How to show that a given function is a polynomial?

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?

Solution 1:

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.

Solution 2:

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 $h(x)=x^5-5 x^4+20 x^3-60 x^2+120 x-120$.

Solution 3:

The polynomials are neither open nor closed in $\mathcal{C}[0,1]$. They are uniformly dense in $\mathcal{C}[0,1]$ [Stone-Weierstrass].

Solution 4:

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: $$g(x)=\sum_{k=0}^{n}a_{k}x^{k}$$