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

[Leopold Infeld]

I thought x^0 was x^(m-n) where m=n, so x^0 = x^m/x^n. If x=0, then
0^0 must be 0/0 = undefined.

Correct me if I am wrong.

 

[master_coda]

I thought x^0 was x^(m-n) where m=n, so x^0 = x^m/x^n. If x=0, then
0^0 must be 0/0 = undefined.

Correct me if I am wrong.

No. x^(m-n) = x^m/x^n doesn't hold for x=0, at least if you want something sensible like 0^1=0 to be true.

 

[matt grime]

It is merely a convention that 0^0 is taken as 1,and not universal.

Think carefully about what it even means to raise numbers to powers.

if x is a whole number and n is a whole number then x^n is straight foward, we can even do this for negative integer x , and negative integer n too, and this extends nicly to x^0=1 when x=/=0 by your above argument. We can even extend to rational x and thence to real x.

We can even take roots occasionally ie we can square root +ve numbers (let's not worry abuot complex ones yet), and so we can raise real numbers to rational powers by, say, for 4/5 rasing to the power 4 then taknig the 5th root. All well and good. But what does it even mean to raise a number to the power sqrt(2)?

To be honest we do not need to consider that.

The most common time we use ^0 is in taylor series, when we want a nice formula, and then it makes sense to define x^0=1 for all x even 0. This makes the function f(x)=x^0 a nice continuous function on the whole of R.

Defining 0^0 to be 1 is useful and consistent with extending x^0 to a continuous function on the whole of R.

 

[Hurkyl]

I thought x^0 was x^(m-n) where m=n,

No... there are some subtleties here -- we're talking about these expressions as strings of symbols in the language of mathematics.

In particular, in this context, x^(m-n) is never x^m / x^n. They're different strings of symbols.

What we have is that, under the right conditions, x^(m-n) = x^m / x^n is a true statement.

Those conditions require x to be nonzero (and positive, if we're talking about the usual exponentiation operation on the reals, and being strict).

This means that trying to substitute x = 0 into this equation to get 0^(m-n) = 0^m / 0^n is meaningless. It makes as much sense as if I were to write "+ 1 & *(".

 

[arildno]

Now, it might be of some interest to provide a rigorous, although tedious method of defining "exponentiation".
1) First, by induction, it is unproblematic to define
x\equiv{x}^{1}, x^{n}=x^{n-1}*x, n\geq{2}, n\in\mathbb{N}, x\in\mathbb{R}

2) Let us now define a function Exp(x) on \mathbb{R}:
Exp(x)=1+\sum_{n=1}^{\infty}\frac{x^{n}}{n!}
Exp(x) can be shown to be larger than zero, and strictly increasing.
We also have Exp(0)=1
Since it is strictly increasing, we may define its inverse, Log(x), defined on the positive half-axis (zero not included).
3) We may now, for any b\in\mathbb{R}, a\in(0,\infty} define:
a^{b}=Exp(b*Log(a))
Thus, we have for any a>0, a^{0}=Exp(0*Log(a))=Exp(0)=1

4) We might extend our idea of exponentiation to give, for example, meaning to the function 0^{x} but the value of the expression 0^{0} will, at best, always be one from a "provisional" definition dictated by convenience, rather than that we may deduce its value from first principles.

 

[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:

<br /> \lim_{x \rightarrow 0} 0^x = 0<br />.




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.

<br /> \lim_{x \rightarrow 0} 0^x = 0<br />.

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]

<br /> \lim_{x \rightarrow 0} 0^x = 0<br />.

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