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I am looking at the following question:

Is the set of all polynomials open in $C[-1,1]$?

I am not sure what functions are considered as polynomial.

For example, let $$f(x) = \frac{x}{1+|x|}.$$ Is $f$ a polynomial?

I think the answer is no as polynomial must be of the form $$\sum_{i=0}^n a_ix^i$$ where $n$ is a natural number. Since $f$ is not of the form given, therefore it is not a polynomial.

However, is $g(x) = |x|$ a polynomial? I think it is because $$g(x)=|x| = sgn(x)x$$ where $sgn(x)$ is the sign function of $x.$

To conclude, I post my question below:

How to show that a given function is a polynomial?

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    $\begingroup$ $g(x)$ is not a polynomial because $a_i$ must be a constant while $\text{sgn}(x)$ is not a constant. $\endgroup$ – user499203 Nov 24 '17 at 1:14
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    $\begingroup$ I was hesitant to post this as an answer, but polynomials eventually differentiate to $\mathbf0$ $\endgroup$ – gen-z ready to perish Nov 24 '17 at 2:54
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    $\begingroup$ @ChaseRyanTaylor: Yeah I basically take this to be the definition. But I'm wondering, do you need to be more precise than that? Wouldn't there need to be some notion of vanishing "uniformly" in some sense? Otherwise I'm imagining what if the derivative at every point vanishes, but you need more and more of them as you go in one direction? Not entirely sure that's possible though, since I'm not sure if it would imply a discontinuity in some derivative somewhere... $\endgroup$ – user541686 Nov 24 '17 at 4:11
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    $\begingroup$ $\dfrac{x}{2+x}$ is not a polynomial either, though $\dfrac{4-x^2}{|2+x|}$ is a polynomial in $C [-1,1]$ and any test needs to distinguish these $\endgroup$ – Henry Nov 24 '17 at 8:55
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    $\begingroup$ @PaulSinclair: I don't think it's possible. If $f^{(n)}$ is identically zero on an open interval, then $f$ is a polynomial of degree less than $n$ on that interval (e.g., by Taylor's theorem with the Cauchy form of the remainder). Now if for every real $x$ there's an $n$ and neighborhood of $x$ on which $f^{(n)}$ vanishes, we must have the same polynomial on overlapping neighborhoods, and a connectedness argument shows that $f$ must be a polynomial globally. $\endgroup$ – Mark Dickinson Nov 24 '17 at 20:22
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A function $f\colon[-1,1]\to\mathbb R$ is a polynomial if and only if it is infinitely differentiable (with one-sided derivatives at the endpoints) and for some $n\in\mathbb N$ the function $f^{(n)}$ is a constant. The minimal such $n$ is the order of the polynomial.

Side remarks: (1) Alternatively, you can demand that $f^{(n)}\equiv0$, but then the minimal $n$ is the order plus 1. (2) The order statement fails when $f\equiv0$, but this is the only exception. Since it is rather uninteresting, I will tacitly exclude the case. (3) To prove this characterization of polynomials, first show that a polynomial has all the required properties. Then, if $f$ is as above, you can compute the indefinite integral of $f^{(n)}$ with the constants $n$ times to get $f(x)$.


Your function $g$ is not a polynomial because it is not differentiable at the origin.

Your function $f$ is smooth outside the origin. It is also differentiable at the origin, and the derivative is $$ f'(x) = \frac{1}{(1+|x|)^2} . $$ Now this $f'$ is not differentiable at the origin, so $f$ cannot be a polynomial. (Alternatively, you could argue that the second derivative of a polynomial would be continuous, but in this case the limits from different sides of the origin are $\pm2$.)

In this particular case, the functions are not polynomials because they are not smooth enough. If you want to show that something is a polynomial, it is often most convenient to use the definition and express the function in an appropriate power sum form.

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Since $f$ is not of the form given, therefore it is not a polynomial.

You have not proved this. The fact that $f$ is not currently written in that form does not necessarily imply that it cannot be written in that form. For example,

Is the function $h:\mathbb{R}\to\mathbb{R}$ defined by $$h(x)=e^{-x}\int_{-\infty}^{x}t^5 e^t\mathrm{d}t$$ a polynomial?

It would be incorrect to say that $h$ is not a polynomial because it isn't written in the form $\displaystyle{\sum_{i=0}^{n}a_ix^i}$. Actually $h$ is a polynomial, and $h(x)=x^5-5 x^4+20 x^3-60 x^2+120 x-120$.

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    $\begingroup$ This is a useful comment on the question's wording, but it doesn't actually seem to answer the question which asks how to show whether something is a polynomial. $\endgroup$ – David Z Nov 24 '17 at 23:32
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The polynomials are neither open nor closed in $\mathcal{C}[0,1]$. They are uniformly dense in $\mathcal{C}[0,1]$ [Stone-Weierstrass].

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    $\begingroup$ It appears that you did not answer his question, but the question that motivated OP's question of the definition of a polynomial $\endgroup$ – Prince M Nov 24 '17 at 1:50
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    $\begingroup$ Dense sets can be open, showing that the polynomials are dense doesn't prove that they aren't open. $\endgroup$ – jwg Nov 24 '17 at 10:25
  • $\begingroup$ Not here; just add $\epsilon\sin(at)$ where $\epsilon$ is small and $a$ is huge. $\endgroup$ – ncmathsadist Nov 24 '17 at 23:12
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If you want to show that $g(x)$ is not a polynomial:

You have to show that for every degree $n$ and every set of coefficients $a_0,a_1,\ldots,a_n$ there exists a value for $x$ such that the equation bellow is not satisfied:

$$ x\cdot\mathrm{sign}(x)=\vert{x}\vert=\sum_{k=0}^{n}a_{k}x^{k}.$$

If you want to show that a given function $h(x)$ is a polynomial (in $\mathbb{R}[x]$) you have to show that there exists:

1) A natural number $n$,

2) A set of scalars $a_0,a_1,\ldots,a_n$ in $\mathbb{R}$,

such that for every $x\in\mathbb{R}$ you have: $$g(x)=\sum_{k=0}^{n}a_{k}x^{k}$$

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    $\begingroup$ Your answer is a bit confusing as to whether you're giving a general or answer or telling the OP how to tell whether |x| in particular is a polynomial. To show that |x| is not a polynomial, all that is necessary is to note that the derivative is undefined at x =0. Another method for showing a function is not a polynomial is to show that it is, on two different subdomains with infinite number of points, equal to two different polynomials. $\endgroup$ – Acccumulation Nov 25 '17 at 19:43
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You gave the right definition of polynomial: it's $\sum_{k=0}^n{a_k x^k}$, where the $a_k$ are constant. $sgn(x)$ is not a constant, so your $g$ is not a polynomial.

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    $\begingroup$ His question is "How to show that a given function is a polynomial?" $\endgroup$ – user499203 Nov 24 '17 at 1:15
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    $\begingroup$ fair point, but the answer is: use the definition you already know. His real question was "why isn't this a polynomial?" $\endgroup$ – dbx Nov 24 '17 at 1:17
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    $\begingroup$ How do you know which is the real question? $\endgroup$ – Ooker Nov 24 '17 at 3:25
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    $\begingroup$ This reasoning is not rigorous here and wrong in general, see math.stackexchange.com/a/2534885/85341. I agree on your intuitive point, though, e.g. it is also easy to see that f'''(x) = sin(x)f(x) + f''(x) is not a linear ODE. $\endgroup$ – ComFreek Nov 24 '17 at 9:31
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$f(x) = \frac{x}{1+|x|}$ is not a polynomial because $\displaystyle \lim_{x \to \infty} f(x) = 1$ but $\displaystyle \lim_{x \to \infty} p(x) = \pm \infty$ for a nonconstant polynomial.

$g(x)=|x|$ is not a polynomial because it is not differentiable at $x=0$ but polynomials are differentiable at all points.

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    $\begingroup$ Unfortunately, the OP ultimately wants to work in $C[-1,1]$ (needs just one (necessarily bounded) non-polynomial), hence any claim about $\lim_{x\to\pm\infty}$ is irrelevant; e.g., the restriction of the non-polynomial $h(x)={\max\{|x|-1,0\}}{}$ to $[-1,1]$ coincides with a polynomial. However, differentiability comes to the rescue for $f$ as well (at $f''(0)$, I suppose) $\endgroup$ – Hagen von Eitzen Nov 24 '17 at 6:04
  • $\begingroup$ @HagenvonEitzen, right, good point! $\endgroup$ – lhf Nov 24 '17 at 10:09
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    $\begingroup$ @HagenvonEitzen wouldn't it be fair to observe that polynomials on $C[-1,1]$ are exactly those functions which are the restrictions of a polynomial on $\mathbb{R}$, and then reason about properties of polynomials on $\mathbb{R}$? $\endgroup$ – jwg Nov 24 '17 at 10:33
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I looked at both Wikipedia and Wolfram, and they define "polynomial" as an expression. According to that definition, there is a distinction between a polynomial and a function that it represents, and whether something is a polynomial can be determined according to its form, without inquiry as to the properties of the function that it represents: a polynomial is an expression in which each term is a monomial. In practice, however, the term "polynomial" is used to refer to what strictly speaking should be referred to as "polynomial function". In this use, there are a variety of tests, such as showing that it is a product of binomials, or showing that it is annihilated by a finite number of differentiation.

Also, it should be "considered to be polynomials", not "considered as polynomial".

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  • $\begingroup$ Polynomials can be formally defined as sequences of coefficients. $\endgroup$ – user370967 Dec 13 '17 at 21:59

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