Question about the formal definition of a polynomial in relation to $\sin(x)$ not being a polynomial

Question

This question has been asked before, but none of the answers seems to satisfy what I'm asking here ..

So, in class, we introduced set of polynomials as a set of sequences of real numbers, where only finitely many elements of a sequence are non zero. Now, I get it that, for example: $p(x) = x^2 + x + 50$ fits the definition. The sequence in definition is actually a sequence of coefficients of polynomial. So, in my example, $p = (50, 1, 2, 0, 0 ...)$, right ?

But how does $\sin(x)$ not fit the definition ? So my guess is that this set contains all the polynomials, but is larger than that. I suppose that condition that is missing is that polynomial is actually linear combination of elements of given sequence and power functions, in natural order $\dots$ Is that correct, or am I wrong? I might have misunderstood the definition or $\sin(x)$ is polynomial ?

Answer 1

We cannot write $\sin(X)$ as a finite combination of the indeterminate $X$ and coefficients $a_i$. The problem is that you probably know that you can express

$$ \sin(x) = \sum_{} ^{\infty} (-1)^{n+1} \frac{x^{2n+1}}{(2n+1)!}$$ In terms of its Taylor series. This is however an infinite series, furthermore it is actually the limit as $n \to \infty$ of the partial sum of the sequence of polynomial functions. That's a mouthful. Basically we need some more tools to properly express $\sin(x)$ as an "infinite polynomial". Notice that my quotes express that this is not the proper way to talk about it, the proper way to talk about Taylor series is in terms of limits of sequences of functions.

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If you were to fit a polynomial of finite degree to $\sin(x)$ we get that this polynomial must, like $\sin(x)$ have infinitely many roots/zeros. We get a contradiction and therefore $\sin(x)$ cannot ever be expressed in terms of a polynomial of finite degree.

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Interestingly, just like with Taylor series being infinite we can define so-called formal power series, which are a generalisation of polynomials as you know them, because they encode infinitely long sequences: https://en.wikipedia.org/wiki/Formal_power_series

Notice that we cannot substitute values into formal power series, or as my combinatorics professor calls them, fops.

Also see the definition on wikipedia https://en.wikipedia.org/wiki/Polynomial#Definition

"That is, a polynomial can either be zero or can be written as the sum of a finite number of non-zero terms."

Answer 2

This is one of these things where context really matters.

Often the first concept of polynomial one deals with is in the case of real (or potentially complex) functions. A polynomial is a function $f : \mathbb{R} \to \mathbb{R}$ that can be written in the form $$f(x) \equiv a_0 + a_1 x + \ldots + a_n x^n,$$ for some $a_0, \ldots, a_n \in \mathbb{R}$ and $n$ a non-negative integer.

It's a handy definition for proving a function is a polynomial, but not so handy for proving a function is not a polynomial. For example, $\sin$ is not a polynomial, but it takes some proving to show that no choice of $n$ and $a_0, \ldots, a_n$ will make the above expression identical to $\sin(x)$. Fortunately, it's not too hard to show; you can examine limits to $\infty$, count roots (polynomials have only finitely many), or take repeated derivatives (polynomials differentiate to $0$).

However, in certain contexts, defining polynomials as functions is not sufficient. For example, to construct $\mathbb{F}_4$, the finite field of order $4$, you start with $\mathbb{F}_2$, which is the ring of integers modulo $2$. You then consider the ring $\mathbb{F}_2[x]$, the ring of "polynomials" whose coefficients are elements of $\mathbb{F}_2$, and quotient out the maximal ideal generated by the irreducible polynomial $x^2 + x + 1$.

However, if you think about this (and are with me so far), this makes no sense if you think of polynomials as functions from $\mathbb{F}_2$ to $\mathbb{F}_2$. As a function from $\mathbb{F}_2$ to $\mathbb{F}_2$, the polynomial $f(x) = x^2 + x + 1$ is equivalent to the constant function $1$, in that it sends both elements of $\mathbb{F}_2$, $0$ and $1$, to $1$. So, thinking about polynomials in this way, in this context, is unhelpful and impedes our way to some vital mathematics.

So, all of this is a long-winded way to say, it depends on context. You can think of them as a sequence that's eventually $0$. You can fit $\sin$ in terms of this definition by considering it as a sequence of coefficients of its Maclaurin Series (which makes it not a polynomial).