A polynomial function is given as,
$P(x)= a_nx^n+a_{n-1}x^{n-1}+.....+a_1x+a_0$
Notice the last but one term $a_1x$. This term is a simplified form of $a_{n-(n-1)}x^{n-(n-1)}$.
Now let us take the last term of the Polynomial. The term $a_0$ is a simplified form of $a_{n-n}x^{n-n}$. Notice that $x^{n-n} = x^0 = 1$ only when $x\neq0$. This is because $0^0$ is indeterminate. It is evident that $x=0$ is clearly not in the domain of $P(x)$. But by definition, the polynomial function given above is defined for all values of $x$, $x\in(-\infty,\infty)$.
Was I right to frame the last term of the polynomial the way I did above? If no, I would like to know why.
9 Answers
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.
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 $\sum_{i=0}^na_iP_i\>$.
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 $0^0=1$?
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.
The general form of a polynomial ($p(x)$) of degree $n$ can indeed be written as:
\begin{align} p(x)=a_nx^n +a_{n-1}x^{n-1} + \cdots +a_1x +a_0 , \qquad a_n \ne 0 \end{align} Now I believe you saw some trends and tried to write it in a more compact way: \begin{align} p(x) = a_{n-0} x^{n-0}+a_{n-1} x^{n-1}+\cdots + a_{n-(n-1)}x^{n-(n-1)} + a_{n-(n)}x^{n-(n)}= \sum_{i=0}^{n} a_{n-i}x^{n-i} \end{align} And you are running into problems with that last term. The first equation above is the proper way to think of a polynomial and the one you proposed is just a compacted way to write it that is almost always equivalent except for the last term.
It is not the right way to frame a polynomial. The reason why you could say is exactly the reason you noticed: it isn't equivalent to the definition of the polynomial (because $x^0$ is not well defined when $x=0$ so our form isn't well defined when $x=0$ however the definition does not have issues when $x=0$)
This is one area where a little computer science can be helpful. Asked to test the value of $0^0$ by implementing the function $x^y$, one person writes this program
def powernat(real x, nat y)
if y = 0 :
return 1
else:
return x * powernat(x,y-1)
and another person writes this program
def powerreal(real x, real y)
return exp(y * log(x))
Here "nat" is a data type for natural numbers and "real" is a data type for real numbers.
We can see immediately that something different will happen with $0^0$. The powernat function will return 1, but the powerreal function will cause an error, because $\log(0)$ is not defined.
The situation in mathematics is not so different - we often define exponentiation for natural numbers as in powernat, and exponentiation for real numbers as in powerreal. But we have no notation to distinguish powernat from powerreal: we write both of them as $x^y$ and rely on context only to tell them apart.
This causes trouble when we write expressions such as $0^0$. If we mean for this to be treated as the powernat function - which is the case in the definition of a power series - then we read $0^0 = 1$. But if we want this to be treated as the powerreal function - which is also used, essentially, to treat complex exponentiation - then $0^0$ is undefined (as is $0^1$, actually...).
For the more basic arithmetical operations, this does not cause any issues. For example $1 + 1 = 2$ is true regardless of whether we think of the numbers $1$ and $2$ as natural numbers or as Dedekind cuts representing real numbers. In each of these cases the "+" means something different, but it causes no confusion. In the case of $x^y$, though, it does matter which definition we use.
The problem is essentially of your own making in that, for your own convenience, you decided to rewrite $a_0$ as $a_0x^0$ and then ran into problems with the case $x=0$. Since the problem is of your own making, the power to solve it is also yours. You're in a context where it makes a huge amount of sense to just adopt the convention that $0^0=1$, so you should do that. This allows you to keep the convenience of writing $a_0x^0$ without the problems associated with evaluating $x^0$ at $x=0$.
When I took an undergraduate course in Abstract Algebra, the text introduced an alternate interpretation of polynomials that you might find more helpful: Forget about the 'x' entirely and use -just- the coefficients! SO: a polynomial P is simply a sequence of values drawn from a field F: { $a_0$, $a_1$, $a_2$, ... } [where all but finitely many of the $a_i$ are non-zero]. Arithmetic of polynomials is defined in the expected manner directly in terms of their coefficient sequences. Evaluating a polynomial is then reintroduced by means of the Evaluation Homomorphism: The map $V_P$: F -> F defined by $V_P(c)$ = $a_0$ + $a_1 c$ + $a_2 c^2$ + ... . The book also suggests that, if you still feel the need for 'x', think of it as the polynomial { 0, 1, 0, 0, 0, ...} (personally I don't find that suggestion too helpful).
Why are you thinking of your polynomial in that way .. Just think of it as sum of n+1 functions $P_0, P_1, P_2 , P_3, ... P_n$
where
$P_0 = a_0$
$P_1 = a_1 x$
. . .
$P_n = a_n x^n$
Domain of all these functions is $R$ Therefore domain of their sum will be $R$ ... That is the domain of your polynomial will be $R$
