

 

[DaveC426913]

So I sketched this up using an online 2D plotter. (I manually overlaid each 2D graph in Photoshop, and gthen offset each layer by a few pixels, to make an ersatz 3D plot.)

This is the graph of y=x ^z, for the values of z=3.0 to -1.0 by increments of 0.2.

That "tail" in the lower left quadrant flips from negative to positive discontinuously every time I cross an even decimal such as n.8 or n.4:

What will happen if I take smaller increments of z, such as .1, or .01? Will the graph ever be continuous, or will it always have these discontinuities? How far down the decimal rabbit hole will the discontinuities go? Is there a fixed number, or does the number depend on the resolution?

 

[jbriggs444]

What will happen if I take smaller increments of z, such as .1, or .01? Will the graph ever be continuous, or will it always have these discontinuities? How far down the decimal rabbit hole will the discontinuities go? Is there a fixed number, or does the number depend on the resolution?

Are you familiar with how exponentiation works with polar coordinates for complex numbers?

There are n complex nth roots of 1. When plotted with cartesian coordinates $(a,b)$ for $a + bi$ these are spread evenly around the unit circle. In polar coordinates $(r, \theta)$, they are each at an angle of $k\frac{2\pi}{n}$ for some $k$. $r$ is the same for all of them.

If n is even then two of the roots will be real numbers. One positive, directly to the right of the origin and one negative, directly to the left of the origin.

If n is odd then only one of the roots will be a real number. The one directly to the right of the origin.

If you try to do complex exponentiation to a rational power, you can will find the finitely many possible results scattered based on whatever number is in the denominator of the rational exponent.

If you try to do complex exponentiation to an irrational power, you will find a countable infinity of possible results spread densely around in a circle.

Your graphs are misleading because they seem to be presented in an isometric manner rather than being properly overlaid with a common origin.

 

[DaveC426913]

Are you familiar with how exponentiation works with polar coordinates for complex numbers?

I am not.

There are n complex nth roots of 1. When plotted with cartesian coordinates $(a,b)$ for $a + bi$ these are spread evenly around the unit circle. In polar coordinates $(r, \theta)$, they are each at an angle of $k\frac{2\pi}{n}$ for some $k$. $r$ is the same for all of them.

If n is even then two of the roots will be real numbers. One positive, directly to the right of the origin and one negative, directly to the left of the origin.

If n is odd then only one of the roots will be a real number. The one directly to the right of the origin.

If you try to do complex exponentiation to a rational power, you can will find the finitely many possible results scattered based on whatever number is in the denominator of the rational exponent.

If you try to do complex exponentiation to an irrational power, you will find a countable infinity of possible results spread densely around in a circle.

Thanks. Good reading there.

Your graphs are misleading because they seem to be presented in an isometric manner rather than being properly overlaid with a common origin.

I don't see why. It is a simple xyz Cartesian graph with the z-axis protruding from the page in increments of .2b (the grey line is the z-axis). The x and y axes have been removed for clarity because I'm only really interested in a qualitative view of the surface in this rough sketch.

There may be other preferred ways of visualizing the data but that doesn't make this visualization misleading. (Albeit the crudeness makes it hard-to-read.)

Here is a similar graph. It has different data and better visualization, but it is fundamentally the same thing:

 

[jbriggs444]

I don't see why. It is a simple xyz Cartesian graph with the z-axis protruding from the page in increments of .2b (the grey line is the z-axis). The x and y axes have been removed for clarity because I'm only really interested in a qualitative view of the surface in this rough sketch.

The z axis is protruding from the page at an angle. If I look at your overlaid drawing, I can identify the curve for $y=x^{2.8}$ and a curve for $y=x^{-1}$. But the origins for the two curves do not coincide.

Be that as it may, let me try to explain what I think you are seeing.

Consider $y=x^{2.8}$. What does that mean? $x^{2.8}$ can be computed as $x^2 \times x^{0.8}$ We treat $x^{0.8}$ as $x^{\frac{4}{5}}$ That is the same thing as the fifth root of $x^4$. Always positive.

This means that $x^{2.8}$ is always positive.

Now consider $y=x^{2.6}$. In the same way, that is $x^2 \times x^{0.6} = x^2 \times x^{\frac{3}{5}}$ The latter term is the fifth root of $x^3$. Not always positive.

The 0.4 power is the same as the $\frac{2}{5}$ power which yields the fifth root of $x^2$. Always positive.

The 0.2 power is the same as the $\frac{1}{5}$ power which yields simply the fifth root of $x$. Not always positive.

If you reduce the step size, you will not get continuity. Which is why we do not usually try to define exponentiation within the reals for negative values of the base and irrational values of the exponent.

Here is one of the first things I Googled up for exponentiation with a negative base and an irrational exponent:

It depends what number system you are working in. In the reals, there is no answer. In the complex numbers, there are many answers.

 

[DaveC426913]

If you reduce the step size, you will not get continuity. Which is why we do not usually try to define exponentiation within the reals for negative values of the base and irrational values of the exponent.

This is what I'm groping toward. I don't usually encounter discontinuous curves like this - unless I'm plotting trig functions, then they're ubiquitous - but at least trig functions' curves make sense. They curve smoothly off toward infinity.

Here is one of the first things I Googled up for exponentiation with a negative base and an irrational exponent:

Yeah. I'm trying to intuit why that's so.

 

[jbriggs444]

Yeah. I'm trying to intuit why that's so.

I will try to give you my take on it.

It is straightforward to define a positive real number $x$ raised to a rational power $\frac{a}{b}$. It is the $b$th root of $x^a$.

By itself, that does not help us in defining $x$ raised to an irrational power. But positive numbers raised to rational powers are continuous. So we can define $x^y$ for irrational $y$ in terms of the limit of $x^q$ as rational $q$ approaches irrational $y$.

If we try the same thing for a negative base, you've already noticed that continuity goes out the window. So the limit would be undefined.