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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

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    $\begingroup$ 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? $\endgroup$ Feb 21 '19 at 15:28
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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

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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.

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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 $x$.

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

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