What does $(-2)^x$ really mean?

Question

I think we can all agree that $(-2)^{-1}=-1/2,(-2)^0=1,(-2)^1=-2,(-2)^2=4$
But what does the function $f=(-2)^x$ really mean? It is defined on the integers based on how most people understand exponents, but on the real numbers it's not so easy...

It means whatever you define it to mean. There are a number of approaches to defining it, or you can leave it undefined. If $x,y$ are complex, then $x^y$ can take infinitely many values, in general, and you often have to pick one. — Thomas Andrews, Mar 31 '16 at 19:19
If you are thinking about functions on the real line, it is undefined for most values of x. If you are thinking of functions on the complex plane, then it looks like a spiral. — Doug M, Mar 31 '16 at 19:22
if $x$ is a real number (not integer) such a term doesn't exist, example $$(-2)^{\sqrt{2}}$$ — Dr. Sonnhard Graubner, Mar 31 '16 at 20:35
@Dr.SonnhardGraubner It does exist, (and surprisingly it is a real number): $$(-2)^{\sqrt2}=-2.665144142690225...$$ — KKZiomek, Mar 31 '16 at 20:42
it is $$- 0.7096088653016609545438498923032672288670855758869730994984249159316629988322227863101748814329390800

2.568939189549122939813366063104304284132809433723442681091269785109717841183284552528775955609159484 ,i $$ — Dr. Sonnhard Graubner, Mar 31 '16 at 20:54
@KKZlomek $-2^\sqrt2 = -2.6651441426902251886502972498731...$, not (-2)^{\sqrt2}$. — sqtrat, Mar 31 '16 at 21:00

score 1 · Answer 1 · answered Apr 02 '16 at 12:51

1

Over the reals, I would use:

$$(-2)^x=2^x(-1)^x$$

$$=2^xe^{(2n-1)\pi ix}$$

Where $n\in\mathbb{Z}$ and $i=\sqrt{-1}$

answered Apr 02 '16 at 12:51

Simply Beautiful Art

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What does $(-2)^x$ really mean?

1 Answers1