# How is 0^0 defined?

1. Jul 12, 2007

### murshid_islam

2. Jul 12, 2007

### matt grime

Please can we not start a very long thread on this one? If people want to see the debates on it then search the forums.

Simply put the most logical definition of 0^0 is that it is equal to 1. This makes 'everything work' without having to make any 'except for 0 when it FOO is equal to 1' statements: partitions, functions, combinatorics, taylor series etc.

3. Jul 12, 2007

### ObsessiveMathsFreak

It isn't. It's an undefined statement like 0/0.

However, a lot of mathematicians like to set it equal to one, but it's really a matter of convenience. So be careful of this one.

4. Jul 12, 2007

### jpr0

5. Jul 12, 2007

### Kummer

I, and most people, define it to be $$0^0$$.
Why? Because it is useful for dealing with infinite series.

But some people do not define it. And what I hate is when a person tells me he does not define it but when he writes the power series he completely overlooks $$0^0$$

The same way we define $$0!=1$$ (but there is actually another reason there).

6. Jul 12, 2007

### Werg22

We can also define in terms of the continuity of the function x^0 or x^x, or (x+x)^x or whatever you want.

7. Jul 13, 2007

### Gib Z

The reason we define 0!=1 has very little relation to this..How many ways can we arrange nothing Kummer? Or if you want, you could take the recursive definition of the factorial function, $n!= n\cdot (n-1)!$ and substituting n=1 gives the desired result.

The reason 0^0 remains undefined is because the limit that represents it does not actually converge. Of course we could somewhat cheat by making some assumptions, eg say that it is the limit:
$$\lim_{x^{+}\to 0} x^x$$, and that is equal to 1, but we assume that the Base and the exponent approach zero at the same rate.

The correct limit is actually:
$$\lim_{x\to 0 , y\to 0} x^y$$, which is multi valued.

8. Jul 13, 2007

### HallsofIvy

What do you mean "define it to be $$0^0$$"? Did you mean to say "define it to be 1"?

9. Jul 13, 2007

### murshid_islam

now i am really confused. is $$0^0 = 1$$ or not?

10. Jul 13, 2007

### Kummer

Thank you.

Yes, that is true. But that does not constitute a formal mathematical proof. The problem is that there is no proof and it is a matter of taste. My preference along with most people is to define it as 1 because it is useful in power series.

Another reason is that the Gamma function evaluated at 1 is equal to 1, and that is a generalization of a factorial. But that is another story.

11. Jul 14, 2007

### HallsofIvy

No, it is an "indeterminate"- like 0/0, if you replace the "x" value in a limit by, say, 0 and get 0^0 the limit itself might have several different values.

To take two obvious examples, if f(x)= x0, then f(0)= 00. For any positive x, f(x)= x0= 1 so the limit as x goes to 0 is 1. If we want to make this a continuous function, we would have to "define" 00= 1.

However, if f(x)= 0x we again have f(0)= 00 but for any positive x, f(x)= 0x= 0 which has limit 0 as x goes to 0. If we want to make this a continuous function, we would have to "define" 00= 0.

Last edited by a moderator: Jul 14, 2007
12. Jul 14, 2007

### murshid_islam

thanks a lot, HallsofIvy. that made it pretty clear to me.

13. Jul 14, 2007

### HallsofIvy

It might be still clearer now that I have edited it to say what I meant!

14. Jul 15, 2007

### murshid_islam

yeah it's clear. 00 cannot be equal to both 0 and 1. so that's why it is indeterminate. am i right?

15. Jul 15, 2007

### HallsofIvy

Yes. Actually, it is possible to alter the limits slightly so as to get ANY number.

16. Jul 16, 2007

### matt grime

Come on people. It is not that 0^0 can be '0 and 1', but that a certain limit, x^y as x and y tend to 0 can be made to be arbitrary. That doesn't say what 0^0 is, just that the function f(x,y)=x^y has a nasty singularity at (0,0). But the symbol 0^0 has a perfectly well understood commonly accepted value as 1 for many other uses.