Are these two expression equal?

Question

My friend insisted that $(-1)^{(-n)}$ is equivalent to $(-1)^n$ for any number of $n$. A quick check in the Wolfram Alpha show false like this:

But then he showed me the working (substituting $n=0$ to $n=3$ in each expression)and indeed they are the same.

So, I am a bit confused on whether those two expressions are really equal. Will they give out the same number every time or am i missing something?

Answer 1

I assume your friend and you both know that the expression $(-1)^x$ only makes sense if $x$ is an integer.

That said, you have

$$(-1)^{(-n)} = \frac{1}{(-1)^n} = \left(\frac{1}{-1}\right)^n = (-1)^n$$

I believe all steps only use equalities you learnt in school:

• For all $a, b$ we have $a^{-b} = \frac{1}{a^b}$
• If $\frac{c}{d}$ is a fraction, then $\left(\frac cd\right)^b = \frac{c^b}{d^b}$
• $\frac{a}{-b} = -\frac{a}{b}$.

Answer 2

Note that $$ (-1)^{-n} = \frac{1}{(-1)^n} = \frac{1^n}{(-1)^n} = \left(\frac{1}{(-1)}\right)^n = \left(\frac{1(-1)}{(-1)(-1)}\right)^n = (-1)^n$$

Answer 3

I guess that mathematica or whatever wolfram thing that gave you this answer was just confused by the definition of n. If it is not an integer, the equality fails. You can convince yourself of this by choosing n to be 1/2 (granted that you know about complex numbers).

Answer 4

$\dfrac{1}{(-1)^n}=(-1)^n \implies 1=(-1)^n(-1)^n \implies 1=(-1)^{2n}$ which is true for all integers, but nothing else.

Answer 5

I suppose that the question in OP is to understand why WolframAlpha gives ''false'' as answer.

This is because WA evaluate the expressions $(-1)^n$ and $(-1)^{-n}$ in the field of complex numbers $\mathbb{C}$. In this case have, for $n=a+ib$ we have: $$ (-1)^n=(e^{i\pi})^{a+ib}=e^{a \pi i}e^{-b \pi} $$ $$ (-1)^{-n}=(e^{i\pi})^{-a-ib}=e^{-a \pi i}e^{b \pi} $$ That are, in general, different.

If $n$ is a real positive number than, using Euler formula $ e^{i \theta}=\cos \theta + i \sin \theta$, we find:

$$ (-1)^{n}=(e^{i\pi})^n=\cos (\pi n)+i\sin(\pi n) $$

$$ (-1)^{-n}=(e^{i\pi})^{-n}=\cos (-\pi n)+i\sin(-\pi n)=\cos (\pi n)-i\sin(\pi n) $$ that are equal iff $n \in \mathbb{Z}$.

So the result for $n$ integer is that $(-1)^{n}=(-1)^{-n}$ and WA agree as you can see here.