Need conceptual help with -1⁰ ...

By Joe A. on Saturday, August 26, 2000 - 07:13 am :

We are told that X⁰ =1 (X not = 0)...
But when X is negative (in particular -1) I am having trouble conceptually. My problem is that if you take the lim(n-> infinity) of -1^1/n this should approach -1⁰ =1, right?

But I can think of an infinite number of situations where -1^1/n =-1 (for all odd n)...

How can this be and maintain the idea of -1⁰ =1?

Please help, I have been racking my brain over this one all day...

By Dan Goodman (Dfmg2) on Tuesday, August 29, 2000 - 06:34 pm :

The answer to this problem is a bit tricky to explain without using complex numbers, do you know about complex numbers? If not, the following won't make much sense.

$(- 1)^{z} = e^{z \log (- 1)} = e^{z (2 m + 1) π i}$ for any integer $m$ . You're considering $z = 1 / n$ , and if $n$ is odd, i.e. $n = 2 k + 1$ , you are choosing the branch of log which gives $m = k$ , in other words, you are taking a different branch of log for different values of $n$ . The correct thing to do is to use the same branch of log ( $m = 0$ is the principal branch), and then you will get the right answer.

Basically, the problem is that there are $n$ solutions to the equation $z^{n} + 1 = 0$ , and you are saying that $(- 1)^{1 / n}$ is a solution to $z^{n} + 1 = 0$ and choosing the solution so that it doesn't get close to 1 at any time.

I'm not sure if that answer will satisfy you, not knowing how much you know about complex numbers and complex functions, etc.

By Joe A. on Wednesday, August 30, 2000 - 04:00 pm :

I know a little about complex numbers, but don't fully understand the explanation. What I understand you are saying is that I am arbitrarily selecting particular values or "a branch" that do not satisfy the result of 1.

Does this mean that the principal branch is all numbers, where I am selecting only odds? If this is the case, then taking successive values for n on the principal branch gives you alternating numbers one that approaches the number 1 and the other that is always -1. So since one branch is always constant at -1 are we really discarding it in favor of the other branch?

By Dan Goodman (Dfmg2) on Monday, September 4, 2000 - 06:57 pm :

OK, quick explanation of branches coming up. A function in mathematics has to take only one value for any particular argument, however there are some "functions" that take more than one value, for instance the square root function can take two values. With these "functions" (sometimes called multivalued functions) we say they have different branches, so for instance sqrt(x) has two branches +sqrt(x) and -sqrt(x), you can extend this to complex numbers as well as real numbers, but it's slightly more difficult. In the above, you are looking at the function

f (z) = (- 1)^{z}

for

z = 1 / n

. This function has an infinite number of branches, but we can work out what they are by using the formula

(- 1)^{z} = e^{z \log (- 1)}

. log is a multivalued function, because if

e^{w} = z

, then

e^{w + 2 m π i} = z

as well, because

e^{2 m π i} = 1

, so there are an infinite number of values for

\log (z)

. To make your question about

(- 1)^{0}

above sensible, you have to choose a particular value of

m

(i.e. a particular branch of the function log), if you do this, and then plug in

z = 1 / n

into the function, and let

n

tend to infinity, you'll find that

f (z)

will tend to 1. I'm in a bit of a hurry right now, if that doesn't make sense, post again and I'll explain further.