how is 0^0 defined?

murshid_islam

i was wondering how 0^0 is defined? can anybody please help?

thanks in advance.

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.

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.

jpr0

I thought 0^0 was nullity?

http://www.bbc.co.uk/berkshire/conte..._feature.shtml

Kummer

I, and most people, define it to be [tex]0^0[/tex].
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 [tex]0^0[/tex]

The same way we define [tex]0!=1[/tex] (but there is actually another reason there).

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.

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, [itex] n!= n\cdot (n-1)![/itex] 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:
[tex]\lim_{x^{+}\to 0} x^x[/tex], 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:
[tex]\lim_{x\to 0 , y\to 0} x^y[/tex], which is multi valued.

HallsofIvy

What do you mean "define it to be [tex]0^0[/tex]"? Did you mean to say "define it to be 1"?

murshid_islam

now i am really confused. is [tex]0^0 = 1[/tex] or not?

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.

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)= 0^x we again have f(0)= 0⁰ but for any positive x, f(x)= 0^x= 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.

murshid_islam

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

HallsofIvy

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

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?

HallsofIvy

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

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.