# Thread: 0 to the power of zero?

1. What is 0 to the power of zero?[/img]  2.

3. 1.  4. 0 to the power 0 is often left undefined. The usual reason to define it is for notational convenience, e.g. to make power series a little simpler to write. In this case it will usually be defined as 1.  5. Thanks. That is what I thought. It is an odd situation though. Can someone please describe why any # to the power of 0 is one?  6. That's how it's defined, x^0=1 when x is non-zero. There's really nothing more to it than that.

Why is this a good or sensible or useful definition? This is another question. The most basic reason is it's required if we want the power laws x^(n+m)=(x^n)*(x^m) to be true when n and m are any integers.  7. it is 1 since anything divided withitself (not 0) is allways 1 x/x=1=x^1/x^1=x^(1-1)=x^0  8. Originally Posted by Zelos
it is 1 since anything divided withitself (not 0) is allways 1 x/x=1=x^1/x^1=x^(1-1)=x^0
Brilliant! Zelos you have just answered an age old question!

U0^U0 = 1U

where U = universe! 8)

All that money wasted on Quantum Theory...  9. Shmoe and I just finished a discussion of this in "power of 4" Originally Posted by shmoe
What would you think if I suggested 0^0 should be 0 based on:

0^1=0
0^0.5=0
0^0.1=0
0^0.001=0

and then went on to suggest 0^0 should be 1 based on:

1^0=1
0.5^0=1
0.1^0=1
0.001^0=1
Shmoe is giving an example of function of two variables which has two different limits at the origin from two different directions, f(x,y) = x^y at (x,y)=(0,0). But I don't see that algebra or consistency with the logarithm will work on this one either.

taking the logarithm of both sides of 0^0 = a
0 log 0 = log a
But log 0 is undefined except as the limit of negative infinity.

Also the other approach of (log a)/log 0 = 0 doesn't help, even when you use the limit of log 0 -> negative infinity for any a solves this.

Therefore I think we must conclude that 0^0 is undefined

Hmmm.... what about a limit from a different angle
like the 45 degree angle using the limit of x^x as x approaches 0

emacs lisp gives me the following progression

(expt .1 .1) 0.7943282347242815
(expt .01 .01) 0.954992586021436
(expt .001 .001) 0.9931160484209338
(expt .0001 .0001) 0.9990793899844618
(expt .00001 .00001) 0.9998848773724686
(expt .000001 .000001) 0.9999861845848758
(expt .0000001 .0000001) 0.9999983881917339

which suggests that the limit is 1.0

I think this means that the limit is 1.0 from all directions except along x=0 where f(x,y) = 0^y = 0

I don't know that might be a case for 1.0 being the best answer if there is any answer at all, other than undefined that is. Originally Posted by shmoe Originally Posted by mitchellmckain
I think this means that the limit is 1.0 from all directions except along x=0 where f(x,y) = 0^y = 0
What if you approach along the curve (x,log(a)/log(x)), where a>0 and x>0 and x->0?
Shmoe is suggesting this because on this curve our two variable function f(x,y) = x^y becomes x^(log base x of a) = a, which means that the limit as x goes to 0 on this curve is a.

For this limit to work (without resorting to analytic continuations of our function) I think that a must not only be greater than 0 but also less than 1.

Now the question I immediately had was at what angle does this approach the origin and as you would expect the tangent to this curve at the origin is the line x=0.

As a result I was correct in my conclusion that approaching the origin at any angle except the line x=0 does give the limit of 1. It is only that when you approach the origin from the angle of the line x=0 that you can get any limit from 0.0 to 1.0 depending on how you take the limit (that is what curve you follow).  10. This smells like a branch cut is involved....?  11. Originally Posted by william
This smells like a branch cut is involved....?
Branch cuts with real valued functions is quite a novel idea Not sure how Mitch got tied into analytic continuations, the function f(x, y) = x<sup>y</sup> has no limit as x, y -> 0. You cant even give it a continous extension

f(x, x^2) -> 1 as x -> 0 but f(2<sup>-1/x</sup>, x) -> 1/2 as x -> 0  12. 0^0 = 0, nothing or all by itself.

0 is the barriar between the real number and the imaginary number. If 0 x 0 or 0^0 equals any real or imaginary number, 0 cannot exist as zero.  13. 0^0 = 0^k/0^k for any integer k except zero

0^k = 0, hence 0^0 = 0/0,

which is generally undefined, but can be defined as a limit for a particular problem, like the limit of f(x)/g(x) as x-> 0, with f(0)=g(0)=0. To arbitrarily define it as 1 is dubious.  14. not to confuse you:

the equal sign in 0^0 = 0/0 should not be taken literally, but as saying both sides are equivalent in the sense that they are both undefined.   15. Originally Posted by shmoe
0 to the power 0 is often left undefined. The usual reason to define it is for notational convenience, e.g. to make power series a little simpler to write. In this case it will usually be defined as 1.
0^0 is undefined. I agree with Zelos. But I never knew it could also be considered equals to 1. Any example in what equation 0^0 = 1?  Bookmarks
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