# Homework Help: What is x^0 when x=0?

1. Feb 1, 2010

### yungman

What is $$x^0$$ when x=0?

My thinking $$0^0=1$$

Am I correct? Why?

Last edited: Feb 1, 2010
2. Feb 1, 2010

Undefined!

3. Feb 1, 2010

### espen180

00 is poorly defined. Its value depends on which zero in the expression in subject to a limit. In you case, you want to take

$$\lim_{x\to0}x^0$$

Note that x never reaches 0, as in your question.

4. Feb 2, 2010

### yungman

I asked because I am working on this:

Defined: $$P_k(x)=\frac{1}{2^k}\sum_{m=0}^M (-1)^m \frac{(2k-2m)!}{m!(k-m)!(k-2m)!} x^{k-2m}$$

$$M=\frac{n}{2}$$ if n is even integer.

Question: Prove $$P_{2n}(0)=(-1)^n\frac{2n)!}{2^{2n}(n!)^2}$$ using formula above.

Since k=2n is even

$$\Rightarrow M=\frac{2n}{2}=n$$

Substitude k=2n

$$\Rightarrow P_{2n}(x)=\frac{1}{2^{2n}}\sum_{m=0}^n (-1)^m \frac{(4n-2m)!}{m!(2n-m)!(2n-2m)!} x^{2n-2m}$$

$$\Rightarrow P_{2n}(0)=\frac{1}{2^{2n}}\sum_{m=0}^n (-1)^m \frac{(4n-2m)!}{m!(2n-m)!(2n-2m)!} 0^{2n-2m}$$

The only way that this is possible is if $$0^{2n-2m}=1 \Rightarrow n=m$$

Also according to Dr. Math:

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

Thanks

5. Feb 2, 2010

### Hurkyl

Staff Emeritus
The problem is that you are mixing up two different kinds of exponentiation. (Alas, the difference is usually not mentioned. )

"xn" the monomial and "xn" the real number are different expressions describing different types of objects. However, monomials can be converted into functions, and expressions in a variable can be converted back and forth with expressions denoting a number, and most of the time it doesn't matter how you interpret things.

Alas, the monomial "x0" is the same as the monomial "1", and so the associated function is f(x)=1 with domain all of R.

But the real number "x0" (with exponentiation interpreted as real exponentiation) is only partially defined -- at best, the variable x must be restricted to nonzero reals.

Generally speaking, though, the only time you would ever encounter 00 is when you were working with monomials, which is why people sometimes adopt a convention that extends real exponentiation so that 00=1.

6. Feb 2, 2010

### yungman

Thanks for your reply. I am not familiar with monomial, what is it? Is it like a "non polynomial" ei. only have single power of x?

In my case,$$x^{2n-2m}$$ is not a real number, it is a monomial. so do you mean if n=m, then $$x^{2n-2m}=x^0$$ and if x=0, then x^0=1? Which in another word $$x^0=1$$ for all x in real number including 0. Am I getting it right?

Thanks

Last edited: Feb 2, 2010
7. Feb 2, 2010

### dextercioby

I would say that $0^0$ is ill-defined. However, one compute the following limit $\lim_{x\rightarrow 0} x^x$ and end up with 1 as a result.

8. Feb 2, 2010

### vela

Staff Emeritus
In the context of power series, just take x0 to mean 1 for all x. The constant term is often written as a0x0 so you don't have to keep breaking out the constant term separately from the rest of the series, but it's understood it's just a constant. You're not actually taking x to the zeroth power there.

9. Feb 2, 2010

### yungman

Thanks guys.

It make sense to me that$$x^0=1$$ for all x including 0.