$y=k^x$ when $k$ is negative

Question

Hi I am learning about exponential functions and it points out that $k$ must be positive in $y=k^x$. I was wondering what would happen if $k$ is negative as I have drawn a rough copy of the graph and it seems quite interesting. Is it a valid graph and can anyone draw a stimulation of it please as I can't find an online graph sketcher that will draw it. Would $y=(-1)^x$ be almost like a oscillating wave?? Also can $k=0$ ? I'm just curious if this sort of graph exists. Thanks

The outputs will only be real for certain values of x. For instance, it’s percectly valid to consider your function for integer values of x, but what happens when x is, say, 1/2? — TM Gallagher, Feb 21 '19 at 15:28

Klaus · Answer 1 · 2019-02-21T15:36:32.817

1

The problem is that $k^x$ is defined as $\exp(x\log(k))$ (at least if $x$ is not rational), which is problematic for negative $k$. Even if you restrict yourself to rational $x$, you need to explain what $k^{1/2} = \sqrt{k}$ is (and similarly for higher roots). It is only unproblematic for integer $x$.

edited Feb 21 '19 at 15:36

answered Feb 21 '19 at 15:27

Klaus

10,578

okay so is the graph not possible because you can't sqr root a negative number? – yt. Feb 21 '19 at 15:31
Do you really mean "The problem is that $k^x$ is defined as $\exp(x\log(k))$"? Perhaps you mean "is equal to"? – user1729 Feb 21 '19 at 15:32
@yt. Yes, basically. And things get worse if you take $x$ irrational. – Klaus Feb 21 '19 at 15:34
@user1729 How would you define $k^x$ for $x$ irrational? – Klaus Feb 21 '19 at 15:34
I don't know. But I also do not know how to define $e^x$ for $x$ irrational. – user1729 Feb 21 '19 at 15:42
@user1729 $e^x = \exp(x)$ is defined by the exponential series for any $x \in \mathbb{R}$. – Klaus Feb 21 '19 at 15:43
Ah, okay, I understand. I didn't know that. Thanks. – user1729 Feb 21 '19 at 15:56

score 1 · Answer 2 · answered Feb 21 '19 at 15:29

In the real plane, it doesn’t make sense for any $x$ that is not an odd integer.

However, if you were to draw it on an Agrand diagram, which is the complex plane, you’ll get a unit circle centered at the origin.

You can write $-1=e^{i\pi}$, so $(-1)^x=e^{i\pi x}$. This traces out the unit circle for $x\in\mathbb{R}$.

For negative $k$, you’ll get a spiral that spirals out anti-clockwise.

Warren Hill · Accepted Answer · 2019-02-21T19:42:10.960

Lets consider $k^x$ when x is an integer and $k<1$.

Provided $x$ is an integer $x \in \mathbb{I}$ we have no issues.

Taking for example $k = -2$ we have:

$$ \begin{array}{c|c|c} x&\text{}&k^x\\ \hline 3&(-2)^3&-8\\ 2&(-2)^2&4\\ 1&(-2)^1&-2\\ 0&(-2)^0&1\\ -1&\dfrac{1}{(-2)^1}&-\dfrac{1}{2}\\ -2&\dfrac{1}{(-2)^2}&\dfrac{1}{4}\\ -3&\dfrac{1}{(-2)^3}&-\dfrac{1}{8} \end{array} $$

Now consider $x = \dfrac{1}{2}$, and $k^x = \sqrt{-2} = ?$

We can solve this with complex numbers $k^x = i \cdot \sqrt{2}$

Where $i = \sqrt{-1}$

Similarly $x = -\dfrac{1}{2}$, and $k^x = \dfrac{1}{\sqrt{-2}} = - \dfrac{i}{\sqrt{2}}$

But our solutions are not real numbers any more $k^x \notin \mathbb{R}$

When $ k \lt 0$ we do not have real solutions for all x

$y=k^x$ when $k$ is negative

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