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

[The Rev]

I've been told that any number raised to the zeroeth power is equal to 1. What about zero raised to the zeroeth power? Is that 1 or 0?

$$0^0=1?$$

The Rev

 

[Popey]

undefined!

 

[hypermorphism]

I've been told that any number raised to the zeroeth power is equal to 1. What about zero raised to the zeroeth power? Is that 1 or 0?

$$0^0=1?$$

The Rev

...and indeterminate. In some cases when a generalization is desired, it is profitable to define it as 1, in others, as 0.

 

[Galileo]

In series, it is always taken to be 1, so that you can write a series expansion compactly:

$$f(x)=\sum_{n=0}^{\infty}a_nx^n$$
so that $f(0)=a_0$

For example, the geometric series:

$$\sum_{n=0}^{\infty}x^n =\frac{1}{1-x}$$
is true at x=0 only if we define 0^0=1.

 

[CRGreathouse]

I don't think that it's undefined; I think that it's 1. 0^0 is an empty product, and empty products are necessarily equal to 1. As Galileo points out above, we need 0^0=1 for series to have compact formulas.

Everyone agrees that $$x^0=1$$ for $$x\neq0$$, but there's no reason to think that it should be different at 0 -- $$0^x$$ is only 0 for x > 0, since it's not defined for negative x.

There's no problem accepting 0^0 as 1, and there are many good reasons to think it shouldn't be undefined or 0.

 

[Popey]

Well, I think that the formula a^0=1 appears when you try to divide a^m by itself:

(a^m)/(a^m) = a^(m-m) = a^0

Since the first part of this equation equals 1, we have a^0=1

But if a=0 we can't do the division (0^m)/(0^m)

 

[Zurtex]

I don't think that it's undefined; I think that it's 1. 0^0 is an empty product, and empty products are necessarily equal to 1. As Galileo points out above, we need 0^0=1 for series to have compact formulas.

Everyone agrees that $$x^0=1$$ for $$x\neq0$$, but there's no reason to think that it should be different at 0 -- $$0^x$$ is only 0 for x > 0, since it's not defined for negative x.

There's no problem accepting 0^0 as 1, and there are many good reasons to think it shouldn't be undefined or 0.

That's not mathematical at all. I've always had a simple way of looking at it:

$$x^0 = \frac{x}{x}$$.

 

[shmoe]

Everyone also agrees that $$0^x=0$$ for all $$x>0$$, so there should be no reason to think it's different at 0, so $$0^0=0$$ right? We have a problem with the limit of $$x^y$$ as $$(x,y)\rightarrow (0,0)$$, it doesn't exist (different values depending on how x and y are approaching zero) so there is no obvious or natural choice of a value for $$0^0$$, so it's usually left as undefined (barring notational convenience).

The reasons to call the symbol 0^0 1 is usually for notational convenience, like empty products, binomial theorem, power series, etc.

 

[master_coda]

Most of the mathematicians I know seem to take the position that 0^0 is technically undefined, but there's nothing wrong with letting it equal 1 for notational convenience.

It's very rare for it to be convenient for 0^0 to be defined as any other value. The standard reason it needs to be something other than 1 is to extend 0^x to x=0, but in practice that doesn't seem to happen very often.


Of course, if you have a limit of the form f(x)^g(x) with f(x),g(x) -> 0 you have to be careful; it won't always converge to one.

 

[Icebreaker]

What about $$x^x$$ as $$x \rightarrow 0$$? I have always found $$x^x$$ fascinating for some obscure reason.

 

[master_coda]

What about $$x^x$$ as $$x \rightarrow 0$$? I have always found $$x^x$$ fascinating for some obscure reason.

x^x -> 1 as x -> 0, of course.

In fact, I believe there's a result that says that if f(x),g(x) -> 0 as x approaches some limit, then f(x)^g(x) -> 1 as long as f and g are analytic.

 

[dextercioby]

COUNTEREXAMPLE:


The functions $\frac{1}{3x+5}$ and $\frac{1}{x^{2}+4}$ are analytical in every point from their domain...

However,

$$\lim_{x\rightarrow +\infty}\left(\frac{1}{x^{2}+4}\right)^{\frac{1}{3x+5}} =-\infty$$

Daniel.

 

[Hurkyl]

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. :tongue: 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.

 

[Zurtex]

What about $$x^x$$ as $$x \rightarrow 0$$? I have always found $$x^x$$ fascinating for some obscure reason.

It is a very nice function though, when I first was thinking about it I tried to think about it in 4D in complex space and was pleasantly surprised when I started graphing it on mathematica recently I had quite a good idea oh how it looked.

 

[CRGreathouse]

That's not mathematical at all. I've always had a simple way of looking at it:

$$x^0 = \frac{x}{x}$$.

Empty products aren't mathematical?

 

[Hurkyl]

A less trivial example is:

$$\lim_{x \rightarrow 0} \left( e^{-1/x^2} \right)^{x^2} = \frac{1}{e}$$

Both the base and the exponent approach 0 from the positive side, but as we can see, the limit is not 1.

(Note the base is not an analytic function of x, despite its infinite differentiability at 0!)

 

[Zurtex]

Empty products aren't mathematical?

No, I meant your approch to the problem.

 

[CRGreathouse]

No, I meant your approch to the problem.

My approach was just stating that it's an empty product, and that empty products are always 1. The rest of my post was descriptive/informative (and perhaps poorly worded).

 

[leon1127]

take log both side...
u get
0 = (log 1)/log(x)
log 1 = 0
0/log(x) as x > 0
= 0/inf
interminate^_^

 

[master_coda]

That's what I get for quoting a theorem from memory.


I was way too general it seems. After looking it up, it seems that the theorem actually requires that:

1) f and g are non-zero
2) f and g are analytic at zero
3) f(x) -> 0 and g(x) -> 0 as x -> 0 from the right
4) f(x) is positive from all positive x <= some value close to zero

Then you get that f(x)^g(x) -> 1 as x -> 0 from the right.


Of course, that result is much less impressive. But I'm pretty sure this one is true.



Edited to remove the counterexample that wasn't really a counterexample.

 

[Hurkyl]

I'm pretty sure the limit dex quoted goes to 1, not -&infin;.

 

[master_coda]

I'm pretty sure the limit dex quoted goes to 1, not -∞.

Yes, it seems that it does. Still, your trivial example of 0^x was a good counterexample.

 

[tongos]

its the hundreth time the same question has been asked

 

[dextercioby]

Yes,you're both right.It goes to 1.

Daniel.

 

[BenGoodchild]

We cannot divide by 0.

Take the classic example:

a = b
a^2 = b^2
a^2 + a^2 = a^2 + ab
2(a^2) = a^2 + ab
2(a^2) - 2ab = 2(a^2) + ab - 2ab
2(a^2) - 2ab = 2(a^2) - ab
2((a^2) - ab) = 1((a^2) - ab)

then divide by (a^2) - ab) to give 1 =2

The fallacy comes right at the end when dividing by 0. So, we cannot divide by zero, and 0^0 is 0/0 therefore it is Mathematically undefined.

Regards,

M

 

[master_coda]

The fallacy comes right at the end when dividing by 0. So, we cannot divide by zero, and 0^0 is 0/0 therefore it is Mathematically undefined.

0^0 isn't 0/0.

 

[BenGoodchild]

Why not?

x^0 = x/x

 

[dextercioby]

Yes,but $$x^{0}=\frac{x}{x} \Leftrightarrow x\neq 0$$.So master coda was right.

Daniel.

 

[BenGoodchild]

Yes,but $$x^{0}=\frac{x}{x} \Leftrightarrow x\neq 0$$.So master coda was right.

I understand that this is true, and is infact what i said.

So, we cannot divide by zero, and 0^0 is 0/0 therefore it is Mathematically undefined.

I say we cannot divide by zero,
and as x^0 is x/x
which becomes, when x = 0, 0/0 cannot be omputed if we cannot divide by 0 it either doesn't exist or as I put it it is mathematically undefined.

Regards,

Ben

 

[master_coda]

I understand that this is true, and is infact what i said.

I say we cannot divide by zero,
and as x^0 is x/x
which becomes, when x = 0, 0/0 cannot be omputed if we cannot divide by 0 it either doesn't exist or as I put it it is mathematically undefined.

Regards,

Ben

x/x is not the definition of x^0. So the fact that x/x is not defined for x=0 doesn't automatically mean that x^0 is not defined for x=0.

 

[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.