The graph of $\ y=(-2)^x$

Question

When I attempt to graph the function $\ y=(-2)^x$, all that is output is a point at (1,0). Why won't my calculator graph anything further? Does this graph exist? What does it look like?

Answer 1

$(-2)^x$ is only a real number when $x$ is an fraction with odd denominator. This is because $(-2)^x = e^{x\log(-2)} = e^{x[\log(2)+(2k+1)i\pi]} = e^{x\log(2)}\cdot e^{(2k+1)i\pi x} = 2^x\left[ \cos((2k+1)\pi x) + i\sin((2k+1)\pi x) \right]$.

This number is real if and only if $\sin((2k+1)\pi x) = 0$, which means that the argument of $\sin$ should be an integer multiple of $\pi$, i.e. $(2k+1)\pi x = n\pi \Rightarrow x = \frac{n}{2k+1}$ for integers $n,k$.

In complex algebra, since there are multiple $p$-th roots, we tend to choose $x^{1/p}$ to be the principal root, which is not a real number when $x$ is real and $p>1$. Thus, in light of this, $(-2)^x$ is only real when $x$ is a fraction whose denominator is $1$, in other words, an integer. Since the function is only real for a (relatively) small subset of real numbered inputs, most graphing utilities won't show a curve.

If you want to see the full curve with real and complex parts, you can input $(-2)^x$ into WolframAlpha.

Answer 2

The fundamental problem here is one of domain. Recall that a function is a way of mapping points or objects in one set (the domain) to points or objects in another set (the codomain). We might write $$ f : X \to Y, $$ which says that $f$ is function with domain $X$ and codomain $Y$. The graph of $f$ is then a subset of the Cartesian product $X\times Y$. Formally, $$ X\times Y := \{ (x,f(x)) : x\in X \}. $$ In English, this is the set of all tuples or ordered pairs $(x,y)$ such that $x$ is an element of the domain of $f$ and $y = f(x)$ is the point in the range of $f$ that $x$ gets sent to. Because human beings have highly evolved visual processing systems, we often draw graphs in an attempt to gain better insight. This is particularly easy to do when the domain and codomain are both the real numbers, because we can draw a pair of vertical axes, and treat the points of the graph as points in the plane.

So, what does this have to do with anything?

When we try to define a function by the formula $f(x) = (-2)^x$, we first have to determine what kinds of objects we can feed to this function—that is, what is the domain?

Typically, we like to work with real numbers. Indeed, unless told to do otherwise, this is likely what you calculator is doing. Unfortunately, this function is not typically defined for most real numbers. It is well defined for integers, and is fine for rational numbers with odd denominators, but starts to have heartburn when you feed it rational numbers with even denominators, and pretty much gives up on irrational numbers.

The problem, more or less, is that if $n$ is an integer, then the function $y \mapsto y^{2n}$ will only output positive numbers. Unfortunately, this means that there are no numbers $x$ and $n$ (with $n$ and integer) such that $-2 = y^{2n}$. Taking inverses, this means that $(-2)^{\frac{1}{2n}} = \sqrt[2n]{-2} = y$ is not defined. By extension, if $n$ is an integer, then $$ f\left( \frac{m}{2n} \right) = (-2)^{\frac{m}{2n}} = \sqrt[2n]{-2}^m $$ is undefined. The problems are even worse for irrational numbers—there isn't even a good way of defining $(-2)^x$ for an arbitrary irrational number without being rather sneaky.

Therefore, as a function with real numbers as its domain, $f(x) = (-2)^x$ is going to be a royal pain to graph. You could, I suppose, graph a large set of points of the form $(x,(-2)^x)$ where $x$ is a rational number with odd denominator, but you wouldn't be able to "connect the dots" in any meaningful way (i.e. the function is quite discontinuous).

On the other hand, if we allow ourselves to work with complex numbers instead of real numbers, there are things that can be done (though not without difficulty). As AlexanderJ93 seems to have addressed this point while I was typing, I will simply refer you to his answer.