Do polynomial expansions have to be in terms of $x$, ie $\sum x^k$?

Question

This seems like a trivial question, but I have not been able to find an answer online. Do power series/polynomial expansions have to be in terms of $x$, ie: $\sum x^k$.

For example, would $\sum_{k=0}^{5} (sin(x))^k$ formally be considered a polynomial expansion? I know that there are ways to $approximate$ this summation of powers of sine functions in terms of powers of $x$, but is the sum itself considered a polynomial expansion?

I am asking because I have a generative problem of the form:

Class $\mathbb{P}$ is generated by $\{\hat{p_i} \rvert \sum_{k=1}^{N}c_k\hat{p}_0^k\}$

Class $\mathbb{Q}$ is generated by $\{\hat{q_i} \rvert \sum_{k=1}^{N}c_k\hat{q}_0^k\}$

I am trying to find a way to classify signals of this form without needing to know $p_0$ and $q_0$. I thought that there may be some transformation or property of a polynomial expansion that could help me with this problem.

The way that this problem is posed is to assume that every signal in $\mathbb{P}$ and $\mathbb{Q}$ is generated by as a summation of powers. Given a new signal $h_i=\sum_{k=1}^{N}c_k\hat{h}_0^k$ belonging to one of these classes, we want to decide whether it is a member of $\mathbb{P}$ or $\mathbb{Q}$, ie: is $h_0=p_0$ or $h_0=q_0$. Ideally, we want to be able to do this without necessarily knowing $p_0$ or $q_0$.

The task is to be able to "undo" the polynomial expansion so that every $p_i$ and $q_i$ have the same form (namely, that every $p_i$ and $q_i$ can be reduced to $p_0$ and $q_0$). This way, when given $h_i$, we can apply our transform which "undoes" the polynomial expansion to compare ${h_i}_{undone}$ to $p_0$ and $q_0$ directly. Then the task becomes very easy.

Answer 1

This question is really about linear algebra. Given a function $\, p(x), \,$ the set $\, P \,$ of all linear combinations of powers of $\, p(x), \,$ is a vector space. It is also known as the set of polynomials of $\, p(x). \,$ The space $\, P \,$ can also be generated by functions of the form $\, a\,p(x) + b\,$ where $\, a, b \,$ are constants and $\, a\ne 0.\,$ If another function $\, q(x) \,$ is not of that form, then it generates a different set $\, Q \,$ of polynomials of $\, q(x). \,$

The classification problem is that, given a functions $\, p(x), f(x), \,$ how to determine if $\, f(x) \,$ can be expressed as a polynomial of $\, p(x). \,$ In practical terms, if we assume that the degree of $\, f(x)\, $ in terms of $\, p(x) \,$ is at most $\, n \,$ then we attempt to solve the equation $\, f(x) = \sum_{k=0}^n c_k\, p(x)^k \,$ for the coefficients $\ \{c_0,\dots,c_n\}. \,$ If a solution (unique) exists then we can express $\, f(x) \,$ as an explicit polynomial of $\, p(x). \,$ If no solution exists then $\, f(x) \,$ is not a polynomial of $\, p(x). \,$

Answer 2

This is not properly a polynomial, but one does say that it is "a polynomial in $\sin x$".

In other words, it is a composition of a polynomial with another function. In your first example, $P(\sin x)$, where $P(x) = 1 + x + x^2 + x^3+x^4+x^5$.

Note too that a polynomial can involve multiple variables (the terms are products of nonnegative integer powers of these variables), such as $x^3 + 3x^2 + xy^4z^3 + xz^2 +7$.