Domain of a Polynomial function

Solution 1:

This is because $0^0$ is indeterminate.

This is an extremely common misconception. There is a vast difference between $0^0$ and the form of a limit, which may be labelled as "$0^0$" (note the quotes!), just as there is a difference between $\frac00$ and the form "$\frac00$" of some limits.

Here are the facts under standard mathematical conventions:

$0^0 = 1$ in contexts where the exponent is a natural number.

"$0^0$" is a label referring to an indeterminate form of some limits.

$\frac00$ is undefined.

"$\frac00$" is a label referring to another indeterminate form of some limits.

Limits with form "$0^0$" or "$\frac00$" may have a value or may not. That is precisely why we call their form indeterminate, because we cannot determine the value so easily by their form alone.

$0^0$ is not a limit, and if the exponent is a natural number (like for rings or in combinatorics or in the binomial theorem or in power series or ...) then its value is always $1$.

If you do not believe this, see the conventional statement of the binomial theorem here and here (equation 4) and the definition of power series here and here.

Solution 2:

On pondering this good question further, I think that part of the problem is that we have no name for the functions $x\mapsto x^n$. A clean way of getting around the difficulty might be the following:

Define functions $P_n$ for nonnegative integers $n$ inductively as follows: for all $x$, $P_0(x)=1$, and for $n\ge0$, define $P_{n+1}(x)=xP_n(x)$. You see that this makes $P_0$ the constant function $1$, and for $n>0$, $P_n(x)=x^n$.

Then your function can be written $\sum_{i=0}^na_iP_i\>$.

Solution 3:

I finished my M.Sc. Mathematics two years ago and in the branches I have studied, I have never encountered the notion that $0^0$ should be undefined.

I always consider $0^0$ to be $1$.

For further explanation, see the question linked to by Hans Lundmark in his comment: Zero to the zero power - is $0^0=1$?

Solution 4:

Formally, you are absolutely correct. $0^0$ is an indeterminate form. But consider a seemingly unrelated case:

$$f(x)=\frac{x}{x}.$$

This used to drive me nuts, because it is clearly just the same as the function $g(x)=1$... right? The answer is no, but only in a way that is disgustingly technical. Similar to your case, $f(0)$ is technically an indeterminate form. The problem is division of $0$ by $0$. So the functions $f(x)$ and $g(x)$ can't really be equal because they have different domains. However, there is a way around this. Consider instead defining a new function $h$ in this way:

$h(x)= \frac{x}{x}$ if $x\neq 0$, and $h(x)=1$ if $x=0$. Now, we have removed the problem with $0$ and defined a function truly equal to $g(x)=1$ everywhere.

In your problem, $a_0$ is not really equal to $a_0x^0$ because those expressions have different domains. Specifically, $0$ is in the domain of the first, but not of the second. However, when people speak of $a_0$ as being "the $0$-order term", they are doing that for reasons that are intuitively helpful, not reasons that are mathematically formal. And it is always helpful to remember that $a_0\neq a_0x^0$ in general, but that $a_0= a_0x^0$ when $x\neq 0$.

Does this help? Please do ask for clarification if you need it, as this is not only an important point but demonstrates great mathematical insight on your part. I enjoyed thinking about it.