Is Zero Raised to the Power of Zero Equal to One?

[Leopold Infeld]

$$0^{0}$$ is what then?
I still think that it is "undefind"

 

[matt grime]

It can be defined and its definition may depend on context so that it is not "well-defined"

if we extend the function f(x)=x^0 to one on all of R continuously it is 1,and this is the rule used for taylor series. If we extend the function g(x)=0^x from the positive reals to include 0 then we would set 0^0=0 for continuiuty, the function h(x)=x^x has no canonical extension to include 0, hence the decision to declare it undefined.

Why must it be absolutely something? It isn't, that's all.

 

[whozum]

Maple says its 1 :\

 

[arildno]

Maple is not a mathematician, and has no authority on these issues.

 

[whozum]

Maple is written by mathematicians, so surely it has some authority?

 

[arildno]

Not the program, no.

 

[dextercioby]

"Scientific Workplace Warning:0^0 is undefined"...So what did you say about mathematicians and math software...?

Daniel.

 

[Zurtex]

I put 0⁰ into mathematica and first it beeps at me with the error:

Power::"indet" "Indeterminate expression 0⁰ encountered

Then gives me the output: "Indeterminate".


Anyway, I've met people who write mathematical software, I wouldn't trust them

 

[dextercioby]

Lol,we've had a least a thread on the difference between "indeterminate" and "undefined"

There it goes again.I'll stay out,though.Enjoy

Daniel.

 

[whozum]

Not the program, no.

So one shouldn't trust the program that claims to allow one to "command the power of a thousand mathematicians"?

Fraudulent advertising, one might say?

If a thousand mathematicians come to the conclusion that 0^0 = 1, I'm not in a position to argue with them. Perhaps we should write to Maple?

 

[master_coda]

The problem is not that Maple is not an authority (it isn't, but neither is a mathematician). The problem is that Maple isn't designed to be a 100% correct, complete representation of our current understanding of mathematics. It's designed to be a convenient software package to work with. So the software may make assumptions that aren't always valid or agreed upon for the sake of convenience.

Of course, the fact that 0^0 returns 1 in Maple does indicate that defining 0^0=1 is probably the most convenient definition of 0^0.

 

[arildno]

whozum:
It ought to have been made abundantly clear to you and others from this thread that however we choose to define $$0^{0}$$ is just a matter of notational convenience, for example in order to make Taylor series look "nice".

We cannot rigorously define an exponentiation process/exponential function in such a manner that the value of $$0^{0$$ pops out in a similar manner as, say, $$\pi^{e}$$.

 

[Zurtex]

If a thousand mathematicians come to the conclusion that 0^0 = 1, I'm not in a position to argue with them. Perhaps we should write to Maple?

It's not a matther of how many mathematicians come up with the same thing, it's a matter of if they worked it out logically.

 

[mathelord]

To me issues like this are controversial,shmoe said 0^x=0,does it imply that log0/log0
gives any number ,since x might be any number,now how do we evaluatelog0?

 

[mathelord]

My friend abia once gave an isight in class ,he posed some kind of theory relating to this,here is what he said ,let 0^0=x,then 0log0=logx,since 0 times a number gives 0then log x=0hereby making x 1.Our high school maths teacher was dumbfounded and ,how true is this theory?

 

[matt grime]

Very much false.

log(0) is not defined. Indeed 0 is an essential singularity of log. In laymans terms, not only does |log(z)| tend to infinity as z tends to zero along the real axis, but it does so in a very very nasty way.

 

[whozum]

Sorry you guys didnt pick it up, there was a heavy tone of sarcasm in my last post.

I compeltely understand what you guys are saying.

 

[The Reverend BigBoa]

An interesting discussion indeed. I thought at least SOMEWHERE there might be something a bit more "conclusive", geez, let's discuss what THAT means too! Hah,,,,

Anyways, here is another aspect, especially what the author brings up regarding gamma functions,,,,

Food for thought anyways, and I'm presuming that's the intent of this thread,,,,,,

http://home.att.net/~numericana/answer/algebra.htm#zeroth

 

[The Reverend BigBoa]

Perhaps this says that everybody's right. At least, given the situation. This guy does agree though, that the majority view is that 0^0 = 1.

http://www.mathforum.org/dr.math/faq/faq.0.to.0.power.html

But isn't it nice when everyone can be correct? :!)

 

[Leopold Infeld]

My calculator symbolically evaluates that limit to 1...

There's a more obvious counterexample:

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




The whole thing about 0^0 depends precisely what ^ means. As an operation on real numbers, 0^0 is undefined. The reason is that (0, 0) is not in the domain of ^.

Okay, so you want the "philosophical" reason. A crucial property about real operations is that they're continuous within their domain.

However, ^ cannot be continuous at (0, 0) -- the classic examples demonstrating this fact are 0^x --> 0 and x^0 --> 1 as x --> 0.


However, there are other meanings to ^. For example, there's a definition of ^ that means repeated multiplication. The empty product is, by definition, 1, so anything to the zero-th power is equal to 1. In polynomials and power series, this definition is what ^ "really" means -- that's why one uses 0^0 = 1.

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

Is this a possible limit?
I thought that the limit from the right isn't equal to that of the left for this case.
cuz 0^0.0000000000000001=0 cause the zillionth root of 0 is still o
but 0^-0.0000000000000001 is undefined if 0^0.0000000000000001=0 is correct.

Dont laugh guys if my arguments and terribly lame because i don't kno anything.

 

[master_coda]

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

Is this a possible limit?
I thought that the limit from the right isn't equal to that of the left for this case.
cuz 0^0.0000000000000001=0 cause the zillionth root of 0 is still o
but 0^-0.0000000000000001 is undefined if 0^0.0000000000000001=0 is correct.

Dont laugh guys if my arguments and terribly lame because i don't kno anything.

That's correct. The limit doesn't exist, only the limit from the right. The limit from the left is just undefined.

 

[arildno]

Note that we may approach $$0^{0}$$ in the following manner, by letting x go to 0 from the positive side:

$$f(x)=(e^{-\frac{1}{x}})^{\alpha{x}}$$

But, evidently, we have $$f(x)=e^{-\alpha}$$
In this case therefore, we have $$\lim_{x\to0}f=e^{-\alpha}$$

 

[mathelord]

What the problem with wat i posed to thye forum concerning wat was done by my friend?I'd love to know so i can let the class be aware of the error

 

[arildno]

See post 51 by matt grime.

 

[mathelord]

but 0 time sany number whatever it is give 0 right

 

[Zurtex]

but 0 time sany number whatever it is give 0 right

What has a^b got to do with multiplying so many times when b is not a natrual number?

 

[arildno]

but 0 time sany number whatever it is give 0 right

There isn't any number log(0)

 

[Leopold Infeld]

log 0 is not a number

 

[leon1127]

i am sure this is right
set y= x^x
y=e^(x Ln[x])
lim[y,x,0]=lim[e^(x Ln[x]),x,0]
since it is continuous, it produces e^(lim[x Ln[x], x, 0]) = e^0 = 1

 

[Hurkyl]

That's (almost) correct; $\lim_{x \rightarrow 0^+} x^x = 1$.

 

[Zurtex]

That's (almost) correct; $\lim_{x \rightarrow 0^+} x^x = 1$.

Although perhaps their proof only covered that it's also true that $\lim_{x \rightarrow 0^-} x^x = 1$

 

[HallsofIvy]

Although perhaps their proof only covered that it's also true that $\lim_{x \rightarrow 0^-} x^x = 1$

What is true is that x^x is not defined for negative x so that doesn't even make sense.

 

[Zurtex]

What is true is that x^x is not defined for negative x so that doesn't even make sense.

$$\left( -\frac{1}{2} \right)^{-\frac{1}{2}} = -i \sqrt{2}$$



A couple years back the function x^x I loved studying, the properties of the function I really enjoy.

 

[Hurkyl]

You're talking about a different x^x function.

 

[BoTemp]

Note that we may approach $$0^{0}$$ in the following manner, by letting x go to 0 from the positive side:

$$f(x)=(e^{-\frac{1}{x}})^{\alpha{x}}$$

But, evidently, we have $$f(x)=e^{-\alpha}$$
In this case therefore, we have $$\lim_{x\to0}f=e^{-\alpha}$$

Interesting, but I'm not sure if f(x) = exp(-a) in the limit as x->0, because there's a 0/0 in the exponent.
I would say 0^0 is undefined, mainly because one can get different answers by attacking from different angles:

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

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

The second I know how to prove with logs, the first I trust the arguments
given. Also, the proof of the second involves l'Hopitals rule, which only
gives limits, not exact values. For instance,

$$\lim_{x \rightarrow 2} (x-2)/(x-2)(x+3) = 1/5$$

but if one were to define that as a function f(x), f(2) is undefined.
In conclusion, I would say treat 0^0 much as 0/0, basically by taking
limits when it comes up.