Why is $(-1)^x=e^{i\pi x}$

Question

I was recently taught exponentials and I decided to play around with negative bases, which they told me were not allowed. The obvious place to start was negative one, and, as expected, the graphing tool did not work. However, after trying Wolfram, it told me it was equaled $e^{i\pi x}.$ I would understand why it oscillates, like the sine and cosine waves (negative base to an even power becomes positive, negative to an odd power is negative) but why are they exactly the same?

Answer 1

Because one logarithm of $-1$ is indeed $i \pi,$ as in $$ e^{i \pi} +1 = 0. $$

Another logarithm is $3 i \pi,$ so a less common assignment for $(-1)^x$ would be $$ e^{3 i \pi x} $$

One application that is not widely known is with the Gelfond-Schneider Theorem, using your expression. Since $-1$ is not $0$ or $1$ and is algebraic (indeed rational and an integer), if $x$ is irrational and algebraic over $\mathbb Q,$ then $$ e^{ i \pi x} = \cos \pi x + i \sin \pi x $$ is transcendental. In turn, that means the real numbers $\cos \pi x$ and $\sin \pi x$ are transcendental.

I used that in an article.

Really.

Answer 2

Usually for $a$ positive, you define

$$a^x := e^{x \ln(a)}$$

Now, for $a$ negativve, $\ln(a)$ is not defined. You can still work around that by noticing that

$$-1 = e^{i\pi}$$

So, it would makes sense to define

$$\ln(-1) := i\pi$$

The problem is that it makes as much sense to define it as

$$\ln(-1) := i(2k+1)\pi$$

So, you could define

$$(-1)^x := e^{x i(2k+1)\pi }$$

And by convention, you choose $k=0$

Answer 3

This has to do with how the exponential function is expanded from the real numbers to the complex plane. This is done by the following basic definition:

Let $x$ be a real number. Then we define $e^{ix}$ as follows:

$e^{ix}=\cos(x)+i\sin(x)$.

With this definition of an exponential of imaginary numbers, we get that:

$e^{i(2k+1)\pi}=\cos((2k+1)\pi)+i\sin((2k+1)\pi)=-1+0i=-1$ for all integers $k$, and thus also in the special case where $k=0$, which yields the result $e^{i\pi}=-1$.

Now we can see why it's natural to reach your identity. Remember, if we have a positive real number $a$, and real numbers $b$ and $x$, we have that

$(a^b)^x=a^{bx}$.

Now if we allow $a$ to be negative, and $b$ to be imaginary, then we would naturally reach the conclusion that

$(-1)^x=(e^{i\pi})^x=e^{i\pi x}$.

It's not really this easy to state the identity, and one needs to be a bit more rigorous in defining the exponentiation of negative numbers than I've been here to make sure no logical contradictions will occur, but this is a basic explanation for your identity.

Answer 4

do you know that cosθ + i sinθ can be written as e^iθ

since -1 = cosπ +i sinπ

       = **e**^iπ

we can write(-1)^x =(e^iπ)^x=e^iπx