Zero to the zero power?

[Guv]

My girl friends's grandson asked me about 0^0, and it was difficult to give him as answer. He is only 9 years old, but knew that the zero power of positive numbers is one. He seemed to understand a simplified explanation based on a variable approaching a limit.

BTW: I can think of a good reason for e^0 = 1, but without more thought, I cannot come up with a good reason for x^0 = one for positive finite values of x.

The question has been rattling around in my head for a few days, and I wonder what some of you might think.

It seems to me that 0^0 is undefined in the same sense that 0 / 0 is undefined.

If you consider x^0 as positive x approaches zero, one looks like a good limit value. x^0 = 1 for all positive values of x, and I think it equals one for all negative values. Why should it not be one when x = 0?

If you consider 0^x as positive x approaches zero, zero looks like a good limit value. 0^x = zero for all positive nonzero values of x, why should it not be zero when x = 0?

I think that 0^x for negative x has infinity as a limit value, so there is a problem with 0^x as negative x approaches zero. Considering 0^x for negative x does not suggest zero or one for 0^0.

Does anybody have some thots on this?

[DerFarm]

I'm not sure, but I seem to remember that 0 is not considered an
integer. Something about it representing a vanishingly small
entity in Rheimanian spaces. Assuming (a BIG assumption) that
I'm even close to being right, this would imply that 0^0 would be
undefined.

My wife studied these silly things for her doctorate, and she once
tried to explain it. It didn't even make to one ear, much less out
the other.

Hope this helps

[chrisf]

think of x^n = (x^(n+1)) / x

so if x^1 =x, then x^0 = (x^1) / x = x / x =1

If x = 0 then: 0^1 = 0, so x ^ 0 = 0 / 0

now x / n heads towards (positive or negative) infinity as n becomes closer to 0.
As infinity is not a number, x / 0 has no numeric solution.

[kedaman]

I remember evaluating the limit for 0^0 using L'Hopitals rule giving 1, since Sin(x)^x approaches 1 at when x approaches 0. I don't remember the details though, will check out later when i come back from school

[thinktank]

you thought, Guv doesn't know this ?

"it was difficult to give him as answer. He is only 9 years old, but knew that the zero power of positive numbers is one. He seemed to understand a simplified explanation based on a variable approaching a limit. "

Try to explain something to a kid and you will know what he meant !

[kedaman]

but there's a flaw in it, i'm sure the limit was 1. But I can't find the flaw

lim x->0 y=x^x
lim x->0, ln(y)=lim x->0, X*ln(x)
=lim x->0, ln(x)/(x^-1) = (-infinity/infinity)= lim x->0, D[ln(x)]/D[x^-1] (L'Hopital II)
=lim x->0, (x^-1)/-(x^-2)
=lim x->0, -X = 0

[kedaman]

never mind, stupid one

ln(y)=0
y=1

[Guv]

As suggested in my original post, it seems to me that you can argue for 0^0 = 1 and also for 0^0 = zero, suggesting that the expression is undefined. It has been so long since I took courses in such subject matter that I am not really sure that the following is correct.

X^0 = 1 for all positive, nonzero values of X.
Hence limit( X^0 ) = 1 as X approaches zero from positive direction. Why should it magically change to something else when X becomes zero?

0^X = 0 for all positive nonzero values of X.
Hence limit( 0^X ) = 0 as X approaches zero from positive direction. Why should it magically change to something else when X becomes zero?

Two reasonable arguments leading to different conclusions.

• If both arguments are valid, 0^0 is undefined.
  
• If both not valid, which one is invalid? Why is it invalid?

A little fooling around with a math evaluator application suggests the x^x approaches 1 as x approaches zero. Using calculations does not constitute a formal proof, and it does not address other functions which might require evaluation of 0^0. For example the following.

(x^3 - 1)^(x -1), x^sin(x), (x - 1)^(x^5 - 1), (sin(x))^tan(x), and many others. Does every such function approach one as the value of 0^0?

[kedaman]

I think X^X is discountinous at X=0 and that the limit should be most useful as 1

[paulw]

X^0 = 1 for all positive, nonzero values of X.

This is the nub of the problem - X^0 = 1 for all positive values - this can be clearly demonstrated by charting the raised values for any positive X. It will always create a smooth curve that passes though 1 at the power 0. This does not hold true for negative numbers (clearly).

Therefore there is a breakdown of our limit assumption. You cannot assume that 0^0 is 1 since all positives to the 0th power do so.

Any thoughts?

Cheers,

P.

[kedaman]

try (-1)^0, what do you get? 1

[chrisf]

because:
-1^1 = -1
-1 ^ 0 = (-1^1) /-1 =1

[nishantp]

VB says that 0^0 = 1. Doesnt mean its right though...