What is the Value of 0^0 in Mathematics?

[yungman]

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

My thinking $$0^0=1$$

Am I correct? Why?

 

[sutupidmath]

Undefined!

 

[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.

 

[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

Please help me on this one.

Thanks

 

[Hurkyl]

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.

 

[yungman]

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.

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

 

[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.

 

[vela]

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.

 

[yungman]

Thanks guys.

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