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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.

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  • $\begingroup$ Just to make sure I understand you correctly, you are asking i) If it is okay to think of the last term as $a_0x^0$ and ii) Why plugging zero doesn't then give an indeterminate form? Am I correct? $\endgroup$
    – The Count
    Commented Mar 13, 2017 at 2:52
  • $\begingroup$ $a_0x^0$ is a term that I arrived at logically by recognizing a pattern. If my logic is right, then $x=0$ shall not be included in the domain. I'm, however, skeptical about my approach because I'm going against the established definition of a polynomial. In summary, If it is OK to think of the last term as $a_0x^0$, plugging zero does give an indeterminate. $\endgroup$
    – R004
    Commented Mar 13, 2017 at 3:12
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    $\begingroup$ This is a serious concern, but not one that greatly bothers people who’ve been in the game for a while. My own attitude is just that when I’m in Polynomial Land, $x^0$ evaluates to $1$ no matter what the value of $x$. $\endgroup$
    – Lubin
    Commented Mar 13, 2017 at 3:25
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    $\begingroup$ math.stackexchange.com/questions/11150/… $\endgroup$
    – Hans Lundmark
    Commented Mar 13, 2017 at 7:16
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    $\begingroup$ The only thing "indeterminate form" means is that if you have a limit $\lim_{x\to x_0} f(x)^{g(x)}$, and it turns out that the limit of both $f(x)$ and $g(x)$ are $0$, then you cannot find the limit of $f(x)^{g(x)}$ by raising the two limits to each other. This only means anything in the context of limits. What it tells you is that the function $x^y$ is not continuous at $x=y=0$, but "not continuous" does not mean "has no value". $\endgroup$
    – hmakholm left over Monica
    Commented Mar 13, 2017 at 13:00

9 Answers 9

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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.

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    $\begingroup$ In combinatorics, the number of ways to do $n$ independent tasks each with $k$ choices is precisely $k^n$. In particular, if $n = 0$ (no tasks) then there is $1$ way (to do nothing). And if $k = 1$ then there is $1$ way (since you have only one choice for each task). And if $k = 0$ then there is no way unless $n = 0$. This property of natural number exponentiation is crucial for decomposing combinatorial problems easily (including recursively), otherwise we would be mired in a myriad of cases simply to side-step what is obviously the natural base case... $\endgroup$
    – user21820
    Commented Mar 13, 2017 at 16:45
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    $\begingroup$ Perhaps indeterminate forms ought to be taught as $({\to}0)^{\to 0}$ and $\frac{\to 0}{\to 0}$ (and likewise $({\to}0)\cdot({\to}\infty)$ and $({\to}\infty)-({\to}\infty)$ etc...) $\endgroup$
    – hmakholm left over Monica
    Commented Mar 13, 2017 at 16:47
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    $\begingroup$ @HenningMakholm: Yes! Yes! I always cringe when people use the term "form" without the quotes. This is especially disastrous in logic, where one had better be extremely clear on what is a symbol in the meta-system and what is a symbol in the system being studied. Your suggestion is very good and I seriously hope some people take it up. $\endgroup$
    – user21820
    Commented Mar 13, 2017 at 16:53
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    $\begingroup$ I think this issue is not so much whether the exponent is an integer, but rather which function is in mind. If we define $x^y = \exp(y \log x)$ - which is a standard definition - then $0^0$ in this function will be undefined. If we define $x^0 = 1$, then $0^0$ in this function will be $1$. So the issue is that $0^0$ can denote a specific value of several functions, so we have to know which one is intended before we can say whether $0^0$ is defined. $\endgroup$
    – Carl Mummert
    Commented Mar 14, 2017 at 0:57
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    $\begingroup$ Another nice answer coming from your end and also the interesting discussion in comments. +1 $\endgroup$
    – Paramanand Singh ♦
    Commented Aug 4, 2017 at 13:31
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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 $\sum_{i=0}^na_iP_i\>$.

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  • $\begingroup$ This definition rules out the possibility of finding an $x^0$ anywhere in the polynomial, thus allowing us to define it at $x=0$. The summation makes perfect sense to me. $\endgroup$
    – R004
    Commented Mar 13, 2017 at 3:54
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    $\begingroup$ What this answer does is to find a pattern from the bottom up (starting with the assumption of a constant term), while the question observes a pattern from the top down (starting at the highest value of the exponent). The question is an insightful observation, but the answer corresponds to the way I think polynomials were meant to be defined. $\endgroup$
    – David K
    Commented Mar 13, 2017 at 19:23
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    $\begingroup$ You know how powers are defined in a monoid (and $\mathbb R$ is a monoid under multiplication!)? Well, $x^0=1$, $x^{n+1}=xx^n$. So your $P_n(x)$ is exactly the function commonly known as power and written as $x^n$. $\endgroup$
    – celtschk
    Commented Mar 14, 2017 at 8:55
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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 $0^0=1$?

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  • $\begingroup$ The problem is when the exponent is an analytic variable. Also, most of the circumstances where $0^0=1$ is applied have an alternate interpretation -- such as "I'm going to use $0^0$ as a shorthand for the evaluation of the polynomial $x^0$ at $0$". $\endgroup$
    – user14972
    Commented Mar 13, 2017 at 12:23
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    $\begingroup$ @Hurkyl: It is true that we would want to distinguish between natural number exponentiation and complex exponentiation, as I state in my answer. However, this provides no argument for not defining $0^0$, since complex exponentiation is incompatible with real exponentiation; cube-roots disagree, and $\sqrt{0}$ is undefined if using the ordinary definition of complex exponentiation with the principal branch. Anyway, I'm sure you'll agree that when the exponent is constrained to be an integer we ought to define $0^0 = 1$, since it is precisely the base case for defining exponentiation in fields. $\endgroup$
    – user21820
    Commented Mar 13, 2017 at 16:51
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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.

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  • $\begingroup$ The domains of $a_0x^0$ and $a_0$ are indeed not the same. And it is indeed true that $a_0x^0=a_0$ for $x\neq0$. Stay with me. $\endgroup$
    – R004
    Commented Mar 13, 2017 at 4:03
  • $\begingroup$ We prefer to not use $a_0x^0$ because that gets us into serious problems. But why should we not use it when logic clearly gets us to it? From this thread, I learnt that polynomials are defined in a safe way. The safe way is provided by Lubin. The definition rules out the possibility of finding $x^0$ anywhere in the polynomial. It is relieving to witness that the definition gets us to the general polynomial without causing us to worry about the function at $x=0$. $\endgroup$
    – R004
    Commented Mar 13, 2017 at 4:12
  • $\begingroup$ The reason behind me running into issues was that I was not aware of a concrete definition of a polynomial. No text book I have read so far has put forward a definition that allowed me to understand the general polynomial. $\endgroup$
    – R004
    Commented Mar 13, 2017 at 4:22
  • $\begingroup$ Well, it looks like you got it sorted out. It seemed to me like you were familiar with the formal definition and just had a small concern with it, but I'm glad you got what you needed either way! :) $\endgroup$
    – The Count
    Commented Mar 13, 2017 at 11:18
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    $\begingroup$ @R004: It is completely safe to consider that $0^0=1$ everywhere. The only potential danger is if you somehow think the function $(x,y)\mapsto x^y$ is continuous everywhere -- which it isn't at $(0,0)$ -- but what you should remember is just the true fact that $x^y$ is not continuous at $(0,0)$ rather than the false superstition that $0^0$ fails to equal $1$. $\endgroup$
    – hmakholm left over Monica
    Commented Mar 13, 2017 at 13:09
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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 $x=0$)

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  • $\begingroup$ The pattern I recognized before arriving at the "compacted" form of a polynomial is, in my opinion, logical. But that gets me into problems. Here is the thing. In sciences, the right logic sometimes takes you to undesired results because at times, that is how things in nature are. This allows us to put forward limitations to the theory. Now, the logic( seemingly right ) I applied leads me to conclude that $x=0$ cannot be included. That's a limitation( as per me ). But universally, there is no such limitation I proposed. Why? $\endgroup$
    – R004
    Commented Mar 13, 2017 at 3:33
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    $\begingroup$ So the difference here is that a polynomial isn't a theory. It is a rigorously defined concept and we use this concept as a tool to communicate theories. It often makes sense to reformulate definitions to get a more intuitive understanding however it must still be equivalent to the original. This is in contrast to a theory in science where we will attempt to extend the theory and find new interesting results and predictions (and sometimes limitations). $\endgroup$
    – Patrick
    Commented Mar 13, 2017 at 3:42
  • $\begingroup$ I appreciate your points. $\endgroup$
    – R004
    Commented Mar 13, 2017 at 4:14
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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.

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  • $\begingroup$ I agree with your answer that we have multiple incompatible definitions for exponentiation and therefore which one we are using will of course affect the answer. The only thing is, I think in practice mathematicians use the simplest definition by default in general, which would be natural number exponentiation if applicable, otherwise integer exponentiation, then real exponentiation, then complex exponentiation. $\endgroup$
    – user21820
    Commented Mar 14, 2017 at 3:06
  • $\begingroup$ This is why we readily understand $x^n$, where $x$ is in any ring and $n$ is natural, or where $x$ is in any field and $n$ is an integer, or where $x$ is a non-negative real number and $n$ is rational, or where $x$ is a real number and $n = \frac{p}{q}$ where $p,q$ are integers and $q$ is odd. Of course when all these do not apply, or in the context of about complex analysis, we typically use complex exponentiation with principal branch cut unless otherwise stated. This is why we don't think twice when someone says that $\sqrt{0} = 0$ or that $( \mathbb{R}\ x \mapsto x^{1/3} )$ is injective. $\endgroup$
    – user21820
    Commented Mar 14, 2017 at 3:09
  • $\begingroup$ I think $0^0$ is somehow a different case - perhaps different from $\sqrt{0}$ even - so that it is less clear which function to use for $0^0$. Perhaps this is because $0^0$ is so often listed as an indeterminate form, which causes students confusion because the indeterminate form claim is for a different function than the $0^0 = 1$ claim. But this quickly becomes a sociological issue, once the fact that there are multiple interpretations for $0^0$ is clear. $\endgroup$
    – Carl Mummert
    Commented Mar 14, 2017 at 3:09
  • $\begingroup$ Yes your last comment is precisely what I was about to say next; $0^0$ is contentious for the wrong reason; people confuse between form and value. If we look solely at the notion of multiple incompatible definitions, the 'problem' with $0^0$ is of the same nature and rather small size as $\sqrt{0}$, in my opinion at least. Anyway thanks for explicitly giving the recursive definition of powernat, which I alluded to in my answer but was too lazy to type out. And I like Python-style too! =) $\endgroup$
    – user21820
    Commented Mar 14, 2017 at 3:12
  • $\begingroup$ I don't think the root confusion is between form and value. For the natural-exponent version $x^y \colon \mathbb{R} \times \mathbb{N} \to \mathbb{R}$, $0^0$ is not an indeterminate form - that function is continuous and locally constant at the origin, and the limit along $y$ is trivial. For the definitions I think of that make $0^0$ indeterminate, $0^0$ is also not defined. $\endgroup$
    – Carl Mummert
    Commented Mar 14, 2017 at 3:14
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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 $x=0$.

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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).

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  • $\begingroup$ That is an interesting pedagogical approach, and I kind of like it; it would surely make explaining finite field arithmetic easier. However, defining arithmetic on polynomials before defining their evaluation would seem to imply that you'd need to defer the main motivation for why polynomial arithmetic works like it does (i.e. the identities $V_{P+Q}(c) = V_P(c)+V_Q(c)$ and $V_{PQ}(c) = V_P(c)V_Q(c)$) until after evaluation is introduced. If you define polynomials in the classical way, as sums of powers of an unknown variable $x$, these arithmetic rules emerge directly from the definition. $\endgroup$
    – Ilmari Karonen
    Commented Mar 13, 2017 at 18:36
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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 $R$

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    $\begingroup$ This is essentially Lubin's answer. $\endgroup$
    – pjs36
    Commented Mar 13, 2017 at 18:07
  • $\begingroup$ Indeed. The only drawback with my approach was that I was not aware of how polynomials were defined in the first place. $\endgroup$
    – R004
    Commented Mar 14, 2017 at 2:34

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