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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 ?

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    ask yourself the question, can I write $\sin(x)$ as a FINITE combination of powers of the indeterminate X and coefficients $a_i$, such that: $$ \sin(x) = a_0 + a_1 X + \dots + a_n X^n $$ –  Dec 17 '18 at 10:58
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    The expansion of $\sin x$ has infinitely many nonzero terms. The given definition says that a polynomial is a sequence with finitely many nonzero terms. – Sujit Bhattacharyya Dec 17 '18 at 11:01
  • I would recommend taking some real-analysis and specifically a section on sequences of functions, this really cleared things up for me to not see Taylor series as an "infinite polynomial". –  Dec 17 '18 at 11:10
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    Just in case you also find this confusing. Looks like the risk for that is low, but still... something that confuses many at first. – Jyrki Lahtonen Dec 17 '18 at 11:14
  • Note that you also consider and infinite sequence of zeros, but the definition often explicitly states we look at a finite number of nonzero coefficients/terms. –  Dec 17 '18 at 11:28
  • So I would say the polynomial $5 + 3X + X^2$ encodes the sequence $(5,3,1)$ –  Dec 17 '18 at 11:28
  • "That is, a polynomial can either be zero or can be written as the sum of a finite number of non-zero terms." –  Dec 17 '18 at 11:29
  • @JyrkiLahtonen I agree, it confused me too at first. But then we started doing finite fields and all became clear, well, mostly. –  Dec 17 '18 at 11:30
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    Also, sine cannot be a polynomial. After all, a polynomial has limits $\pm\infty$ when $x\to\pm\infty$ (think about the leading term), but sine is bounded. The only bounded (real) polynomial functions are the constants. – Jyrki Lahtonen Dec 17 '18 at 11:30
  • @JyrkiLahtonen Had no idea about difference of two terms in link you left, actually .. This cleared things up a bit. –  Dec 17 '18 at 12:52

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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.

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.

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."

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    This one answers question precisely. Thanks .. Nice way to see it with infinitely many roots. But am I right about the additional condition for definition, though ? Of course, from definition one should intuitively get the idea that polynomial should be linear combination of a sequence with power functions, but person who never seen example of polynomial can't get a clear picture of one, right ? If definition is just a set of sequences alone, then many things are considered to be polynomial, right ? –  Dec 17 '18 at 12:46
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    I have only been taught that the coefficients of the polynomial of degree $n$ and a sequence of length $n$ of the form $(a_0, a_1 \dots a_n )$ are isomorphic, so any of the two will do. We started out with the definition of a polynomial in $X$ over a field/ring $R$, where $a_i \in R$. Then the elements of R[X] are represented by: $$P= a_0 +a_ 1 X \dots a_n X^n$$ The structure where we consider all such expression we called a polynomial ring. –  Dec 17 '18 at 12:54
  • Again, finite sequences and finite polynomials. –  Dec 17 '18 at 12:56
  • I would say that $p$ can also be represented by $(50, 1, 1)$ if we take: $$p(X)= 50 +X + X^2 $$ Of course you can add infinitely many zeros and then you can get infinitely long sequences, but then you get infinitely long series in correspondence with this. –  Dec 17 '18 at 12:57
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    Well yes, so the definition in question is not really complete and precise, right ? –  Dec 17 '18 at 13:01
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    Well, yes. Your professor should really state something like "where any polynomial is a finite linear combination of power of the indeterminate and coefficients from the field/ring $R$" Notice that this sum runs up to $n$ and we are not saying something like : $$ \lim_ {n \to \infty } \sum_{i=0}^n a_i x^i= \sum_{i=0}^\infty a_i x^i$$ –  Dec 17 '18 at 13:03
  • The right thing to do in your example is to observe that if every polynomial has at most degree $n$, we write the sequence: $$ (50,1,1, 0 \dots 0) $$ Where we have $n-2$ zeros. (because a polynomial of degree $n$ has $n+1$ coefficients) –  Dec 17 '18 at 13:07
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    Oh, okay, thats the thing that was bothering me. Thanks! –  Dec 17 '18 at 13:08
  • No problem, feel free to ask any time ^^ –  Dec 17 '18 at 13:09
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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).

Theo Bendit
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