# Is 0^0 defined?

1. Jun 2, 2006

### pivoxa15

Is it defined? if not why not?

2. Jun 2, 2006

### TD

I'm sure you'll find one or more topics about that if you use the search function.

When defined, it is usually defined as 1 (and not as 0) for practical reasons (making certain summations and other formula/notations easier because you don't have the consider the case with 0 separately). Looking at the graph y = x^x for x approaching 0 will make it clear why this is the more logical choice (above 0). Try taking the limit of x^x for x->0 for example.

It is often left undefined though.

3. Jun 2, 2006

### Hurkyl

Staff Emeritus
The most common use of 0^0 = 1 is a result of de-abstraction.

In a polynomial ring, or power series ring, one might have a symbol x. x does not stand for a number; it is a number in its own right! And since x is not zero, or a zero divisor, we would properly say that x^0 = 1.

One of the typical uses of a polynomial ring, or a power series ring, is that we can substitute a number for x. I.E. if we have the polynomial f = x^2 + x + 1, we can make a substitution f(2) = 2^2 + 2 + 1.

Now, if f = x^0 = 1, then f(0) = 1 as well, but formally it looks like we said f(0) = 0^0 = 1!

When exponents can only refer to repeated multiplication, then 0^0 = 1 is usually right.

But when we are working with real numbers, none of the above applies. A very important thing about exponentiation is that it's continuous, and exponentiation cannot be continuous at 0^0, therefore it is undefined there.

4. Jun 2, 2006

### NEWO

it seems also similar to the 0/0 paradox. theory states that anything divided by zero is infinte, theory also states that anything to the power of 0 is 1. Therefore, 0^0 =1.

5. Jun 2, 2006

### arildno

No, it doesn't. Neither is there a paradox here.
No, it doesn't and isn't.

6. Jun 2, 2006

### NEWO

no offense, im sure your credentials are much better than mine but that is how i've been taught and it works for me.

7. Jun 2, 2006

### NEWO

i suppose because zero is strange that 0^0 is not defined.

8. Jun 2, 2006

### AlphaNumeric

As with so many subjects in education, there are lies everywhere. In physics you're told "electrons spin around the nucleus like planets around the sun" only to then find out they form 'distributions'. You are often given the impression at high school level that all functions can be differentiated or integrated nicely, then you try to integrate $$e^{-x^{2}}$$ indefinitely.

Particularly at high school level the notion of proof is often missing so 'just because it works for you' doesn't mean there aren't more powerful results and concepts which should be applied :)

9. Jun 2, 2006

### matt grime

Who taught you that "0/0 is a paradox"? What does thet even mean? Theory does not state that anything divided by zero is infinite. Theory does state that the operation x --->1/x is not a well defined function on R, but merely on R\{0}. Theory states that there are extensions of R (compactifications) where 1/x does extend to a function whose domain includes 0, and the symbol 1/0 is commonly labelled $\infty$. It still does not provide a way to define 0/0 though, as can be readily demonstrated, but it is not a paradox.

Last edited: Jun 2, 2006
10. Jun 2, 2006

### arildno

this is a good thought on your part!
It is precisely because of the properties of zero (which you perceive as "strange"), and because of what we want exponentiation to mean ,that 0^0 is left undefined (usually).

11. Jun 2, 2006

### JCCol

From this am i supposed to gather that 1/0 is infinity or is that wrong?

12. Jun 2, 2006

### matt grime

You can gather what you will. 1/0=$\infty$ is true in the extended real numbers or the extended complex plane. The phrase 1/x tends to infinity as x tends to zero is true in the real numbers, where 'tends to infinity' has a very specific meaning which is often abbreviated in a convenient shorthand to 1/0=[itex]\infty[/tex].

13. Jun 2, 2006

### pivoxa15

Why can't exponentiation be continuous at 0^0?

14. Jun 2, 2006

### Hurkyl

Staff Emeritus
Because 0^x = 0 and x^0 = 1 for positive real numbers x. If exponentiation was continuous at 0^0, then it would be equal to the limit of both of those expressions as x goes to zero.

15. Jun 2, 2006

### rbj

often the question is what is

$$\lim_{x \rightarrow \ 0} x^x \$$

and the answer to that is

$$\lim_{x \rightarrow \ 0} x^x = 1 \$$

16. Jun 2, 2006

### pivoxa15

Also I realised that the function x^x is discontinous at x=0 because there is a limit going from positive x to 0 but no limit going from negative x to 0. So limit as x->0 doesn't exist.

So if a function is continous at a point, we define the value of the function to be that value at that point. If f is not continous there, we defne f to be the limit at that point. But if f dosen't have a limit there than we are in trouble like 0^0. Although the next best thing would be to define f by one side of the limit if it exists which is 1 or 0 in this case. Hence we define 0^0 to be 0 or 1 at different times depending on which is more suitable but its safer to leave it undefined?

17. Jun 2, 2006

### Hurkyl

Staff Emeritus
You've caught a trick there -- depending on what you're doing, you might not even want to define 0^x and x^0 for that very reason!

But at least if we restrict ourselves to the closed first quadrant, but with the origin removed, x^y would be everywhere continuous, so it is still reasonable to leave x^0 and 0^x defined for positive x.

Minor nitpick -- if f is continuous at a point, then its value at that point is the limit, and if is discontinuous there, then its value is something other than the limit.

But I think your idea is right -- you're talking about making a new function whose value at that point is equal to the limit. (And, by abuse of notation, calling that function f too)

But for 0^0, in practice I've only ever really seen it defined when there's something sneaky lurking in the background.

18. Jun 2, 2006

### pivoxa15

This usually occurs in power series? For example, the Taylor expansion of cos(x) about x=0 http://mathworld.wolfram.com/Cosine.html where the first term would be 0^0 when expanding cos(0). But in that case they have assumed it to equal 1 without explicitly stating it. Is it usually the case that if 0^0 occurs but they don't mention its value, just assume it equals 1?

19. Jun 2, 2006

### Gokul43201

Staff Emeritus
Ooh, cos(0) is a veritable playground for the "zeroists" - the Taylor expansion even has the famous 0! in it. And seeing as how it's the lim(x->0) sinx/x, you can throw in the more famous 0/0 into the ring.

Now back to your serious discussion...

20. Jun 2, 2006

### shmoe

You shouldn't have to assume it's anything- any text should explain this notation if they use it. I don't know if Mathworld explicitly mentions them sticking to this usage, but they do mention it http://mathworld.wolfram.com/ExponentLaws.html and the context it comes up in makes it clear what they mean. Some texts will avoid this completely, writing power series instead as

$$a_0+\sum_{n=1}^{\infty}a_{n}x^n$$

21. Jun 3, 2006

### arildno

The mathworld power series definition of cos(x) is incomplete; it ought to have had, say, the following definition appended to it:
$$0^{0}\equiv{1}$$
It should be obvious why such a local definition of the symbol is needed.

Last edited: Jun 3, 2006
22. Jun 3, 2006

### Universe_Man

I asked a simular question a week or two ago. Here's my interpretation (I apologize if it is incorrect, I am but a mere HS student):

As I understand it:

$$n^-1 = 1/n$$ and $$n^1= n$$

and

$$(n^-1)(n^1) = n^0 = (1/n)(n/1) = 1$$

correct?

so if n=0:

$$(0^-1)(0^1) = (1/0)(0/1) = 1 = 0^0$$

So given that the following is true, I would come to the conclusion that $$0^0$$ is defined. But there is something weird about that $$1/0$$.

23. Jun 3, 2006

### Hurkyl

Staff Emeritus
Nope. That identity is false. You get one guess as to why. The actual identity has another clase you left out.

24. Jun 3, 2006

### Universe_Man

Really? I didn't know that. I was always taught that n^-1= 1/n. I asked my teachers why of course (they just said, "that's the way it is, we can't discuss it now, I have material to cover"). But my calculator came out with an agreeable answer when I plugged a number in for n. So, please enlighten me if you will (Your mathematic credentials are very likey far better than mine), as I have no earthly idea, and probably could not give you a rational guess.

25. Jun 3, 2006

### Universe_Man

Oh, well it would be undefined of course raising 0 to a negative exponent, I just assumed that for the sake of argument, you could notate it as 1/0, but obviously my assumption is incorrect. thank you for your information.